From Fibonacci Quilts to Benford s Law through Zeckendorf Decompositions

Transcription

1 From Fibonacci Quilts to Benford s Law through Zeckendorf Decompositions Steven J. Miller (sjm1@williams.edu) AMS Special Session on Difference Equations, San Antonio, January 10, 2015

2 Introduction

3 Goals of the Talk Generalize Zeckendorf decompositions Analyze gaps (in the bulk and longest) Patterns and new recurrences Some open problems (if time permits)

4 Collaborators and Thanks Collaborators: Gaps (Bulk, Individual, Longest): Olivia Beckwith, Amanda Bower, Louis Gaudet, Rachel Insoft, Shiyu Li, Philip Tosteson. Kentucky Sequence, Fibonacci Quilt: 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 DMS , DMS , DMS and DMS , AIM and Williams College.

5 5 Intro Gaps Kentucky and Quilts Benfordness in Interval Random + Zeck Decomposition References Generalizations Previous Results Fibonacci Numbers: F n+1 = F n + F n 1 ; F 1 = 1, F 2 = 2, F 3 = 3, F 4 = 5,... Zeckendorf s Theorem Every positive integer can be written uniquely as a sum of non-consecutive Fibonacci numbers. Example: 2014 = = F 16 + F 13 + F 8 + F 4 + F 1. Lekkerkerker s Theorem (1952) The average number of summands in the Zeckendorf n decomposition for integers in [F n, F n+1 ) tends to ϕ n, where ϕ = is the golden mean.

6 Old Results Central Limit Type Theorem As n, the distribution of number of summands in Zeckendorf decomposition for m [F n, F n+1 ) is Gaussian Figure: Number of summands in [F 2010, F 2011 ); F

7 7 Intro Gaps Kentucky and Quilts Benfordness in Interval Random + Zeck Decomposition References Generalizations Benford s law Definition of Benford s Law A dataset is said to follow Benford s Law (base B) if the probability of observing a first digit of d is ( log B 1+ 1 ). d More generally probability a significant at most s is log B (s), where x = S B (x)10 k with S B (x) [1, B) and k Z. Find base 10 about 30.1% of the time start with a 1, only 4.5% start with a 9.

8 Gaps Joint with Olivia Beckwith, Amanda Bower, Louis Gaudet, Rachel Insoft, Shiyu Li, Philip Tosteson

9 Distribution of Gaps For F i1 + F i2 + +F in, the gaps are the differences i n i n 1, i n 1 i n 2,...,i 2 i 1. Example: For F 1 + F 8 + F 18, the gaps are 7 and 10.

10 Distribution of Gaps For F i1 + F i2 + +F in, the gaps are the differences i n i n 1, i n 1 i n 2,...,i 2 i 1. Example: For F 1 + F 8 + F 18, the gaps are 7 and 10. Let P n (g) be the probability that a gap for a decomposition in [F n, F n+1 ) is of length g.

11 Distribution of Gaps For F i1 + F i2 + +F in, the gaps are the differences i n i n 1, i n 1 i n 2,...,i 2 i 1. Example: For F 1 + F 8 + F 18, the gaps are 7 and 10. Let P n (g) be the probability that a gap for a decomposition in [F n, F n+1 ) is of length g. Bulk: What is P(g) = lim n P n (g)?

12 Distribution of Gaps For F i1 + F i2 + +F in, the gaps are the differences i n i n 1, i n 1 i n 2,...,i 2 i 1. Example: For F 1 + F 8 + F 18, the gaps are 7 and 10. Let P n (g) be the probability that a gap for a decomposition in [F n, F n+1 ) is of length g. Bulk: What is P(g) = lim n P n (g)? Individual: Similar questions about gaps for a fixed m [F n, F n+1 ): distribution of gaps, longest gap.

13 New Results: Bulk Gaps: m [F n, F n+1 ) and φ = m = k(m)=n j=1 F ij, ν m;n (x) = k(m) 1 δ ( x (i j i j 1 ) ). k(m) 1 j=2 Theorem (Zeckendorf Gap Distribution) Gap measures ν m;n converge to average gap measure where P(k) = 1/φ k for k Figure: Distribution of gaps in [F 2010, F 2011 ); F

14 New Results: Longest Gap Fair coin: largest gap tightly concentrated around log n/ log 2. Theorem (Longest Gap) As n, the probability that m [F n, F n+1 ) has longest gap less than or equal to f(n) converges to elog n f(n) logφ Prob(L n (m) f(n)) e µ n = ( log ) φ 2 φ 2 n +1) logφ + γ logφ Small Error. If f(n) grows slower (resp. faster) than log n/ logφ, then Prob(L n (m) f(n)) goes to 0 (resp. 1).

15 Main Results Theorem (Distribution of Bulk Gaps (SMALL 2012)) Let H n+1 = c 1 H n + c 2 H n 1 + +c L H n+1 L be a positive linear recurrence of length L where c i 1 for all 1 i L. Then 1 ( a 1 C Lek )(2λ a 1 1 3) : g = 0 P(g) = λ 1 1 ( 1 C Lek )(λ 1 (1 2a 1 )+a 1 ) : g = 1 ( (λ 1 1) 2 a1 λ g 1 : g 2. C Lek )

16 Main Results Theorem (Longest Gap (SMALL 2012)) As n, the probability that m [F n, F n+1 ) has longest gap less than or equal to f(n) converges to elog n f(n) logφ Prob(L n (m) f(n)) e

17 Kentucky Sequence and Quilts with Minerva Catral, Pari Ford, Pamela Harris & Dawn Nelson

18 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 from the first s bins to the left or the first s to the right.

19 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 from the first s bins to the left or the first s to the right. Fibonaccis: These are (s, b) = (1, 1). Kentucky: These are (s, b) = (1, 2). [1, 2], [3, 4], [5, 8], [11, 16], [21, 32], [43, 64], [85, 128].

20 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 from the first s bins to the left or the first s to the right. Fibonaccis: These are (s, b) = (1, 1). Kentucky: These are (s, b) = (1, 2). [1, 2], [3, 4], [5, 8], [11, 16], [21, 32], [43, 64], [85, 128]. a 2n = 2 n and a 2n+1 = 1 3 (22+n ( 1) n ): a n+1 = a n 1 + 2a n 3, a 1 = 1, a 2 = 2, a 3 = 3, a 4 = 4.

21 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 from the first s bins to the left or the first s to the right. Fibonaccis: These are (s, b) = (1, 1). Kentucky: These are (s, b) = (1, 2). [1, 2], [3, 4], [5, 8], [11, 16], [21, 32], [43, 64], [85, 128]. a 2n = 2 n and a 2n+1 = 1 3 (22+n ( 1) n ): a n+1 = a n 1 + 2a n 3, a 1 = 1, a 2 = 2, a 3 = 3, a 4 = 4. a n+1 = a n 1 + 2a n 3 : New as leading term 0.

22 Gaussian Behavior Figure: Plot of the distribution of the number of summands for 100,000 randomly chosen m [1, a 4000 ) = [1, ) (so m has on the order of 602 digits).

23 Gaps Figure: Plot of the distribution of gaps for 10,000 randomly chosen m [1, a 400 ) = [1, ) (so m has on the order of 60 digits).

24 Gaps Figure: Plot of the distribution of gaps for 10,000 randomly chosen m [1, a 400 ) = [1, ) (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 even and odd gaps.

25 The Fibonacci (or Log Cabin) Quilt: Work in Progress a n+1 = a n 1 + a n 2, non-uniqueness (average number of decompositions grows exponentially). In process of investigating Gaussianity, Gaps, K min, K ave, K max, K greedy.

26 Average Number of Representations d n: the number of FQ-legal decompositions using only elements of {a 1, a 2,...,a n}. c n requires a n to be used, b n requires a n and a n 2 to be used. n d n c n b n a n Table: First few terms. Find d n = d n 1 + d n 2 d n 3 + d n 5 d n 9, implying d FQ;ave (n) C n.

27 Greedy Algorithm h n : number of integers from 1 to a n+1 1 where the greedy algorithm successfully terminates in a legal decomposition. n a n h n ρ n Table: First few terms, yields h n = h n 1 + h n and percentage converges to about

28 Benfordness in Interval Joint with Andrew Best, Patrick Dynes, Xixi Edelsbunner, Brian McDonald, Kimsy Tor, Caroline Turnage-Butterbaugh and Madeleine Weinstein

29 Benfordness in Interval Theorem (SMALL 2014): Benfordness in Interval The distribution of the summands in the Zeckendorf decompositions, averaged over the entire interval [F n, F n+1 ), follows Benford s Law.

30 Benfordness in Interval Theorem (SMALL 2014): Benfordness in Interval The distribution of the summands in the Zeckendorf decompositions, averaged over the entire interval [F n, F n+1 ), follows Benford s Law. Example Looking at the interval [F 5, F 6 ) = [8, 13) 8 = 8 = F 5 9 = = F 5 + F 1 10 = = F 5 + F 2 11 = = F 5 + F 3 12 = = F 5 + F 3 + F 1

31 Preliminaries for Proof Density of S For a subset Sof the Fibonacci numbers, define the density q(s, n) of S over the interval [1, F n ] by q(s, n) = #{F j S 1 j n}. n Asymptotic Density If lim n q(s, n) exists, define the asymptotic density q(s) by q(s) = lim n q(s, n).

32 Needed Input Let S d be the subset of the Fibonacci numbers which share a fixed digit d where 1 d < B. Theorem: Fibonacci Numbers Are Benford ( q(s d ) = lim q(s d, n) = log n B 1+ 1 ). d Proof: Binet s formula, Kronecker s theorem on equidistribution of nα mod 1 for α Q.

33 Random Variables Random Variable from Decompositions Let X(I n ) be a random variable whose values are the the Fibonacci numbers in [F 1, F n ) and probabilities are how often they occur in decompositions of m I n : P{X(I n ) = F k } := F k 1 F n k 2 µ nf n 1, if 1 k n 2 1 µ n, if k = n 0, otherwise, where µ n is the average number of summands in Zeckendorf decompositions of integers in the interval [F n, F n+1 ).

34 Approximations Estimate for P{X(I n ) = F k } P{X(I n ) = F k } = 1 ( µ n φ 5 + O φ 2k +φ 2n+2k). Constant Fringes Negligible For any r (which may depend on n): r<k<n r P{X(I n ) = F k } = 1 r O ( ) 1. n

35 Estimating P{X(I n ) S} Set r := log n logφ. Density of S over Zeckendorf Summands We have P{X(I n ) S} = nq(s) µ n φ + o(1) q(s). 5

36 Remark Stronger result than Benfordness of Zeckendorf summands. Global property of the Fibonacci numbers can be carried over locally into the Zeckendorf summands. If we have a subset of the Fibonacci numbers S with asymptotic density q(s), then the density of the set S over the Zeckendorf summands will converge to this asymptotic density.

37 Benfordness of Random and Zeckendorf Decompositions Joint with Andrew Best, Patrick Dynes, Xixi Edelsbunner, Brian McDonald, Kimsy Tor, Caroline Turnage-Butterbaugh and Madeleine Weinstein

38 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 F 1 + F 2 + F 3 + F 4 + F 5 + F 6 + F 7 + F 8 + F 9 + F 10 = = 120

39 Choosing a Random Decomposition Select a random subset A of the Fibonaccis as follows: Fix q (0, 1). Let A 0 :=. For n 1, if F n 1 A n 1, let A n := A n 1, else { A n 1 {F n } with probability q A n = with probability 1 q. Let A := n A n. A n 1

40 Main Result Theorem With probability 1, A (chosen as before) is Benford. Stronger claim: For any subset S of the Fibonaccis with density d in the Fibonaccis, S A has density d in A with probability 1.

41 Preliminaries Lemma The probability that F k A is p k = q 1+q + O(qk ). Using elementary techniques, we get Lemma Define X n := #A n. Then nq E[X n ] = 1+q + O(1) Var(X n ) = O(n).

42 Expected Value of Y n Define Y n,s := #A n S. Using standard techniques, we get Lemma E[Y n ] = Var(Y n,s ) = o(n 2 ). nqd 1+q + o(n).

43 Expected Value of Y n Define Y n,s := #A n S. Using standard techniques, we get Lemma E[Y n ] = Var(Y n,s ) = o(n 2 ). nqd 1+q + o(n). Immediately implies with probability 1 + o(1) Y n,s = nqd Y n,s + o(n), lim = d. 1+q n X n Hence A S has density d in A, completing the proof.

44 Zeckendorf Decompositions and Benford s Law Theorem (SMALL 2014): Benfordness of Decomposition If we pick a random integer in [0, F n+1 ), then with probability 1 as n its Zeckendorf decomposition converges to Benford s Law.

45 Proof of Theorem Choose integers randomly in [0, F n+1 ) by random decomposition model from before. Choose m = F a1 + F a2 + +F al [0, F n+1 ) with probability { q l (1 q) n 2l if a l n p m = q l (1 q) n 2l+1 if a l = n. Key idea: Choosing q = 1/ϕ 2, the previous formula simplifies to { ϕ n if m [0, F n ) p m = ϕ n 1 if m [F n, F n+1 ), use earlier results.

46 References

47 References References Bower, Insoft, Li, Miller and Tosteson, The Distribution of Gaps between Summands in Generalized Zeckendorf Decompositions, preprint. Beckwith, Bower, Gaudet, Insoft, Li, Miller and Tosteson: The Average Gap Distribution for Generalized Zeckendorf Decompositions: The Fibonacci Quarterly 51 (2013), Kologlu, Kopp, Miller and Wang: On the number of summands in Zeckendorf decompositions: Fibonacci Quarterly 49 (2011), no. 2, Miller and Wang: From Fibonacci numbers to Central Limit Type Theorems: Journal of Combinatorial Theory, Series A 119 (2012), no. 7, Miller and Wang: Survey: Gaussian Behavior in Generalized Zeckendorf Decompositions: To appear in CANT 2011 Proceedings.

48 Generalizations

49 Positive Linear Recurrence Sequences This method can be greatly generalized to Positive Linear Recurrence Sequences: linear recurrences with non-negative coefficients: H n+1 = c 1 H n (j1 =0) + c 2 H n j2 + +c L H n jl. Theorem (Zeckendorf s Theorem for PLRS recurrences) Any b N has a unique legal decomposition into sums of H n, b = a 1 H i1 + +a ik H ik. Here legal reduces to non-adjacency of summands in the Fibonacci case.

50 Messier Combinatorics The number of b [H n, H n+1 ), with longest gap < f is the coefficient of x n s in the generating function:

51 Messier Combinatorics The number of b [H n, H n+1 ), with longest gap < f is the coefficient of x n s in the generating function: +x t 1 1 ( c1 1+c 2 x t 2 + +c L x t ) L 1 x [ ((c1 1)x t 1 + +(c L 1)x t )( x s+1 x f L k 0 ( x s+t 2 t 1 +1 x f 1 x 1 x ) + ) ( + +x t L 1 x s+t L t L x f )] k. 1 x

52 Messier Combinatorics The number of b [H n, H n+1 ), with longest gap < f is the coefficient of x n s in the generating function: +x t 1 1 ( c1 1+c 2 x t 2 + +c L x t ) L 1 x [ ((c1 1)x t 1 + +(c L 1)x t )( x s+1 x f L k 0 ( x s+t 2 t 1 +1 x f 1 x A geometric series! 1 x ) + ) ( + +x t L 1 x s+t L t L x f )] k. 1 x

53 Generalized Generating Function Let f > j L. The number of x [H n, H n+1 ), with longest gap < f is given by the coefficient of s n in the generating function where and F(s) = 1 s j L M(s)+s f R(s), M(s) = 1 c 1 s c 2 s j 2+1 c L s j L+1, R(s) = c j1 +1s j 1 + c j2 +1s j 2 + +(c jl +1 1)s j L. and c i and j i are defined as above.

54 What are the extra obstructions? The coefficients in the partial fraction expansion might blow up from multiple roots.

55 What are the extra obstructions? The coefficients in the partial fraction expansion might blow up from multiple roots. Theorem (Mean and Variance for "Most Recurrences") For x in the interval [H n, H n+1 ), the mean longest gap µ n and the variance of the longest gap σ 2 n are given by µ n = ( ) R( 1 ) λ log 1 n G( 1 ) λ 1 logλ 1 + γ logλ Small Error+ǫ 1(n), 55 and σ 2 n = π2 6 logλ Small Error+ǫ 2(n), where ǫ i (n) tends to zero in the limit, and Small Error comes from the Euler-Maclaurin Formula.