From the Kentucky Sequence to Benford's Law through ......probability of observing a ﬁrst digit of...

Intro Previous Work Kentucky Sequence Other Rules Benfordness in Interval Random + Zeck Decomposition References

From the Kentucky Sequence to Benford’sLaw through Zeckendorf Decompositions.

Steven J. Miller ([email protected])http://www.williams.edu/Mathematics/sjmiller/public_html

AMS Special Session on Difference EquationsGeorgetown University, Washington, DC, 3/7/15

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http://www.williams.edu/Mathematics/sjmiller/public_html


Introduction

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Collaborators and Thanks

Collaborators:Kentucky Sequence: Joint with Minerva Catral, Pari Ford,Pamela Harris & Dawn Nelson.Benfordness: Andrew Best, Patrick Dynes, Xixi Edelsbunner,Brian McDonald, Kimsy Tor, Caroline Turnage-Butterbaugh &Madeleine Weinstein.

Supported by:NSF Grants DMS1265673, DMS0970067, DMS1347804 andDMS0850577, AIM and Williams College.

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Previous Results

Fibonacci Numbers: Fn+1 = Fn + Fn−1;First few: 1,2,3,5,8,13,21,34,55,89, . . . .

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Previous Results


Zeckendorf’s TheoremEvery positive integer can be written uniquely as a sum ofnon-consecutive Fibonacci numbers.

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Previous Results



Example: 51 =?

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Previous Results



Example: 51 = 34 + 17 = F8 + 17.

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Previous Results



Example: 51 = 34 + 13 + 4 = F8 + F6 + 4.

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Previous Results



Example: 51 = 34 + 13 + 3 + 1 = F8 + F6 + F3 + 1.

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Previous Results



Example: 51 = 34 + 13 + 3 + 1 = F8 + F6 + F3 + F1.

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Previous Results



Example: 51 = 34 + 13 + 3 + 1 = F8 + F6 + F3 + F1.Example: 83 = 55 + 21 + 5 + 2 = F9 + F7 + F4 + F2.Observe: 51 miles ≈ 82.1 kilometers.

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Old Results

Central Limit Type Theorem

As n → ∞, the distribution of number of summands inZeckendorf decomposition for m ∈ [Fn,Fn+1) is Gaussian.

500 520 540 560 580 600

0.005

0.010

0.015

0.020

0.025

0.030

Figure: Number of summands in [F2010,F2011); F2010 ≈ 10420.

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Benford’s law

Definition of Benford’s LawA dataset is said to follow Benford’s Law (base B) if theprobability of observing a first digit of d is

logB

(

1 +1d

)

.

More generally probability a significant at most s is logB(s),where x = SB(x)10k with SB(x) ∈ [1,B) and k an integer.

Find base 10 about 30.1% of the time start with a 1, only4.5% start with a 9.

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Previous Work

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Equivalent Definition of the Fibonaccis

Fibonaccis are the only sequence such that each integer canbe written uniquely as a sum of non-adjacent terms.

1,

15




1, 2,

16




1, 2, 3,

17




1, 2, 3, 5,

18




1, 2, 3, 5, 8,

19




1, 2, 3, 5, 8, 13 . . . .

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1, 2, 3, 5, 8, 13 . . . .

Key to entire analysis: Fn+1 = Fn + Fn−1.

View as bins of size 1, cannot use two adjacent bins:

[1] [2] [3] [5] [8] [13] · · · .

Goal: How does the notion of legal decomposition affectthe sequence and results?

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Generalizations

Generalizing from Fibonacci numbers to linearly recursivesequences with arbitrary nonnegative coefficients.

Hn+1 = c1Hn + c2Hn−1 + · · · + cLHn−L+1, n ≥ L

with H1 = 1, Hn+1 = c1Hn + c2Hn−1 + · · ·+ cnH1 + 1, n < L,coefficients ci ≥ 0; c1, cL > 0 if L ≥ 2; c1 > 1 if L = 1.

Zeckendorf: Every positive integer can be written uniquelyas

∑

aiHi with natural constraints on the ai ’s (e.g. cannotuse the recurrence relation to remove any summand).

Central Limit Type Theorem

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Example: the Special Case of L = 1, c1 = 10

Hn+1 = 10Hn, H1 = 1, Hn = 10n−1.

Legal decomposition is decimal expansion:∑m

i=1 aiHi :ai ∈ {0,1, . . . ,9} (1 ≤ i < m), am ∈ {1, . . . ,9}.

For N ∈ [Hn,Hn+1), first term is anHn = an10n−1.

Ai : the corresponding random variable of ai . The Ai ’s areindependent.

For large n, the contribution of An is immaterial.Ai (1 ≤ i < n) are identically distributed random variableswith mean 4.5 and variance 8.25.

Central Limit Theorem: A2 + A3 + · · · + An → Gaussianwith mean 4.5n + O(1) and variance 8.25n + O(1).

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Kentucky Sequencewith Minerva Catral, Pari Ford, Pamela Harris & Dawn Nelson

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Kentucky Sequence

Rule: (s,b)-Sequence: Bins of length b, and:

cannot take two elements from the same bin, and

if have an element from a bin, cannot take anything fromthe first s bins to the left or the first s to the right.

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Kentucky Sequence




Fibonaccis: These are (s,b) = (1,1).

Kentucky: These are (s,b) = (1,2).

[1, 2], [3, 4], [5,

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Kentucky Sequence






[1, 2], [3, 4], [5, 8],

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Kentucky Sequence






[1, 2], [3, 4], [5, 8], [11,

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Kentucky Sequence






[1, 2], [3, 4], [5, 8], [11, 16],

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Kentucky Sequence






[1, 2], [3, 4], [5, 8], [11, 16], [21,

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Kentucky Sequence






[1, 2], [3, 4], [5, 8], [11, 16], [21, 32], [43, 64], [85, 128].

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Kentucky Sequence






[1, 2], [3, 4], [5, 8], [11, 16], [21, 32], [43, 64], [85, 128].

a2n = 2n and a2n+1 = 13(2

2+n − (−1)n):an+1 = an−1 + 2an−3,a1 = 1,a2 = 2,a3 = 3,a4 = 4.

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Kentucky Sequence






[1, 2], [3, 4], [5, 8], [11, 16], [21, 32], [43, 64], [85, 128].

a2n = 2n and a2n+1 = 13(2

2+n − (−1)n):an+1 = an−1 + 2an−3,a1 = 1,a2 = 2,a3 = 3,a4 = 4.

an+1 = an−1 + 2an−3: New as leading term 0.33


What’s in a name?

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What’s in a name?

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Theorem: Gaussian Behavior

620 640 660 680 700 720

0.005

0.010

0.015

0.020

0.025

0.030

0.035

Figure: Plot of the distribution of the number of summands for100,000 randomly chosen m ∈ [1, a4000) = [1, 22000) (so m has on theorder of 602 digits).

Proved Gaussian behavior.36


Theorem: Geometric Decay for Gaps

Figure: Plot of the distribution of gaps for 10,000 randomly chosenm ∈ [1, a400) = [1, 2200) (so m has on the order of 60 digits).

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Theorem: Geometric Decay for Gaps

Figure: Plot of the distribution of gaps for 10,000 randomly chosenm ∈ [1, a400) = [1, 2200) (so m has on the order of 60 digits). Left(resp. right): ratio of adjacent even (resp odd) gap probabilities.

Again find geometric decay, but parity issues so break into evenand odd gaps.

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Other Rules(Coming Attractions)

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Tilings, Expanding Shapes

1

2

4 6

8

16

24

32

64 96

1, 2, 4, 6, 8, 16, 24, 32, 64, 96, ...

Figure: (left) Hexagonal tiling; (right) expanding triangle covering.

Theorem:A sequence uniquely exists, and similar to previous work candeduce results about the number of summands and thedistribution of gaps.

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Fractal Sets

Figure: Sierpinski tiling.41


Upper Half Plane / Unit Disk

Figure: Plot of tesselation of the upper half plane (or unit disk) bythe fundamental domain of SL2(Z), where T sends z to z + 1 and Ssends z to −1/z.

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Benfordness in IntervalJoint with Andrew Best, Patrick Dynes, Xixi Edelsbunner, Brian

McDonald, Kimsy Tor, Caroline Turnage-Butterbaugh andMadeleine Weinstein

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Benfordness in Interval

Theorem (SMALL 2014): Benfordness in Interval

The distribution of the summands in the Zeckendorfdecompositions, averaged over the entire interval [Fn,Fn+1),follows Benford’s Law.

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Benfordness in Interval

Theorem (SMALL 2014): Benfordness in Interval

The distribution of the summands in the Zeckendorfdecompositions, averaged over the entire interval [Fn,Fn+1),follows Benford’s Law.

Example

Looking at the interval [F5,F6) = [8,13)

8 = 8 = F5

9 = 8 + 1 = F5 + F1

10 = 8 + 2 = F5 + F2

11 = 8 + 3 = F5 + F3

12 = 8 + 3 + 1 = F5 + F3 + F1

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Preliminaries for Proof

Density of S

For a subset S of the Fibonacci numbers, define the densityq(S,n) of S over the interval [1,Fn] by

q(S,n) =#{Fj ∈ S | 1 ≤ j ≤ n}

n.

Asymptotic Density

If limn→∞ q(S,n) exists, define the asymptotic density q(S) by

q(S) = limn→∞

q(S,n).

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Needed Input

Let Sd be the subset of the Fibonacci numbers which share afixed digit d where 1 ≤ d < B.

Theorem: Fibonacci Numbers Are Benford

q(Sd) = limn→∞

q(Sd ,n) = logB

(

1 +1d

)

.

Proof: Binet’s formula, Kronecker’s theorem on equidistributionof nα mod 1 for α 6∈ Q.

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Random Variables

Random Variable from Decompositions

Let X (In) be a random variable whose values are the Fibonaccinumbers in [F1,Fn) and probabilities are how often they occurin decompositions of m ∈ In:

P (X (In) = Fk ) :=

Fk−1Fn−k−2µnFn−1

if 1 ≤ k ≤ n − 2

1µn

if k = n

0 otherwise,

where µn is the average number of summands in Zeckendorfdecompositions of integers in the interval [Fn,Fn+1).

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Approximations

Estimate for P (X (In) = Fk)

P (X (In) = Fk ) =1

µnφ√

5+ O

(

φ−2k + φ−2n+2k)

.

Constant Fringes Negligible

For any r (which may depend on n):

∑

r<k<n−r

P (X (In) = Fk ) = 1 − r · O(

1n

)

.

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Estimating P (X (In) ∈ S)

Set r :=⌊

log nlogφ

⌋

.

Density of S over Zeckendorf Summands

We have

P (X (In) ∈ S) =nq(S)

µnφ√

5+ o(1) → q(s).

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Remark

Stronger result than Benfordness of Zeckendorfsummands.

Global property of the Fibonacci numbers can be carriedover locally into the Zeckendorf summands.

If we have a subset of the Fibonacci numbers S withasymptotic density q(S), then the density of the set S overthe Zeckendorf summands will converge to this asymptoticdensity.

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Benfordness of Random and Zeckendorf DecompositionsJoint with Andrew Best, Patrick Dynes, Xixi Edelsbunner, Brian

McDonald, Kimsy Tor, Caroline Turnage-Butterbaugh andMadeleine Weinstein

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Random Decompositions

Theorem 2 (SMALL 2014): Random Decomposition

If we choose each Fibonacci number with probability q,disallowing the choice of two consecutive Fibonacci numbers,the resulting sequence follows Benford’s law.

Example: n = 10

F1 + F2 + F3 + F4 + F5 + F6 + F7 + F8 + F9 + F10

= 2 + 8 + 21 + 89

= 120

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Choosing a Random Decomposition

Select a random subset A of the Fibonaccis as follows:

Fix q ∈ (0,1).

Let A0 := ∅.

For n ≥ 1, if Fn−1 ∈ An−1, let An := An−1, else

An =

{

An−1 ∪ {Fn} with probability q

An−1 with probability 1 − q.

Let A :=⋃

n An.

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Main Result

TheoremWith probability 1, A (chosen as before) is Benford.

Stronger claim: For any subset S of the Fibonaccis withdensity d in the Fibonaccis, S ∩ A has density d in A withprobability 1.

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Preliminaries

LemmaThe probability that Fk ∈ A is

pk =q

1 + q+ O(qk ).

Using elementary techniques, we get

LemmaDefine Xn := #An. Then

E [Xn] =nq

1 + q+ O(1)

Var(Xn) = O(n).

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Expected Value of Yn

Define Yn,S := #An ∩ S. Using standard techniques, we get

Lemma

E[Yn] =nqd

1 + q+ o(n).

Var(Yn,S) = o(n2).

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Expected Value of Yn

Define Yn,S := #An ∩ S. Using standard techniques, we get

Lemma

E[Yn] =nqd

1 + q+ o(n).

Var(Yn,S) = o(n2).

Immediately implies with probability 1 + o(1)

Yn,S =nqd

1 + q+ o(n), lim

n→∞

Yn,S

Xn= d .

Hence A ∩ S has density d in A, completing the proof.58


Zeckendorf Decompositions and Benford’s Law

Theorem (SMALL 2014): Benfordness of Decomposition

If we pick a random integer in [0,Fn+1), then with probability 1as n → ∞ its Zeckendorf decomposition converges to Benford’sLaw.

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Proof of Theorem

Choose integers randomly in [0,Fn+1) by randomdecomposition model from before.

Choose m = Fa1 + Fa2 + · · · + Faℓ∈ [0,Fn+1) with

probability

pm =

{

qℓ(1 − q)n−2ℓ if aℓ ≤ n

qℓ(1 − q)n−2ℓ+1 if aℓ = n.

Key idea: Choosing q = 1/ϕ2, the previous formulasimplifies to

pm =

{

ϕ−n if m ∈ [0,Fn)

ϕ−n−1 if m ∈ [Fn,Fn+1),

use earlier results.60


References

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References

Beckwith, Bower, Gaudet, Insoft, Li, Miller and Tosteson: The Average GapDistribution for Generalized Zeckendorf Decompositions: The FibonacciQuarterly 51 (2013), 13–27.http://arxiv.org/abs/1208.5820

Best, Dynes, Edelsbrunner, McDonald, Miller, Turnage-Butterbaugh andWeinstein, Benford Behavior of Generalized Zeckendorf Decompositions, toappear in the Fibonacci Quarterly. http://arxiv.org/pdf/1409.0482

Best, Dynes, Edelsbrunner, McDonald, Miller, Turnage-Butterbaugh andWeinstein: Benford Behavior of Generalized Zeckendorf Decompositions,preprint. http://arxiv.org/pdf/1412.6839

Bower, Insoft, Li, Miller and Tosteson, The Distribution of Gaps betweenSummands in Generalized Zeckendorf Decompositions, preprint. http://arxiv.org/pdf/1402.3912

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http://arxiv.org/abs/1208.5820

http://arxiv.org/pdf/1409.0482





References

Catral, Ford, Harris, Miller and Nelson, Generalizing Zeckendorf’s Theorem: TheKentucky Sequence, submitted August 2014 to the Fibonacci Quarterly.http://arxiv.org/pdf/1409.0488.pdf

Kologlu, Kopp, Miller and Wang: On the number of summands in Zeckendorfdecompositions: Fibonacci Quarterly 49 (2011), no. 2, 116–130. http://arxiv.org/pdf/1008.3204

Miller and Wang: From Fibonacci numbers to Central Limit Type Theorems:Journal of Combinatorial Theory, Series A 119 (2012), no. 7, 1398–1413.http://arxiv.org/pdf/1008.3202

Miller and Wang: Survey: Gaussian Behavior in Generalized ZeckendorfDecompositions: Combinatorial and Additive Number Theory, CANT 2011 and2012 (Melvyn B. Nathanson, editor), Springer Proceedings in Mathematics &Statistics (2014), 159–173. http://arxiv.org/pdf/1107.2718

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http://arxiv.org/pdf/1409.0488.pdf





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