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

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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 (sjm1@williams.edu)http://www.williams.edu/Mathematics/sjmiller/public_html

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

Introduction

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.

Previous Results

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

Previous Results

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

Previous Results

Example: 51 =?

Previous Results

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

Previous Results

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

Previous Results

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

Previous Results

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

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.

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

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

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

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.

Previous Work

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, 2, 3,

1, 2, 3, 5,

1, 2, 3, 5, 8,

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?

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

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).

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

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.

Kentucky Sequence

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

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

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

Kentucky Sequence

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

Kentucky Sequence

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

Kentucky Sequence

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

Kentucky Sequence

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

Kentucky Sequence

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

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.

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?

Theorem: Gaussian Behavior

620 640 660 680 700 720

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).

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.

Other Rules(Coming Attractions)

Tilings, Expanding Shapes

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.

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.

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

McDonald, Kimsy Tor, Caroline Turnage-Butterbaugh andMadeleine Weinstein

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.

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

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}

Asymptotic Density

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

q(S) = limn→∞

q(S,n).

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

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

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

if k = n

0 otherwise,

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

Approximations

Estimate for P (X (In) = Fk)

P (X (In) = Fk ) =1

µnφ√

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(

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).

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.

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

McDonald, Kimsy Tor, Caroline Turnage-Butterbaugh andMadeleine Weinstein

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

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−1 ∪ {Fn} with probability q

An−1 with probability 1 − q.

Let A :=⋃

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.

Preliminaries

LemmaThe probability that Fk ∈ A is

1 + q+ O(qk ).

Using elementary techniques, we get

LemmaDefine Xn := #An. Then

E [Xn] =nq

1 + q+ O(1)

Var(Xn) = O(n).

Expected Value of Yn

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

E[Yn] =nqd

1 + q+ o(n).

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

Expected Value of Yn

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

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→∞

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.

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

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

ϕ−n if m ∈ [0,Fn)

ϕ−n−1 if m ∈ [Fn,Fn+1),

use earlier results.60

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

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

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

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