Distribution of summands in generalized Zeckendorf decompositions

Transcription

1 Distribution of summands in generalized Zeckendorf decompositions Amanda Bower (UM-Dearborn), Louis Gaudet (Yale University), Rachel Insoft (Wellesley College), Shiyu Li (Berkeley), Steven J. Miller and Philip Tosteson (Williams College) AMS Session on Number Theory, I Room 12, Mezzanine Level, San Diego Wednesday January 9, 2013, 3:30p.m.

2 Introduction

3 Goals of the Talk Explain consequences of combinatorial perspective. Perspective important: misleading proofs. Highlight techniques. Some open problems. Joint work at SMALL (Undergraduate REU Program) at Williams College in 2010, 2011 and 2012.

4 Previous Results Fibonacci Numbers: F n+1 = F n + F n 1 ; F 1 = 1, F 2 = 2, F 3 = 3, F 4 = 5,...

5 5 Intro Gaussian Behavior Gaps (Bulk) References 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.

6 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: 2013 = = F 16 + F 13 + F 8 + F 4.

7 7 Intro Gaussian Behavior Gaps (Bulk) References 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: 2013 = = F 16 + F 13 + F 8 + F 4. 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.

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

9 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 almost surely to average gap measure where P(k) = 1/φ k for k Figure: Distribution of gaps in [F 1000, F 1001 ); F

10 New Results: Longest Gap 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 Immediate Corollary: If f(n) grows slower or faster than logφ log n, then Prob(L n (m) f(n)) goes to 0 or 1, respectively.

11 Gaussian Behavior

12 From The Cookie Problem to Gaussian Behavior Reinterpreting the Cookie (or Stars and Bars) Problem The number ) of solutions to x 1 + +x P = C with x i 0 is. ( C+P 1 P 1

13 From The Cookie Problem to Gaussian Behavior Reinterpreting the Cookie (or Stars and Bars) Problem The number ) of solutions to x 1 + +x P = C with x i 0 is. ( C+P 1 P 1 Let p n,k = # {N [F n, F n+1 ): the Zeckendorf decomposition of N has exactly k summands}.

14 From The Cookie Problem to Gaussian Behavior Reinterpreting the Cookie (or Stars and Bars) Problem The number ) of solutions to x 1 + +x P = C with x i 0 is. ( C+P 1 P 1 Let p n,k = # {N [F n, F n+1 ): the Zeckendorf decomposition of N has exactly k summands}. For N [F n, F n+1 ), the largest summand is F n. N = F i1 + F i2 + +F ik 1 + F n, 1 i 1 < i 2 < < i k 1 < i k = n, i j i j 1 2.

15 From The Cookie Problem to Gaussian Behavior Reinterpreting the Cookie (or Stars and Bars) Problem The number ) of solutions to x 1 + +x P = C with x i 0 is. ( C+P 1 P 1 Let p n,k = # {N [F n, F n+1 ): the Zeckendorf decomposition of N has exactly k summands}. For N [F n, F n+1 ), the largest summand is F n. N = F i1 + F i2 + +F ik 1 + F n, 1 i 1 < i 2 < < i k 1 < i k = n, i j i j 1 2. d 1 := i 1 1, d j := i j i j 1 2 (j > 1). d 1 + d 2 + +d k = n 2k + 1, d j 0.

16 From The Cookie Problem to Gaussian Behavior Reinterpreting the Cookie (or Stars and Bars) Problem The number ) of solutions to x 1 + +x P = C with x i 0 is. ( C+P 1 P 1 Let p n,k = # {N [F n, F n+1 ): the Zeckendorf decomposition of N has exactly k summands}. For N [F n, F n+1 ), the largest summand is F n. N = F i1 + F i2 + +F ik 1 + F n, 1 i 1 < i 2 < < i k 1 < i k = n, i j i j 1 2. d 1 := i 1 1, d j := i j i j 1 2 (j > 1). d 1 + d 2 + +d k = n 2k + 1, d j 0. Cookie counting p n,k = ( n 2k+1 + k 1) ( k 1 = n k k 1).

17 Generalizations Generalizing from Fibonacci numbers to linearly recursive sequences with arbitrary nonnegative coefficients. H n+1 = c 1 H n + c 2 H n c L H n L+1, n L with H 1 = 1, H n+1 = c 1 H n + c 2 H n 1 + +c n H 1 + 1, n < L, coefficients c i 0; c 1, c L > 0 if L 2; c 1 > 1 if L = 1. Zeckendorf: Every positive integer can be written uniquely as a i H i with natural constraints on the a i s (e.g. cannot use the recurrence relation to remove any summand). Lekkerkerker Central Limit Type Theorem

18 Example: the Special Case of L = 1, c 1 = 10 H n+1 = 10H n, H 1 = 1, H n = 10 n 1. Legal decomposition is decimal expansion: m i=1 a ih i : a i {0, 1,..., 9} (1 i < m), a m {1,...,9}. For N [H n, H n+1 ), m = n, i.e., first term is a n H n = a n 10 n 1. A i : the corresponding random variable of a i. The A i s are independent. For large n, the contribution of A n is immaterial. A i (1 i < n) are identically distributed random variables with mean 4.5 and variance Central Limit Theorem: A 2 + A 3 + +A n Gaussian with mean 4.5n+O(1) and variance 8.25n + O(1).

19 Far-difference Representation Theorem (Alpert, 2009) (Analogue to Zeckendorf) Every integer can be written uniquely as a sum of the ±F n s, such that every two terms of the same (opposite) sign differ in index by at least 4 (3). Example: 1900 = F 17 F 14 F 10 + F 6 + F 2. K : # of positive terms, L: # of negative terms. Generalized Lekkerkerker s Theorem As n, E[K] and E[L] n/10. E[K] E[L] = ϕ/ Central Limit Type Theorem As n, K and L converges to a bivariate Gaussian. corr(k, L) = (21 2ϕ)/(29+2ϕ).551, ϕ = K + L and K L are independent

20 Gaps in the Bulk

21 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.

22 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.

23 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 (k) be the probability that a gap for a decomposition in [F n, F n+1 ) is of length k.

24 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 (k) be the probability that a gap for a decomposition in [F n, F n+1 ) is of length k. What is P(k) = lim n P n (k)?

25 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 (k) be the probability that a gap for a decomposition in [F n, F n+1 ) is of length k. What is P(k) = lim n P n (k)? Can ask similar questions about binary or other expansions: 2012 =

26 Main Results Theorem (Base B Gap Distribution) For base B decompositions, P(0) = (B 1)(B 2) B 2, and for k 1, P(k) = c B B k, with c B = (B 1)(3B 2) B 2. Theorem (Zeckendorf Gap Distribution) For Zeckendorf decompositions, P(k) = φ(φ 1) φ k φ = the golden mean. for k 2, with

27 Main Results H n : H n+1 = c 1 H n + c 2 H n 1 + +c L H n+1 L a positive linear recurrence of length L where c i 1 for all 1 i L. λ 1 > 1: largest root (in absolute value) of characteristic polynomial of H n. Generalized Binet: H n = a 1 λ n 1 +. Theorem Notation as above, probability of a gap of length j is 1 ( a 1 C Lek )(λ n+2 1 λ n λ a 1 1 3) j = 0 λ 1 1 ( 1 C Lek )(λ 1 (1 2a 1 )+a 1 ) j = 1 ( (λ 1 1) 2 a1 λ j 1 j 2 C Lek )

28 Proof of Fibonacci Result Lekkerkerker total number of gaps F n 1 n φ Let X i,j = #{m [F n, F n+1 ): decomposition of m includes F i, F j, but not F q for i < q < j}. P(k) = lim n n k i=1 X i,i+k n F n 1 φ 2 +1.

29 Proof sketch of almost sure convergence m = k(m) j=1 F i j, ν m;n (x) = 1 k(m) k(m) 1 j=2 δ( x (i j i j 1 ) ). µ m,n (t) = x t dν m;n (x). Show E m [µ m;n (t)] equals average gap moments, µ(t). Show E m [(µ m;n (t) µ(t)) 2 ] and E m [(µ m;n (t) µ(t)) 4 ] tend to zero. Key ideas: (1) Replace k(m) with average (Gaussianity); (2) use X i,i+g1,j,j+g 2.

30 References

31 References References Beckwith, Bower, Gaudet, Insoft, Li, Miller and Tosteson: The Average Gap Distribution for Generalized Zeckendorf Decompositions, to appear in the Fibonacci Quarterly, Kologlu, Kopp, Miller and Wang: Gaussianity for Fibonacci case, 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, Gaussian Behavior in Generalized Zeckendorf Decompositions (with Yinghui Wang), to appear in the conference proceedings of the 2011 Combinatorial and Additive Number Theory Conference,