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J A N L U B I N A
jaa
n
n
bilu
A P P L E P I E O R D E R
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A P P L E P I E O R D E RIt doesn‘t fit the margin, but it does go in my book.
„The scientist dose not study Nature because it is useful; hi studies it because hi delights in it,and he delights in it because it is beautiful. If Nature were not beautiful, it would not be worthknowing, and if Nature were not worth knowing, life would not be worth living.‖
Jules Henri Poincaré.
―Mathematicians do not study objects, but relations between objects. Thus, they are free toreplace some objects by others so long as the relations remain unchanged. Content to them is
irrelevant: they are interested in form only‖.
Jules Henri Poincaré.
For the memory of
Katharina LubinaMarch 7, 2009
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"Sed omnia in mensura, et numero, et pondere diposuisti.‖ Sapientia 11, 21.
World and all what oneself it on him finds it carries mathematical structures. So God createdhim with mathematical point of sight. Given the man the strength to him from God of reason,plan of his building can the discoveries. It is the mathematics so the key to understanding of
world. In peer with her development, she went the change of aims what her was placed. Itdoes not serve the mathematics the endeavor to better perception only and the understandingof nature, but it has to permit her to master.
Both points of sight, chief place of mathematics, strength of granted her certainty and theincontestability, which are useful different disciplines scientific character, as also change of this, what oneself it under this notion understands and what it the thanks were wanted was toreach her, they gave the beginning my trials they would solve problem of primes. So beganmy adventure with primes.
Human spirit and human culture they unrolled such formal system of thinking, to can
formulae recognize, to classify and to use. We call him mathematics, because it ismathematician the science of formulae.The only right of existence for mathematician, the desire of discovery of new formulae is andthe inherent in rights of nature regularities, as and announcing this what it will happen.Though looking for formulae and structures it is mathematician's activity, then proper his task is formulation their in irrefutable proofs.Numbers are the simplest mathematical object, and the simplest formulae of nature arenumerical, because perfect relations between numbers reign.
The basic theorem of algebraically theory of numbers sounds: All numbers descend from one..
"O M N I A E X U N O‖
Theorem this be leaning on system of certainties, what Italian mathematician Giuseppe Peanoin 1889 r. submitted on unquestionable truth the undemanding proof "parental power‖ of number one, giving the same bearing foundation theorem taking out from one all naturalnumbers.The forcible model of principle "all of one‖/ OMNIA EX UNO / is the draught of naturalnumbers 1, 2, 3, 4,.. in which number one, it is for all numbers the "point of exit‖. One isreally only corner stone the whole draught of numbers on which is bases here. In gathering of natural numbers the number one is the class alone for me, the "Unity‖ is called also from here.
One is only number, which does not change when oneself it divide her by her, or it increases.Geometric he be introduced as point, by what his elusiveness be expressed. Point's the lack of length, width and height, upper or bottom side, any color, and even the position.It was cannot say even, points are round, because taking at all closely they do not widen. Thisborders on with miracle directly, that attributes number these essential and necessary featureswithout which the whole draught of natural numbers would not can exist. Then she is the"Point‖ of reference, what to which all natural numbers graphic be co-ordinated, introducedon two co-ordinates the a and the b. She is the "source‖ even and odd units also from which itcomes into being whole row of prime and folded numbers. Exists such "Unity‖, from whichthe whole wealth of world results, as one axiom will suffice as foundation the fine edifice of arithmetic.
"It exists such number 1 possessing property, which treats to every number - n‖:
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n · 1 = n = n + 0 1 · p = p
Really comparative size with 2 enters in life, in support about which , all differentmeasurable can pit .She beyond this is with nature the number of "unification‖ from twounit's make one number.
1 =
312
2 =3
12 +
3
12
p = a + b b = p - a a =3
1 p
p = )
3
1(
3
1
p p
pp = a + ( p - a )
Prime numbers this "building blocks‖, from which be built all different natural numbers. Notwe will find them however in multiplication table, because number first cannot be the result"sensible‖ operation of multiplication, but only addition. Every prime numbers is the sum two components defining her place in draught of naturalnumbers p = a + b.
Component a =3
1 pthen they came into being with divisible numbers even quotient by 3.
Component b = p - a then difference among prime number, and even quotient.
It number 2 is only even prime number and across her principle "larger about one‖ it willbecometransferred on next natural numbers, guaranteeing contact and progress in draught.
1 =3
12
2 = 1 + 11
3 = 1 + 2 = 1 + ( 3 - 1 )
14 = 2 + 2 1
5 = 2 + 3 =3
15 + ( 5 -
3
15)
16 = 3 + 3
7 = 2 + 5 =3
17 + ( 7 -
3
17 )
1
8 = 3 + 5
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All prime numbers precede or they follow after divisible number by 3 eg. 2, 3, 17, 18, 19,23, 24, for except 3 even. Eureka!
p n21 17 + 1 = 18 = 19 – 1 because 3(p n2)1 3(7-1) = 18
p n21 5 + 1 = 6 = 7 – 1 because 3(p n2)1 3(3-1) = 6
5 11 17 23 29 41 47 59 71 83 101 107 113 131
6 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96, 102, 108, 114, 120, 126, 132,
7 13 19 31 37 43 61 67 73 79 97 103 109 127
Prime number is about one larger or smaller from previous or following even divisiblenumber by 3.
p = 2n 1 e.g. 1999 = 1998 + 1
Odd numbers, as this results from every multiplication table, are product of prime numbers,odd and almost prime.
3
n= b n = 3 b (2 b + b)
9 = 2 (3) + 3In odd numbers the relation of even components to odd is always 2 : 1,we can from here write n = 2b + b e.g. 15 = 2(5) + 5 21 = 2(7) + 7
If decomposes the sum of units of number on the components the being in relation expressedin equation n = 2b + b, then it is surely then the odd number.
Triangle of numbers.
"Number is collection of units‖, Euclid defines her in book "Elements‖ so.
"Tria juncta in uno" / Three join in one / In triangle of numbers the Principle "larger aboutone" the links units in integers.
If decomposes the sum of individuals of number on the components the being in relationexpressed in equation p = a + ( p - a) this is surely then prime number.
If every number natural larger from one, can be written in aspect of the sum of unity or the
sum primes, and ―unity ‖ is quotient of the sum of prime and ―unity‖ by next number prime,then the infinite sum of natural numbers is equals infinite sum ―unity‖, e. g. 4 = 1 + 1 + 1 + 1 N 1
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Since natural numbers is infinitely many, then and primes is infinitely many, because alldifferent with them consist, and what with this goes also pair of twin primes. This is yetcompletely comprehensible! And simultaneously not natural in natural numbers.
All natural numbers which carry in me principle ―larger about one‖, can be written as the sumof ones, or primes 2 and 3.
2k = p + p… 2k = )2( 1 =312 n = p‘ + p‘ n = )3(
1 + 1 = 2 = 1(2)1 + 1 + 1 = 3 = 1(3)
1 + 1 + 1 + 1 = 4 = 2 + 21 + 1 + 1 + 1 + 1 = 5 = 2 + 3
1 + 1 + 1 + 1 + 1 + 1 = 6 = 2 + 2 + 21 + 1 + 1 + 1 + 1 + 1 + 1 = 7 = 2 + 2 + 3
1 + 1 + 1 + 1 + 1 + 1+ 1 + 1 = 8 = 2 + 2 + 2 + 21 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 9 = 3 + 3 + 3
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 =10= 2 + 2 + 2 + 2 + 2
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 =11= 2 + 2 + 2 + 2 + 31 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 =12= 2 + 2 + 2 + 2 + 2 + 2
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 =13= 2 + 2 + 2 + 2 + 2 + 31 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 =14= 2 + 2 + 2 + 2 + 2 + 2 + 2
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 =15= 3 + 3 + 3 + 3 + 315 + 1 =16= 2 + 2 + 2 + 2 + 2 + 2 + 2 + 216 + 1 =17= 7(2) + 1(3) p = n(p) + p‘ 17 + 1 =18= 9(2)18 + 1 =19= 8(2) + 1(3)19 + 1 =20= 10(2)24 + 1 =25=5(2) + 5(3) „p‖= n(p) + n(p‘) 34 + 1 =35=7(2) + 7(3)+ 1 =
k + k = n 1 + 1 = 2 + 1 = 3, 4 = 2 + 2, 5 = 2 + 3, 6 = 4 + 2, 7 = 4 + 3, 8 = 6 + 2, 9 = 6 + 3,...
12
34
56
78 9 10
1112
1314
1516
1718
1920
2122
2324
2526
2728
2930
3132
3334
3536
3738
3940
4142
4344
4546
4748 49
5051
5253
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 223 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3
24
6 68 8 1010
12121414
16161818
20202222
24242626
28283030
32323434
36363838
40404242
44444646
484850
3
50
2
4
0
10
20
30
40
50
60
1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53
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Theorem: Every number prime 3 consists from ternary and number even diminish aboutthree.
N = 2k – 1 p = 3 + [(2k – 1) – 3] 2k – 1 – 3 = 2k – 4 = 2k – 2 k > 2
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 +
\ 3 / + \ 2 / + \ 2 / + \ 2 / + \ 2 / + \ 2 / + \ 2 / + \ 2 / + \ 2 / + \ 2 / + \ 2 / 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
p = )2(3 n p = 3 + 2k – 323 = 10(2) + 1(3) 37 = 17(2) + 1(3)29 = 13(2) + 1(3) 41 = 19(2) + 1(3)31 = 14(2) + 1(3) 43 = 20(2) + 1(3)47 = 22(2) + 1(3) 67 = 32(2) + 1(3)53 = 25(2) + 1(3) 71 = 34(2) + 1(3)59 = 28(2) + 1(3) 73 = 35(2) + 1(3)
61 = 34(2) + 1(3) 79 = 38(2) + 1(3)83 = 40(2) + 1(3) 89 = 43(2) + 1(3)97 = 47(2) + 1(3) 107 = 52(2) + 1(3)
101 = 49(2) + 1(3) 109 = 53(2) + 1(3)103 = 50(2) + 1(3) 113 = 55(2) + 1(3)
The whole infinite file of natural numbers consists from infinite quantity 2 and 3, which are―units ‖ all numbers.
N = (2) + (3) = (1)
N = 1 = 3
12
Alone meanwhile ―units‖ they are even and odd multiplicity ―unity‖ 1 =3
12 1(2), 1(3).
In this way was proved mathematically indirectly that all numbers descend from one, becausethey consist from ―units ‖.
P = 13
12
3
1
p pe. g. 179 = 1
3
11792
3
1179
p = [ 2(k) – 2] + 3 727 = [ 2(363) – 2] + 3 = (726 – 2) + 3
As to that indivisibility. Euler announced, that possesses algebraically proof on existence
God. His form looked so: xn
ban
, hence God exists. If in place of algebraically signs to
substitute three first numbers, then for mathematician equation this can to be proof on
indivisibility number 3 13
123
. Philosopher can tell, that only plurality can to get unite.
Theologian meanwhile it will say: Father and Son with triple only Holy Spirit it is indivisibleHoly trinity, hence exists one God in three persons. And all are right, because plurality is theform of unity.
See this on example primes, which despite that they consist from many individuals, they existas individual indivisible numbers.
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9
6 7
2 3 4 5 1 =
3
121 =
3
12
1 =
3
121 =
3
121 =
3
121 =
3
12
1 =
3
121 =
3
121 =
3
121 =
3
121 =
3
121 =
3
12
0 - - - - - - - - - - - 1 =
3
12------------- 1 =
3
12--------------------- 1 =
3
12
1 =
3
121 =
3
121 =
3
121 =
3
121 =
3
121 =
3
12
1 =
3
121 =
3
121 =
3
121 =
3
12
1 =
312 1 =
312
Three, as all odd numbers possesses symmetry creator "centre‖. Centre of three is two,quintuple the three, seven the five etc, hence with 2 and 3 consist all natural numbers and thethree the state "centre‖ of all odd numbers, and all natural numbers are the quotient of three.
Every plurality is the plurality of „units‖.
N = 1
3
12
From those "units" be folded the whole cosmos, world of minerals, plants, animals and human
existences. "Man was created on range and similarity of only God's‖, it tells us so many belief, "all consists from the smallest and indivisible particles‖ - natural sciences teach so.Both on their way struggle about formulating one and the same truth: Such Unity is, fromwhich whole plurality descends.
0 : 1 = 03
0
13
3
3
12
3
12
3
12
= 23
6
312
312 + 3
39
312
3
12
3
12
+
3
12
3
12
= 4
3
12
3
12
3
12
+
3
12
3
12
+ 5
3
15
3
12
3
12
3
12
+
3
12
3
12
+
3
12
3
12
= 6
3
18
3
12
3
12
+
3
12
3
12
+
3
12
3
12
+ 7
3
21
3
12
3
12
3
12 +3
12
3
12 +3
12
3
12 +3
12
3
12 = 83
24
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10
3
12
3
12
+
3
12
3
12
+
3
12
3
12
+
3
12
3
12
+ 9
3
27
3
12
3
12
3
12
+
3
12
3
12
+
3
12
3
12
+
3
12
3
12
+
3
12
3
12
= 10
3
30
If sum two following numbers prime form n and n + 2, it is divisible by 12, then they are
surely then twin numbers.
p + (p`+ 2) =12
)2( p p
12 24 36 60 84 / \ / \ / \ / \ / \
5 + 7 11 + 13 17 + 19 29 + 31 41 + 43"Twin ‖ call pair of numbers prime between which steps out the even number divisible by 3,e.g. 5-6-7, 11 -12- 13, 17-18-19, 29-30- 31, 41 -42- 43, 59-60- 61,. it but not pair 131 -132-133, or 10 000 037 -10 000 038- 10 000 039, it because 2,3,5 number can was take apart onprime factors 133 = 7(19), 10 000 039 = 7(1 428 577), 10 000 037 = 43(232 559).
Divide the sum of twin pair by 12, we will find out near which following even number
divisible by 3, came into being numbers prime. 502312
3013930137
because 5023 · 6
=30138/3During when sequence of the reciprocal of primes is divergent / with reason of growing space
(n)6 / pprim p
1, sequence of the reciprocal of all twin numbers is convergent / because they
near mutually on distance 2/
prim p p p2 2
11< ∞, and his exact value be well-known!
The six- wide array further helps to demonstrate the otherwise still unproven conjecture thatthere must be infinitely many twin primes.In the six- wide rectangular array, the consecutive multiples of each number higher than threelay on a straight line from zero to that number and beyond, and on periodic parallels to thatline further ―down‖ if we begin writing the numbers from the ―top‖ of the array. Soon after this ―factor line‖ leaves the array rectangle on one side, a parallel to it re - enters it on the otherside, farther down in the array at the next such multiple. Each so broken factor line thuscascades in evenly spaced stripes down the layers of the array. Whenever the factor lines fromall the primes above a given layer in the six- wide array happen to miss the two spaces beforeand after the 6n column in that layer, the entries there are not multiples of any among those
prior primes. They are therefore primes themselves and from a pair of twin primes, asillustrated in following table. This approach to the way Euclid suggested to multiply all the
primes, up to a supposedly ―largest‖ one, with each other. He imagined this equally unfeasiblemultiplication to show that the result plus or minus one is either a prime, or else the productof two or more primes larger than the previously ―largest‖. By this method, he proved thatthere always exists a prime larger than any allegedly ―largest‖ one, and that there must thus bean infinite quantity of them.
It is from in pairs twin numbers similarly. Always the foundling oneself the larger pair of twinnumbers from allegedly "largest‖, and by then sequence their has not the end.
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11
0 1 2 3
4 5 6 7 8 9
10 11 12 13 14 15
16 17 18 19 20 21
22 23 24 25 26 27
28 29 30 31 32 33
34 35 36 37 38 39
40 41 42 43 44 45
46 47 48 49 50 51
52 53 54 55 56 57
58 59 60 61 62 63
64 65 66 67 68 69
70 71 72 73 74 75
76 77 78 79 80 81 82 83 84 85 86 87
88 89 90 91 92 93
94 95 96 97 98 99
100 101 102 103 104 105
106 107 108 109 110 111
112 113 114 115 116 117
118 119 120 121 122 123
124 125 126 127 128 129
130 131 132 133 134 135 136 137 138 139 140 141
142 143 144 145 146 147
148 149 150 151 152 153
154 155 156 157 158 159
160 161 162 163 164 165
166 167 168 169 170 171
172 173 174 175 176 177
178 179 180 181 182 183
184 185 186 187 188 189
190 191 192 193 194 195
196 197 198 199 200 201
202 203 204 205 206 207
208 209 210 211 212 213
214 215 216 217 218 219
The sum the pair of twin numbers equals sum of first three successive the pair as thetriangular multiplicities number 12, and the next different multiplicities in dependence fromthis, which they in turn are the pair with infinite set of numbers.
)2(
)(12)6,3,1(12 p p
N n n p p 12
)2(
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Theorem: They twin numbers prime, place oneself before and after even number divisibleby 3, when sum of their ciphers of units equal 4, 10 or 16.
11 + 12/3 + 13 17 + 18/3 + 19 29 + 30/3 + 31 2087 + 2088/3 + 20891 + 3 = 4 7 + 9 = 16 9 + 1 = 10 7 + 9 = 16
Only primes, which even components are even, create the not only that is to say, of twinnumber e.g. 5 and 7, 11 and 13, form n and n + 2, but once even number ‖triplets‖: 3, 5, 7,form n and n + 2 and n + 4, in which this even components are even : -1 -3 -5 =2It exist also one peer of successive prime 2 and 3 which are not "twins‖ yet only "successive‖.
p, (p +2), 5, (5 + 2), 11, (11 + 2), 29, (29 + 2), 107, (107 + 2)
511
1319
31
37
41
43
47
61
59
67
71
7379
97
101
107
109
127
131
137
139
151
157167
181
179
2
3 7
17
2329 53 8389 103113149 163 173
1 2 3
7
13
19
31
37
43
61
67
735
11
17
23
29
41
47
53
59
71
6
12
18
24
30
36
42
48
54
60
66
72
0%
20%
40%
60%
80%
100%
1
2 3 4 56
78
910
11
12
13
14
15
16
17
18
19
20
21
22
2324
25
26
27
2829
3031
3233
343536373839404142
4344
4546
47
48
49
50
5152
53
54
55
56
57
58
59
60
61
62
63
6465
6667
6869
70 71 72 73
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13
12 24 36 60 84
/ \ / \ / \ / \ / \ pd =12
)2( nn
5 + 7 11 + 13 17 + 19 29 +31 41 + 43
p, (p +2), 5, (5 + 2), 11, (11 + 2), 29, (29 + 2), 107, (107 + 2)
511
1319
31
37
41
43
47
61
59
67
71
7379
97
101
107
109
127
131
137
139
151
157167
181
179
2
3 7
17
2329 53 8389 103113149 163 173
_1 + _3 = 4 _7 + _9 = 16 _9 + _1 = 10
1237
13
19
31
3743
61
67
73
5 11
17
23
29
41
47
53
59
71
6 12
18
24
30
3642
48
54
60
66
72
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All natural numbers congruent to me according to module n‘ – n 0 mod 6.
Crossing through prism light, it appears as rainbow of colors. Goes out with unity of numbernatural put on shape six waves about length 6. Congruent to me according to module 6numbers they divide on three groups of even numbers and odd / 2, 4, 6 / 3 -2- 5 -2- 7 / keeping among me solid space 2 and 6 in every group 2/8, 3/9, 4/10, 5/11, 6/12, 7/13.
(p - 1) + (p + 1) /2 = p (53 - 1) + (53 + 1) /2 = 53
0
1
2
3
4
5
6
7
8
10
11
12
13
14
16
17
18
19
20
22
23
24
26
28
29
30
31
32
34
36
37
38
40
41
42
43
44
46
47
48
50
52
53
54
n' - n = 0 mod 6
1 2 2 2 2 24 4 4 4 428
1420
2632
3
915
2127
33
4
10
16
22
28
34
5
11
17
23
29
35
6
12
18
24
30
36
7
13
19
25
31
37
0
50
100
150
200
250
Serie8 7 13 19 25 31 37
Serie7 6 12 18 24 30 36
Serie6 5 11 17 23 29 35
Serie5 4 10 16 22 28 34
Serie4 3 9 15 21 27 33
Serie3 2 8 14 20 26 32
Serie2 4 4 4 4 4
Serie1 1 2 2 2 2 2
1 2 3 4 5 6 7 8 9 10 11
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78 + 80 /2 = 79 7(8) /3 - 7(9) - 8(0) /17/ 82 + 84 /2 = 83 8(2) - 8(3) - 8(4) /3 /9/ 96 + 98 /2 = 97
9(6)/3 - 9(7) - 9(8) /21/
83
89
101
107
131
137 84
90
96
102
108
114
120
126
132
138
144
113
79
97
103
109
127
139
150 + 152 / 2 = 151 /3/ 17(9) - 18(0) / 3 - 18(1) /10/ 19(1) - 19(2) / 3 - 19(3) /6/
167
173
179
191
197
156
162
168
174
180
186
192
198
204
210
216
151
157
163
181
193
199
211
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From first ten numbers prime rise for them four characteristic the number of unity
n n + 2 n + 6 n + 8k + 1 k + 3 k + 7 k + 9
11 13 17 19
222 + 224 /2 = 223 /9/ 22(7) - 22(8) /3 - 22(9) /24/ 23(9) - 24(0) /3 - 24(1) /10/
227
233
239
251
257
263
269
281
228
234
240
246
252
258
264
270
276
282
288
223
229
241
271
277
283
306 + 308 /2 = 307 /21/ 31(1) - 31(2) /3 - 31(3) /6/ 358 + 360 /2 = 359 /17/
311
317
347
353
359
312
318
324
330
336
342
348
354
360307
313
331337
349
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and they step out in tens which number after deduction 1 is divisible by 3 e.g. 10-1 = 9:
Every almost prime numbers we can introduce as sum of 2 and 3 keeping definite proportions.
In almost prime numbers, which are multiplicity of number 5 the relation of 2 to 3 amount 1 1because 5 = 3 + 2
25 = 3 + 3 + 3 + 3 + 3 + 2 + 2 + 2 + 2 + 2
25 = 5(2) + 5(3) 5(5) „p― = )3()2( nn 55 = 11(2) + 11(3) 5(11)
65 = 13(2) + 13(3) 5(13) 85 = 17(2) + 17(3) 5(17)95 = 19(2) + 19(3) 5(19) 115 = 23(2) + 23(3) 5(23)
125 = 25(2) + 25(3) 5(25) 145 = 29(2) + 29(3) 5(29)155 = 31(2) + 31(3) 5(31) 175 = 35(2) + 35(3) 5(35)185 = 37(2) + 37(3) 5(37) 205 = 41(2) + 41(3) 5(41)625 = 125(2) + 125(3) 5(5)(25) 875 = 175(2) + 175(3) 5(7)(25)
In almost prime numbers, which are multiplicity of number 7 the relation of 2 to 3 amount 2 1because 7 = 2(2) +3
35 = 3 + 3 + 3 + 3 + 3 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2
35 = 10(2) + 5(3) 7(5) „p― = )3()2(2 nn 49 = 14(2) + 7(3) 7(7)
77 = 22(2) + 11(3) 7(11) 91 = 26(2) + 13(3) 7(13)119 = 34(2) + 17(3) 7(17) 133 = 38(2) + 19(3) 7(19)161 = 46(2) + 23(3) 7(23) 203 = 58(2) + 29(3) 7(29)
2401 = 686(2) + 343(3) 7(7)(49) 2695 = 770(2) + 385(3) 7(7)(55)
In almost prime numbers, which are multiplicity of number 11 the relation of 2 to 3 amount
4 1 because 11 = 4(2) + 3
121 = 44(2) + 11(3) 11(11) „p― = )3()2(4 nn 143 = 52(2) + 13(3) 11(13)
275 = 100(2) + 25(3) 11(25) 385 =140(2) + 35(3) 11(35)
In almost prime numbers, which are multiplicity of number 13 the relation of 2 to 3 amount
5 1 because 13 = 5(2) + 3
169 = 65(2) + 13(3) 13(13) „p― = )3()2(5 nn 221 = 85(2) + 17(3) 13(17)
637 = 245(2) + 49(3) 13(49) 715 = 275(2) + 55(3) 13(55)
In almost prime numbers, which are multiplicity of number 17 the relation of 2 to 3 amount
7 1 because 17 = 7(2) + 3
289 = 119(2) + 17(3) 17(17) „p― = )3()2(7 nn 323 = 133(2) + 19(3) 17(19)
1105 = 455(2) + 65(3) 17(65) 1309 = 539(2) + 77(3) 17(77)
In almost prime numbers, which are multiplicity of number 19 the relation of 2 to 3 amount
8 1 because 19 = 8(2) + 3
361 = 152(2) + 19(3) 19(19) „p― = )3()2(8 nn 437 = 184(2) + 23(3) 19(23)
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In almost prime numbers, which are multiplicity of number 23 the relation of 2 to 3 amount
10 1 because 23 = 10(2) + 3
529 = 230(2) + 23(3) 23(23) „p‖ = 10n(2) + n(3) 575 = 250(2) + 25(3) 23(25)
In almost prime numbers, which are multiplicity of number 29 the relation of 2 to 3 amount13 1 because 29 = 13(2) + 3
841 = 377(2) + 29(3) 29(29) „p‖ = 13n(2) + n(3) 899 = 403(2) + 31(3) 29(31) 841 = 754 + 87 899 = 806 + 93
In almost prime numbers, which are multiplicity of number 31 the relation of 2 to 3 amount
14 1 because 31 = 14(2) + 3
In almost prime numbers, which are multiplicity of number 7 the relation of 2 to 3 amount 2 1because 7 = 2(2) +3
961 = 434(2) + 31(3) 31(31) „p‖ = 14n(2) + n(3) 1147 = 518(2) + 37(3) 31(37)961 = 868 + 93 1147 = 1036 + 111
In almost prime numbers, which are multiplicity of number 37 the relation of 2 to 3 amount
17 1 because 37 = 17(2) + 3 1369 = 629(2) + 37(3) 37(37) „p‖= 17n(2) + n(3) 25271 = 11611(2) + 683(3) 37(683)1369 = 1258 + 111 25271 = 23222 + 2049
It the whole infinite set of natural numbers consists with infinite quantity of 2 and 3, whichare "units‖ of all numbers.
N = (2) + (3) = (1)Proof N = 1 =
3
12
Alone meanwhile "units‖ they are even and odd multiplicity "unity‖ 1(2), 1(3), 1 =3
12
In this way was proved indirectly the basic theorem of algebraically theory of numbers, thatall numbers descend from one, because they consist from "units‖. Only plurality can to get unite, and primes as only they possess this ability, because they areindivisible.Why a number is prime? Because could be written as two smaller numbers multipliedtogether. That is, it is not possible to represent a prime as the product of two integers a x b with a, b > 1. Let q and r be the quotient and remainder of the division of n by d . That is, foreach n and d, let n = d q + r, where r and q are positive integers and 0 ≤ r < d .
Because all prime numbers contain in me one 3, it was not possible divide here by two.Superiority meanwhile 2 it causes, that they don´t divide by three a lso. So they are indivisibleby all different numbers, and on this depends the complete primality certificate! p = n(2) + 3
2 = 1(2) + 0 3 = 0(2) + 3 5 = 2 + 3 7 = 2(2) + 3 11 = 4(2) + 3 13 = 5(2) + 3
17 = 7(2) + 3 19 = 8(2) + 3 23 = 10(2) + 3 233 = 115(2) + 3 251 = 124(2) + 3
p = 1312
31
p p e.g. 179 = 1
311792
31179
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p = [ 2(k) – 2] + 3 727 = [ 2(363) – 2] + 3 = (726 – 2) + 3
2 1127 = 170 141 183 460 469 231 731 687 303 715 884 105 727
3170 141 183 460 469 231 731 687 303 715 884 105 724
34 279 974 696 877 740 253 374 607 431 768 211 4573
34 279 974 696 877 740 253 374 607 431 768 211 454
The natural numbers in scheme of 2 and 3.
If p ≥ 2 and p‘ ≠ 0, are whole numbers not having common divisor, than such arithmeticalsequence contains in me all natural numbers.
2, 3, n(2), 2 + 3, n(3), 3 + n(2), n(2), n(3), n(2), ... n(2) + n(3)
2, 3, 4, 5, 6, 7, 8, 9, 10, .... 10 + 15 = 25
P(n) = p, p‘, n(p), p + p‘, n(p‘), p‘+ n(p), .... n(p) + n(p‘),
n(2) + n(3) = N
2 2
3 3
2(2) 4
2 3 5
2(3) 6
2(2) 3 7
4(2) 8
3(3) 9
5(2) 10
4(2) 3 11
4(3) 12
5(2) 3 13
7(2) 14
5(3) 15
8(2) 16
7(2) 3 17
6(3) 18
8(2) 3 19
10(2) 20
7(3) 21
11(2) 2210(2) 3 23
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8(3) 24
5(2) 5(3) 25
13(2) 26
9(3) 27
14(2) 28
13(20 3 29
10(3) 30
14(2) 3 31
16(2) 32
11(3) 33
17(2) 34
7(2) 7(3) 35
12(3) 36
17(2) 3 37
19(2) 38
13(3) 3920(2) 40
19(2) 3 41
14(3) 42
20(2) 3 43
22(2) 44
15(3) 45
23(2) 46
And here how with two primes 2 and 3 come into being all natural numbers.
1
2 2
3 3
4 2
5 3 2
6 27 3 4
8 2
9 3 6
10 2
11 3 8
12 2
13 3 10
14 2
15 3 12
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16 2
17 3 14
18 2
19 3 16
20 2
21 3 18
22 2
23 3 20
24 2
25 3 22
26 2
27 3 24
28 2
29 3 26
30 2
31 3 28
32 2
33 3 30
34 2
35 3 32
p = 3 + n(2) n (2) =n= n (3) "p" = n (2) + n (3)
12 3 4 5 67
89
10
11
12
13
14
15
1617
18192021
2223
24
25
26
27
28
29
30
31
32
33
34
35
36
2 22
222
22222
2222
22 2 3
33
3
3
33
3333
3
3
3
33
3
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The scheme of natural numbers
And so harmoniously develop natural numbers in support about principle "larger about one‖on the base of 2 and 3 in 360 ° the circle.
The proprieties of natural numbers repeat oneself periodically, what six numbers according to
pattern of primes.Proof: 1 + 2 + 3 = 6 p + 6 = p‘ n + 6 = n‘ „p‘― – „p‖= 6
012345678910
1112
1314
1516
17
18
19
20
2122
23 24 25 2627
28
29
30
31
32
33
343536
p + 6 = p'
2 1719
2329
31
37
41
43
47
53
5961
67
71
73
79
83
891311753
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With discovery of regularity in sequence of primes, that what 6 numbers repeat oneself thesame proprieties, was decoded together pattern how be distributed primes and the basing onhim periodicity of natural numbers.
1 =3
12
Two first numbers / 1 + 2 / added to me and divided by third next number / 3 /, it equals / 1 / that is to say, again the same first number from three taking part in this working. Three firstnext numbers added to me give perfect and triangular number 6, defining length of period inwhat will repeat oneself the same proprieties in whole sequence of natural numbers.
Tres faciunt collegium, then it means three numbers they decide about whole scheme of natural numbers. It 2 (3) = 6, was can introduce all numbers from here, as sum of ones (+ 1),the twos (+ 2) and the threes (+ 3). The periodical scheme of natural numbers is so perfect, asperfect is first perfect number 6, him untouched basis.
1 + 2 + 3 = 6 = 2 · 3Ranked according to propriety natural numbers create 6 groups. Propriety of numbers in fourcentral groups repeat oneself in turn periodically, what 6 numbers. Primes create here two therows the complementary to two rows of group sixth the almost prime numbers.
Periodical scheme of natural numbers.
n 1 n 2 n 3 n 4 n 5 n 6
2
3 4
5 6
7 8 9 10
11 12
13 14 15 16
17 18
19 20 21 22
23 24 26 25
27 28
29 30
31 32 33 34
36 35
37 38 39 40
41 4243 44 45 46
47 48 50 49
51 52
53 54 56 55
57 58
59 60
61 62 36 64
66 65
67 68 69 70
71 72
73 74 75 7678 77
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79 80 81 82
83 84 86 85
87 88
89 90
91 92 93 94
96 9597 98 99 100
101 102
103 104 105 106
107 108
109 110 111 112
113 114 116 115
117 118 119
120 122 121
123 124
126 125
Sieve of Eratosthenes.
In the six- wide rectangular array, the consecutive multiples of each number higher than threelay on a straight line from zero to that number and beyond, and on periodic parallels to thatline further ―down‖ if we begin writing the numbers from the ―top‖ of the array. In six groups of numbers we have 3 group of even numbers (II, IV, VI), and 3 odd (I, III, V).Her multiplicities for prime number 5 on left have lain cascade, until after number almostprime 25 = 5(5).Next multiplicities for prime number 7 on right have lain her cascade, among which we havesecond almost prime number 35 = 7(5). Parallel line by her runs factor 5 falling on left in pit,
until to fourth almost prime number 55 = 5(11).The parallel line factor 7 falls from the multiplicity number 7(7) = 49 in right, until to lying inV group of almost prime number 77 = 7(11).Parallel line factor 5 falling on left in pit it crosses out their 13(5) = 65 and 15(5) = 85multiplicity.In this way they the parallel lines factors 5 and 7 cross out all almost prime numbers in I andV the group of numbers.
So of the sieve Eratosthenes is situated less than 100 numbers 25 primes.
2 3 5 7 9 1113 15 17 19 21 2325 27 29 31 33 3537 39 41 43 45 4749 51 53 55 57 5961 63 65 67 69 7173 75 77 79 81 8385 87 89 91 93 95 97 99
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I II III IV V VI
All natural numbers congruent to me according to module.n‘ – n 0 mod. 6
2n = p – 3
(2n - 1)
=p(n) 2n = p – 3
p = 2n + 3
"p"= n(2)+n(3)
2n = p(n)
= p±1
p = 2n + 3
"p"= n(2)+n(3)
5 - 3 = 2 3 7 - 3 = 4 2 +3 = 5 3(2) = 6 2(2) + 3 = 7
11 - 3 = 8 3(3) = 9 13 - 3 = 10 2(4) + 3 = 11 3(4) = 12 2(5) + 3 = 13
17 -3 = 14 3(5) = 15 19 - 3 = 16 2(7) + 3 = 17 3(6) = 18 2(8) + 3 = 19
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23 - 3 = 23 3(7) = 21 22 2(10)+3 = 23 3(8) = 24 5(2 + 3) = 25
29 - 3 = 26 25 + 2 = 27 31 - 3 = 28 2(13)+3 = 29 3(10) = 30 2(14)+3 = 31
32 35 - 2 = 33 34 7(2 + 3) = 35 3(12) = 36 37
41 - 3 =38 3(13) = 39 43 - 3 = 40 41 3(14) = 42 43
47 - 3 = 44 3(15) = 45 46 47 48 14(2)+7(3)=4953 - 3 = 50 49 + 2 = 51 52 53 54 11(2 + 3) = 55
59 - 3 = 56 55 + 2 = 57 61 - 3 = 58 59 60 61
62 65 - 2 = 63 64 13(2 + 3) = 65 66 67
68 3(23) = 69 70 71 72 73
74 75 + 2 = 77 76 22(2)+11(3)=77 78 79
80 3(27) = 81 82 83 84 17(2 + 3) = 85
86 85 + 2 = 87 88 89 90 26(2)+13(3)=91
92 95-2=91+2 94 19(2 + 3) = 95 96 97
98 3(33) = 99 100 101 102 103
2n - 1 = 6n - 3 9 = 6(2) - 3 15 = 6(3) - 3 21 = 6(4) - 3 27 = 6(5) - 3 33 = 6(6) -3 39 = 6(7) - 3 45 = 6(8) - 3 2n = 6n - 4 2 = 6(1) - 4 8 = 6(2) - 4 14 = 6(3) - 4 20 =6(4) - 4 26 = 6(5) - 4 32 = 6( 6)- 4 38 = 6(7) - 4 p = 6n - 7 5 = 6(2) - 7 11 = 6(3) - 717 = 6(4) - 7 23 = 6(5) - 7 29 = 6(6) - 7 p = 6n - 5 7 = 6(2) - 5 13 = 6(3) - 52n = 6n - 6 6 = 6(2) - 6 12 = 6(3) - 6 18 = 6(4) - 6 24 = 6(5) - 6 30 = 6(6) - 6 36 = 6(7) - 642 = 6(8) - 62n = 6n - 8 4 = 6(2) - 8 10 = 6(3) - 8 16 = 6(4) - 8 22 = 6(5) - 8 28 = 6(6) - 834 = 6(7) - 8 40 = 6(8) - 8And all runs according to pattern of prime numbers which seems, that they be scattered howsavagely growing weeds among natural numbers, but only there where they create fertile soilgiving the infinite quantity of natural numbers.
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
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From first ten prime numbers can possess four characteristic ends: - 1, - 3, - 7, - 9, resultingwith rhythm 2/4 in what 11 +(2), 13 +(4), 17 + (2), 19 step out and they repeat oneself what21 places in tens divisible by 3.
Table of tens in which prime numbers step out.
I II III IV V VI VII VIII IX X XI XII XIII XIV ..|E – 1 | - 1 : 3 | : 3 | - 1 : 3 | : 3 | - 1 : 3 | - 1 : 3 | : 3 | -1 : 3 | : 3 | - 1 : 3 |: 3 | -1 : 3 | : 3 | - 1 : 3 || | 1 | 3 | 4 | 6 | 7 | 10 | | 13 |15 | | 18 | 19 | 21 | || | | 24 | 25 | 27 | 28 | 31 | 33 | | | | | 40 | 42 | 43 || 21x2 | | | 46 | | 49 | 52 | 54 | | 57 | | 60 | | 63 | 64 || | | 66 | 67 | 69 | 70 | 73 | 75 | 76 | | | 81 | 82 | | || | | | 88 | | 91 | 94 | | 97 | 99 | |102| 103 |105 | 106 || E- 7 | - 1 : 3 | : 3 | - 1 : 3 | : 3 | : 3 |- 1 : 3 | : 3 |- 1 : 3 | : 3 | - 1 : 3 | : 3 | - 1 : 3 | : 3 | - 1 : 3 |
| | 1 | 3 | 4 | 6 | 9 | 10 | 12 | 13 | 15 | 16 | | 19 | | 22 || | | | 25 | 27 | 30 | 31 | 33 | 34 | 36 | | 39 | | | || 21x2| | 45 | 46 | 48 | | | 54 | 55 | 57 | 58 | 60 | 61 | 63 | 64 || | | | 67 | | 72 | | 75 | | 78 | 79 | | 82 | | .| | | 87 | 88 | 90 | 93 | 94 | 96 | 97 | | | | | | |
. I II III IV V VI VII VIII IX X XI XII XIII XIV XV .|E- 3 | - 1 : 3 |+1: 3|-1 : 3|+1: 3|-1 : 3 |+1: 3|-1: 3|+1: 3|-1: 3|+1: 3|-1: 3|+1: 3|-1: 3|+1:3 |- 1: 3 || | 1 | 2 | 4 | 5 | 7 | 8 | 10 | 11 | | | 16 | 17 | 19| | 22 || | | 23 | 25 | 26 | 28 | 29 | 31 | | | 35 | 37 | 38 | | | 43 ||21x3 | | 44 | 46 | | | 50 | 52 | | | 56 | | 59 | 61| | 64 |
| | | 65 | 67 | 68 | | | 73 | 74 | | 77 | | | 82| | 85 || | | 86 | 88 | | 91 | | | 95 | | 98 | | 101| 103| | 106 ||E- 9 | - 1 : 3 |+1: 3 |-1 : 3|+1:3 |-1 : 3 |+1:3 |-1: 3|+1: 3|-1: 3|+1: 3|-1: 3|+1: 3|-1: 3|+1:3 |- 1: 3 || | 1 | 2 | | 5 | 7 | 8 | 10 | | 13 | 14 | | 17 | 19 | 20 | 22 || | | 23 | | 26 | | | | | 34 | 35 | 37 | 38 | 40 | 41 | 43 ||21x3| | 44 | | 47 | 49 | 50 | | | | 56 | | 59 | 61 | | || | | 65 | | | 70 | 71 | 73 | | 76 | | | 80 | 82 | 83 | || | | | | | 91 | 92 | | | | | 100 | 101| 103| 104 | 106 |
The table of tens in which step out prime numbers betrays us sure regularity what it reigns insequence of prime numbers. Not accidentally has written down in this table of ending of
prime numbers in this way 1 - 7 = 6 = 3 - 9. This shows that the regularity what 6 numbersfrom what can step out prime numbers, crosses over on the whole sequence the naturalnumbers, which of propriety what they 6 numbers repeat oneself in six groups. Primes with ending 1 - 7 create XIV ranks, in which their endings repeat oneself what 21 and42 place, and with ending 3 - 9 create XV ranks, in which their endings repeat oneself what21, 42 or 63 places and they in both cases are then divisible numbers by 7, which will befurther great meaning.Prime, even and odd numbers they create "twelve segmental cycles‖.
5 + 7 = 12 = 2 + 4 + 6 = 12 = 3 + 9
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Periodical scheme of prime numbers results with principle the "twelve of segmental cycles‖ in360 numbers which be comprises 30. Multiply thirty by unitary length of period (7) primes inwhat step out 30 · 7 = 210 - receive decimal length of period of prime numbers.
3
1'
3
1''210
3
1
3
1'
p p p p
11 - 3931 i 17 - 4217
+
1/3
- 1
/3
+
1/3
-
1/3
+
1/3
+
1/3 - 1/3
+
1/3
-
1/3
+
1/3
-
1/3
+
1/3
-
1/3
+
1/3
11 31 41 61 71 101 131 151 181 191 211
241 251 271 281 311 331 401 421 431
461 491 521 541 571 601 631 641
661 671 691 701 731 751 761 811 821
881 911 941 971 991 1021 1031 1051 1061
1091 1151 1181 1201 1231
1291 1301 1321 1361 1381 1451 1471 1481
1511 1531 1571 1601 1621
1721 1741 1801 1811 1831 1861 1871 1901
1931 1951 2011 2081 2111
2131 2141 2161 2221 2251 2281 2311
2341 2351 2371 2381 2411 2441 2521 2531
2551 2591 2621 2671 2711 2731 2741
2791 2801 2851 2861
2971 3001 3011 3041 3061 3121
3181 3191 3221 3251 3271 3301 3331 3361 3371
3391 3461 3491 3511 3541 3571 3581
3631 3671 3691 3701 3761
3821 3851 3881 3911 3931
17 37 47 67 97 107 127 137 157 167 197 227
257 277 307 317 337 347 367 397
457 467 487 547 557 577 587 607 617 637 647
677 727 757 787 797 827
877 887 907 937 947 967 977 997
1087 1097 1117 1187 1217 1237 1277
1297 1307 1327 1367 1427 1447 1487
1567 1597 1607 1627 1637 1657 1667 1697
1747 1777 1787 1847 1867 1877 1907
1987 1997 2017 2027 2087
2137 2207 2237 2267 2287 2297
2347 2357 2377 2417 2437 2447 2467 2477
2557 2617 2647 2657 2677 2687 2707
2767 2777 2797 2837 2857 2887 2897 2917 2927 2957
3037 3067 3137 3167
3187 3217 3257 3407 3347
3407 3457 3467 3517 3527 3547 3557
3607 3617 3637 3677 3697 3797
3847 3877 3907 3917 3947 3967 40074027 4057 4127 4157 4177 4217
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13 - 3793 i 19 - 4409
+1/3 -1/3 +1/3 -1/3 +1/3 -1+ 1
: 3- 1
: 3+ 1
: 3- 1
: 3+ 1
: 3- 1
: 3+ 1
: 3- 1
: 3
13 23 43 53 73 83 103 113 163 173 193 223
233 253 263 283 293 313 353 373 383 433
443 463 503 523 563 593 613 643
653 673 683 733 743 773 823 853
863 883 913 953 983 1013 1033 1063
1093 1103 1123 1153 1163 1193 1213 1223
1283 1303 1373 1423 1433 1453 1483
1493 1523 1543 1553 1583 1613 1663 1693
1723 1733 1753 1783 1823 1873
1913 1933 1973 1993 2003 2053 2063 2083 21132143 2203 2213 2243 2273 2293
2333 2383 2393 2423 2473 2503
2543 2593 2633 2663 2683 2693 2713
2753 2803 2833 2843 2903 2953
2963 3023 3083 3163
3253 3313 3323 3343 3373
3413 3433 3463 3533 3583
3593 3613 3623 3643 3673 3733 3793
19 29 59 79 89 109 139 149 179 199 209 229
239 269 349 359 379 389 409 419 439
449 479 499 509 569 599 619659 709 719 739 769 809 829 839
919 929 1009 1019 1039 1049 1069
1109 1129 1229 1249 1259 1279
1289 1319 1399 1429 1439 1459 1489
1499 1549 1559 1579 1609 1619 1669 1699
1709 1759 1789 1879 1889
1949 1979 1999 2029 2039 2069 2089 2099
2129 2179 2239 2269 2309
2339 2389 2399 2459 2539
2549 2579 2609 2659 2689 2699 2719 2729 2749
2789 2819 2879 2909 2939
2969 2999 3019 3049 3079 3089 3109 3119 3169
3209 3229 3259 3299 3319 3329 3359
3389 3469 3499 3529 3539 3559
3659 3709 3719 3739 3769 3779
3889 3919 3929 3989
4019 4049 4079 4099 4129 4139 4159 4219
4229 4259 4289 4339 4349 4409
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Spiral twelve segmental cycles of primes.
2 5 11 17 23 29 41 47 53 59 71
3 7 13 19 31 37 43 61 67 73 79
2 83 89 101 107 113 131 137 149
2 97 103 109 127 1393 151 157 163 191 197
3 167 173 179 181 193 199 211
4 227 233 239 251 257 263 269 281 293
4 223 229 241 271 277 283
5 307 313 331 337 349 367 373 379
5 311 317 347 353 359 383
6 397 409 421 433 439 457 463
6 389 401 419 431 443 449 461
7 487 499 503 509 521
7 467 479 491 523 541
8 563 569 587 593 599 6178 547 571 577 601 607 613 619
9 641 647 653 659 677 683
9 631 643 661 673 691
10 701 719 743 761 773
10 709 727 733 739 751 757 769
11 797 809 821 827 839
11 787 811 823 829 853
12 857 863 881 887 911 929
12 859 877 883 907 919
13 941 947 953 971 977 983
13 937 967 991 997 1009
2 + 3 = 5 + 7 = 12
23571113 83 151
227223
307
311
389
467
547
701
857859
937
171989 157
233
229
313
317397631
709
787
863 941
23
97
163
167239
401
479
94729
31
101 103
173
241
409487563
641
643
719
797
877
953
37
107 109
179
251
331
491
569
571647
727
881
88341 43
113
257
337
421
419499
577653
733
809
811
887967
47
181
263
347
503
659
661
739
971
53
127
191
269
349
353433
431
509
587
743
821
823
977
59
61197
193
271
359
439
593
673
751
827
829
907
983
67131
199
281
277
443
521
523599
601
677
757
911
991
71
73
137 139
283
367
449
607
683
761
839
919997
79
211
293
373
457
613
691
769
149
379
383
463
461541617619
773
853
929
1009
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Spiral scheme of natural numbers.
With spiral arrangement of primes and almost prime result spiral arrangement of all naturalnumbers, what we see in following table.
2 3 45 6 7 8 9 10 11
12 13 14 15 16 17 18
19 20 21 22 23 24 25
26 27 28 29 30 31 32
33 34 35 36 37 38 39
40 41 42 43 44 45 46
47 48 49 50 51 52 53
54 55 56 57 58 59 60
61 62 63 64 65 66 67
68 69 70 71 72 73 74
75 76 77 78 79 80 81
82 83 84 85 86 87 88
89 90 91 92 93 94 95
96 97 98 99 100 101 102
103 104 105 106 107 108 109
110 111 112 113 114 115 116
117 118 119 120 121 122 123
124 125 126 127 128 129 130
131 132 133 134 135 136 137
138 139 140 141 142 143 144
145 146 147 148 149 150 151152 153 154 155 156 157 158
159 160 161 162 163 164 165
166 167 168 169 170 171 172
173 174 175 176 177 178 179
180 181 182 183 184 185 186
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The spiral sequence of natural numbers / primes and almost prime /.
Spiral of primes.
Spiral of almost prime.
a = b mod 17
0 1 2345
67
891011
12
13
14
1516
17 1819
20
21
22
23
242526
27
28
29
30
31
32
3334 35
36
37
38
39
40
41
424344
45
46
47
48
49
5051
52
53
54
55
56
57
58
5960
61
62
63
64
65
66
6768
69
70
71
72
73
74
75
7677
78
79
80
81
82
83
84
8586
87
88
89
90
91
92
9394
95
96
97
98
99
100
101
p' - p= 0 mod 17 = 19 - 2
0 23
5
711
13
1719
23
29
31 37
41
43
47
53
59
61
67
71
73
79
83
89
97
101
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It is true in spirals primes and almost prime congruent according to different modules,however difference between them is common module all natural numbers 23 -17 = mod 6, what show above mentioned graphs. Module 40 = 17 + 23 arranges natural numbers ininfinite spiral.
Congruence according to module 6 shine numbers with all colors of rainbow.
"p" - p = mod 23 = 25 - 2
0 225
35 55
65
7785
91
95
115
119
121
125
133 143
145
155
161
169
175
185
187
203
205 209
215
217
221
235
245
247
253
259
265
275
287
289
295
299
301
305
323
325
329
335
341
343
355
361
365
371
377
385
391
395
403
407
413
415
425
427
437
445
451
455
473
475
481
485
493
497
2 + 3 + 5 + 11 + 19 = mod 40
012 345678910
11121314
151617181920212223242526
272829
303132333435
3637 3839404142 4344
4546
47484950
5152
5354
5556
57585960616263
6465
6667
68
69
70
71
72
73
7475
7677
78 79 80 81 8283
8485
86
87
88
89
90
91
92
93
94
95
9697
9899100101102
103104
105
106
107
108
109
110
111
112
113
114
115
116117
118119 120 121 122
123124
125
126
127
128
129
130
131
132
133
134
135
136
137138
139140141142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157158
159 160 161162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178179180181182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198199
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The cross of numbers is bases on number 6, appointed by primes 2 and 3, which squeeze outhis brand on whole scheme of natural numbers.
a = b mod 6
06121824303642485460
667278
84
90
96
102
108114
120126132138
144
150
156
162
168
174
171319253137434955
6167
73
79
85
91
97
103
109
115121127
133
139
145
151
157
163
169
175
281420263238445056
6268
74
80
86
92
98
104
110
116122128
134
140
146
152
158
164
170
176
3915212733394551
5763
69
75
81
87
93
99
105
111
117123129
135
141
147
153
159
165
171
177410162228344046
5258
64
70
76
82
88
94
100
106
112
118
124130
136
142
148
154
160
166
172
178
51117232935 4147
53
59
65
71
77
83
89
95
101
107
113
119
125131
137
143
149
155
161
167
173
179
3 + 3 = 6 = 2 x 3
123
56
7
1112
13
1718
19
2324
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It comes from structures of cross of natural numbers from congruence of primes and almostprime according to module 2(2)2.
+
0 235
7
1113
17
19
23 25
29
31
3537
41
43
47 49
53
55
5961
65
67
71 73
77
79
8385
89
91
95 97
101
103
107109
113
115
119 121
125
127
131133
137
139
143
+
0
2
35
7
1113
17
19
23 25
29
31
3537
41
43
47 49
53
55
5961
65
67
71 73
77
79
8385
89
91
95 97
101
103
107109
113
115
119 121
125
127
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Sequence of primes and almost prime in five groups about the same endings are possible thethanks their congruence according to module 5.
3 + 2(2)2 = 11 + 2(2)2 = 19 + 2[2(2)2] = 35 + 2(2)2 = 43 + 2[2(2)2] = 59 + 2(2)2 = 67
0
17 25 41 49 65 73 89 97 113 121
2
311
19
35
43
59
67
83
91
107
1155
13
29
37
53
61
77
85
101
109
125
7 23 31 47 55 71 79 95 103 119 127
0%
10%
20%
30%
40%
50%
60%
70%
80%
90%
100%
p ' + p
( p ) p = p
"
Serie8 7 23 31 47 55 71 79 95 103 119 127
Serie7Serie6 5 13 29 37 53 61 77 85 101 109 125
Serie5
Serie4 3 11 19 35 43 59 67 83 91 107 115
Serie3 2
Serie2 17 25 41 49 65 73 89 97 113 121
Serie1 0
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
- 9 - 3 - 7 - 1 - 5
0
23
5
1317
19
23
25
29
35
3743
47
49
53
55
59
65
7377
79
83
85
89 91
95
97
7
1131
41
61
67
71
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So looks clocks of primes measuring it in rhythm 2/4 (6).
According to this rhythm of primes flow away us days in four times year.
(p-1) + (p+1) / 2 = p2 3
7
13
19
31
37
43
61677379
91
97
103
109
127
5
11
17
23
29
41
47
53
5971
83
89
101
107
113
131139137
2 + 3 = 5 x 73 = 3652 3
7
13
19
31
3743
61
67
735
11
17
23
29
41
47
53
59
71 46
89
10
12
14
15
16
18
20
21
22
24
25
26
27
28
30
323334353638394042
4445
46
48
49
50
51
52
54
55
56
57
58
60
62
63
64
65
66
6869
7072
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Binary and Ternary Goldbach's Conjecture, equation Pythagoras and great Fermat'stheorem.
Creative process in mathematics begins from conjecture. Mathematical conjecture really thenit becomes theorem, when we have on his truth irrefutable proof.
Theorem: Even numbers are "larger about 1‖ from one's odd, prime or almost primepredecessor, and so they are duplication different natural number.
Proof: (2n – 1) 1 \ „p‖ 1 = 2n
p 1 /
(2n – 1) + 1 = 2n 1 + (2n – 1) = 2n = 3p - p = p + p‘7 – 5 mod 2 5 – 3 p‘- p = n/2 p‘ + p = 2n 6 – 4 mod 2 10 – 8
3(2) – 2 = 4 3(3) – 3 = 6 3 + 5 = 8 3(5) – 5 = 10 5 + 7 = 12 3(7) – 7 = 14 11 + 5 = 16
This theorem proves the just truth Binary Goldbach's Conjecture, that every even largernumber than 2 is the sum two primes. Both even numbers how and primes congruent to meaccording to modules 2, that is to say differences between them divisible they are by 2. Fromhere simple conclusion, if differences this and sum two primes divisible they are by 2, as evennumbers.
2 + 3 = 5 x 73 = 365
237
13
19
31
3743
61
67
73
5 11
17
23
29
41
47
53
59
71
4 6 891012141516
18202122
24252627
28
3032
3334
353638394042
444546
484950
5152
54
55
56
57
58
60
62
63
64
65
66
6869
70
72
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It will permit then us on formulating polynomial describing the solution of Binary Goldbach‘sConjecture.
P(2n) = (p + p‘) (2+2) (3 + 3) (3 + 5) (5 + 5) (5 + 7) (7 + 7) (5 + 11) (5 + 13) (7 + 13) (5 + 17)
2p = 2n = p + p', 2 + 2 = 4, 3 + 3 = 6, 3 + 5 = 8, 5 + 5 = 10, 5 + 7 = 12, 7 + 7 = 14, 5 + 11 = 16,
7 + 11 = 18, 7 + 13 = 20, 5 + 17 = 22, 7 + 17 = 24, 13 + 13 = 26, 11 + 17 = 28, 13 + 17 = 30, 13 +
19 = 32, 17 + 17 = 34
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 1819 20 21 22 23 24 25 26 27 28 29
24
68 10 12
1416
1820
2224
2628
3032
3436
3840
4244
4648
5052
5456
58
23
5
7
11
13
17
19
23
29
0
20
40
60
80
100
120
140
Serie3 2 3 5 7 11 13 17 19 23 29
Serie2 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58
Serie1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29
p + p' = 2n 2 + 2 = 4 3 + 3 = 6 3 + 5 = 8 17 + 19 = 36 37 + 41 =78 59 + 61 = 120 67 +
71 = 138 101 + 103 = 204 109 + 113 = 222 163 + 167 = 330 193 + 197 = 390 227 + 229 = 456
197; 193
113; 109
71; 67
349; 359
307; 317
167; 163
41; 37
269; 257
227; 229
101; 103
59; 61
17; 19
3; 5
311; 313
-100
-50
0
50
100
150
200
250
300
350
400
450
-5 0 5 10 15 20 25 30 35
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Theorem: Every odd number larger than 5, is sum three primes, because difference amongodd and even number is always 3.
Proof: 2n = p + p (2n – 1) – (p + p) = p 2n – 1 = p + p + p‖
4 = 2 + 2 7-
(2 + 2) = 3 7 = 2 + 2 + 3
2n = p + p' 4 = 2 + 2 6 = 3 + 3 8 = 3 + 5 10 = 5 + 5 12 = 5 + 7 14 = 7 + 7 16 = 5 + 11
2 3 35 5
75
7 7
11 1113
1113 13
17 1719
2
35
57
711
1113
1113
1317
1719
1719
19
4
6
8
10
12
14
16
18
20
22
24
26
28
3032
34
36
38
0
10
20
30
40
50
60
70
80
Serie3 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38
Serie2 2 3 5 5 7 7 11 11 13 11 13 13 17 17 19 17 19 19
Serie1 2 3 3 5 5 7 5 7 7 11 11 13 11 13 13 17 17 19
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
p' + (p + p) = 2n - 1 2n = p + p (2n - 1) - (p + p) = p 3 + (2 + 2) = 7 4 = 2 + 2 7 - (2 + 2) = 3
3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 34
68
1012
14 16 1820
2224
2628
3032
3436
3840
4244
4648
79 11
1315
1719
2123
2527
2931
3335
3739
4143
4547
4951
0
20
40
60
80
100
120
Serie3 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51
Serie2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48
Serie1 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23
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Theorem: Difference among two successive square numbers is always odd number.a(a) - b(b) = (2n - 1) 25 - 16 = 2(5) - 1
Next numbers from infinite file of odd numbers added to square minuend create alwayssquare subtrahend.
p + p' = p" 2 + 3 = 5 p + p = 2n 5 + 3 = 8 p'' + p' + p = 2n - 1 7 + 5 + 3 = 15
12
34
56
78
910
1112
1314
1516
1718
1920
2122
2324
2526
2728
29
23
23
23
23
23
23
23
23
23
23
23
23
23
23
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 10
5
10
15
20
25
30
35
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29
a(a) - b(b) = (a - b)(a + b) 5(5) - 3(3) = (5 - 3)(5 + 3) 25 - 9 = 2(8)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 253 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49
1
4964
100121
144
169
196225
256
289
324
361
400
441
484
529
576
625
81
36251694
0
100
200
300
400
500
600
700
1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49
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z 222 x y
That is to say, that equation x nnn z y from great theorem Fermat‘s, only near n = 2 hassolution, because when add only odd number to square, we receive next square number.
1 and 3 are not square numbers, and above mentioned theorem despite this and on themchecks because primes are the multiplicity of number one and only me also. p = 1(p)
In other words, equation for n > 2 in infinite file of natural numbers does not possess nosolution, because only the square numbers create the ternary Pythagoras.
z 222 y x 25 – 9 = 16 ( y 2mod0)
2
16 : 2 = 8 Rest 0
x x‘ 22 x x 2
2 2 ! = 24
2 2 ! + 1 2 = 5 2 3+5+7+9 + 1 = 25
First ternary Pythagoras comes into being, when the sum of differences among successive
square numbers reaches value of faculty 2 2 !.It number 2 modulates so formation so square numbers how and ternary Pythagoras, that is tosay that differences among odd numbers and squares in ternary Pythagoras they are divisibleby 2, therefore squares how and ternary Pythagoras are product 2 factors.
3(3) + 4(4) = 5(5) [2(5) – 1] + 2(8) = 2(13) – 1 3[2(2) – 1] + 2(2)2(2) = 5[2(3) – 1]
1 + 3 = 4 + 5 = 9 + 7 = 16 + 9 = 25 + 11 = 36 + 13 = 49 + 15 = 64 + 17 = 81 + 19 = 100 + 21.. \ 2 /\ 2 /\ 2 /\ 2 /\ 2 /\ 2 /\ 2 /\ 2 /\ 2 /\ 2 /
25 – 9 = 2(8) 169 – 25 = 2(72) 289 – 225 = 2(32) 625 – 49 = 2(288) 841 – 441 = 2(200)
It because 2 is the solid value of differences among two the successive odd numbers becomesshe the modules of differences with them square numbers in ternary Pythagoras, where withdifference among horizontal the and vertical length the side of triangle square creates, beingsimultaneously the proof on truth of equation the Pythagoras and the Fermat's conjecture.
z 22
24 z x
xy
z 2 = 2xy + x 2 - 2xy + y 2 z 2 = x 2 + y 2
1 + 1 1
+ 3
2 4+ 5
3 9
+ 7
4 16
+ 9
5 = 24 25
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If the product of hypotenuse to be equal to sum two products of legs and product of theirdifference, that is to say, that the square of hypotenuse is the sum of squares of legs.
Theorem: Difference among every odd square in triples Pythagoras even square is, which
congruent according to modules y 2mod2
Proof: z 222 y x 25 – 9 = 16 ( y 2mod0)
2
16 : 2 = 8 Rest 0
[2(5) – 1] + 2(8) = 2(13) – 1The congruence the even square y 2mod0
2 signifies, that in quadrate of hypotenuse(z 2 ) 4triangles of the same hypotenuse be comprise (z), replenished about quadrate came into beingwith differences between horizontal and vertical length the sides 4 triangles. e.g.
y 2 + x 2 = z 2 y – x = n 42
yx+ n 2 = z 2
e.g. 4 2 + 3 2 = 5 2 4 – 3 = 1 4(6) + 1 2 = 25
12 2 + 5 2 = 13 2 12 – 5 = 7 4(30) + 7 2 = 169
8 2 + 15 2 = 17 2 15 – 8 = 7 4(60) + 7 2 = 289
24 2 + 7 2 = 25 2 24 – 7 = 17 4(84) + 17 2 = 625
20 2 + 21 2 = 29 2 21 – 20 = 1 4(210) + 1 2 = 84112 2 + 35 2 = 37 2 35 – 12 = 23 4(210) + 23 2 = 1369
40 2 + 9 2 = 41 2 40 – 9 = 31 4(180) + 31 2 = 1681
28 2 + 45 2 = 53 2 45 – 28 = 17 4(630) + 17 2 = 2809
60 2 + 11 2 = 61 2 60 – 11 = 49 4(330) + 49 2 = 3721
56 2 + 33 2 = 65 2 56 – 33 = 23 4(928) + 23 2 = 4225
84 2 + 13 2 = 85 2 84 – 13 = 71 4(546) + 71 2 = 7225
72 2 + 65 2 = 97 2 72 – 65 = 7 4(2340) + 7 2 = 9409
144 2 + 17 2 = 145 2 144 – 17 = 127 4(1224) + 127 2 =21025
180
2
+ 19
2
= 181
2
180 – 19 = 161 4(1710) + 161
2
= 37261
4xy/2+(x - y)^= z^ z^- 4xy/2= (x - y)^2xy+(x-y)(x-y)=z(z) 2xy+x(x)-2xy+y(y)=z(z)
x(x) = x^ y(y) = y^ z(z) = z^ ^ = 2
Y
X
Z
Y
X
+ =
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Ternary Pythagoras this square equation, and how there are all quadratic functions as graph of function is a parabola. The running by vertex axis of symmetry be shifted in them about 2 in
direction on line - x, and about (y – x) 2 in direction on line - y.
Number 2 in every semi stabile elliptic curve over rational numbers modular is.
(1+3+5+7+9)25 + (11+13+15+17+19+21+23+25)144 = 169 = 49(1+3+5+7+9+11+13) + \2/\2/\2/\2/ \2/ \2/ \2/ \2/ \2/ \2/ \2/ 120(15 +17+19+21+23+25)
y = (y - x)^ = 1 x^ + y^ = z^ 9 + 16 = 25 (1 + 3 + 5) + (7 + 9) = 25 - 1 = 24 - 3 = 21 - 5 = 16 - 7 = 9 -
9 = 0
25
1
25
0
5
10
15
20
25
30
0 0,5 1 1,5 2 2,5 3 3,5
y = (y - x)^ =(12 - 5)^ = 49 5^ +12^ =13^ 25+144=169 (1+3+5+7+9) + (11+13+15+17+19+21+23+25)
=169 -1=168 - 3 =165 - 5=160 -7=153 - 9 =144 -11=133 -13=120 -15 = 105 -17 = 88 -19 = 69 -21= 48 -
23=25 - 25 = 0
169
49
169
0
20
40
60
80
100
120
140
160
180
0 0,5 1 1,5 2 2,5 3 3,5
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Theorem: The square of hypotenuse is equal the sum of squares of legs,when it is sum of such quantity of odd numbers how degree of square hypotenuse.
Proof: x(2n – 1) = x 2 y(2n ) = y 2 z(2n – 1) = z 2 x + y = z y = z – x x +(z – x) = z
x2
+ y2
= z2
z2
= z(2n – 1) z2
- z(2n – 1) = 0
x(2n - 1) + y(2n - 1) = z^ - z(2n -1) = 0 (1 + 3 + 5) + (7 + 9) = 25 - 1 = 24 - 3 = 21 - 5 = 16 - 7 = 9 - 9 = 0
3^ + 4^ = 5^ x^ - [2xy + (y-x)^] + y^ = 0 9 - (24 + 1) + 16 = 0
9
7
5
3
1
3
5
7
9
-11
0
9
16
21
2425
24
21
16
9
0
-11
-15
-10
-5
0
5
10
15
20
25
30
Serie1
Serie2 9 7 5 3 1 3 5 7 9
Serie3 -11 0 9 16 21 24 25 24 21 16 9 0 -11
1 2 3 4 5 6 7 8 9 10 11 12 13
2xy +(y - x)^ = z^ 2(3)4+(4 -3)^ = 25 2(5)12+(12 -5)^ =169 2(15)8+(15 -8)^ =289 2(7)24+(24 -7)^ =625
2(21)20 + (21 - 20)^ =841
25 25169 169
625 625
841 841
2809 2809
149
289289 289
0
500
1000
1500
2000
2500
3000
3500
4000
4500
5000
1 2 3
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3[2(2) – 1] + 2(2)2(2) = 5[2(3) – 1]
1 + 3 + 5 = 9 = 3(3) 1 + 3 + 5 + 7 = 16 = 4(4) 1 + 3 + 5 + 7 + 9 = 25 = 5(5)
(1 + 3 + 5) + (1 + 3 + 5 + 7) = (1 + 3 + 5 + 7 +9) Σ(2n - 1) = n(a + z)/2 = n(n)
3(1 +5)/2 = 3(3) 4(1 + 7)/2 = 4(4) 5(1 + 9)/2 = 5(5) 3(3) + 4(4) = 5(5) 9 + 16 = 25
1 + 3 +
1 + 3 5
1 + + +
3 + + 5 + = 7 +
5 = 9 7 = 16 9 = 25
Because the square of hypotenuse is sum of such quantity of successive odd numbers, as
degree of square of hypotenuse, equation Pythagoras was can write as fraction:
2
2
2
2
2
2
z
z
z
y
z
x
The common square denominator confirmed that the square of hypotenuse is the sum of thesquares of legs.
3^+4 =̂ 5(2n - 1) =5^= (1+3+5)+(7+9) = 25 13(2n - 1) = 13^=(1+3+5+7+9)+(11+13+15+19+21+23+25)
=169 15^+8 =̂17(2n - 1) = 17 =̂(1+3+5+7+9+11+13+15+17+19+21+23+25+27+29)+(31+33)=289
25(2n - 1) = 25^= 625 29(2n - 1) = 29^=841
1 3 5 7 9
251 3 5 7 9
11
13 15 17 19 21 23 25
169
1 3 5 7 9
11
13 15 17 19 21 23 25
27
29 31 33
2891 3 5 7 911
13 15 17 19 21 23 25
27
29 31 33
35
37 39 41 43 45 47 49
625
1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57
841
2n - 1 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53 55 57 841
2n - 1 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 625
2n - 1 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 289
2n - 1 1 3 5 7 9 11 13 15 17 19 21 23 25 169
2n - 1 1 3 5 7 9 25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
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Theorem: If the product of difference and the sum two numbers square number is, then she issquare difference two square numbers.
Proof: (z – y)(z + y) = xn
= (znn
y ) n = 2
-10
0
9
16
21
24 25 24
21
16
9
0
-10
0 2 4 6 8 10 12 14
(1 + 3 + 5) + (1 + 3 + 5 + 7) = (1 + 3 + 5 + 7 +9) Σ(2n - 1)= n(a + z)/2 = n(n)
3(1 +5)/2 = 3(3) 4(1 + 7)/2 = 4(4) 3(3) + 4(4) = 5(5) 9 + 16 = 25
0 0 00 0 00 0 00 0 0625
0625
289
1681 1681
961
3721 3721
2401
7225 7225
5041
1 2 3
2xy +(y-x)^= z^ 2(7)24+(24-7)^=25^=625 2(9)40+(40-9)^=41^=1681
2(11)60+(60-11)^=61^=3721 2(13)84+(84-13)^=85^=7225
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z2
(z – y)(z + y) = x2
x2
25 (5 – 4)(4 + 5) = 9 9
169 (13 – 12)(12 + 13) = 25
289 (17 – 15)(17 + 15) = 64
625 (25 – 24)(25 + 24) = 49
841 (29 – 21)(29 + 21) = 4001369 (37 – 35)(37 + 35) = 144
1681 (41 – 40)(41 + 40) = 81
2809 (53 – 45)(53 + 45) = 784
3721 (61 – 60)(61 + 60) = 121
4225 (65 – 33)(65 + 33) = 3136
4225 (65 – 63)(65 + 63) = 256
5329 (73 – 55)(73 + 55) = 2304
7225 (85 – 77)(85 + 77) = 1296
7225 (85 – 84)(85 + 84) = 169
7921 (89 – 39)(89 + 39) = 6400
9409 (97 – 65)(97 + 65) = 518410201 (101 –99)(101 +99)= 400
11881 (109-91)(109+91) = 3600
12769 (113-112)(113+112) = 225
15625 (125-117)(125+117) = 1936
18769 (137-105)(137+105) = 7744
21025 (145-143)(145+143) = 576
21025 (145-144)(145+144) = 289
22201 (149-51)(149+51) = 19600
Also prime numbers except 2 can introduce as product of difference and sum two naturalnumbers and they are then prime difference two square numbers.
3 = (2 – 1)(2 + 1) 5 = (3 – 2)(3 + 2) 7 = (4 – 3)(4 + 3) 11 = (6 – 5)(6 + 5)
13 = (7 – 6)(7 + 6) 17 = (9 – 8)(9 + 8) 19 = (10 – 9)(19 + 9) 23 = (12 – 11)(12 + 11).
p = ))((]2
1[]
2
1[
22 baba p
p p
3 = )12)(12(]2
133[]
2
13[
22
5 =)23)(23(]
2
155[]
2
15[
22
7 =)34)(34(]
2
177[]
2
17[
22
This is Great Fermat's theorem for all values of n proved, because he is for all prime values of n valid.Looking closer at the following graph, you will see that half of the following sums of twoprimes on a straight line parallel to the y - axis with real part ½ y lie. This means that thelinear Diophantine equation ax + by - c = 0, with given integer pairs not have common divisorCoefficient a, b, c, always in prime x, y is solvable.
1(2) + 1(3) – 5 = 0 1(3) + 1(7) – 10 = 0 1(5) + 1(13) – 18 = 0 1(11) + 1(19) – 30 = 0
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Natural numbers, divisibility and primes.
Central notion within of natural numbers concerns divisibility, and more far order of primes -natural number larger than 1, which has not natural divisor, that is to say, no different divisorexcept 1 or me alone. Sequence of primes has begun since 2, 3, 5, 7, 11, 13, 17, 19,
23,…Already Euclid proved before over 2 000 years, that this sequence does not end, and sothere is no the largest prime. Beyond 2 all primes are odd with characteristic endings - 1 - 7 -3 - 9. From second side is in force the main theorem of arithmetic: every natural number willgive oneself unambiguously to introduce as product of primes. Primes gain by this onmeaning for mathematics, as contribution to construction of all different numbers. The everynumber, which is not prime, will give oneself with these indivisible factors to to put together.Prime numbers 2 and 3 are components of all natural numbers really. Why, for example, can‘tall numbers be built simply by multiplying and adding together different combinations of theprimes 2 and 3. e.g. 4 = 2(2),5 = 2 + 3, 6 = 3 + 3, 7 = 2(2) + 3, 8 = 2(2)2, 9 = 3(3), 10 = 2(2) + 3 + 3, 11 = 2(2)2 + 3,12 = 3(3)+ 3, 13 = 2(2)+ 3(3), 14 = 2(2)2 + 3 + 3, 15 = 3(3)+ 3 + 3, 16 = 2(2)2(2),
17 =2(2)2+3(3), 18 =2(2)2(2)+2, 19 =2(2)2(2)+3, 20 =2(2)2(2)+2(2), 21 =2(2)2(2)+2+322 = 2(2)2(2)+ 3 + 3, 23 = 2(2)2(2)+ 2(2)+ 3, 24 = 2(2)2(2)+ 2(2)2, 25 = 2(2)2(2)+ 3(3),26 =2(2)2(2)+2(2)2+2, 27 =3(3)3, 28 =2(2)2(2)+3+3(3), 29 =2+3(3)3, 30 =3+3(3)3.All prime be built according to simple formula: p = n(2)+3, 5 =2+3, 7 =2(2)+3 11 =4(2)+3,13 = 5(2) + 3, 17 = 7(2) + 3, 19 = 8(2) + 3, 23 = 10(2) + 3, 29 = 13(2) + 3. Formulathis permits us to divide primes on two classes: they class of basic primes (2, 3, 5, 7), whichalone for me are the building material and these, which are already the multiplicity of number7. e.g.11 = 7+(4) 13 = 7+(6) 17 = 2(7)+(3) 19 = 2(7)+(5) 23 = 3(7) + (2) 29 = 4(7) + (1)And so we write new formula: p = n(7) + The rest (1,2,3,4,5,6)
0 0 02
0
32,5
3
75
5
13
9
11
0
19
15
2y+3y=5y/2=2+0,5=3-0,5 3y+7y=10y/2=3+2=7-2
5y+13y=18y/2=5+4=13-4 11y+19y=30y/2=11+4=19-4
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It was can sequence of primes and write so: 2, 3, 5,-2- 7,-4- 11,-2- 13,-4- 17,-2- 19,-4- 23. Inspaces among numbers notice hidden formula: - 2 - 4 - 2 - 4. These two last formulae, theywill play further decisive part.Are there formulas that produce some of the prime? Here you are! p = n(2) + 3
2 = 1(2) + 0 3 = 0(2) + 3 5 = 1(2) + 3 7 = 2(2) + 3 11 = 4(2) + 3 13 = 5(2) + 317 = 7(2) + 3 19 = 8(2) + 3 23 = 10(2) + 3 13(2) + 3 = 29 14(2) + 3 =31 233 = 115(2) + 3251 = 124(2) + 3.
Irrefutable proof.
Mathematicians knew, however, that proving the Riemann Hypothesis would be of far greatersignificance for the future of mathematics than knowing that Fermat‘s equation has nosolutions when n is bigger than 2. The Riemann Hypothesis seeks to understand the mostfundamental objects in mathematics – prime numbers.
The primes are those indivisible numbers that cannot be written as two smaller numbersmultiplied together. The primes are the jewels studded throughout the vast expanse of theinfinite universe of numbers that mathematicians have explored down the centuries.
Their importance to mathematics comes from their power to build all other numbers. Everynumber that is not a prime can be constructed by multiplying together these prime buildingblocks /2 and 3/. Mastering these building blocks offers the mathematician the hope of discovering new ways of charting a course through the vast complexities of the mathematicalworld.
Yet despite their apparent simplicity and principal character, prime numbers remain the most
mysterious objects studied by mathematicians. They question about distribution of primes
1 2 35
711
13
1719
2325
2931
3537
4143
4749
53
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 22 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 324
8 10
1416
2022
2628
3234
3840
4446
50
1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53
p = 3 + n(2)
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belonged to the most difficult. They were the long time then the question of plain theoreticalnature, however today found primes in different realms the use.
Suddenly the economic interest appears also the question, or proof the Riemann's hypothesiscans you something say about distribution of primes in world of numbers. If centuries of
searching had failed to unearth some unknowing formula which would generate the list of prime numbers, perhaps it was time to adopt a different strategy. Look through a list of primenumbers, and you‘ll find that it‘s impossible to predict when the next prime will appear. Thelist seems chaotic, random, and offers no clues as to how to determine the next number. Canyou find a formula that generates the numbers in this list, some unknowing rule that will tellyou what the 10 000 000th prime number is?
Not the question about quantity of primes in given interval of numbers, but the observation of spaces between two primes, she directed me on sure regularity from what they appear. 2, 3,5,-2- 7,-4- 11,-2- 13,-4- 17,-2- 19,-4- 23 and so 2, 4, 2, 4, then the smallest space is amongtwo primes and the decisive structure, recognizable in whole does not end sequence of primes.
It after 23 number first 29 comes however in space 6 (23,- 2 -25,-4-29), because placebetween them is for first product of primes, number almost prime 25 = 5(5). Since then allalmost prime numbers, as product of primes will take free place in sequence of primes,keeping spaces - 2 - 4 - 2 - 4. Generations have sat listening to the rhythm of the primenumber drum as it beats out its sequence of numbers: two beats, followed by three beats, five,seven, eleven. As the beat goes on, it becomes easy to believe that random white noise,without any inner logic, is responsible. At the centre of mathematics, the pursuit of order,mathematicians could only hear the sound of chaos.
I do realize, that prime and almost prime numbers appear in interval two and fourth. If itwalks about finding formulae and order, then primes are not more unequalled challenge.Knowing in what space sequent prime or relatively prime will appear, we can easily wholetheir list take down. And when we to this have yet the hand, as to qualify in sequence thesequent number, or it is prime or almost prime numbers, then and list of primes does notappear us as chaotic and accidental. The List of primes is, the heartbeat of mathematics, but apulse wired regular in rhythm by multiplicity of seven in two – by – four steps.Fractions are the numbers whose decimal expansions have a repeating pattern. For example
1/7 = 0,142 857 142 857
At bases of distribution of primes in sequence of numbers, lies decomposition of their
products on prime factors. According to Fermat small theorem numbers to power ( p - 1)minus one, they are divisible without the rest by prime. e.g. 106 - 1 = 999 999/7 = ↓- 142 857
857 142Proof:
−1−1
≅ 0 if a ≠ p p ≥ 3 a ≥ 2 26 = 64 – 1 = 63/7 36 = 729 – 1 = 728/7
Similarly by fractions: 1/7 = 0,142 857 142 857 1 …2/7 = 0,2857 142 857 14 …3/7 = 0,42857 142 857 1 …
4/7 = 0,57 142857 142857 1..5/7 = 0,7 142587 142587 1..
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6/7 = 0,857 142587 142587..8/7 = 1,142 857 142857 9/7 = 1,2857 142 857 14..
10/7 = 1,42857 142 857 …11/7 = 1,57 142 857 1428 …
12/7 = 1,7 142 857 14285 …13/7 = 1,857 142 857 142 …
where the quotient in decimal expansion from some place after comma begins repeating ininfinity six - digits numbers since 1, and finishing on 7. In practice this marks that every thethe six - digit combination of numbers e.g. (x x x x x x)/ 7, (x y x y x y)/ 7, (y x y x y x)/ 7,(xyz xyz)/ 7, (zxy zxy)/ 7, (yzx yzx)/ 7, (zyx zyx)/ 7, (yxz yxz)/ 7, (xzy xzy)/ 7, and theirmultiplicities divide without the rest by
7. 111 111 111 111 111 111 / 7 = 15 873 015 873 015 873
0 1 2 3 4 5 6 7 1 2 3 4 5 6 7 8
2 3 4 5 6 7 8 9
3 4 5 6 7 8 9 0
4 5 6 7 8 9 0 1
5 6 7 8 9 0 1 2
6 7 8 9 0 1 2 3
7 8 9 0 1 2 3 4
This gets from here, that all numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, 9,.. they are congruent to meaccording to module 7, as this shows following radar graph
0 1
2
3
4
5
6
7
1 2
3
4
5
6
7
8
23
4
5
6
7
8
9
34
5
6
7
8
9
0
4
5
6
7
8
9
0
1
5
6
7
8
9
0
1
2
6
7
8
9
01
2
3
7
8
9
0
12
3
4
a - 7/7
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If difference among given number a, and prime is divisible by prime, then given number be
complex. pV p
pa
e.g. 10
17
17187
Only difference among two primes divisible it is by 7, because p - (2,3,5,11,13,29) = n(7), and
all primes be distributed according to multiplicity of number 7 (23 - 2)/ 7 = 3 (17 - 3)/ 7 = 2(19 - 5)/ 7 = 2 (53 - 11)/ 7 = 6 (41 - 13)/ 7 = 4 (43 - 29)/ 7 = 2
Primes and almost prime follow after me in rhythm on two fourth.
1. 3. + 2 = 5.- 2 – 7 – 4 – 11 – 2 – 13 – 4 – 17- 2 – 19 – 4 – 23 - 2 – 25 – 4 – 29 – 2 – 31 -
Theorem: Odd numbers, the cannot be written as product of two smaller numbers a and b witha, b> 1, are prim and congruent to me modulo 7.
Proof: p p`mod p 11 53 mod 7 when 11 =1(7)+ 4 and 53 = 7(7)+ 4 then p and p`congruent according to mod p, and difference p`- p is multiplicity p. 53 - 11 = 42/7 = 6
This proof gives mathematics to instruction very quick procedure on qualification of primesabout any quantity of places. p - (2,3,5,11,13,29) = n7
p = n(7) + R(1,2,3,4,5,6)
0 1 2 3 4 5 6
7 11 13
17 19
23 25
29 31
35 37 41
43 47
49 53 55
59 61
65 67
71 73
77 79 83
85 89
91 95 97
101 103 107 109
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113 115
119 121 125
127 131
133 137 139
143 145
149 151
155 157
161 163 167
169 173
175 176 177 178 179 180 181
182 183 184 185 186 187 188
189 190 191 192 193 194 195
196 197 198 199 200 201 202
203 204 205 206 207 208 209
210 211 212 213 214 215 216 217 218 219 220 221 222 223
224 225 226 227 228 229 230
231 232 233 234 235 236 237
238 239 240 241 242 243 244
245 246 247 248 249 250 251
252 253 254 255 256 257 258
259 260 261 262 263 264 265
266 267 268 269 270 271 272
273 274 275 276 277 278 279 280 281 282 283 284 285 286
287 288 289 290 291 292 293
294 295 296 297 298 299 300
301 302 303 304 305 306 307
308 309 310 311 312 313 314
315 316 317 318 319 320 321
322 323 324 325 326 327 328
329 330 331 332 333 334 335
336 337 338 339 340 341 342
343 344 345 346 347 348 349
350 351 352 353 354 355 356
357 358 359 360 361 362 363
364 365 366 367 368 369 370
371 372 373 374 375 376 377
378 379 380 381 382 383 384
385 386 387 388 389 390 391
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Discovery of rhythm beating the heart of mathematics storm the safety of system the RSA,any business selling prime numbers could realistically in support about this proof peddle theirwares under the banner ―satisfaction guaranteed or your money back‖, without too much fear of going bust. And so it turns out, that it 64 numerical factor with 129 numerical code is not
number prime, because divisible it is by 7.
3 490 529 510 847 650 949 147 844 619 903 898 133 417 764 638 493 387 843 990820577:7= 498 647 072 978 235 849 878 263 517 129 128 304 773 966 376 927 626 834 855 831 511
But second 65 numerical factor is prime (32769132993 266 709 549 961 988 190 834 461413177 642 967 992 942 539 798 288 533 – 5):7 = 4 681 304 713 323 815 649994569741547780201 882 520 423 998 991 791 399 755 504
Prime numbers are numbers that are divisible only by one and themselves. They are the atomsof arithmetic, for any number is either a prime or a product of primes. The first few primes are
2, 3, 5, 7, 11, 13, 17, 19, but despite their simple definition the prime numbers appear to bescattered randomly amid the integers.
There is simple way to tell if a number is prime – than they cannot be written as product of two smaller numbers a and b, with a, b>1, and that is the basis for most modern encryptionschemes.
Solving the Riemann Hypothesis could lead to new encryption schemes and possibly providetools that would make existing schemes, which depend on the properties of prime numbers,more vulnerable.
0
10
20
30
40
50
60
70
8090
100
1 2 3 4 5 6 7 8 9 10 11 12 13 14
07
0 0 0
35
0
49
0 0 0
77
0
91
0 0 0 0
29
0
43
0 00
71
0
85
0
20 0
23
0
37
0 0 0
65
0
79
0 0
30
17
0
31
0 0 0
59
0
73
0 0 0
0
11
0
25
0 0 0
53
0
67
0 0 0
95
0
19
0 0 0
47
0
61
0 0 0
89
0
0
13
0 0 0
41
0
55
0 0 0
83
0
97
p'- p = n(7) 19 - 5 = 2(7) 47 - 19 = 4(7) 61 - 47 = 2(7) 89 - 61 = 4(7)
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Primes do not possess except 1 and only number no factors, but number almost prime arealmost so good, because they have only two factors prime. 23 is prime, but 25 (5 · 5) it isalmost prime. So alone numbers 35(5·7), 49(7·7), 55(5·11), 65(5·13), 77(7· 11), 85(5 · 17)
01
2
34
5
6
7
11
13
17
19 23
25
29
31
35
37
41 43
47
49
53
55
59
61 65
67
71
73
77
79
83 85
89
91
95
97
p, p', p" = 2n + p' 2, 3, 5 = 2 + 3, 7 = 4 + 3, p'" = n(7) + Rest(1,2,3,4,5,6), 11 = 7 + 4, 13 = 7 +
6, 17 = 2(7) + 3, 19 = 2(7) + 5, 23 = 3(7) + 2, 29 = 4(7) + 1, 2, 3, 5, -2- 7, -4- 11, -2- 13, -4- 17, -2- 19,
-4- 23,
2337
79
107
149163
1731
5973
101
157
11
5367
109
137
151
179
1947
61
89
103
131
173
13
41
83
97
139
167
181
0
50
100
150
200
250
300
350
400
13 41 83 97 139 167 181
5 19 47 61 89 103 131 173
11 53 67 109 137 151 179
3 17 31 59 73 101 157
2 23 37 79 107 149 163
7 29 43 71 113 127
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Number almost prime built with prime numbers larger than three, they develop how splendidfan in infinity.
Sequences of numbers almost prime.
(10)(20)
25
35 (14)(28)49
"p" = n(3) + (1,2) 25 = 8(3) + 1 35 = 11(3) + 1 49 =16(3) + 1 55 = 18(3) +1 65 = 21(3) + 2
77 = 25(3) + 2 85 = 28(3) + 1 91 = 30(3) + 1
275
215
299
235
319
91
133
175
217
259
301
155
323
221
305
115
325
265
119
161
203
245
287
329
143
185
121
205247
289
145
187
125
209251
293
335
9585
169253
295
5,7,11,13,17,19,23,29,31,37,41,43,(p) = "p"
5 711
13
1719
23252931
35
37
4143 25 35
55
65
85
95
115125145
155
175
185
205
215
4977
91
119
133
161175203
217
255
259
287
301
121
143
187
209
253
275319
341
385
407
451
473
169
221
247
299
325377
403
455
481
559
289
323
391
425493
361
437
475
0
100
200
300
400
500
600
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58
55
65
77
85
95 91
115 119 (22)(44)
125 121
133
145 143
155 (52)(26)
161 169
175
185 187
205 203 209
215 217221
235
245 247
259 253
265
275 (34)(68)
287 289
295 299
305 301
319
325 329 323
335
343 341
355 (76)(38)
365 361
371 377
385
395 391
407 403
415 413
425 427
437
445
455 451
469
475 473
485 481
497 493
505
515 511 517 (46)(92)
527 529
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In interval what 30 numbers (10-40) on three numbers with ending 5 (15, 25, 35), two of themare almost prime. Primes and almost prime follow after me in interval what 2(p) and 4(p).
25 – 2(5) – 35 – 4(5) – 55 49 – 4(7) – 77 – 2(7) – 91 121 – 2(11) - 143 – 4(11) – 187
17 - 4(5) – 37 – 2(5) – 47 – 4(5) – 67- 2(5) – 77 – 4(5) – 97 – 2(5) – 107 – 4(5) – 127 – 2(5)..
Triangle of almost prime.
5 7 11 13 17 19 23 25 29 31 35 37 41 43 47 49 53 55 59
5 25 35 55 65 85 95 115 125 145 155 175 185 205 215 235 245 265 275 295
7 49 77 91 119 133 161 175 203 217 245 259 287 301 329 343 371 385 413
11 -22 -44 121 143 187 209 253 275 319 341 385 407 451 473 517 539 583 605 649
13 -52 -26 169 221 247 299 325 377 403 455 481 533 559 611 637 689 715 767
17 -34 -68 289 323 391 425 493 527 595 629 697 731 799 833 901 935 1003
19 -76 -38 361 437 475 551 589 665 703 779 817 893 931 1007 1045 1121
23 -46 -92 529 575 667 713 805 851 943 989 1081 1127 1219 1265 1357
25 100 -50 625 725 775 875 925 1025 1075 1175 1225 1325 1375 1475
29 -58 116 841 899 1015 1073 1189 1247 2363 1421 1537 1595 1711
31 124 -62 961 1085 1147 1271 1333 1457 1519 1643 1705 1829
35 -70 140 1225 1295 1435 1505 1645 1715 1855 1925 2065
37 148 -74 1369 1517 1591 1739 1813 1961 2035 2183
41 -82 164 1681 1763 1927 2009 2173 2255 2419
43 172 -86 1849 2021 2107 2279 2365 2537
47 -94 188 2209 2303 2491 2585 2773
49 196 -98 2401 2597 2695 2891
53 106 212 2809 2915 3127
55 220 110 3025 3245
59 118 236 3481
"p" + 2(p) "p' " + 4(p) 25 + 2(5), 35 + 4(5), 55 + 2(5).. 49 + 4(7), 77 + 2(7), 91 + 4(7), 119 +
2(7), 133 + 4(7), 121 + 2(11), 143 + 4(11)
23 43 53 73 83 103 113 163 173
17
37 47 67 97
107
127
137
157 167
31
41
61
71 101
151
25 55 85 17535 65 95 125 15549
77 91
133 161121 143 169
181
131
115
145119
0%
20%
40%
60%
80%
100%
2 5 11 169
2 5 11 121 143
2 5 11 49 77 91 119 133 161
2 5 11 35 65 95 125 155
2 5 11 25 55 85 115 145 175
2 5 11 31 41 61 71 101 131 151 181
3 7 13 17 37 47 67 97 107 127 137 157 167
3 13 23 43 53 73 83 103 113 163 173
19 29 59 79 89 109 139 149 179
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25 + 2(5) = 35 + 4(5) =55 + 2(5) = 65,… 49 + 4(7) = 77 + 2(7) = 91 + 4(7) = 119,..
25 35
55 65 49
85 95 77 91
115 125 121 119145 155 133 143
175 185 169 161
205 215 203 187 209
235 245 221 217
265 275 289 253 247
295 305 319 287 301 299
325 335 323 329
355 365 361 343 341
385 395 377 391 371
415 425 407 403 413
445 455 437 451 427475 485 481 473
505 515 497 511 493 517
535 545 527 533 539
565 575 553 551
595 605 601 583 581
625 635 623 611 629
655 665 649 637
685 695 679 671 667 689
715 725 707 721 703 713 697
745 755 737 731 749
775 785 767 781 763 779
805 815 799 793 803 791
835 845 841 833 817
865 875 871 851 847 869
895 905 889 901 893 899
925 935 917 931 913 923
955 965 949 961 943 959
985 995 979 973 989
1015 1025 1007 1003 1001
1045 1055 1037 1043 1027
1075 1085 1067 1081 1073 1057 1079
1105 1115 1099 1111
1135 1145 1127 1141 1133 1121 1139
1165 1175 1159 1157 1147 1169
1195 1205 1189 1183 1177 1199
1225 1235 1219 1211 1207
1255 1265 1247 1261 1243 1253 1241
1285 1295 1273 1271 1267
1315 1325 1309 1313
1345 1355 1339 1337 1351 1333 1343 1331 1349
1375 1385 1369 1363 1357 1379
1405 1415 1397 1411 1393 1403 1391 1387
1435 1445 1441 1421 1417
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1465 1475 1457 1463 1469
1495 1505 1501 1493 1477
1525 1535 1519 1517 1513 1507 1529
1555 1565 1547 1561 1541 1537
1585 1595 1577 1591 1573 1589
1615 1625 16031645 1655 1639 1651 1633 1643 1631 1649
1675 1685 1681 1673 1661 1679
1705 1715 1711 1703 1691 1687
1735 1745 1729 1727 1717 1739
1765 1775 1757 1771 1751 1769
1795 1805 1793 1781 1799
5 7 11 13 17 19
5 29 23 25
37 31 35
7 43 47 41 49
59 53 55
67 61 65
11 79 73 71 77
89 83 85
13 97 95 91
109 103 107 101
17 113 115 119
11 127 125 121
19 139 137 131 133
13 149 145 143
157 151 155
13 23 163 167 161 169
179 173 175
17 181 185 187
199 193 197 191
19 29 205 203 209
31 211 215 217
17 229 223 227 221
239 233 23519 241 245 247
23 37 257 251 259 253
269 263 265
277 271 275
17 41 283 281 287 289
23 293 295 299
43 307 305 301
29 313 317 311 319
19 47 325 329 323
337 331 335
31 49 349 347 343 341359 353 355
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19 367 365 361
29 53 379 373 371 377
389 383 385
23 397 395 391
31 37 409 401 407 403
59 419 415 41361 421 425 427
23 439 433 431 437
449 443 445
41 457 455 451
67 463 467 461 469
43 479 475 473
37 487 485 481
29 71 499 491 497 493
509 503 505
47 73 515 511 517
Theorem: Odd numbers, the can be written as product of two primes, are almost prime andcongruent to me modulo 3.
Proof: "p" "p`" mod p 49 85 mod 3 when 49 =16(3)+1 and 85=28(3)+1, then "p" and "p`"congruent according mod p, and difference "p`"-"p" is multiplicity p. 85 - 49 = 36/3 = 12
„p‖- (5,7,11,13) = n(3)
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
250
49 55
0 0
85 91
0
115121133145
0
169175
0 0
205
0
217
0
235247253265
0
289295 301
0
35
0 0
65 77
0
95
0
119125
0
143155161
0
185
0
203209215221
0
245259
0
275287299305
"p" ≡ "p" mod 3, 91 - 49 = 14(3), 119 - 77 = 14(3), 161 - 143 = 6(3), 169 -
133 = 12(3), 299 - 287 = 4(3), 301 - 289 = 4(3),
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350 351 352 353 354 355 356 355 (76)(38)
357 358 359 360 361 362 363 361
364 365 366 367 368 369 370 365
371 372 373 374 375 376 377 377
378 379 380 381 382 383 384
385 386 387 388 389 390 391 391
392 393 394 395 396 397 398 395
399 400 401 402 403 404 405 403
406 407 408 409 410 411 412 407
413 414 415 416 417 418 419 415
420 421 422 423 424 425 426 425
427 428 429 430 431 432 433
434 435 436 437 438 439 440 437
441 442 443 444 445 446 447 445
448 449 450 451 452 453 454 451
455 456 457 458 459 460 461
462 463 464 465 466 467 468
469 470 471 472 473 474 475 475 473
476 477 478 479 480 481 482 481
483 484 485 486 487 488 489 485
490 491 492 493 494 495 496 493
497 498 499 500 501 502 503
504 505 506 507 508 509 510 505
511 512 513 514 515 516 517 515 517
518 519 520 521 522 523 524 (46)(92)
525 526 527 528 529 530 531 527 529
532 533 534 535 536 537 538 535 533
539 540 541 542 543 544 545 539
546 547 548 549 550 551 552
553 554 555 556 557 558 559
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It will suffice only to look on table of primes and almost prime, to oneself about this toconvince, what order and rules reign here. This would border about absurdity, these justfundamental elements of well ordered world of mathematics, if would they behave savagelyand unforeseeable.
2617 2615 2611
2621 2627 2623 26292633 2635 2639
2647 2645 2641
2659 2657 2651 2653
2663 2665 2669
Above mentioned table shows, that primes and almost prime ranked according to 4 basicnumbers unity 9 – 3 – 7 – 1, divide number almost prime about number unity 5 on twocomplementary parts. Fact this is the denial so far general opinion, about irregularities of occurrence of primes and them the alleged decreasing on further of up growth of sequencenumbers. Smaller or larger gaps among primes 2-4-6-8-14-18-20-22-24-26-30-34-36-44-52-
60-72-86-96-112-114-118-132-148-154-180-210-220-222-234-248-250-282-288-292-320-336-354-382-384-394-456-464-468-474-486-490-500-514-516-532-534-540-582-588-602-652-674-716-766-778-804-806-906, … be full with numbers almost prime, so as appearingprimes in interval 2 and 4.Among primes 1327 and 1361 his place occupies 10 successive numbers divisible by primes,that is almost primes.
1327 + 4 = 1331/11 + 2 = 1333/31 + 4 = 1337/7 + 2 = 1339/13 + 4 = 1343/17 + 2 = 1345/5 +4 = 1349/19 + 2 = 1351/7 + 4 = 1355/5 + 2 = 1357/23 + 4 = 1361 – 1327 = 34
Similarly is among primes 8467 and 8501. Among 370261 and 370373 we have gap about
length 112. For p < N the largest at present well-known maximal gap equal m =1442, p =804 212 830 686 677 669.
"p"+n(7)="p' " 35+ 2(7)=49+ 4(7)= 77 + 2(7) = 91 + 4(7) = 119 + 2(7) = 133 + 4(7) = 161 + 2(7) = 175 +
4(7) = 203 + 2(7) = 217 + 4(7) = 245 + 2(7) = 259 + 4(7) = 287
25
3549
55
77
6585
91
95
119133
125
121
145
175
169
209
205
217
215 221
247
265
275
287
295
299
7
115143
161
155
187
185
203
235
245
253
259
289
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10 000 019+2 = 10 000 021/97+4 = 10 000 025/5+2 = 10 000 027/37+4 = 10 000 031/227 + 2
=10 000 033/397+4 = 10 000 037/43+2 = 10 000039/7+4 = 10 000 43/2089+2 = 10 000 045/5
+ 4=10000049/47+2 =10 000 051/73+4 =10 000 055/5+2=10 000 57/79+4= 10 000 061/19+2
=10 000 063/17+4 = 10 000 067/7+2 = 10 000 069/181+4 = 10 000 073/31+ 2 = 10 000 075/5
+ 4 = 10 000 079 – 10 000 019 = 60
2677 2671 2675
2689 2683 2687 2681
2699 2693 2695
2707 2705 2701
2719 2713 2711 2717
2729 2725 2723
2731 2735 27372749 2741 2747 2743
2753 2755 2759
2767 2765 2761
n p p n P P
2 3 5 292 1453168141 1453168433
4 7 11 320 2300942549 2300942869
6 23 29 336 3842610773 3842611109
8 89 97 354 4302407359 4302407713
14 113 127 382 10726904659 10726905041
18 523 541 384 20678048297 20678048681
20 887 907 394 22367084959 22367085353
22 1129 1151 456 25056082087 25056082543
30 13063 13093 464 42652618343 42652618807
34 1327 1361 468 127976334671 127976335139
36 9551 9587 474 182226896239 182226896713
44 11633 11677 486 241160624143 141160624629
52 19609 19661 490 297501075799 297501076289
60 100000019 100000079 500 303371455241 303371455741
72 31397 31469 514 304599508537 304599509051
86 155921 156007 516 416608695821 416608696337
96 360653 360749 532 461690510011 461690510543
112 370261 370373 534 614487453523 614487454057
114 492113 492227 540 738832927927 738832928467
118 1349533 1349651 582 1346294310749 1346294311331
132 1357201 1357333 588 1408695493609 1408695494197
148 2010733 2010881 602 1968188556461 1968188557063
154 4652353 4652507 652 2614941710599 2614941711251 180 17051707 17051887 674 7177162611713 7177162612387
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Theorem about congruence odd number permits faultlessly to distinguish primes from anotherdivisible numbers, that is almost prime. e. g. prime confirms the legitimacy of formula:
p = 1 + n(7) p = 2 + n(7) p = 3 + n(7) p = 4 + n(7) p = 5 + n(7) p = 6 + n(7)
2 89 - 1 = 618 970 019 642 690 137 449 562 111- 3
618 970 019 642 690 137 449 562 108/7 = 88 424 288 520 384 305 344 937 444
(3 203 000 719 597 029 781 – 3) : 7 = 457 571 531 371 004 254
(810 433 818 265 726 529 159 – 5) : 7 = 115 776 259 752 246 647 022 andalmost prime with numerous iterations inside, as and in quotient of formula ―p‖= 2 + n(3)
7 · 20408163265306122449 = 142 857 142 857 142 857 143- 2
142 857 142 857 142 857 141/3 = 476 190 476 190 476 190 47
We happen in second factor of following expression sure unusual prime:
10 31+ 1 = 11· 909 090 909 090 909 090 909 090 909 091 =10 000 000 000 000 000 000 000 000 000 001
Decomposition on primes her product, it lies at bases of iteration in this number.From this, that 1001 = 7 · 143 = 11 · 91 = 13 · 77 and 10 001 = 73 · 137 create followingiterations. Products:
7 · 1001 = 7007 11 · 1001 = 11011 13 · 1001 = 13013 77 · 1001 = 7707791 · 1001 = 91091 143 · 1001 = 143143 73 · 1001 = 73073 137 · 1001 = 137137and 999 multiplicity 1001 e.g. 323 · 1001 = 323 323 and number 10 001, 43 ·10001= 43004329 · 430 043 = 124 7 124 7 3 · 12 471 247 = 37 41 37 41
We see noteworthy iterations in prime 9 090 909 091 and her square, and so number almostprime 826 644 628 100 826 446 281 and prime 82 644 628 099 173 553 719, in which exceptiteration see two peers of numbers in reflection mirror.
210 20831323 20831533 716 13829048559701 13829048560417
220 47326693 47326913 766 19581334192423 19581334193189
222 122164747 122164969 778 42842283925351 42842283926129
234 189695659 1899695893 804 90874329411493 90874329412297
248 191912783 191913031 806 171231342420521 171231342421327
250 387096133 387096383 906 218209405436543 218209405437449
282 436273009 436273291 1132 1693182318746371 1693182318747503
288 1294268491 1294268779 1308 749565457554371299 749565457554372607
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On radar graph number almost prime are visible on black background.
9999907 9999901 9999905
9999913 9999911 9999917 9999919
9999929 9999925 9999923
9999931 9999937 9999935
9999943 9999949 9999947 9999941
9999955 9999959 9999953
9999961 9999965 9999967
9999973 9999971 9999977 9999979
9999985 9999983 9999989
9999991 9999997 9999995
10000003 10000001 10000007 10000009
10000019 10000015 10000013
10000021 10000025 10000027
10000039 10000037 10000031 10000033
10000045 10000043 10000049
10000051 10000055 10000057
10000069 10000067 10000061 10000063
10000079 10000073 10000075
10000085 10000081 10000087
10000097 10000091 10000099 10000093
Let‘s apply so well-known us a formula to constructing successive primes and almost prime,that could generate this kind of pattern.
29
79
139
149199
229239
269
43
53
113
193
223
283
293
313
7
17
3747
127
137157
197
257
277
337
5 11
41
71
131
251
281
25
35
55
65
85
95
125
145
155175
205
235
265
275
295
305
355
91
121
161
221
301
341
77
187217
247
287
133
143
253
343
49
119
169209
259
289
299
319
329
179
109
89
59
349
19359 2
233
83
173
163
353
263
73
23
3
103
13
167
227
347
367
307
317
97
67
107
151
31
61
271
191
181
241
211
331
311
101
325
335
115245
215185
365
361
203
323
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(1,2,3,4,5,6) + n(7) = p 2 + 3(7) = 23 (1,2) + n(3) = „p― 1 + 8(3) = 25
9 999 901 = 1 428 557(7) + 2 9 999 905 = 3 333 301(3) + 29 999 907 = 1 428 558(7) + 1 9 999 913 = 3 333 304(3) + 1
9 999 929 = 1 428 561(7) + 2 9 999 923 = 3 333 307(3) + 29 999 931 = 1 428 561(7) + 4 9 999 925 = 3 333 308(3) + 19 999 937 = 1 428 562(7) + 3 9 999 935 = 3 333 311(3) + 29 999 943 = 1 428 563(7) + 2 9 999 949 = 3 333 316(3) + 19 999 971 = 1 428 567(7) + 2 9 999 977 = 3 333 325(3) + 29 999 973 = 1 428 567(7) + 4 9 999 985 = 3 333 328(3) + 19 999 991 = 1 428 570(7) + 1 9 999 997 = 3 333 332(3) + 1
10 000 019 = 1 428 574(7) + 1 10 000 015 = 3 333 338(3) + 110 000 079 = 1 428 582(7) + 5 10 000 085 = 3 333 361(3) + 2
9999907 9999901 9999905
9999911 9999917 9999913 9999919
9999929 9999925 9999923
9999937 9999931 9999935
9999943 9999941 9999947 9999949
9999955 9999953 9999959
9999965 9999961 9999967
9999973 9999971 9999977 9999979
9999985 9999983 9999989
9999991 9999995 9999997
10000001 10000007 10000003 10000009
10000019 10000015 10000013
10000025 10000021 10000027
10000031 10000037 10000033 10000039
10000045 10000043 10000049
10000055 10000051 10000057
10000061 10000067 10000063 10000069
10000079 10000075 10000073
10000085 10000081 10000087
10000091 10000097 10000093 10000099
Here are the primes amongst the 100 numbers either side of 10 000 000. For example in the100 numbers immediately before 10 000 000 since 9 999 901 to 9 999 991 there are 9 primes,but look now at how few there are in the 100 numbers above 10 000 000: only 2 primessince10 000 001 to 10 000 099.
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Arithmetical sequence of primes and almost prime are sequence line and helical growing.
Helical sequence of primes and almost prime.
3 2
7 5
11 13
f(p"p") = p, p', p' + d, p" + 2d, d = 2, f(p"p") = 2, 3, 3 + 2, 5 + 2, 7 + 4, 11 + 2, 13 + 4, 17 + 2,
12
35
7
11
13
1719
2325
2931
3537
4143
47
49
53
2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 223 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 33
0
10
20
30
40
50
60
1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39 41 43 45 47 49 51 53
P(p) = (2,3,5,11,13,29) + n(7) 59 = 3 + 8(7) P("p") = (5,7,11,13) + n(3) 55 = 7 + 16(3)
1 2 2 2 2 2 2 2 2 2 24 4 4 4 4 4 4 4 425
11
17
2329
35
41
47
53
59
3
7
13
19
25
31
37
43
49
55
61
0
20
40
60
80
100
120
140
2
, 3 , - 2 - 5 , - 2 - 7 , - 4 - 1 1 , - 2 - 1 3 , -
4
- 1 7 , . .
Serie4 3 7 13 19 25 31 37 43 49 55 61
Serie3 2 5 11 17 23 29 35 41 47 53 59
Serie2 4 4 4 4 4 4 4 4 4
Serie1 1 2 2 2 2 2 2 2 2 2 2
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
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Theorem: It sequence of primes and relatively primes is the twin sequence frominitial term p = 2 and constantly difference d = 6 (- 2 – 4).
Therefore though in Riemann‘s conjecture function of location of primes π (x), is the gradualfunction about high irregularity, as helical arithmetical twin sequence of primes and almostprimes, which the difference the d = 6 is constant, it shows the amazing smoothness.
From here sequence of prime numbers is not similar to accidental sequence of numbers, but towell ordered structure. So basic numbers does not be definite per nature the method of accidental throw with coin. Accident and chaos they are for mathematician simply cruelty.
2 + 3 = 5 -2- 7 -4- 11 -2- 13 -4- 17 -2- 19 -4- 23 -2- 25
2
3
51117
2329
31
35
37
41
43
47
49
53
55 7131925
1
7
13
19
31
37
43
25
811
1417
20 23
2629
32
3841
4447
5053
6
12
1824
30
36
42
48
54
3
9
15
21
27
33
39
45
51
4
10
1622
28
34
40
46
52
25
35
49
55
0
10
20
30
40
50
60
Serie6 25 35 49 55
Serie5 4 10 16 22 28 34 40 46 52
Serie4 3 9 15 21 27 33 39 45 51
Serie3 6 12 18 24 30 36 42 48 54
Serie2 2 5 8 11 14 17 20 23 26 29 32 38 41 44 47 50 53
Serie1 1 7 13 19 31 37 43
1 2 3 4 5 6 7 8 91
0
1
1
1
2
1
3
1
4
1
5
1
6
1
7
1
8
1
9
2
0
2
1
2
2
2
3
2
4
2
5
2
6
2
7
2
8
2
9
3
0
3
1
3
2
3
3
3
4
3
5
3
6
3
7
3
8
3
9
4
0
4
1
4
2
4
3
4
4
4
5
4
6
4
7
4
8
4
9
5
0
5
1
5
2
5
3
5
4
5
5
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Twin sequences of prime and almost prime numbers a congruent to me according to algebraicmodule 72.
Distribution of primes and almost prime according to rules of congruence of modules 7 and 3is the reason, why these folded with two and threes numbers be place on straight line of line,which confirms the legitimacy of the Riemann hypothesis. The uniformity from what rises thegraph of primes e.g.: by 100 000, he owes not quantity of primes to number N what can
express with logarithmic function, but proportionate distributing, resulting from congruenceof according to modules 7.
P(p) = (2, 3 + R2 + R4), 2, 3+ 2 = 5, 3 + 4 = 7= 5 + 2, 7 + 4 = 11 + 2 = 13 + 4 = 17 + 2 = 19 + 4 = 23 + 2 =
25 + 4 = 29 + 2 = 31 + 4 = 35 + 2 = 37 + 4 = 41 + 2 = 43 + 4 = 47 + 2 = 49 + 4 = 53 + 2 = 55
2 35
11
17
23
29
41
47
53
59
71
83
89
101
107
113
131
24
24
24
24
24
24
24
24
24
24
24
24
24
24
24
24
24
24
24
24
24
2
7
13
19
31
37
43
61
67
73
79
97
103
109
25
35
55
65
85
95
115
125
49
77
91
119121
127
0
20
40
60
80
100
120
140
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45
p'="p" mod p 65 = 13(2) + 13(3) 137 = 67(2) + 3 p' - "p" = n/p 137 - 65 = 72/2
71
73
143
145215
217
287
289
359
36177
79
149
151221
223
293
295
365
367
83
85
155
157227
229
299
301
371
373
17 19 89
91
161
163233
235
305
307
377
379
2325 95
97
167
169
239
241
311
313
383
385
2931
101
103
173
175
245
247
317
319
389
391
3537
107
109
179
181
251
253
323
325
395
397
41
43
113
115
185
187
257
259
329
331
361
363 47
49
119
121
191
193
263
265
335
337
367
36953
55
125
127
197
199271
341
343
373
375
59
61
131
133
203
205277
347
349
379
381
67
137
139
209
211283
353
355
385
387
2 3
751311
269275281
65
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„Upon looking at these numbers, one has the feeling of being in the presence of theinexplicable secrets of creation.‖ /D. Zagier/ Are the primes distributed chaotically or can we find some means for computing them?Will it ever be possible to predict with arbitrary accuracy where the next one lies? Yes, hereyou are! p + 6(7) = p‘, or almost prime ―p‖
65
107
149
191
233
275
31747
89
131
173215
257
299
67
109
151
193
235
277
319
49
91
133
175217
259
301
71
113
155
197
239
281
323
53
95
137
179221
263
305
73
115
157
199
241
283
325
55
97
139
181223
265
307
77
119
161
203
245
287
329
59
101
143
185227
269
311
79
121
163
205
247
289
331
61
103
145
187229
271
313
83
125
167
209
251
293
335
25
29
3135
37
4147
49
53
55
59
61
67
71
7377
79
83
89
91
95
97
101
103
109
113
115119
121
125
131
133
137
139
143
145
151
155
157161
163
167
173
175
179
181
185
187
193
197
199203
205
209
215
217
221
223
227
229
235
239
241245
247
251
257
259
263
265
269
271
277
281
283287
289
293
299
301
305
307
311
313
319
323
325329
331
335
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The prime numbers are distributed not chaotically. All prime and almost prime numbers to becongruent modulo 7.Because the smallest gap between their equal 2 + 4 = 6, and 6(7) = 42than is possible to predict with arbitrary accuracy that the next one lies what 42 gap.
Primes and almost prime can settle according to their quantity. Such in a row creates fourteenthe vertical groups and the innumerable amount of horizontal rows / periods / primes andalmost prime. Length period 42 = 7(6) it is product of length of period all natural numbers and
seven units about what grow primes and almost prime. In third and eighth group exceptingprime 7, have only almost prime numbers.I II III IV V VI VII VIII IX X XI XII XIII XIV
5 7 11 13 17 19 23
25 29 31 35 37 41 43
47 49 53 55 59 61 65
67 71 73 77 79 83 85
89 91 95 97 101 103 107
109 113 115 119 121 125 127
131 133 137 139 143 145 149
151 155 157 161 163 167 169
173 175 179 181 185 187 191
193 197 199 203 205 209 211
215 217 221 223 227 229 233
235 239 241 245 247 251 253
257 259 263 265 269 271 275
277 281 283 287 289 293 295
299 301 305 307 311 313 317
319 323 325 329 331 335 337341 343 347 349 353 355 359
p + 7(6) = p' 5 -42- 47 -42- 89 -42- 131 -42- 173 -84- 257 -126- 383 -84-467 -42- 509 -,,, "p"+ 7(6)
= "p' " 35 -42- 77 -42- 119 -42- 161 -42- 203 - 42- 245 -42- 287 -42- 329 -42- 371 -42- 413 -42- 455 -
42- 497 -42- 539 ..
0 23
5
711
13
17
19 23
25
2931
35
37
41 43
49
53
55
59
61 65
67
7173
77
79
83 85
91
9597
101
103 107
109
113115
119
121
125 127
133
137
139
143
145 149
151
155
157
161
163
167 169
175
179
181
185
187 191
193
197
199
203
205
209 211
217
221
223
227
229 233
235
239
241
245
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251 253
257
259
263
265
269
271 275
277
281
283
287
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293 295
301
305
307
311
313 317
319
323
325
329
331
335 337
341
343
347
349
353
355 359
361
365
367
371
373
377 379
385
389
391
395
397 401
403
407
409
413
415
419 421
425
427
431
433
437
439 443
445
449
451
455
457
461 463
469
473
475
479
481 485
487
491
493
497
499
503 505
511
515
517
521
523 527
529
533
535
539
541
545
4789 131173215 299 383 467509
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Number 19, is prime in that case the sum 19 + 42 = 61 is prime too. Number 9 091, is primein that case the sum 9 091 + 42 = 9 133 is prime too. 9 091 – 19 = 9 072 : 42 = 216Number 909 091, is prime in that case the sum 909 091 + 42 = 909 133 is prime too.Number 909 090 909 090 909 090 909 090 909 091, is prime so I can predict with arbitrary
accuracy that the next one lies in gap 42 = 909 090 909 090 909 090 909 090 909 133 – 43 = 909 090 909 090 909 090 909 090 909 090 : 42 = 21 645 021 645 021 645 021 6 45 021 645
909 1, is prime + 42 = 9 133 too,
909 091, is prime + 42 = 909 133 too,
9 090 909 091 = 11 · 23 · 4093 · 8779
909 090 909 091 = 859 · 1 058 313 049
9 090 909 090 909 091 = 103 · 4013 · 21 993 833 369
909 090 909 090 909 091, is prime
9 090 909 090 909 090 909 091, is prime
909 090 909 090 909 090 909 091, is prime
9 090 909 090 909 090 909 090 909 091 = 59(154 083 204 930 662 557 781 201 849)
909 090 909 090 909 090 909 090 909 091, is prime too. They are 4,6,18, 22,24, and 30 digitsprimes. One from 100 and 1000 million digits prime are 9.090909091e99 999 999and 9.090909133e999 999 999.
8 264 462 809 917 355 371 900 826 446 281 They are 32 digits 90 909 090 909 090 909 090 909 090 909 091 is dividable by 11 and
e38―p― = e26 + e10 + e2 105 831 304 899 989 415 869 510 001 058 313 049 38 digits numbers 90 909 090 909 090 909 090 909 090 909 090 909 091 : 859
e32 „p― = e22 + e9 + e1 8 264 462 809 917 355 371 900 826 446 281
+ 82 644 628 099 173 553 719 008 264 462 8190 909 090 909 090 909 090 909 090 909 091
e99 999 998 ―p― = e4 545 454(22) + e9 + e1 = 9.090909091e99 999 997
e1000 000 000 „p― = e38 461 538(26) + e10 + e2 = 9.090909091e999 999 999
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2 + 3 + 5 + 7 + 11,..+ 29 + 31 + 37 = 222 35 + 41 + 43 + 47 + 49,..+ 71 + 73 + 79 = 798
77 + 83 + 85,.. + 113 + 115 + 121 = 1386 119 + 125 + 127,.. + 155 + 157 + 163 = 1986
222 – 48(12) – 798 – 49(12) -1386 – 50(12) – 1986 ,.. 14n + [14n + n(12)],..
They in this way grow with 14 of primes and almost prime built-up terms, arrange inexquisite mosaic illustrating their row in intervals 2 and 4 in arrangement of sevens.There are two facts about the distribution of prime numbers of which I hope to convince youso overwhelmingly that they will be permanently engraved in your hearts. The first is that,despite their simple definition and role as the building blocks of the natural numbers, theprime numbers same for me a balding blocks, that is to say every prime bigger than 3 the sumtheir predecessor 2, 3, 5, 11, 13, and 29 is, and n-the multiplicity of prime 7. They grow notlike weeds among the natural numbers, seeming to obey no other law than that of chance, andnobody can predict where the next one will sprout. The second fact is even more astonishing,for it states just the opposite: that the prime numbers exhibit stunning regularity, that there are
laws governing their behavior, congruence laws modulo 7, and that they obey these laws withmilitary precision. To support the first of these claims, let me begin by showing you a list of the prime up to 100. I hope you will agree that there is apparent reason why one number isprime and another not.2, 3, 2 + 3 = 5 5 + 6(7) = 47
5 + 2 = 7 11 + 6(7) = 532(2) + 7 = 11 3 + 8(7) = 592(3) + 7 = 13 5 + 8(7) = 613 + 2(7) = 17 11 + 8(7) = 675 + 2(7) = 19 29 + 6(7) = 712 + 3(7) = 23 3 + 10(7) = 73
1 + 4(7) = 29 2 + 11(7) = 793 + 4(7) = 31 13 + 10(7) = 83
p + 2(7) = p' + 4(7) = p" 3 + 2 = 5 + 2(7) = 19 + 4(7) = 47 + 2(7) = 61 + 4(7) = 89 + 2(7) = 103 + 4(7) =
131 + 2(7) = 145 + 4(7) = 173
519
47 61 89 103 131 145 173
1341 55 83 97 125 139 167
7
35
49
77
91
119
133
161
175
29
43
71
85
113
127
155
169
23
37
65
79
107
121
149
16317
31 59 73 101 115 143 15711
25 53 67 95 109 137 151 179
2
181
3
185
0%
20%
40%
60%
80%
100%
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
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2 + 5(7) = 37 5 + 12(7) = 8913 + 4(7) = 41 13 + 12(7) = 9729 + 2(7) = 43
I II III IV V VI VII VIII IX X XI XII XIII 2 3 5 7 4 6 8 9
11 13 10 12 14 15
17 19 16 18 20 21
23 25 22 24 26 27
31 29 28 30 32 33
37 35 34 36 38 39
41 43 40 42 44 45
47 49 46 48 50 51
53 55 52 54 56 57 59 61 58 60 62 63
67 65 64 66 68 69
73 71 70 72 74 75
79 77 76 78 80 81
83 85 82 84 86 87
89 91 88 90 92 93
97 95 94 96 98 99
101 103 100 102 104 105
107 109 106 108 110 111
113 115 112 114 116 117
121 119 118 120 122 123
127 125 124 126 128 129
The numbers 2 and 3 are building blocks all natural numbers. Even indivisible by 2 and 3prime and almost prime numbers can you from n(2) and n(3) to put together e.g. 2 + 3 = 52(2) + 3 = 7 4(2) + 3 = 11 5(2) + 3 = 13 7(2) + 3 = 17 8(2) + 3 = 19 10(2) + 3 = 235(2) + 5(3) = 25 9(3) = 27The periodical table of natural numbers distinguishes 13 groups of even and odd numbers. Incolumns I - VII we have prime numbers appearing what n(7). e.g. 5 + 2(7) = 19 + 4(7) = 47 +
2(7) = 61 + 4(7) = 89 + 2(7) = 103 + 4(7) = 131 + 6(7) = 173 …In VI column except 7 we have free places on stepping out what n(7) almost prime numbers.e.g. 35 + 2(7) = 49 + 4(7) = 77 + 2(7) = 91 + 4(7) = 119 + 2(7) = 133 + 4(7) = 161 which arein VIII and IX column.25 + 2(5) = 35 + 4(5) = 55 + 2(5) = 65 + 4(5) = 85 + 2(5) = 95 + 4(5) = 115 + 2(5) = 125 …121 + 2(11) = 143 + 4(11) = 187 + 2(11) = 209 + 4(11) = 253 + 2(11) = 275 + 4(11) = 319 …
In tenth and twelfth column we have even numbers, and in XI and XIII column even and oddnumbers divisible by 3, following what 2(3).
101, 1 001=11(91), 100 001=11(9091), 10 000 001=11(909 091), 1.000 001E+99 999 999
103, 1 003=17(59), 100 003, 1 000 003, 1.000 003E+12,+18,+19,+99 999 999,+999 999 999
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107, 1 007=19(53), 100 007, 1 000 007, 1.000 007E+99 999 999, E+999 999 999
109, 1 009, 10 009, 100 009, 1 000 009, 1.000 009E+99 999 999, E+999 999 999
113, 1 013, 10 013, 100 013, 1 000 013, 1.000 013E+99 999 999, E+999 999 999
115, 1 015, 10 015, 100 015, 1 000 015, 1.000 015E+99 999 999, E+999 999 999
119, 1 019, 10 019, 100 019, 1 000 019, 1.000 019E+99 999 999, E+999 999 999
121, 1 021, 10 021, 100 021, 1 000 021, 1.000 021E+99 999 999, E+999 999 999
125, 1 025, 10 025, 100 025, 1 000 025, 1.000 025E+99 999 999, E+999 999 999
127, 1 027=13(79), 100 027, 1 000 027, 1.000 027E+99 999 999, E+999 999 999
131, 1 031, 10 031, 100 031, 1 000 031, 1.000 031E+99 999 999, E+999 999 999
133, 1 033, 10 033, 100 033, 1 000 033, 1.000 033E+99 999 999, E+999 999 999
137, 1 037=17(61), 10 037, 100 037, 1 000 037, 1.000 037E+14, E+16, E+99 999 999
139, 1 039, 10 039, 100 039, 1 000 039, 1.000 039E+13, E+99 999 999, E+999 999 999
2, 3, 5, 11, 13, 29 + n(7) = p
2 + 15(7) = 107 3 + 14(7) = 101 5 + 14(7) = 103
11 + 14(7) = 109 13 + 18(7) = 139 29 + 12(7) = 113
3 + 148(7) = 1 039 29 + 1430(7) = 10 039 5 + 142 862(7) = 1 000 039
5 + 142 857 142 862(7) = 1 000 000 000 039
5 + 142 857 142 862e99 999 999(7) = 1.000 000 039E+100 000 000
5 + 142 857 142 862e999 999 999(7) = 1.000 000 039E+1000 000 000
3 + 1(7) = 10
2 + 14(7) = 100
6 + 142(7) = 1 000
4 + 1 428(7) = 10 000
5 + 14 285(7) = 100 000
1 + 142 857(7) = 1,00E+06
3 + 1 428 571(7) = `1,00E+07
2 + 14 285 714(7) = 1,00E+08
6 + 142 857 142(7) = 1,00E+09
4 + 1 428 571 428(7) = 1,00E+10
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5 + 14 285 714 285(7) = 1,00E+11
1 + 142 857 142 857(7) = 1,00E+12
3 + 1 428 571 428 571(7) = 1,00E+13
2 + 14 285 714 285 714(7) = 1,00E+14
6 + 142 857 142 857 142(7) = 1,00E+15
4 + 1 428 571 428 571 428(7) = 1,00E+16
5 + 14 285 714 285 714 285(7) = 1,00E+17
4 + 1,428 571 428e99(7) = 1,00E+100
4 + 1,428 571 428e999(7) = 1,00E+1000
4 + 1,428 571 428e99 999 999(7) = 1,00E+100 000 000
4 + 1,428 571 428e999 999 999(7) = 1,00E+1000 000 000
Factorization of almost primes in prime factors.
Factorise large numbers on factors prime, it was in last 2000 years difficult problem. Majoritymathematicians‘ is opinion, that factorisation numbers is fundamental extraordinarycomputational problem. One of main reasons, why the factorise numbers is so difficult; shewas alleged fortuity of occurrence of primes.We know since, that primes and almost prime are present not accidentally, but according torules of congruence of modules 7 and 3, we have also the way on factorise their products.It with theorems about congruence odd numbers results brightly, or number is prime or almostprime, and we for help of binomial formula easily will take apart every odd number on factors
prime. We know, that difference among two successive square numbers state always oddnumber, we have such number from here to write down in some way as difference twosquares and take out the common factor.Difference of 2 squares:
a(a) – b(b) = (a – b)(a + b)p is common to both terms. Put this common factor outside the brackets.
―p‖ = p(p) ―p‖ = p(p + p`) 25 = 5(2 + 3)„p― = p(p´) = [(p + p‗)/2 – {(p + p‗)/2 – p}][(p + p‗)/2 + {(p + p‗)/2 – p}]
147 573 952 589 676 412 927 = 193 707 721(761 838 257 287) =[(193 707 721 + 761 838 257 287)/2 – {(193 707 721 + 761 838 257 287)/2 – 193 707 721}][(193 707 721 + 761 838 257 287)/2 + {(193 707 721 + 761 838 257 287)/2 – 193 707 721}]
(381 015 982 505 – 380 822 274 783)( 381 015 982 505 + 380 822 274 783)
35 = 6(6) – 1(1) = (6 – 1)(6 + 1) = 5(7) 55 = 8(8) – 3(3) = (8 – 3)(8 + 3) = 5(11)
143 = 12(12) – 1(1) = (12 – 1)(12+ 1) = 11(13) 221 = 15(15)-2(2) =(15-2)(15 + 2) = 13(17)
253 = 17(17) – 6(6) = (17 – 6)(17+ 6) = 11(23) 247 = 16(16)-3(3) =(16-3)(16 + 3) = 13(19)If the difference between one number a, and a prime number is divisible by a prime number,then the number is complex.
)1'( p p pa e. g. 40
77287 p(p‘) = (p‘- 1)p + p 7(41) = (41 – 1)7 + 7
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The straight is dividing by 3 given number and to subtract from her rounded quotient withoutthe rest. Then we level rounded quotient to the closest third multiplicity of prime numbers oralmost prime. About the same number we round off received previously the difference to n -the multiplicity of the same prime numbers or almost prime. And so we receive prime factorson what factorizing is number almost prime.
319 : 3 = 106 - 19 = 87 : 3 = 29 343 : 3 = 114 + 33 = 147 : 3 = 49
319 - 106 = 213 + 19 = 232 : 8 = 29 343 - 11 4 = 229 - 33 = 196 : 4 = 49
319 = 11(29) 343 = 7(49)
8051 : 3 = 2683 – 2392 = 291 : 3 = 978051 – 2683 = 5368 + 2392 = 7760 : 80 = 97
8051 = 83(97)
9 090 909 091 : 3 = 3 030 303 030 – 550 964 187 = 2 479 338 843 : 3 = 826 446 2819090909091 – 3030303030 =6060606061 + 550964187 = 6611570248 : 8 = 826446281
9 090 909 091 = 11(826 446 281)
909 090 909 091 : 3 = 303 030 303 030 – 299 855 363 883 = 3174939147 : 3 = 1 058 313 049909090909091 – 303030303030 = 606060606061 + 299855363883 = 905 915 969 944 : 856=1 058 313 049 909 090 909 091 = 859(1 058 313 049)
9090909090909091 : 3 = 3030303030303030 – 2765519270373639 = 264783759929391 : 3=88261253309797
9090909090909091 – 3030303030303030 = 6060606060606061 + 2 765 519 270 373 639 =8 826 125 330 979 700 : 100
9 090 909 090 909 091 = 103(88 261 253 309 797)
9 090 909 090 909 090 909 090 909 091 : 3 = 3 030 303 030 303 030 303 030 303 030- 2 568 053 415 511 042 629 686 697 483
462 249 614 791 987 673 343 605 547/ 3 = 154 083 204 930 662 557 781 201 849
9090909090909090909090909091 – 3030303030303030303030303030 = 6 060 606 060 606 060 606 060 606 061
+ 2 568 053 415 511 042 629 686 697 483
8 628 659 476 117 103 235 747 303 544 : 56 = 154 083 204 930 662 557 781 201 8499 090 909 090 909 090 909 090 909 091 = 59(154 083 204 930 662 557 781 201 849)
8 051 = 90(90) – 7(7) = (90 – 7)(90 + 7) = 83(97) 493=23(23)-6(6)=(23-6)(23+6) = 17(29)
341 = 21(21)-10(10) = (21-10)(21+10) = 11(31) 391 = 20(20)-3(3) =(20-3)(20+3)= 17(23)
529 = 23(20+3) 497 = 17(68+3) 1105 = 17(62 + 3) 1309 = 17(74 + 3) 1147 = 31(34 + 3)
1369 = 37(34 + 3) 25271 = 37(680 + 3) 734 591 = 11(66778 + 3)
8453= 79(107) 11111 = 41(271) 120481 = 211(571) 526313=281(1873) 322577= 163(1979)
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434779=197(2207) 353357=307(1151) 10 000 043=2089(4787) 10 000 127= 167(59881)
370 267 = 479(773) 370 283 = 379(977) 370 289 = 349(1061) 370 297 = 353(1049)
370 303 = 367(1009) 370 319 = 547(677) 370 327 = 107(3461) 370 339 = 199(1861)
370 351 = 179(2069) 370 361 = 383(967) 370 309 = 67(5527) 370 313 = 47(7879)
370 273 = 43(8611) 370 301 = 29(12769) 370 333 = 37(10009) 370 369 = 23(16103)
370 243 = 17(21779) 370 249 = 11(33659) 370 253 = 13(28481) 370 271 = 11(33661)
370 277 = 17(21781) 370 291 = 19(19489) 370 331 = 13(28487) 370 337 = 11(33667)
370 379 = 17(21787) 370 343 = 59(6277) 370 279 = 7(52897) 370 381 = 11(33671)
370 307 = 7(52901) 370 321 = 7(52903) 370 349 = 7(52907) 370 363 = 7(52909)
9 999 913 = 7(1428559) 9 999 917 = 23(434779) 9 999 941 = 7(1428563)
9 999 947 = 19(526313) 9 999 949 = 31(322579) 9 999 971 =13(769229)
10 000 001 = 11(909 091) 10 000 003 = 13(769 231) 10 000 007 = 941(10 627)
10 000 009 = 23(434 783) 10 000 013 = 421(23 753) 10 000 021 = 97(103 093)
10 000 027 = 37(270 271) 10 000 031 = 227(44 053) 10 000 037 = 43(232 559)
10 000 039 = 7(1 428 577) 10 000 033 = 397(25 189) 10 000 043 = 2 089(4 787)
10 000 049 = 47(212 767) 10 000 061 = 19(526 319) 10 000 067 = 7(1 428 581)
10 000 081 = 7(1 428 583) 10 000 091 = 251(39 841) 10 000 093 = 53(188 681)
10 000 097 = 17(588 241) 10 000 099 = 19(526 321) 10 000 111 = 11(909 101)
10 000 123 = 7(1 428 589) 10 000 127 = 167(59 881) 10 000 133 = 11(909 103)
10 000 129=89(112361) 10 000 171 = 271(36901) 10 000 187 = 41(243907) 4 294 967 297=6 700 417(638+3) 1000001=101(9901) 8 547 008 547(13) = 111 111 111 111
7 709 321 041 217 = 25 271(305 065 927)
7 709321041217=(152 545 599-152520 328)(152 545 599+152 520 328)=25271(305065927)
We have with same the also fast way on qualification of primes, necessary to construction of
code the RSA. She in end was found hidden behind primes and almost prime full secrets
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structure, since ages in demand throughout mathematicians, and her music can write inaddition in infinity.
Who knows this basic interval two four, two four, knows also where what note will come withprime or almost prime numbers. We cannot already more now tell about their fortuity, but
more about them timeless and universal character.
No perceptible order and Riemann‘s Hypothesis.
Mathematicians since centuries listened intently in sound primes, and they heard unsettledtones only. These numbers resemble accidentally spilled notes on mathematical notes paper,without recognizable melody. Riemann sinusoidal waves what created right away zero Zetathey - showed scenery hidden harmony.
Mathematicians despite all could with sure probability to estimate, how many prime numbers
is in given interval. Only four in first ten are (2, 3, 5 and 7). It in first hundred is them 25, infirst thousand 168 their part comes down from 40 by 25 on 16,8 percentage.
Among smaller numbers from billion, 5% is the only just. To describe this down come of frequency of an occurrence in approximation the simple formula. From this however satisfiedmathematicians are not. They want to know how far real occurrence numbers first deviatesfrom counted frequency. Riemann in one's famous eight page paper ―On the Number of Prime Numbers less than a Given Quantity / "Über die Anzahl der Primzahlen unter einer gegebenen Größe" / he wrote: "The known approximating expression F( x) = Li(x) is therefore
valid up to quantities of the order x 2
1
and gives somewhat too large a value; But also the
increase and decrease in the density of the primes from place to place that is dependent on theperiodic terms has already excited attention, without however any law governing thisbehavior having been observed. In any future count it would be interesting to keep track of the influence of the individual periodic terms in the expression for the density of the primenumbers.‖ Real quantity of prime numbers differs from them counted frequency so aloneoften, as eagle near repeated throw with coin will fall out tails. Differently saying that isRiemann supposed, that occurrence prime numbers be subject to the rights of case. And hemade a mistake here, because prime numbers be subject to the rights of congruence of according to module p‘≡ p (mod.7).
He has written: ―One now finds indeed approximately this number of real roots within theselimits, and it is very probable that all roots are real. Certainly one would wish for a stricter
proof here.‖
Riemann Hipothesis.
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Riemann zeta- function for s = 0,5 + i * t.
The Riemann hypothesis (also called the Riemann zeta-hypothesis), along with suitablegeneralizations, is considered by many mathematicians to be the most important unresolvedproblem in pure mathematics. First formulated by Bernhard Riemann in 1859, it haswithstood concentrated efforts from many outstanding mathematicians for 150 years (as of 2009).
The Riemann hypothesis (RH) is a conjecture about the distribution of the zeros of theRiemann zeta-function ζ(s). The Riemann zeta-function is defined for all complex numbers s ≠ 1. It has zeros at the negative even integers (i.e. at s = −2, s = −4, s = −6, ...). These are
called the trivial zeros. The Riemann hypothesis is concerned with the non-trivial zeros, andstates that:
The real part of any non-trivial zero of the Riemann zeta function is ½.
Thus the non-trivial zeros should lie on the so-called critical line, ½ + it , where t is a realnumber and i is the imaginary unit. The Riemann zeta-function along the critical line issometimes studied in terms of the Z-function, whose real zeros correspond to the zeros of thezeta-function on the critical line.
The Riemann hypothesis implies a large body of other important results. Most mathematicians
believe the Riemann hypothesis to be true, A $1,000,000 prize has been offered by the ClayMathematics Institute for the first correct proof.
Unsolved problems in mathematics: Does every non-trivial zero of the Riemann zeta function have real part ½?
Riemann mentioned the conjecture that became known as the Riemann hypothesis in his 1859paper On the Number of Primes Less Than a Given Magnitude, but as it was not essential tohis central purpose in that paper, he did not attempt a proof. Riemann knew that the non-trivial zeros of the zeta-function were symmetrically distributed about the line s = ½ + it , andhe knew that all of its non-trivial zeros must lie in the range 0 ≤ Re(s) ≤ 1.
In 1896, Hadamard and de la Vallée-Poussin independently proved that no zeros could lie onthe line Re(s) = 1. Together with the other properties of non-trivial zeros proved by Riemann,
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this showed that all non-trivial zeros must lie in the interior of the critical strip 0 < Re(s) < 1.This was a key step in the first proofs of the prime number theorem.
In 1900, Hilbert included the Riemann hypothesis in his famous list of 23 unsolved problems — it is part of Problem 8 in Hilbert's list, along with the Goldbach conjecture. When asked
what he would do if awakened after having slept for five hundred years, Hilbert said his firstquestion would be whether the Riemann hypothesis had been proven (Derbyshire 2003:197).The Riemann Hypothesis is one of the Clay Mathematics Institute Millennium PrizeProblems.
In 1914, Hardy proved that an infinite number of zeros lie on the critical line Re(s) = ½.However, it was still possible that an infinite number (and possibly the majority) of non-trivial zeros could lie elsewhere in the critical strip. Later work by Hardy and Littlewood in1921 and by Selberg in 1942 gave estimates for the average density of zeros on the criticalline.
The Riemann hypothesis and primes
The zeta-function has a deep connection to the distribution of prime numbers. Riemann gavean explicit formula for the number of primes less than a given number in terms of a sum overthe zeros of the Riemann zeta function. Helge von Koch proved in 1901 that the Riemannhypothesis is equivalent to the following considerable strengthening of the prime numbertheorem: for every ε > 0, we have
where π( x) is the prime-counting function, ln( x) is the natural logarithm of x, and the Landaunotation is used on the right-hand side.[5] A non-asymptotic version, due to LowellSchoenfeld, says that the Riemann hypothesis is equivalent to
The zeros of the Riemann zeta-function and the prime numbers satisfy a certain dualityproperty, known as the explicit formulae, which shows that in the language of Fourier
analysis the zeros of the Riemann zeta-function can be regarded as the harmonic frequenciesin the distribution of primes.
The Riemann hypothesis can be generalized by replacing the Riemann zeta-function by theformally similar, but much more general, global L-functions. In this broader setting, oneexpects the non-trivial zeros of the global L-functions to have real part 1/2, and this is calledthe generalized Riemann hypothesis (GRH). It is this conjecture, rather than the classicalRiemann hypothesis only for the single Riemann zeta-function, which accounts for the trueimportance of the Riemann hypothesis in mathematics. In other words, the importance of 'theRiemann hypothesis' in mathematics today really stems from the importance of thegeneralized Riemann hypothesis, but it is simpler to refer to the Riemann hypothesis only in
its original special case when describing the problem to people outside of mathematics.
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For many global L-functions of function fields (but not number fields), the Riemannhypothesis has been proven. For instance, the fact that the Gauss sum, of the quadraticcharacter of a finite field of size q (with q odd), has absolute value
is actually an instance of the Riemann hypothesis in the function field setting.
Consequences and equivalent formulations of the Riemann hypothesis
The practical uses of the Riemann hypothesis include many propositions which are stated tobe true under the Riemann hypothesis, and some which can be shown to be equivalent to theRiemann hypothesis. One is the rate of growth in the error term of the prime number theoremgiven above.
Riemann was particularly interested in feeding imaginary numbers into functions. Usually we
can draw a graph of a function where the input runs along the horizontal and the output is theheight of the graph. But a graph of an imaginary function consists of a landscape where theoutput is represented by the height above any point in the world of imaginary numbers.
An imaginary landscape
Riemann had found one very special imaginary landscape, generated by something called the zeta function, which he discovered held the secret to prime numbers. In particular, the pointsat sea-level in the landscape could be used to produce these special harmonic waves whichchanged Gauss's graph into the genuine staircase of the primes. Riemann used the coordinates
of each point at sea-level to create one of the prime number harmonics. The frequency of eachharmonic was determined by how far north the corresponding point at sea-level was, and howloud each harmonic sounded was determined by the east-west frequency.
Riemann‘s sinus – waves what created with zero place Zeta – topography, they showedhidden harmony.
A prime number is a positive whole number greater than one which is divisible only by itself
and one. The first few are shown above. If the definition doesn‘t mean much to you, think of prime numbers as follows:
If you are presented with a pile of 28 stones, you will eventually deduce that the pile can bedivided into 2 equal piles of 14, 4 equal piles of 7, 7 equal piles of 4, etc. However, if onemore stone is added to the pile, creating a total of 29, you can spend as long as you like, butyou will never be able to divide it into equal piles (other than the trivial 29 piles of 1 stone).In this way, we see that 29 is a prime number, whereas 28 is non-prime or composite.All composites break down uniquely into a product of prime factors: i.e. 28 = 2 x 2 x 7. Notethat 2 is the only even prime - all other even numbers are divisible by 2. 1 is neither prime nor
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composite by convention.
Here the sequence of primes is presented graphically in terms of a step function or counting
function which is traditionally denoted . (Note: this has nothing to do with the value
=3.14159...) The height of the graph at horizontal position x indicates the number of primes less than or equal to x. Hence at each prime value of x we see a vertical jump of oneunit. Note that the positions of primes constitute just about the most fundamental, inarguable,nontrivial information available to our consciousness. This transcends history, culture, andopinion. It would appear to exist 'outside' space and time and yet to be accessible to anyconsciousness with some sense of repetition, rhythm, or counting. The explanation in theprevious page involving piles of stones can be used to communicate the concept of primenumbers without the use of spoken language, or to a young child
By zooming out to see the distribution of primes within the first 100 natural numbers, we seethat the discrete step function is beginning to suggest a curve.
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Zooming out by another factor of 10, the suggested curve becomes even more apparent.Zooming much further, we would expect to see the "granular" nature of the actual graph
vanish into the pixilation of the screen.
Now zooming out by a factor of 50, we get the above graph. Senior Max Planck Institutemathematician Don Zagier, in his article "The first 50 million primes" [ Mathematical
Intelligencer , 0 (1977) 1-19] states:"For me, the smoothness with which this curve climbs is one of the most astonishing facts inmathematics."(Note however that you are not looking at a smooth curve. Sufficiently powerfulmagnification would reveal that it was made of unit steps. The smoothness to which Zagierrefers is smoothness in limit .)The juxtaposition of this property with the apparent 'randomness' of the individual positionsof the primes creates a sort of tension which can be witnessed in many popular-mathematicalaccounts of the distribution of prime numbers. Adjectives such as "surprising", "astonishing",
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"remarkable", "striking", "beautiful", "stunning" and "breathtaking" have been used. Zagiercaptures this tension perfectly in the same article:
In 1896, de la Valee Poussin and Hadamard simultaneously proved what had been suspectedfor several decades, and what is now known as the prime number theorem:
In words, the (discontinuous) prime counting function is asymptotic to the (smooth)
logarithmic function x /log x. This means that the ratio of to x /log x can be made
arbitrarily close to 1 by considering large enough x. Hence x /log x provides an approximationof the number of primes less than or equal to x, and if we take sufficiently large x thisapproximation can be made as accurate as we would like (proportionally speaking - verysimply, as close to 100% accuracy as is desired).The original proofs of the prime number theorem suggested other, better approximations. In
the above graph we see that x /log x, despite being asymptotic to , is far from being the
smooth function which suggests when we zoom out - there is plenty of room forimproving the approximation. These improvements turn out to be greatly revealing.
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The first improvement on x/log x we consider is the logarithmic integral function Li(x). Thisis defined to be the area under the curve of the function 1/log u between 2 and x, as illustratedin the left-hand figure. Gauss arrived at this from the empirical fact that the probability of
finding a prime number at an integer value near a very large number x is almost exactly 1/log x.
l'Hopital's rule can be used to show that the ratio of x /log x to tends to 1
as x approaches infinity. Thus we may use either expression as an approximation to inthe statement of the prime number theorem.
In the right-hand figure we see that this function provides an excellent fit to the function. As Zagier states, "within the accuracy of our picture, the two coincide exactly."
Zagier goes on to state:
"There is one more approximation which I would like to mention. Riemann's research onprime numbers suggests that the probability for a large number x to be prime should be evencloser to 1/log x if one counted not only the prime numbers but also the powers of primes,counting the square of a prime as half a prime, the cube of a prime as a third, etc. This leads tothe approximation
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or, equivalently,
The function on the right side . . . is denoted by R( x), in honor of Riemann. It represents an
amazingly good approximation to as the above values show."To be clear about this, it should be pointed out that the explicit definition for the the functionR( x) is
where are the Möbius numbers. These are defined to be zero when n is divisible by asquare, and otherwise to equal (-1)k where k is the number of distinct prime factors in n. As 1
has no prime factors, it follows that (1) = 1.
It seems, then, that the distribution of prime numbers 'points to' or implies Riemann's functionR( x). This function can be thought of as a smooth ideal to which the actual, jagged, primecounting function clings. The next layer of information contained in the primes can be seen
above, which is the result of subtracting from R( x). This function relates directly to thelocal fluctuations of the density of primes from their mean density.
In their article "Are prime numbers regularly ordered?", three Argentinian chaos theoristsconsidered this function, treated it as a 'signal', and calculated its Liapunov exponents. Theseare generally computed for signals originating with physical phenomena, and allow one todecide whether or not the underlying mechanism is chaotic. The authors conclude"...a regular pattern describing the prime number distribution cannot be found. Also, from aphysical point of view, we can say that any physical system whose dynamics is unknown butisomorphic to the prime number distribution has a chaotic behavior."A physicist shown the above graph might naturally think to attempt a Fourier analysis - i.e. tosee if this noisy signal can be decomposed into a number of periodic sine-wave functions. Infact something very much like this is possible. To understand how, we must look at the
Riemann zeta function.
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We encountered this graph earlier. Recall that it shows us the deviations of the prime counting
function from the smooth approximating function R(x). It was hinted that this noisyfunction might somehow decompose into fairly simple component functions. Indeed, this isthe case.The usual process of Fourier analysis essentially decomposes "signals" such as this into
(periodic) sine wave functions. In this case, the component functions are quasi-periodic, basedon sine waves but with a particular kind of logarithmic deformation.Remarkably, the functions in question, the sum of which produces the function seen above,are intimately connected with the nontrivial zeros of the zeta function which we've just seen.
The difference function R( x) - seen earlier can be expressed as the infinite sum over the
set of zeros (both trivial and nontrivial) of the Riemann zeta function:
This sum separates into sums over the trivial and nontrivial zeros respectively. The former is
the relatively simple function R( x-2
) + R( x-4
) + R( x-6
) + ...
The sum over the nontrivial zeros can be expressed as the sum of the sequence of functions {-Tk ( x)} where Tk is defined as follows:
where the and are the k th pair of nontrivial zeta zeros, which we know must becomplex conjugates. The first four functions T1( x), T2( x),T3( x), and T4( x) are pictured above.
Our first apparent obstacle is that and are complex numbers. However, the function xk
can be meaningfully extended from real k to complex k in a fairly straightforward way. Thismeans that the and are also complex- valued.
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This also initially seems like a problem, as the Riemann function R defined earlier as an
approximation of was clearly intended to act on real values only. However, by the sameprocess of analytic continuation discussed earlier, R can be extended to the entire complexplane, taking the form given by the Gram series:
Here ln x is the usual extension of the logarithm function to . Also note the role of theRiemann zeta function.
So we see that and can be given precise meanings, and will yield complexnumbers. Usefully, the imaginary parts of this pair of complex numbers can be shown tocancel, so that their sum which is Tk ( x) will always be a real-valued function.
Some numbers have the special property that they cannot be expressed as the product of twosmaller numbers, e.g., 2, 3, 5, 7, etc. Such numbers are called prime numbers, and they playan important role, both in pure mathematics and its applications. The distribution of suchprime numbers among all natural numbers does not follow any regular pattern; however theGerman mathematician G.F.B. Riemann (1826 - 1866) observed that the frequency of primenumbers is very closely related to the behavior of an elaborate function
ζ (s) = 1 + 1/2s
+ 1/3s
+ 1/4s
+...
called the Riemann Zeta function. The Riemann hypothesis asserts that all interesting solutions of the equation
ζ(s) = 0
lie on a certain vertical straight line. This has been checked for the first 1,500,000,000solutions. A proof that it is true for every interesting solution would shed light on many of themysteries surrounding the distribution of prime numbers.
Primes seem to be, at the same time very irregularly distributed among all numbers, and yet – if squinted at from a sufficiently far distance – they reveal an astoundingly elegant pattern.
Over 2,300 years ago Euclid proved that the number of primes is infinite, so two possiblequestions come to mind:
1. How many primes are there less than the number x?2. There are infinitely many primes, but how big of an infinity?
This document will focus on the first question.1.1. π(x) is the number of primes less than or equal to xLet x be a positive real number. The question "how many primes are there less than x?" hasbeen asked so frequently that its answer has a name:
[π (x) using the Greek letter pi] = π(x) = the number of primes less than or equal to x.
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The primes under 25 are 2, 3, 5, 7, 11, 13, 17, 19 and 23 so π(3) = 2, π(10) = 4 and π25) = 9.
Table 1. Values of π(x)
x π(x) reference
10 4100 251,000 16810,000 1,229100,000 9,5921,000,000 78,49810,000,000 664,579100,000,000 5,761,4551,000,000,000 50,847,53410,000,000,000 455,052,511100,000,000,000 4,118,054,8131,000,000,000,000 37,607,912,01810,000,000,000,000 346,065,536,839100,000,000,000,000 3,204,941,750,802 [LMO85]1,000,000,000,000,000 29,844,570,422,669 [LMO85]10,000,000,000,000,000 279,238,341,033,925 [LMO85]100,000,000,000,000,000 2,623,557,157,654,233 [DR96]1,000,000,000,000,000,000 24,739,954,287,740,860 [DR96]10,000,000,000,000,000,000 234,057,667,276,344,607100,000,000,000,000,000,000 2,220,819,602,560,918,8401,000,000,000,000,000,000,000 21,127,269,486,018,731,928
10,000,000,000,000,000,000,000 201,467,286,689,315,906,290100,000,000,000,000,000,000,000 1,925,320,391,606,803,968,923
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
2 5 110
31 41
0
61 71
0 0
101
0 0
131
0
151
0 0
181 191
3 7 13 23
0
43 53
0
73 83
0
103 113
0 0 0 0
163 173
0
193
0 017
0
37 47
0
67
0 0
97 107
0
127 137
0
157 167
0 0
197
0 019 29
0 0
59
0
79 89
0
109
00
139 149
0 0
179
0
199
π(10) = 4 = 7(0,57143), π(100) = 25 = 7(3,57143), π(1 000) = 168
=7(0,57143)6(7)
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In 1859 the German mathematician Bernard Riemann proposed a way of understanding andrefining that pattern. Its main result is a suggestion, not rigorously proved, for a perfectlyprecise formula giving the number of primes less than a given quantity. His hypothesis haswide – ranging implications, and this day after 150 years of careful research and exhaustive
study we know it is correct.
N π(x) π(N) π(x) = 4(n)7 4 = 8(0,5)
1,00E+03 168 5,9524 168 = 8(0,5)6(7)
7
1,00E+06 78 498 12,739 78 498 = 8(0,5) 2803,5(7)
7
1,00E+09 50 847 534 19,667 50 847 534 = 8(0,5)1 815 983,3514286(7)
7
1,00E+12 37 607 912 018 26,59 37 607 912 018 = 8(0,5)1 343 1139 714,92857(7)
7
1,00E+15 29 844 570 422 669 33,507 29 844 570 422 669 = 8(0,5)1 065 877 515 095,32(7)
7
1,00E+18 24 739 954 287 740 860 40,42 24739954287740860 = 8(0,5)883569795990,745(7)
7
1,00E+21 21127269486018731928 47,332 21127269486018731928=4(754545338786383283,143)7
There are 4 primes up to 10 (2, 3, 5, 7), because those they cannot be expressed as the productof two smaller numbers (4 = 2(2), 8 = 4(2), 9 = 3(3), 10 = 5(2). Between 1 and 100 there are
25 primes, and 168 primes up to 1 000. Why 168? Is there a rule, a formula, to tell me howmany primes there are less than a given number?The formula is simple: The ratio of half a given number by a given number N, is directlyproportional to the quotient of quantity prime numbers by its dual quantity.
½N : N = πx : 2(πx) πx ∝ ½N ½N(2 πx) = N(πx) πx/(2 πx) = y = ½
2πx(½) = πx
N πx
1,0 E+3 168
1,0 E+6 78498
1,0 E+9 508475341,0 E+12 37607912018
1,0 E+15 29844570422669
1,0 E+18 24739954287740860
1,0 E+21 2112726946018731928
A two – column table like this is an illustration of a function. The main idea of a function isthat some number depends on some other number according to some fixed rule or procedure.Another way to say the same thing is: a function is a way to turn (―maps‖) a number in toanother number.
The function πx ∝ ½N turns, or maps, the number 1000 in to the number 168 – again, by wayof some definite procedure. 500(336) = 1000(168)
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Therefore primes there are less than a given number sure do thin out, but are directlyproportional to the half a given number.We know that all prime numbers be congruent to me modulo 7, and this seven tell me howmany primes there are less than a given number. We show in table, that quantity primes thereare less than a given number is always product number 4 = 7(0,57143), and n – the multiple
number 7. If so the formula is:π(x) = 4(n)7, 4 = 7(0,57143), 168 = 4(6)7
Ultimately, it is in the Riemann Hypothesis about the multiplicative basic building blocks of natural numbers to understand: the primes. Can their distribution in the sea of natural numbersmean? How long do you calculate until the next prime coming? Why is the next primenumber, such as accidental times already after a few steps, sometimes on the other hand, onlyafter great distance? Is there perhaps a hidden pattern?
2 0 3 0 5
7 9 11 13 15 17
19 21 23
25 27 29
31 33 35
37 39 41
43 45 47
49 51 53
55 57 59
61 63 65 67 69 71
0
100
200
300
400
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
Serie1 2 11 23 41 59 73 97 109 137 157 179 197 227 241 269 283
Serie2 3 13 29 43 61 79 101 113 139 163 181 199 229 251 271 293
Serie3 5 17 31 47 67 83 103 127 149 167 191 211 233 257 277 307
Serie4 7 19 37 53 71 89 107 131 151 173 193 223 239 263 281 311
7(0,57143) = 4 = π (10), 4(6)7 = 168 = π (1 000)
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73 75 77
79 81 83
85 87 89
91 93 95
97 99 101
103 105 107
109 111 113
115 117 119
121 123 125
127 129 131
133 135 137
139 141 143
145 147 149
151 153 155
157 159 161
163 165 167
169 171 173
175 177 179
181 183 185
187 189 191
193 195 197
199 201 203
205 207 209
211 213 215 217 219 221
223 225 227
229 231 233
235 237 239
241 243 245
247 249 251
253 255 257
259 261 263
265 267 269
271 273 275
277 279 281
283 285 287
289 291 293
295 297 299
301 303 305
307 309 311
313 315 317
319 321 323
325 327 329
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589 591 593
595 597 599
601 603 605
607 609 611
613 615 617
619 621 623
625 627 629
631 633 635
637 639 641
643 645 647
649 651 653
655 657 659
661 663 665
667 669 671
673 675 677
679 681 683
685 687 689
691 693 695
697 699 701
703 705 707
709 711 713
715 717 719
721 723 725
727 729 731 733 735 737
739 741 743
745 747 749
751 753 755
757 759 761
763 765 767
769 771 773
775 777 779
781 783 785
787 789 791
793 795 797
799 801 803
805 807 809
811 813 815
817 819 821
823 825 827
829 831 833
835 837 839
841 843 845
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847 849 851
853 855 857
859 861 863
865 867 869
871 873 875
877 879 881
883 885 887
889 891 893
895 897 899
901 903 905
907 909 911
913 915 917
919 921 923
925 927 929
931 933 935
937 939 941
943 945 947
949 951 953
955 957 959
961 963 965
967 969 971
973 975 977
979 981 983
985 987 989 991 993 995
997 999 1001
All zeros of ζ(s) in the half-plane Re(s) > 0 have real part ½.
2 0 3
0 7 0
5 0 9
0 13 0
11 0 15
0 19 0
17 0 21
0 25 0
23 0 27
0 31 0
29 0 33
0 37 0
35 0 39
0 43 0 41 0 45
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100
0 49 0
47 0 51
0 55 0
53 0 57
0 61 0
59 0 63
0 67 0
65 0 69
0 73 0
71 0 75
0 79 0
77 0 81
0 85 0
83 0 87
0 91 0
89 0 93
0 97 0
95 0 99
0 103 0
[p + (p+8)]/2 = 2n – 1 then 2(2n – 1) = [p + (p+8)] p = [2(2n – 1) – 8]/2[5 + (5 + 8)]/2 = 9 2(9) = [5 + (5 + 8)] 5 = [2(9) – 8]/2
[11 + (11 + 8)]/2 = 15 2(15) = [11 + (11 + 8)] 11 = [2(15) – 8]/2
The Riemann Hypothesis is about the distribution of primes in the sea of natural numbers.This sea is defined over the sum, because of numbers will always be number 1 add – just thenormal process of counting. The primes, however, are about the multiplication defined, theyare about the factorization the prime multiplicative components of the natural numbers.The distribution of primes and the Riemann Hypothesis says something about the changingrelationship between addition and multiplication of natural numbers. Both are not problemsfor themselves, but both together are incredibly complex and still not fully penetrated, such asthe lack of evidence for the Riemann Hypothesis impressive displays.All these ideas are based on an analogy, which is easier to describe something like this lets:The primes are ―elementary particles‖, which are about the multiplication in interaction occur
and so the composite numbers up. At the same time, ―the particles‖ are arranged through theaddition. In the zeta functions are now in the form of a sum – relatively product formula bothaspects (additive/natural numbers and multiplicative/primes) linked.
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2 + 3 = 5 + 2 = 2(2) + 3 = 7 + 3 + 2 = 12 = 6(2)2 + 3 = 5 + 4(2) = 13 = 5(2) + 3 + 5 = 18 = 9(2)
4(2) + 3 = 11 + 4(2) = 19 = 8(2) + 3 + 11 = 30 = 15(2)7(2) + 3 = 17 + 4(2) = 25 = 5(2 + 3) + 17 = 42 = 21(2)
10(2) + 3 = 23 + 4(2) = 31 = 14(2) + 3 + 23 = 54 = 27(2)13(2) + 3 = 29 + 4(2) = 37 = 17(2) + 3 + 29 = 66 = 33(2)
2
3 7 12
5 13 18
11 19 30
17 25 42
23 31 54
29 37 66
43 35 78 41 49 90
47 55 102
53 61 114
59 67 126
73 65 138
71 79 150
85 77 162
83 91 174
89 97 186
103 95 198
0
50
100
150
200
250
1 2 3 4 56 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
2 3 5 11 17 23 29
0
41 47 53 59
0
71
0
83 89
0
101 107 113
0 7 13 19 31 37 43
0 0
61 67 73 79
0 0
97 103 109
0 0
0 0 0 025
0 0 0
49 55
0 0 0 0
85 91
0 0 0
115 121
0 0 0 0 0 0 0
35
0 0 0 0
65
0
77
0 0
95
0 0 0
0 12 18 30 42 54 66 78 90 102 114 126 138 150 162 174186
198210
222234
2 + 3 = 5 + 4(2) = 13 = 5(2) + 3 + 5 = 18 = 9(2)4(2) + 3 = 11 + 4(2) = 19 = 8(2) + 3 + 11 = 30 = 15(2)
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359 367 726
373 365 738
379 371 750
377 385 762
383 391 774
389 397 786
403 395 798
401 409 810
407 415 822
421 413 834
419 427 846
433 425 858
431 439 870
437 445 882
443 451 894
449 457 906
463 455 918
461 469 930
467 475 942
481 473 954
487 479 966
493 485 978
491 499 990
505 497 1002 503 511 1014
509 517 1026
523 515 1038
521 529 1050
535 527 1062
541 533 1074
547 539 1086
553 545 1098
559 551 1110
557 565 1122
563 571 1134
577 569 1146
583 575 1158
589 581 1170
587 595 1182
593 601 1194
599 607 1206
Theorem: The quotient of half a given magnitude ½N, by a given magnitude N, is directlyproportional to the quotient of quantity prime numbers, by its dual quantity.
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Proof: ½N : N = πx : 2(πx) πx ∝ ½N ½N(2 πx) = N(πx) πx/(2 πx) = Re(s) = ½
5/10 = 4/8 = 3/6 = 2/4 = 1/2 8(1/2) = 4 2πx(½) = πx
N
x N 22
1
In the interval of 10 numbers of 5 consecutive odd numbers is always one divisible by 3, sothat it can act like this at most 4 primes, so double large interval of 8 numbers claim ( 4·2 = 819 - 11 = 8 11, 12, 13, 14, 15, 16, 17 , 18, 19). Generally in the interval of 10 numbers eacheven and odd number 2 number interval claims, including in 4(2), 3(2), 2(2), 1(2), primes. If the interval first 10 numbers come before 4 primes, then in the interval of 100 numbers 25prime numbers occur, and each number in the interval, the proportion ½ is to keep. In otherwords, all zeros of ζ(s) in the half-plane Re(s) > 0 have real part ½ .
In mathematics, two quantities are said to be proportional if they vary in such a way that oneof the quantities is a constant multiple of the other, or equivalently if they have a constantratio. Proportion also refers to the equality of two ratios. In proportional quantities is thedoubling (tripling, halved) one quantity is always a double (triple, halve) connected to theother quantities.
2πx πx N ½ N 2 2 2 2
8 4 10 5 2 3 5 7
16 8 20 10 11 13 17 19
20 10 30 15 0 23 0 29
24 12 40 20 31 0 37 0
30 15 50 25 41 43 47 0
34 17 60 30 0 53 0 59
38 19 70 35 61 0 67 0
44 22 80 40 71 73 0 79
48 24 90 45 0 83 0 89
50 25 100 50 0 0 97 0
58 29 110 55 101 103 107 109
The proportion of ½ means, that is involved in the creation of a half-block of numbers twicethe amount of prime numbers. 5/10 = 4/8 50/100 = 25/50 500/1000 = 168/336
1 = 3 – 2 5 = 3 + 2 7 = 5 + 2 9 = 7 + 2
11 = 9 + 2 13 = 11 + 2 15 = 13 + 2 17 = 15 + 2 19 = 17 + 2 21 = 19 + 2 23 = 21 + 225 = 23 + 2 27 = 25 + 2 29 = 27 + 2 31 = 29 + 2 33 = 31 + 2 35 = 33 + 2 37 = 35 + 239 = 37 + 2 41 = 39 + 2 43 = 41 + 2 45 = 43 + 2 47 = 45 + 2 49 = 47 + 2 51 = 49 + 253 = 51 + 2 55 = 53 + 2 57 = 55 + 2 59 = 57 + 2 61 = 59 + 2 63 = 61 + 2 65 = 63 + 267 = 65 + 2 69 = 67 + 2 71 = 69 + 2 73 = 71 + 2 75 = 73 + 2 77 = 75 + 2 79 = 77 + 281 = 79 + 2 83 = 81 + 2 85 = 83 + 2 87 = 85 + 2 89 = 87 + 2 91 = 89 + 2 93 = 91 + 295 = 93 + 2 97 = 95 + 2 99 = 97 + 2
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Assuming proportional functions graphically in a coordinate system, so you can see thatproportional functions are monotonically increasing. The properties of the zeroes out in thecomplex plane determine the properties of the primes! Riemann conjectured that all the
relevant zeroes have real part ½ . The statement that the equation πx/2(πx) = y = ½, is valid
for every x with real part equal ½, with the quotient on the right hand side converging, isequivalent to the Riemann hypothesis.
2 3 5 711 13
17 19 23
29 31
3741 43
4753
59 6167
71 73
7983
89
97
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
πx ∝ ½N ½N(2πx) = N(πx) 2πx(½) = πx 8 · 0,5 = 4 50 · 0,5 = 25
y = 0,5x
R² = 1
0
5
10
15
20
25
30
35
0 10 20 30 40 50 60 70
Y
X
πx = 2πx(0,5) 4 = 8(0,5) 25 = 50(0,5) 168 = 336(0,5)
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Riemann for help of total numbers translated distribution of prime numbers in mathematicalscenery on two-dimensional plane (so called zeta-function). It topography of this scenerycontains near this general knowledge about prime numbers. It will suffice, so to know onlevel of sea points (zero places), to can reconstruct whole scenery.
2
3
5
7 11
13
17
1923
29
31
37
41
4347
53
59
61
6771
73
79 83
8997
101
103
πx ∝ ½N ½N(2πx) = N(πx) 2πx(½) = πx 8 · 0,5 = 4 50 · 0,5 = 25
20
50
11
0
17
0
23
0
29
0
35
0
41
0
47
0
53
0
59
0
65
0
71
0
77
0
83
0
89
00
7
0
13
0
19
0
25
0
31
0
37
0
43
0
49
0
55
0
61
0
67
0
73
0
79
0
85
0
91
0
97
3
0
9
0
15
0
21
0
27
0
33
0
39
0
45
0
51
0
57
0
63
0
69
0
75
0
81
0
87
0
93
0
0 5 10 15 20 25 30 35
[p + (p + 8)]/2 = 2n - 1 [5 + (5 + 8)]/2 = 9 [11 + (11 + 8)]/2 = 15 [ 17 +
(17 + 8)]/2 = 21 [23 + (23 + 8)]/2 = 27
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Because zero places contain all information about distribution of prime numbers. Riemanncreated concrete formula, to right away zero to regain distribution of prime numbers. Nearwhat every zero place is how source for spreading wave we which can introduce me asacoustic sound. Sounds of all zero places overlap on me in distribution of prime numbers.Near what zero place is about so many louder, if it lies further eastwards (in right from axis -
y), and her sound is about so many higher, if it lies further north (over axis - x).
They fill with the same gap in thousands theorems basing on legitimacy Riemann‘shypothesis. Because many mathematicians be obliged for its results such presumption toaccept. Primes betrayed their secret, and by this was proved Riemann‘s Hypothesis
A solution of the Riemann Hypothesis are huge implications for many other mathematicalproblems. The transformation of hypothesis the Riemann in theorem, suddenly it proves allthe not proved results.
Riemann Hypothesis admits to receive, so that really every from infinitely of many, of zeroplaces lies on this straight line then it means, that all sounds in music of prime numbers arealike loud. This would mean, it that was can distribution of prime numbers to me reallyintroduce how even throw dice. Hexahedron dice after line of natural numbers rolls, whichwhat second and fourth wall shows next prime number or almost prime.
5_7__11_13__17_19__23_25=5·5__29_31__35=5·7_37__41_43__47_49=7·7__53_55=11·5
__59_61__65 = 13·5_67__71_73__77 = 11·7_79__83_85 = 17·5__89_91 = (13·7)_95 = 19·5
y = 0,5x
R² = 1
-5E+17
0
5E+17
1E+18
1,5E+18
2E+18
2,5E+18
-2E+18 0 2E+18 4E+18 6E+18
y
x
πx ∝ ½N ½N(2πx) = N(πx) 8 · 0,5 = 4 50 · 0,5 = 25
336 · 0,5 = 168 πx/(2 πx) = y = ½ 2πx(½) = πx
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½ N = πx + [½ N – (πx + N/6)] + N/6 50 = 25 + [50 – (25 +16)] + 16
2
0
5
0
11
0
17
0
23
0
29
0
35
0
41
0
47
0
53
0
59
0
65
0
71
0
77
0
83
0
89
0
95
0
101
0
0
3
0
9
0
15
0
21
0
27
0
33
0
39
0
450
510
570
630
690
750
810
870
930
990
105
7
0
13
0
19
0
25
0
31
0
37
0
43
0
49
0
55
0
61
0
67
0
73
0
79
0
85
0
91
0
97
0
103
0
109
0
0%
10%
20%
30%
40%
50%
60%
70%
80%
90%
100%
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
2
0
30
5
7
9
11
13
15
17
19
21
23 25
27
29
31
3335
37
39
41
43
45
47 49
51
53 55
57
59
61
6365
67
69
71
73
7577
79
81
83 85
87
89 91
9395
97
99
101
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Riemann hypothesis:The non-trivial zeros (lie zeros in the strip right next to the y-axis with real part of s from 0 to 1)are all on a line parallel to the y-axis with real part ½ .
Proof:
547 89 131 173 215 257 299 341 383 425 467 509 551 593 635 677 719 761 803 845 887 929971
7
49 91 133 175 217 259 301 343 385 427 469 511 553 595 637 679 721 763 805 847 889 931973
11
5395 137 179 221 263 305 347 389 431 473 515 557 599 641 683 725 767 809 851 893 935977
13
5597 139 181 223 265 307 349 391 433 475 517 559 601 643 685 727 769 811 853 895 937979
17
59101 143 185 227 269 311 353 395 437 479 521 563 605 647 689 731 773 815 857 899 941983
19
61103 145 187 229 271 313 355 397 439 481 523 565 607 649 691 733 775 817 859 901 943985
23
65107 149 191 233 275 317 359 401 443 485 527 569 611 653 695 737 779 821 863 905 947989
25
67109 151 193 235 277 319 361 403 445 487 529 571 613 655 697 739 781 823 865 907 949991
29
71113 155 197 239 281 323 365 407 449 491 533 575 617 659 701 743 785 827 869 911 953995
31
73115 157 199 241 283 325 367 409 451 493 535 577 619 661 703 745 787 829 871 913 955997
35
77119 161 203 245 287 329 371 413 455 497 539 581 623 665 707 749 791 833 875 917 9591001
37
79 121 163 205 247 289 331 373 415 457 499 541 583 625 667 709 751 793 835 877 919 96110034183 125 167 209 251 293 335 377 419 461 503 545 587 629 671 713 755 797 839 881 923 9651007
43 85 127 169 211 253 295 337 379 421 463 505 547 589 631 673 715 757 799 841 883 925 9671009
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
1/2N = πx + [1/2N - (πx + N/6)] + N/6 500 = 168 + [500 - (168 + 166)] +
166
0 2 33 7513111917
2523
3129
3735
4341
4947
5553
6159
6765
7371
79
1 2 3
y^- 5y + 6 = 0 x^= 1 y = x+ 1 = 2 y = 4x - 1 = 3+ 4= 7 y= 8 x- 3= 5+ 8 = 13
y=8x+3 = 11 + 8= 19 y=8x+9= 17
y=8x+15=23 + 8=31 y=8x+21=29+8=37 y= 8x+27=35 +8=43 y=8x+33=41
y=8x+39=47 y= 8x+45=53+8=61
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What is the solution of the equation y² - 5y + 6 = 0? The solution is 2 und 3. Another way to say,
this is that 2 and 3 are the zeros of the function y ² - 5y + 6 = 0.
y y‘ ½ y
y² - 10y + 21 = 0 Solution 3 and 7 5
y² - 18y + 65 = 0 - „ - 5 and 13 9
y² - 30y + 209 = 0 - „ - 11 and 19 15
y² - 42y + 425 = 0 - „ - 17 and 25 = 5(5) 21
y² - 54y + 713 = 0 - „ - 23 and 31 27
y² - 66y +1073 = 0 - ― - 29 and 37 33
y² - 78y +1505 = 0 5(7 )= 35 and 43 39
y² - 90y +2009 = 0 - ― - 41 and 49 = 7(7) 45
y² - 102y +2585 = 0 47 i 55 = 5(11) 51
q.e.d.
2 0 00 2,50 03
05
0
7
05
0
9
0
13
0
11
0
15
0
19
0
17
0
21
0
25
0
23
0
27
0
31
0
29
0
33
0
37
0
35
0
39
0
43
0
41
0
45
0
49
0
47
0
51
0
55
0
53
0
57
0
61
0
59
0
63
0
67
0
65
0
69
0
73
0
71
0
75
0
79
0
77
0
81
0
85
0
83
0
87
0
91
0
89
0
93
0
97
0
95
1 2 3
(y + y')/2 = 1/2 y (2+3)/2=2,5 (3+7)/2=5 (5+13)/2=9 (11+19)/2=15(17+25)/2=21 (23+ 31)/2=27 (29+37)/2=33
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A zero off the critical line would induce a pattern in the distribution of the primes.
2
19
29
0 0
59
0
79
89
0
109
313
23
0
43
53
0
73
83
0
103
7 17
0
37
47
0
67
0 0
97
107
5 11
0
31
41
0
61
71
0 0
101
0 0 0 0 0 0 0 0 0 0 0
0 2 4 6 8 10 12
p + 5(6) = p' 7 + 5(6) = 37 + 5(6) = 67 + 5(6) = 97 + 5(6) = 127 + 5(6) = 157
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114
97 101 103
107 109
115 119 113
121 125 127
133 131
143 137 139
145 149 151
155 157
161 163 167
169 173
175 179 181
187 185 191
193 197 199
205 203
209 211
217 215 223
221 227 229
235 233 239
247 245 241
253 251
259 257 263
265 269 271
275 277
287 281 283 289 293
295 299
301 305 307 311
319 313 317
325 323
329 331
343 335 337
341 347 349
355 353 359
361 365 367
The Riemann Hypothesis had been proved, and we are able, to answering the severity of theproblem of Goldbach to go, whether each grade number as the sum of two primes is representable. If proportionality factor all primes in a given quantity ½ is, but this means that the
equation πx/2πx = ½N/N is the answer to the problem of Goldbach. She says that every evennumber is composed of two primes.
Theorem: The quotient of quantity prime numbers by its dual quantity, is directly proportionalto the quotient of quantity even numbers by a given magnitude.
Proof: πx/2πx = 2n/N 25/50 = 50/100 = ½
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The proportion of ½ in the case of even numbers means that all even numbers in a block madeup of two primes. 2 + 2 = 4 3 + 3 = 6 3 + 5 = 8 5 + 5 = 10 That is to say 50 even numbers in a block of 100 numbers, is the sum of 4(25) primes, asshown in the diagram below.
-2
410 16 22 28 34 40 46 52 58 64 70 76 82 88 94 100
0
612 18 24 30 36 42 48 54 60 66 72 78 84 90 96 102
2
8 14 20 26 32 38 44 50 56 62 68 74 80 86 92 98 104
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
2n/N = πx/2πx 50/100 = 25/50
Serie1 Serie2 Serie3
2 55
11 11 17 17 23 23 2923 29 29
41 41 41
47
25 11 11 17 17 23 23 29 29 41 41 47 41 47 53
53
4
10 16 22 28 34 40 46 52 58 64 70 76 82 88 94
100
35 7
11 13 17 19 17 23 29 29 31 37 41 43 433
7 11 13 17 19 23 29 31 31 37 41 41 43 47 53
6
12 18 24 30 36 42 48 54 60 66 72 78 84 90 963
7 713 13
1913 19 19
31 31 37 37 4331 31
57 13 13 19 19 31 31 37 31 37 37 43 43 61 67
8 14 20 26 32 38 44 50 56 62 68 74 80 86 92 98
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
πx/2πx = 1/2N/N 2n = p + p 25/50 = 50/100 4 = 2 + 2
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How we see on above diagram, sum two prime numbers lies always on parallel straight line toaxis - y and it is even numbers that is consisting with two prime numbers.
0 002
02
02
0
4
0
3
0
3
0
6
0
3
0
5
0
8
0
5
0
5
0
10
0
5
0
7
0
12
0
7
0
7
0
14
0
5
0
11
0
16
0
7
0
11
0
18
0
7
0
13
0
20
0
11
0
11
0
22
0
11
0
13
0
24
0
13
0
13
0
26
0
11
0
17
0
28
0
11
0
19
0
30
0
13
0
19
0
32
0
17
0
17
0
34
0
17
0
19
0
36
0
19
0
19
0
38
0
17
0
23
0
40
0
19
0
23
0
42
0
13
0
31
0
44
0
23
0
23
0 0
0 0,5 1 1,5 2 2,5 3 3,5
n/p + p = 1/2 2n = p + p' 2+2=4 3+3=6 3+5=8 5+5=10 5+7=12
7+7=14 5+11=16 7+11=18 7+13=20...
2 23
0
7
0
3 3 3
0
9
0
3 3
5
0
11
0
3
5 5
0
13
0
5 5 5
0
15
0
5 5
7
0
17
0
5
7 7
0
19
0
5 5
11
0
21
0
5 5
13
0
23
0
7 7
11
0
25
0
7 7
13
0
27
0
1 2 3
2n-1=p+p´+p" 7= 2+2+3 9 = 3+3+3 11=3+3+5 13=3+5+5 15=5+5+5
17=5+ 5+7 19=5+7+7 19-16=17-14=15-12=13-10=11-8=9-6=7-4=3
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Legitimacy "strong‖ hypothesis Goldbacha proves legitimacy "weak‖ hypothesis Goldbacha -because 3 have will sufficed from given odd larger number since 7 to subtract and tointroduce received even number with strong hypothesis Goldbacha peaceably.
(2n - 1) - 3 = 2n = p + p' → 2n - 1 = p + p + p
Also, as you can see, every odd integer greater than 5 is the sum of 3 primes, because thedifference between odd and even numbers always of prime numbers 3 is.
4
0
4
0
7
0
6
0
6
0
9
0
8
0
8
0
11
0
10
0
10
0
13
0
12
0
12
0
15
0
14
0
14
0
17
0
16
0
16
0
19
0
18
0
18
0
21
0
20
0
20
0
23
0
22
0
22
0
25
0
24
0
24
0
27
0
26
0
26
0
29
0
28
0
28
31
0 0,5 1 1,5 2 2,5 3 3,5
2n + p = 2n - 1 = p + p + p' 2 + 2 + 3 = 7 3 + 3 + 3 = 9 3 + 5 + 3 = 11
5 + 5 + 3 = 13 5 + 7 + 3 = 15
0
2
4
6
8
10
12
14
16
17
16
14
12
10
8
6
4
2
0
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 0
p+p=2n+p=2n-1 2+2=4+3=7 3+3=6+3=9 3+5=8+3=11 5+5=10+3=13
5+7=12+3 =15 7+7=14+3=17
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In addition to the familiar question of whether there are infinitely many prime pairs withdifference 2 there. The six- wide array further helps to demonstrate the otherwise stillunproven conjecture that there must be infinitely many twin primes.Here are the reasons for this: if there are infinite primes, then twin pairs, with even number
divisible by 3 shares.
5 70
60
11 13
0
12
0
17 19
0
18
0
29 31
0
30
0
41 43
0
42
0
59 61
0
60
0
71 73
0
72
0
101 103
0
102
0
107 109
0
108
0
137 139
0
138
0
149 151
0
150
0
1 2 3
p + (p +2)/2 = 2n/3 5+7/2=6/3 11+13/2=12/3 17+19/2=18/3
29+31/2=30/3 41+43/2=42/3
0 0 0
2
0
32,5
3
7
5
5
13
9
11
0
19
15
2y+3y=5y/2=2+0,5=3-0,5 3y+7y=10y/2=3+2=7-2
5y+13y=18y/2=5+4=13-4 11y+19y=30y/2=11+4=19-4
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Looking closer at the above graph, you will see that half of the following sums of two primeson a straight line parallel to the y - axis with real part ½ y lie. This means that the linear Diophantine equation ax + by - c = 0, with given integer pairs not have common divisorCoefficient a, b, c, always in prime x, y is solvable.1(2) + 1(3) – 5 = 0 1(3) + 1(7) – 10 = 0 1(5) + 1(13) – 18 = 0 1(11) + 1(19) – 30 = 0
Still, then, we go on listing to that mysterious prime numbers beat: 2, 3, 5, 7, 11, 13, 17, 19.The primes stretch out into the far reaches of the universe of numbers, never running dry. Dowe really have to accept that, despite our desire for order and explanation, these fundamentalnumbers might forever remain out of reach?
Sale long we reflect upon with perspective Gauß and Riemann‘s and we should earlier alreadylook for different possibilities, to better to get know these full of secrets numbers. The primesbetrayed in end their secret, and remain not an unanswered riddle. I‘m who made the primessing.
The system of numbers is not the man‘s invention, because in distribution of primes and
almost primes uncover coded plan of building of not only nature, but whole universe. Thereality of existence of timeless plan motivates consideration, or also for events in time andspace that is our history does not hide transcendent guidance.
Number reveals divine think and order. It permits simultaneously to get to know the basicstructures of reality. Number assures insight in God‘s internal secret and the world‘s. Who knows definite number, possesses power. Calculating man makes something like howonly God, when full arranging power over things: he distinguishes and it assigns, he definesborders and it unites together. He places also qualitative standards: what it lays on first place,is most valuable, important, and significant.
Then our perception called with science would bring us to confession with belief for Book Wisdom 11, 21: ―But you settled all according to measure, number and weight‖, and under -standing his deep meaning. Seemingly disarrays are regulated, for what God let will be thanksthat we need not what the least million years wait on understanding of mysterious nature of primes.
FOR THE GREATER GLORY OF GOD.
„AD MAJOREM DEI GLORIAM
SOLI DEO HONOR ET GLORIA.
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2,3,5,11,13,29, + n(7) = p p(p) + 6(7) = p‗(p―) 7(7) + 6(7) = 13(7) 5(7) + 6(7) = 11(7)
2 3 5 7
17 19 13 11
23 29 25
37 31 35
47 41 43 49
59 53 55
61 67 65
79 73 71 77
89 83 85
97 91 95
107 101 103 109
113 119 115
127 121 125
131 139 137 133
149 143 145
157 151 155
163 167 169 161
173 179 175
181 185 187
191 199 193 197
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121
205 203 209
211 215 217
227 229 223 221
233 239 235
241 247 245
257 251 259 253
269 263 265
271 277 275
283 281 289 287
293 299 295
307 305 301
317 311 313 319
325 323 329
331 337 335
349 347 341 343359 353 355
367 361 365
373 379 377 371
383 389 385
397 391 395
401 409 403 407
419 415 413
421 425 427
439 433 431 437
443 449 445
457 455 451
467 461 463 469
479 475 473
487 485 481
499 491 497 493
509 503 505
511 515 517
521 523 527 529
535 533 539541 547 545
557 559 551 553
569 563 565
577 571 575
587 589 581 583
593 599 595
607 601 605
619 613 617 611
625 623 629
631 637 635
647 643 641 649
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122
653 659 655
661 667 665
677 673 679 671
683 685 689
691 697 695
709 701 703 707
719 715 713
727 721 725
733 739 731 737
743 745 749
751 757 755
761 769 763 767
773 775 779
787 781 785
797 793 791 799809 805 803
811 817 815
821 829 823 827
839 835 833
841 845 847
857 859 853 851
863 865 869
877 871 875
887 881 883 889
895 893 899
907 901 905
919 911 913 917
929 925 923
937 931 935
947 941 943 949
953 955 959
967 961 965
971 977 973 979
983 985 989
997 991 995
1009 1007 1001 1003
1013 1019 1015
1021 1025 1027
1031 1039 1033 1037
1049 1045 1043
1051 1057 1055
1069 1063 1061 1067
1075 1073 1079
1087 1081 1085
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123
1097 1091 1093 1099
1109 1103 1105
1117 1111 1115
1129 1123 1121 1127
1135 1133 1139
1141 1145 1147
1151 1153 1157 1159
1163 1165 1169
1171 1177 1175
1181 1187 1183 1189
1193 1195 1199
1201 1207 1205
1213 1217 1219 1211
1223 1229 1225
1237 1231 12351249 1243 1241 1247
1259 1255 1253
1261 1265 1267
1277 1279 1271 1273
1283 1289 1285
1297 1291 1295
1307 1301 1303 1309
1319 1315 1313
1321 1327 1325
1333 1331 1337 1339
1345 1343 1349
1351 1355 1357
1367 1361 1369 1363
1373 1375 1379
1381 1387 1385
1399 1393 1391 1397
1409 1405 1403
1411 1415 1417
1423 1427 1429 14211433 1439 1435
1447 1441 1445
1451 1459 1453 1457
1465 1463 1469
1471 1477 1475
1487 1489 1483 1481
1493 1499 1495
1501 1505 1507
1511 1513 1517 1519
1523 1525 1529
1531 1537 1535
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124
1549 1543 1541 1547
1559 1553 1555
1567 1565 1561
1571 1579 1573 1577
1583 1585 1589
1597 1591 1595
1601 1609 1607 1603
1619 1613 1615
1627 1621 1625
1637 1633 1631 1639
1645 1643 1649
1657 1651 1655
1669 1663 1667 1661
1675 1673 1679
1681 1685 16871697 1699 1693 1691
1709 1705 1703
1711 1715 1717
1721 1723 1729 1727
1733 1735 1739
1741 1747 1745
1759 1753 1751 1757
1765 1769 1763
1777 1771 1775
1787 1783 1789 1781
1795 1799 1793
1801 1807 1805
1811 1813 1817 1819
1823 1825 1829
1831 1837 1835
1847 1843 1841 1849
1855 1853 1859
1867 1861 1865
1871 1879 1873 18771889 1885 1883
1891 1895 1897
1907 1901 1909 1903
1913 1915 1919
1921 1925 1927
1931 1933 1937 1939
1949 1943 1945
1951 1957 1955
1963 1961 1967 1969
1979 1973 1975
1987 1985 1981
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125
1997 1993 1999 1991
2003 2009 2005
2011 2017 2015
2029 2027 2023 2021
2039 2035 2033
2041 2045 2047
2053 2051 2057 2059
2063 2069 2065
2071 2075 2077
2081 2089 2083 2087
2099 2095 2093
2101 2105 2107
2113 2111 2117 2119
2129 2123 2125
2137 2131 21352141 2143 2149 2147
2153 2155 2159
2161 2167 2165
2179 2173 2171 2177
2185 2183 2189
2191 2195 2197
2207 2203 2201 2209
2213 2219 2215
2221 2225 2227
2239 2237 2231 2233
2243 2245 2249
2251 2257 2255
2267 2269 2263 2261
2273 2275 2279
2287 2281 2285
2293 2297 2299 2291
2309 2303 2305
2311 2315 2317
2329 2321 2327 23232333 2339 2335
2347 2341 2345
2357 2351 2353 2359
2363 2369 2365
2371 2377 2375
2389 2383 2381 2387
2399 2393 2395
2407 2405 2401
2417 2411 2413 2419
2423 1429 2425
2437 2435 2431
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126
2441 2447 2443 2449
2459 2455 2453
2467 2461 2465
2473 2477 2471 2479
2485 2483 2489
2491 2495 2497
2503 2501 2507 2509
2519 2513 2515
2521 2525 2527
2539 2531 2537 2533
2543 2549 2545
2557 2551 2555
2569 2561 2567 2563
2579 2573 2575
2581 2585 25872593 2591 2597 2599
2609 2603 2605
2617 2611 2615
2621 2623 2627 2629
2633 2635 2639
2647 2641 2645
2659 2657 2653 2651
2663 2665 2669
2677 2671 2675
2683 2687 2689 2681
2693 2699 2695
2707 2701 2705
2711 2719 2713 2717
2729 2725 2723
2731 2737 2735
2749 2741 2743 2747
2753 2755 2759
2767 2761 2765
2777 2773 2771 27792789 2785 2783
2791 2797 2795
2803 2801 2809 2807
2819 2815 2813
2821 2825 2827
2837 2833 2831 2839
2843 2845 2849
2851 2857 2855
2861 2863 2867 2869
2879 2875 2873
2887 2881 2885
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127
2893 2891 2899
2903 2909 2905
2917 2915 2911
2927 2929 2921 2923
2939 2933 2935
2941 2945 2947
2957 2953 2951 2959
2963 2969 2965
2971 2977 2975
2989 2981 2987 2983
2999 2995 2993
3001 3007 3005
3019 3011 3013 3017
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11161 11165 11167
11177 11173 11171 11179
11183 11189 11185
11197 11195 11191
11207 11201 11209 11203
11213 11219 11215
11225 11227 11221
11239 11237 11231 11233
11243 11249 11245
11251 11257 11255
11261 11267 11263 11269
11279 11273 11275
11287 11285 11281
11299 11291 11297 11293
11303 11309 11305
11317 11311 11315
11321 11329 11323 1132711335 11333 11339
11341 11345 11347
11353 11351 11357 11359
11369 11363 11365
11371 11375 11377
11383 11381 11387 11389
11399 11393 11395
11401 11405 11407
11411 11413 11417 11419
11423 11425 11429
11437 11431 11435
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146
11447 11443 11441 11449
11455 11453 11459
11467 11461 11465
11471 11473 11477 11479
11489 11483 11485
11497 11491 11495
11503 11501 11507 11509
11519 11513 11515
11527 11525 11521
11539 11537 11531 11533
11549 11545 11543
11551 11555 11557
11569 11567 11561 11563
11579 11575 11573
11587 11581 1158511597 11593 11599 11591
11605 11603 11609
11617 11611 11615
11621 11623 11627 11629
11633 11639 11635
11647 11645 11641
11659 11651 11653
11669 11663 11665
11677 11675 11671
11681 11689 11687 11683
11699 11693 11695
11701 11705 11707
11719 11717 11711 11713
11729 11723 11725
11731 11735 11737
11743 11741 11747 11749
11753 11759 11755
11761 11765 11767
11777 11779 11771 11773 11783 11789 11785
11791 11795 11797
11807 11801 11809 11803
11813 11819 11815
11821 11827 11825
11839 11833 11831 11837
11845 11849 11843
11857 11855 11851
11867 11863 11861 11869
11879 11873 11875
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12319 12311 12317 12313
12329 12323 12325
12331 12335 12337
12343 12347 12341 12349
12353 12359 12355 12361 12365 12367
12379 12377 12373 12371
12385 12383 12389
12391 12397 12395
12409 12401 12403 12407
12413 12415 12419
12421 12427 12425
12437 12433 12439 12431
12445 12443 12449
12451 12457 12455
12469 12467 12461 12463
12479 12473 12475
12487 12481 12485
12497 12491 12493 12499
12503 12509 12505
12511 12517 12515
12527 12529 12521 12523
12539 12535 12533
12547 12541 12545
12553 12559 12557 12551
12569 12565 12563
12577 12571 12575
12589 12583 12581 12587
12595 12593 12599
12601 12607 12605
12619 12613 12611 12617
12625 12629 12623
12637 12631 12635
12647 12641 12643 12649
12659 12653 12655
12667 12665 12661
12671 12673 12677 12679
12689 12685 12683
12697 12691 12695
12703 12709 12707 12701
12713 12715 12719
12721 12727 12725
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12739 12733 12731 12737
12743 12745
12757 12751 12755
12763 12769 12761 12767
12775 12773 12779 12781 12787 12785
12791 12799 12793 12797
12809 12805 12803
12817 12815 12811
12829 12823 12821 12827
12839 12833 12835
12841 12845 12847
12853 12857 12851 12859
12863 12869 12865 12877 12875 12871
12889 12887 12881 12883
12899 12893 12895
12907 12905 12901
12917 12911 12919 12913
12923 12929 12925
12935 12931 12937
12941 12947 12943 12949
12959 12953 12955
12967 12965 12961
12973 12979 12971
12983 12989 12977 12985
12995 12997 12991
13001 13009 13007 13003
13013 13019 13015
13025 13027 13021
13037 13033 13031 13039
13043 13049 13045 13055 13051 13057
13063 13067 13061 13069
13079 13073 13075
13085 13081 13087
13099 13093 13097 13091
13109 13103 13105
13115 13117 13111
13127 13121 13123 13129
13139 13133 13135 13147 13145 13141
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150
13151 13159 13157 13153
13163 13169 13165
13177 13171 13175
13183 13187 13181 13189
13199 13193 13195 13205 13201 13207
13219 13217 13211 13213
13229 13223 13225
13235 13231 13237
13249 13241 13247 13243
13259 13253 13255
13267 13265 13261
13277 13271 13273 13279
13289 13283 13285
13291 13297 13295
13309 13301 13307 13303
13313 13319 13315
13327 13325 13321
13337 13331 13339 13333
13343 13349 13345
13355 13351 13357
13367 13361 13363 13369
13373 13375
13381 13379 13385 13387
13399 13397 13391 13393
13403 13409 13405
13417 13411 13415
13421 13427 13423 13429
13433 13439 13435
13441 13445 13447
13457 13451 13453 13459
13463 13469 13465
13477 13475 13471
13487 13481 13483 13489
13499 13493 13495
13505 13501 13507
13513 13517 13511 13519
13523 13529 13525
13537 13535 13531
13541 13547 13543 13549
13553 13559 13555
13567 13565 13561
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13577 13571 13573 13579
13583 13585
13597 13591 13589 13595
13607 13609 13601 13603
13613 13619 13615 13627 13621 13625
13633 13637 13631 13639
13649 13645 13643
13651 13655 13657
13669 13667 13663 13661
13679 13675 13673
13687 13681 13685
13697 13693 13691 13699
13709 13703 13705 13711 13715 13717
13629 13723 13721 13727
13739 13735 13733
13741 13745 13747
13757 13751 13759 13753
13763 13765 13769
13771 13775 13777
13781 13789 13783 13787
13799 13795 13793
13807 13801 13805
13817 13813 13811 13819
13829 13825 13823
13831 13837 13835
13841 13843 13847 13849
13859 13853 13855
13867 13865 13861
13877 13879 13873 13871
13883 13889 13885
13897 13895 13891
13907 13903 13901 13909
13913 13919 13915
13921 13927 13925
13933 13931 13939 13937
13943 13945 13949
13955 13951 13957
13967 13963 13961 13969
13973 13979
13981 13985 13987
13999 13997 13993 13991
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14009 14003 14005
14011 14015 14017
14029 14021 14027 14023
14033 14039 14035
14045 14047 14041 14051 14057 14059 14053
14063 14069 14065
14071 14075 14077
14087 14083 14081 14089
14093 14095 14099
14107 14101 14105
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E X P O S É. Jan Lubinaul. Korfantego 51/10Pl. 43-200 Pszczynae-mail: [email protected]
Personal informationI was born in Katowice/Poland on 01.05.1947. I'm no mathematician, but mathematics hasbecome increasingly intriguing to me. I find it extremely fascinating. As my earliermathematical research had not involved number theory.Mathematics is often referred to as the search for patterns; it is the language of nature.Everything around us can be represented and understood through numbers, and primenumbers are the building block of all numbers, the DNA of arithmetic.But the famous mathematician Euler had to say about the primes: "There are some mysteriesthat the human mind will never penetrate. To convince ourselves we have only to cast aglance at tables of primes and we should perceive that there reigns neither order nor rule."And a Hungarian mathematician Paul Erdös, who spent his career investigating primenumbers, said that it will be another million years, at least, before we understand the primes.Despite Euler´s and Erdös pessimism, I have found ways to understand the primes, and theirdistribution.Are there formulas that produce some of the prime? Here you are! p = n(2) + 32 = 1(2) + 0 3 = 0(2) + 3 5 = 1(2) + 3 7 = 2(2) + 3 11 = 4(2) + 3 13 = 5(2) + 3
17 = 7(2) + 3 19 = 8(2) + 3 23 = 10(2) + 3 233 = 115(2) + 3 251 = 124(2) + 3Why a number is prime? Because could be written as two smaller numbers multipliedtogether. That is, it is not possible to represent a prime as the product of two integers a x b with a, b > 1. Let q and r be the quotient and remainder of the division of n by d . That is, foreach n and d, let n = d q + r, where r and q are positive integers and 0 ≤ r < d .
Because all prime numbers contain in me one 3, it was not possible divide here by two.Superiority meanwhile 2 it causes, that they don´t divide by three also. So they are indi visibleby all different numbers, and on this depends the complete primality certificate!"God does not play dice with the universe." /A. Einstein/, also not with the prime numbers.The nature used not a dice to decide if the number N is prime, but rule of congruence modulo
p‘≡ p mod 7. p = n(7) + (1, 2, 3, 4, 5, 6) 11 = (7) + 4 13 = (7) + 6 17 = 2(7) + 3
19 = 2(7) + 5 23 = 3(7) + 1 29 = 3(7) + 4(2) 31 = 4(7) + 3 37 = 5(7) + 2 p‘ – p = n(7) 19 – 5 = 2(7) 47 – 19 = 4(7) 61 – 47 = 2(7) 89 – 61 = 4(7) 103 – 89 = 2(7)So the next question is, can we understand how the primes are distributed? Can the primes befitted into a pattern in the way that the elements can be organized in the Periodic Table? Theanswer is yes!
And I find a good model for the way the primes are distributed. It looks like they have beenchosen with 6 sides dice on one side what two and four space is painted successive prime 2, 3,_5_7__11_13__17_19__23_25__29_31__35_37__41_43__47_49__53_55__59_61__65
_67__71_73__77_79__83_85, and almost prime / 25, 35, 49, 55, 65, 77, 85 /. The primenumbers are distributed not chaotically. All prime and almost prime numbers to be congruent
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modulo 7.Because the smallest gap between their equal 2 + 4 = 6, and 6(7) = 42 than ispossible to predict with arbitrary accuracy that the next one lies what 42 gap.
I II III IV V VI VII VIII IX X XI XII XIII XIV
5 7 11 13 17 19 23
25 29 31 35 37 41 43
47 49 53 55 59 61 65
67 71 73 77 79 83 85
89 91 95 97 101 103 107
109 113 115 119 121 125 127
131 133 137 139 143 145 149
151 155 157 161 163 167 169
173 175 179 181 185 187 191
193 197 199 203 205 209 211
215 217 221 223 227 229 233
235 239 241 245 247 251 253
257 259 263 265 269 271 275
277 281 283 287 289 293 295
299 301 305 307 311 313 317
319 323 325 329 331 335 337
341 343 347 349 353 355 359
Some numbers have the special property that they cannot be expressed as the product of two
smaller numbers, e.g., 2, 3, 5, 7, etc. Such numbers are called prime numbers, and they playan important role, both in pure mathematics and its applications. The distribution of suchprime numbers among all natural numbers does not follow any regular pattern; however theGerman mathematician G.F.B. Riemann (1826 - 1866) observed that the frequency of primenumbers is very closely related to the behavior of an elaborate function
ζ (s) = 1 + 1/2s
+ 1/3s
+ 1/4s
+...
called the Riemann Zeta function. The Riemann hypothesis asserts that all interesting solutions of the equation
ζ(s) = 0
lie on a certain vertical straight line. This has been checked for the first 1,500,000,000solutions. A proof that it is true for every interesting solution would shed light on many of themysteries surrounding the distribution of prime numbers.
Primes seem to be, at the same time very irregularly distributed among all numbers, and yet – if squinted at from a sufficiently far distance – they reveal an astoundingly elegant pattern. In1859 the German mathematician Bernard Riemann proposed a way of understanding andrefining that pattern. His hypothesis has wide – ranging implications, and this day after 150
years of careful research and exhaustive study, we know it is correct.
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N π(x) π(N) π(x) = 4(n)7 4 = 8(0,5)
1,00E+03 168 5,9524 168 = 8(0,5)6(7)
7
1,00E+06 78 498 12,739 78 498 = 8(0,5) 2803,5(7)
7
1,00E+09 50 847 534 19,667 50 847 534 = 8(0,5)1 815 983,3514286(7)
7
1,00E+12 37 607 912 018 26,59 37 607 912 018 = 8(0,5)1 343 1139 714,92857(7)
7
1,00E+15 29 844 570 422 669 33,507 29 844 570 422 669 = 8(0,5)1 065 877 515 095,32(7)
7
1,00E+18 24 739 954 287 740 860 40,42 24739954287740860 = 8(0,5)883569795990,745(7)
7
1,00E+21 21127269486018731928 47,332 21127269486018731928=4(754545338786383283,143)7
There are 4 primes up to 10 (2, 3, 5, 7), because those they cannot be expressed as the productof two smaller numbers (4 = 2(2), 8 = 4(2), 9 = 3(3), 10 = 5(2). Between 1 and 100 there are25 primes, and 168 primes up to 1 000. Why 168? Is there a rule, a formula, to tell me howmany primes there are less than a given number?We know that all prime numbers be congruent to me modulo 7, and this seven tell me howmany primes there are less than a given number. We show in table, that quantity primes thereare less than a given number is always product number 4 = 7(0,57143), and n – the multiplenumber 7: π(x) = 4(n)7, 168 = 4(6)7.
Theorem: The quotient of half a given magnitude ½N, by a given magnitude N, is directly proportional to the quotient of quantity prime numbers, by its dual quantity.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21
2 5 110
31 41
0
61 71
0 0
101
0 0
131
0
151
0 0
181 191
3 7 13 23
0
43 53
0
7383
0
103 113
0 0 0 0
163 173
0
193
0 017
0
37 47
0
67
0 0
97 107
0
127 137
0
157 167
0 0
197
0 019 29
0 0
59
0
79 89
0
109
0 0
139 149
0 0
179
0
199
π(10) = 4 = 7(0,57143), π(100) = 25 = 7(3,57143), π(1 000) = 168
=7(0,57143)6(7)
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Proof: ½N : N = πx : 2(πx) πx ∝ ½N ½N(2 πx) = N(πx) πx/(2 πx) = Re(s) = ½
5/10 = 4/8 = 3/6 = 2/4 = 1/2 8(1/2) = 4 2πx(½) = πx
N
x N 22
1
In the interval of 10 numbers of 5 consecutive odd numbers is always one divisible by 3, sothat it can act like this at most 4 primes, so double large interval of 8 numbers claim (4·2 = 819 - 11 = 8 11, 12, 13, 14, 15, 16, 17 , 18, 19). Generally in the interval of 10 numbers eacheven and odd number 2 number interval claims, including in 4(2), 3(2), 2(2), 1(2), primes.If the interval first 10 numbers come before 4 primes, then in the interval of 100 numbers 25
prime numbers occur, and each number in the interval, the proportion ½ is to keep. In other words, all zeros of ζ(s) in the half-plane Re(s) > 0 have real part ½ .
In mathematics, two quantities are said to be proportional if they vary in such a way that oneof the quantities is a constant multiple of the other, or equivalently if they have a constantratio. Proportion also refers to the equality of two ratios. In proportional quantities is thedoubling (tripling, halved) one quantity is always a double (triple, halve) connected to theother quantities.
2πx πx N ½ N 2 2 2 2
8 4 10 5 2 3 5 7
16 8 20 10 11 13 17 19
20 10 30 15 0 23 0 29
24 12 40 20 31 0 37 0
30 15 50 25 41 43 47 034 17 60 30 0 53 0 59
38 19 70 35 61 0 67 0
44 22 80 40 71 73 0 79
48 24 90 45 0 83 0 89
50 25 100 50 0 0 97 0
58 29 110 55 101 103 107 109
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Riemann hypothesis:
The non-trivial zeros (lie zeros in the strip right next to the y-axis with real part of s from 0 to 1)
are all on a line parallel to the y-axis with real part 1 / 2.
Proof:
2
0
5
0
11
0
17
0
23
0
29
0
35
0
41
0
47
0
53
0
59
0
65
0
71
0
77
0
83
0
89
0
95
0
101
0
0
3
0
9
0
15
0
21
0
27
0
33
0
39
0
450
510
570
630
690
750
810
870
930
990
105
7
0
13
0
19
0
25
0
31
0
37
0
43
0
49
0
55
0
61
0
67
0
73
0
79
0
85
0
91
0
97
0
103
0
109
0
0%
10%
20%
30%
40%
50%
60%
70%
80%
90%
100%
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
0 2 33 7513111917
2523
3129
3735
4341
4947
5553
6159
6765
7371
79
1 2 3
y^- 5y + 6 = 0 x^= 1 y = x+ 1 = 2 y = 4x - 1 = 3+ 4= 7 y= 8 x- 3= 5+ 8 = 13
y=8x+3 = 11 + 8= 19 y=8x+9= 17
y=8x+15=23 + 8=31 y=8x+21=29+8=37 y= 8x+27=35 +8=43 y=8x+33=41
y=8x+39=47 y= 8x+45=53+8=61
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What is the solution of the equation y² - 5y + 6 = 0? The solution is 2 und 3. Another way to say,
this is that 2 and 3 are the zeros of the function y ² - 5y + 6 = 0.
y y‘ ½ y
y² - 10y + 21 = 0 Solution 3 and 7 5
y² - 18y + 65 = 0 - „ - 5 and 13 9
y² - 30y + 209 = 0 - „ - 11 and 19 15
y² - 42y + 425 = 0 - „ - 17 and 25 = 5(5) 21
y² - 54y + 713 = 0 - „ - 23 and 31 27
y² - 66y +1073 = 0 - ― - 29 and 37 33
y² - 78y +1505 = 0 5(7 )= 35 and 43 39
y² - 90y +2009 = 0 - ― - 41 and 49 = 7(7) 45
y² - 102y +2585 = 0 47 i 55 = 5(11) 51
2 0 00 2,50 03
05
0
7
05
0
9
0
13
0
11
0
15
0
19
0
17
0
21
0
25
0
23
0
27
0
31
0
29
0
33
0
37
0
35
0
39
0
43
0
41
0
45
0
49
0
47
0
51
0
55
0
53
0
57
0
61
0
59
0
63
0
67
0
65
0
69
0
73
0
71
0
75
0
79
0
77
0
81
0
85
0
83
0
87
0
91
0
89
0
93
0
97
0
95
1 2 3
(y + y')/2 = 1/2 y (2+3)/2=2,5 (3+7)/2=5 (5+13)/2=9 (11+19)/2=15
(17+25)/2=21 (23+ 31)/2=27 (29+37)/2=33
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A zero off the critical line would induce a pattern in the distribution of the primes.
2
19
29
0 0
59
0
79
89
0
109
313
23
0
43
53
0
73
83
0
103
7 17
0
37
47
0
67
0 0
97
107
5 11
0
31
41
0
61
71
0 0
101
0 0 0 0 0 0 0 0 0 0 0
0 2 4 6 8 10 12
p + 5(6) = p' 7 + 5(6) = 37 + 5(6) = 67 + 5(6) = 97 + 5(6) = 127 + 5(6) = 157
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The second graph shows the values of the zeta function in the field The
x and the y -axis corresponds to real- and imaginary part of functional value. The coloring and the- axis give real and imaginary archetype of the back. The black curve corresponds to the critical
line
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163
00
2
30
5
0
7
0
9
0
11
0
13 0 15
0
17
0
19
0
21
0
23
0
25
0
270
29
0
31
033
0
35
0
370
39
0
41
0
43
0
45
047
0
49
0
510
53
0
55
0
57
0
59
061
0
63
0
650
67
0
69
0
71
0
73
075
0
77
0
790
81
0
83
0
85
0
87
089
0
91
0
930
95
0
97
0
99
0
101
0103
0
105
0
1070
109
0
111
65
107
149
191
233
275
31747
89
131
173215
257
299
67
109
151
193
235
277
319
49
91
133
175217
259
301
71
113
155
197
239
281
323
53
95
137
179221
263
305
73
115
157
199
241
283
325
55
97
139
181223
265
307
77
119
161
203
245
287
329
59
101
143
185227
269
311
79
121
163
205
247
289
331
61
103
145
187229
271
313
83
125
167
209
251
293
335