Cookie Monster Meets the Fibonacci Numbers. Mmmmmm Theorems!

Transcription

1 Cookie Monster Meets the Fibonacci Numbers. Mmmmmm Theorems! Steven J. Miller (MC 96) Yale University, April 14, 2014

2 Introduction

3 Goals of the Talk You can join in: minimal background needed! Ask questions: lots of natural problems ignored. Look for right perspective: generating fns, partial fractions. End with open problems. Joint with Olivia Beckwith, Amanda Bower, Louis Gaudet, Rachel Insoft, Shiyu Li, Philip Tosteson.

4 Pre-requisites: Probability Review Let X be random variable with density p(x): p(x) 0; p(x)dx = 1; Prob(a X b) = b a p(x)dx. Mean: µ = xp(x)dx. Variance: σ 2 = (x µ)2 p(x)dx. Gaussian: Density (2πσ 2 ) 1/2 exp( (x µ) 2 /2σ 2 ).

5 5 Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs Pre-requisites: Combinatorics Review n!: number of ways to order n people, order matters. n! k!(n k)! = nck = ( n k) : number of ways to choose k from n, order doesn t matter. Stirling s Formula: n! n n e n 2πn.

6 Previous Results Fibonacci Numbers: F n+1 = F n + F n 1 ;

7 7 Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs Previous Results Fibonacci Numbers: F n+1 = F n + F n 1 ; F 1 = 1, F 2 = 2, F 3 = 3, F 4 = 5,...

8 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.

9 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: 2012 = = F 16 + F 13 + F 8 + F 3 + F 1.

10 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: 2012 = = F 16 + F 13 + F 8 + F 3 + 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.

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

12 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

13 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 Prob(L n (m) f(n)) e elog n f(n)/ logφ. Immediate Corollary: If f(n) grows slower or faster than log n/ logφ, then Prob(L n (m) f(n)) goes to 0 or 1, respectively.

14 Preliminaries: The Cookie Problem The Cookie Problem The number of ways of dividing C identical cookies among P distinct people is ( ) C+P 1 P 1.

15 Preliminaries: The Cookie Problem The Cookie Problem The number of ways of dividing C identical cookies among P distinct people is ( ) C+P 1 P 1. Proof : Consider C + P 1 cookies in a line. Cookie Monster eats P 1 cookies: ( C+P 1) P 1 ways to do. Divides the cookies into P sets.

16 Preliminaries: The Cookie Problem The Cookie Problem The number of ways of dividing C identical cookies among P distinct people is ( ) C+P 1 P 1. Proof : Consider C + P 1 cookies in a line. Cookie Monster eats P 1 cookies: ( C+P 1) P 1 ways to do. Divides the cookies into P sets. Example: 8 cookies and 5 people (C = 8, P = 5):

17 Preliminaries: The Cookie Problem The Cookie Problem The number of ways of dividing C identical cookies among P distinct people is ( ) C+P 1 P 1. Proof : Consider C + P 1 cookies in a line. Cookie Monster eats P 1 cookies: ( C+P 1) P 1 ways to do. Divides the cookies into P sets. Example: 8 cookies and 5 people (C = 8, P = 5):

18 Preliminaries: The Cookie Problem The Cookie Problem The number of ways of dividing C identical cookies among P distinct people is ( ) C+P 1 P 1. Proof : Consider C + P 1 cookies in a line. Cookie Monster eats P 1 cookies: ( C+P 1) P 1 ways to do. Divides the cookies into P sets. Example: 8 cookies and 5 people (C = 8, P = 5):

19 Preliminaries: The Cookie Problem: Reinterpretation Reinterpreting the Cookie Problem The number ) of solutions to x 1 + +x P = C with x i 0 is. ( C+P 1 P 1

20 Preliminaries: The Cookie Problem: Reinterpretation Reinterpreting the Cookie 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}.

21 Preliminaries: The Cookie Problem: Reinterpretation Reinterpreting the Cookie 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.

22 Preliminaries: The Cookie Problem: Reinterpretation Reinterpreting the Cookie 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.

23 Preliminaries: The Cookie Problem: Reinterpretation Reinterpreting the Cookie 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).

24 Gaussian Behavior

25 Generalizing Lekkerkerker: Erdos-Kac type result Theorem (KKMW 2010) As n, the distribution of the number of summands in Zeckendorf s Theorem is a Gaussian. Sketch of proof: Use Stirling s formula, n! n n e n 2πn to approximates binomial coefficients, after a few pages of algebra find the probabilities are approximately Gaussian.

26 (Sketch of the) Proof of Gaussianity The probability density for the number of Fibonacci numbers that add up to an integer in [F n, F n+1 ) is ( ) f n(k) = n 1 k /F k n 1. Consider the density for the n + 1 case. Then we have, by Stirling f n+1 (k) = = ( ) n k 1 k F n (n k)! 1 = (n 2k)!k! F n 1 2π (n k) n k+ 1 2 k (k+ 1 2 ) (n 2k) n 2k F n plus a lower order correction term. Also we can write F n = 1 5 φ n+1 = φ 5 φ n for large n, whereφ is the golden ratio (we are using relabeled Fibonacci numbers where 1 = F 1 occurs once to help dealing with uniqueness and F 2 = 2). We can now split the terms that exponentially depend on n. Define f n+1 (k) = ( 1 2π (n k) k(n 2k) )( 5 φ n (n k) n k ) φ k k (n 2k) n 2k. N n = Thus, write the density function as 1 (n k) 5 n k 2π k(n 2k) φ, Sn = (n k) φ n k k (n 2k) n 2k. f n+1 (k) = N ns n where N n is the first term that is of order n 1/2 and S n is the second term with exponential dependence on n.

27 (Sketch of the) Proof of Gaussianity Model the distribution as centered around the mean by the change of variable k = µ + xσ whereµ and σ are the mean and the standard deviation, and depend on n. The discrete weights of f n(k) will become continuous. This requires us to use the change of variable formula to compensate for the change of scales: Using the change of variable, we can write N n as f n(k)dk = f n(µ + σx)σdx. N n = = = = 1 n k φ 2π k(n 2k) k/n 2πn 1 2πn 1 2πn (k/n)(1 2k/n) 5 φ 1 (µ + σx)/n 5 ((µ + σx)/n)(1 2(µ + σx)/n) φ 1 C y 5 (C + y)(1 2C 2y) φ where C = µ/n 1/(φ + 2) (note that φ 2 = φ + 1) and y = σx/n. But for large n, the y term vanishes since σ n and thus y n 1/2. Thus N n 1 1 C 5 2πn C(1 2C) φ = 1 (φ + 1)(φ + 2) 5 2πn φ φ = 1 5(φ + 2) 1 = 2πn φ 2πσ 2 since σ 2 φ = n 5(φ+2).

28 (Sketch of the) Proof of Gaussianity For the second term S n, take the logarithm and once again change variables by k = µ + xσ, ( ) log(s n) = log φ n (n k) (n k) k k (n 2k) (n 2k) = n log(φ) + (n k) log(n k) (k) log(k) (n 2k) log(n 2k) = n log(φ) + (n (µ + xσ)) log(n (µ + xσ)) (µ + xσ) log(µ + xσ) (n 2(µ + xσ)) log(n 2(µ + xσ)) = n log(φ) +(n (µ + xσ)) ( ( log(n µ) + log 1 xσ n µ ( ( (µ + xσ) log(µ) + log 1 + xσ )) µ )) ( ( (n 2(µ + xσ)) log(n 2µ) + log 1 xσ )) n 2µ = n log(φ) +(n (µ + xσ)) ( ( n log ( (µ + xσ) log 1 + xσ µ ) µ 1 ) ( + log 1 xσ )) n µ ( ( ) ( n (n 2(µ + xσ)) log µ 2 + log 1 xσ )). n 2µ

29 (Sketch of the) Proof of Gaussianity Note that, since n/µ = φ + 2 for large n, the constant terms vanish. We have log(s n) ( n = n log(φ) + (n k) log ( (µ + xσ) log 1 + xσ µ ) ( n µ 1 (n 2k) log ) (n 2(µ + xσ)) log ) µ 2 ( 1 xσ n 2µ = n log(φ) + (n k) log(φ + 1) (n 2k) log(φ) + (n (µ + xσ)) log ( + (n (µ + xσ)) log 1 xσ n µ ) ( 1 xσ ) n µ ( (µ + xσ) log 1 + xσ ) ( (n 2(µ + xσ)) log 1 xσ ) µ n 2µ ( = n( log(φ) + log φ 2) log(φ)) + k(log(φ 2 ) + 2 log(φ)) + (n (µ + xσ)) log ( (µ + xσ) log 1 + xσ ) ( (n 2(µ + xσ)) log µ ( = (n (µ + xσ)) log 1 xσ n µ ( (n 2(µ + xσ)) log xσ 1 2 n 2µ ) ( (µ + xσ) log 1 + xσ µ ). ) xσ 1 2 n 2µ ) ( 1 xσ ) n µ )

30 (Sketch of the) Proof of Gaussianity Finally, we expand the logarithms and collect powers of xσ/n. log(s n) = ( (n (µ + xσ)) xσ n µ 1 ( ) xσ ) 2 n µ ( xσ (µ + xσ) µ 1 ( ) xσ ) 2 µ ( xσ (n 2(µ + xσ)) 2 n 2µ 1 ( ) xσ ) 2 n 2µ 2 = (n (µ + xσ)) xσ n (φ+1) 1 xσ 2 n (φ+1) +... (φ+2) (φ+2) 2 (µ + xσ) xσ 1 xσ +... n 2 n φ+2 φ+2 2 (n 2(µ + xσ)) 2xσ n φ 1 2xσ 2 n φ +... φ+2 φ+2 = ( ( xσ n n 1 1 ) ( (φ + 2) φ + 2 (φ + 1) ) ) φ + 2 φ + 2 φ 1 ( ) xσ 2 n( 2 φ n φ φ + 2 ) (φ + 2) (φ + 2) + 4φ φ + 1 φ +O (n(xσ/n) 3)

31 (Sketch of the) Proof of Gaussianity log(s n) = ( xσ n n φ + 1 ) φ + 2 φ + 2 φ φ φ + 2 φ + 2 φ 1 ( ) xσ 2 n(φ + 2)( 1 2 n φ ) φ ( ( ) ) xσ 3 +O n n = 1 (xσ) 2 ( ) ( ( ) ) 3φ + 4 xσ 3 (φ + 2) 2 n φ(φ + 1) O n n = 1 (xσ) 2 ( ) ( ( ) ) 3φ φ + 1 xσ 3 (φ + 2) + O n 2 n φ(φ + 1) n = 1 2 x2 σ 2 ( 5(φ + 2) φn ) ( + O n(xσ/n) 3).

32 (Sketch of the) Proof of Gaussianity But recall that σ 2 = φn 5(φ + 2). ( ) ( Also, since σ n 1/2, n xσn 3 ( ) ) n 1/2. So for large n, the O n xσn 3 term vanishes. Thus we are left with log S n = 1 2 x2 S n = e 1 2 x2. Hence, as n gets large, the density converges to the normal distribution: f n(k)dk = N ns ndk = 1 2πσ 2 e 1 2 x2 σdx = 1 2π e 1 2 x2 dx.

33 Generalizations Generalizing from Fibonacci numbers to linearly recursive sequences with arbitrary nonnegative coefficients. H n+1 = c 1 H n + c 2 H n 1 + +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

34 Generalizing Lekkerkerker Generalized Lekkerkerker s Theorem The average number of summands in the generalized Zeckendorf decomposition for integers in [H n, H n+1 ) tends to Cn+d as n, where C > 0 and d are computable constants determined by the c i s. L 1 C = y (1) y(1) = m=0 (s m + s m+1 1)(s m+1 s m )y m (1) 2 L 1 m=0 (m+1)(s. m+1 s m )y m (1) s 0 = 0, s m = c 1 + c 2 + +c m. y(x) is the root of 1 L 1 sm+1 1 m=0 j=s m x j y m+1. y(1) is the root of 1 c 1 y c 2 y 2 c L y L.

35 Central Limit Type Theorem Central Limit Type Theorem As n, the distribution of the number of summands, i.e., a 1 + a 2 + +a m in the generalized Zeckendorf decomposition m i=1 a ih i for integers in [H n, H n+1 ) is Gaussian

36 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 A n Gaussian with mean 4.5n + O(1) and variance 8.25n + O(1).

37 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 =

38 Generating Function (Example: Binet s Formula) Binet s Formula F 1 = F 2 = 1; F n = 1 5 [( 1+ ) n ( ) n ] 5. 2

39 Generating Function (Example: Binet s Formula) Binet s Formula F 1 = F 2 = 1; F n = 1 5 [( 1+ ) n ( ) n ] 5. 2 Recurrence relation: F n+1 = F n + F n 1 (1)

40 Generating Function (Example: Binet s Formula) Binet s Formula F 1 = F 2 = 1; F n = 1 5 [( 1+ ) n ( ) n ] 5. 2 Recurrence relation: F n+1 = F n + F n 1 (1) Generating function: g(x) = n>0 F nx n.

41 Generating Function (Example: Binet s Formula) Binet s Formula F 1 = F 2 = 1; F n = 1 5 [( 1+ ) n ( ) n ] 5. 2 Recurrence relation: F n+1 = F n + F n 1 (1) Generating function: g(x) = n>0 F nx n. (1) n 2 F n+1 x n+1 = n 2 F n x n+1 + n 2 F n 1 x n+1

42 Generating Function (Example: Binet s Formula) Binet s Formula F 1 = F 2 = 1; F n = 1 5 [( 1+ ) n ( ) n ] 5. 2 Recurrence relation: F n+1 = F n + F n 1 (1) Generating function: g(x) = n>0 F nx n. (1) n 2 n 3 F n+1 x n+1 = n 2 F n x n = n 2 F n x n+1 + n 2 F n x n+1 + n 1 F n x n+2 F n 1 x n+1

43 Generating Function (Example: Binet s Formula) Binet s Formula F 1 = F 2 = 1; F n = 1 5 [( 1+ ) n ( ) n ] 5. 2 Recurrence relation: F n+1 = F n + F n 1 (1) Generating function: g(x) = n>0 F nx n. (1) n 2 n 3 n 3 F n+1 x n+1 = n 2 F n x n = n 2 F n x n = x n 2 F n x n+1 + n 2 F n x n+1 + n 1 F n x n + x 2 n 1 F n x n+2 F n 1 x n+1 F n x n

44 Generating Function (Example: Binet s Formula) Binet s Formula F 1 = F 2 = 1; F n = 1 5 [( 1+ ) n ( ) n ] 5. 2 Recurrence relation: F n+1 = F n + F n 1 (1) Generating function: g(x) = n>0 F nx n. (1) n 2 n 3 F n+1 x n+1 = n 2 F n x n = n 2 F n x n+1 + n 2 F n x n+1 + n 1 F n x n+2 F n 1 x n+1 F n x n = x F n x n + x 2 F n x n n 3 n 2 n 1 g(x) F 1 x F 2 x 2 = x(g(x) F 1 x)+x 2 g(x)

45 Generating Function (Example: Binet s Formula) Binet s Formula F 1 = F 2 = 1; F n = 1 5 [( 1+ ) n ( ) n ] 5. 2 Recurrence relation: F n+1 = F n + F n 1 (1) Generating function: g(x) = n>0 F nx n. (1) n 2 n 3 F n+1 x n+1 = n 2 F n x n = n 2 F n x n+1 + n 2 F n x n+1 + n 1 F n x n+2 F n 1 x n+1 n 3 F n x n = x n 2 F n x n + x 2 n 1 F n x n g(x) F 1 x F 2 x 2 = x(g(x) F 1 x)+x 2 g(x) g(x) = x/(1 x x 2 ).

46 Partial Fraction Expansion (Example: Binet s Formula) Generating function: g(x) = n>0 F nx n = x 1 x x 2.

47 Partial Fraction Expansion (Example: Binet s Formula) Generating function: g(x) = n>0 F nx n = x 1 x x 2. Partial fraction expansion:

48 Partial Fraction Expansion (Example: Binet s Formula) Generating function: g(x) = n>0 F nx n = x 1 x x 2. Partial fraction expansion: g(x) = ( x 1 x x = x ) x x x 2 2

49 Partial Fraction Expansion (Example: Binet s Formula) Generating function: g(x) = n>0 F nx n = x 1 x x 2. Partial fraction expansion: g(x) = ( x 1 x x = x ) x x x 2 2 Coefficient of x n (power series expansion): [( F n = 1 1+ ) n ( ) n ] 5 - Binet s Formula! 2 1 (using geometric series: = 1+r + r 2 + r 3 + ). 1 r

50 Differentiating Identities and Method of Moments Differentiating identities Example: Given a random variable X such that Pr(X = 1) = 1 2, Pr(X = 2) = 1 4, Pr(X = 3) = 1 8,... then what s the mean of X (i.e., E[X])? Solution: Let f(x) = 1x + 1x 2 + 1x 3 + = x/2 f (x) = x + 3 1x f (1) = = E[X] Method of moments: Random variables X 1, X 2,... If l th moments E[Xn l ] converges those of standard normal then X n converges to a Gaussian. Standard normal distribution: 2m th moment: (2m 1)!! = (2m 1)(2m 3) 1, (2m 1) th moment: 0.

51 New Approach: Case of Fibonacci Numbers p n,k = # {N [F n, F n+1 ): the Zeckendorf decomposition of N has exactly k summands}. Recurrence relation: N [F n+1, F n+2 ): N = F n+1 + F t +, t n 1. p n+1,k+1 = p n 1,k + p n 2,k +

52 New Approach: Case of Fibonacci Numbers p n,k = # {N [F n, F n+1 ): the Zeckendorf decomposition of N has exactly k summands}. Recurrence relation: N [F n+1, F n+2 ): N = F n+1 + F t +, t n 1. p n+1,k+1 = p n 1,k + p n 2,k + p n,k+1 = p n 2,k + p n 3,k +

53 New Approach: Case of Fibonacci Numbers p n,k = # {N [F n, F n+1 ): the Zeckendorf decomposition of N has exactly k summands}. Recurrence relation: N [F n+1, F n+2 ): N = F n+1 + F t +, t n 1. p n+1,k+1 = p n 1,k + p n 2,k + p n,k+1 = p n 2,k + p n 3,k + p n+1,k+1 = p n,k+1 + p n 1,k.

54 New Approach: Case of Fibonacci Numbers p n,k = # {N [F n, F n+1 ): the Zeckendorf decomposition of N has exactly k summands}. Recurrence relation: N [F n+1, F n+2 ): N = F n+1 + F t +, t n 1. p n+1,k+1 = p n 1,k + p n 2,k + p n,k+1 = p n 2,k + p n 3,k + p n+1,k+1 = p n,k+1 + p n 1,k. y 1 y xy 2. Generating function: n,k>0 p n,kx k y n = Partial fraction expansion: ( ) y 1 y xy = y 1 2 y 1 (x) y 2 (x) y y 1 (x) 1 y y 2 (x) where y 1 (x) and y 2 (x) are the roots of 1 y xy 2 = 0. Coefficient of y n : g(x) = k>0 p n,kx k.

55 New Approach: Case of Fibonacci Numbers (Continued) K n : the corresponding random variable associated with k. g(x) = k>0 p n,kx k. Differentiating identities: g(1) = k>0 p n,k = F n+1 F n, g (x) = k>0 kp n,kx k 1, g (1) = g(1)e[k n ], (xg (x)) = k>0 k 2 p n,k x k 1, (xg (x)) x=1 = g(1)e[k 2 n], ( x (xg (x)) ) x=1 = g(1)e[k 3 n], Similar results hold for the centralized K n : K n = K n E[K n ]. Method of moments (for normalized K n ): E[(K n )2m ]/(SD(K n ))2m (2m 1)!!, E[(K n) 2m 1 ]/(SD(K n)) 2m 1 0. K n Gaussian.

56 New Approach: General Case Let p n,k = # {N [H n, H n+1 ): the generalized Zeckendorf decomposition of N has exactly k summands}. Recurrence relation: Fibonacci: p n+1,k+1 = p n,k+1 + p n,k. sm+1 1 j=s m p n m,k j. General: p n+1,k = L 1 m=0 where s 0 = 0, s m = c 1 + c 2 + +c m. Generating function: Fibonacci: y 1 y xy 2. General: n L p n,kx k y n L 1 sm+1 1 m=0 1 L 1 m=0 j=s m x j y m+1 n<l m p n,kx k y n sm+1. 1 j=s m x j y m+1

57 New Approach: General Case (Continued) Partial fraction expansion: ( y Fibonacci: y 1 (x) y 2 (x) General: 1 sl 1 j=s L 1 x j L i=1 B(x, y) = L 1 p n,k x k y n n L 1 1 y y 1 (x) y y 2 (x) ). B(x, y) (y y i (x)) j i (y j(x) y i (x)). m=0 s m+1 1 j=s m x j y m+1 n<l m y i (x): root of 1 L 1 sm+1 1 m=0 j=s m x j y m+1 = 0. p n,k x k y n, 57 Coefficient of y n : g(x) = n,k>0 p n,kx k. Differentiating identities Method of moments: implies K n Gaussian.

58 Gaps in the Bulk

59 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.

60 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.

61 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.

62 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)?

63 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 =

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

65 Main Results Theorem 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 P(j) = 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 ) a 1 C Lek λ j 1 j 2

66 Proof of Fibonacci Result Lekkerkerker total number of gaps F n 1 n. φ 2 +1

67 Proof of Fibonacci Result Lekkerkerker total number of gaps F n 1 n. φ 2 +1 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}.

68 Proof of Fibonacci Result Lekkerkerker total number of gaps F n 1 n. φ 2 +1 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}. n k P(k) = lim n i=1 X i,i+k n F n 1 φ 2 +1.

69 Calculating X i,i+k How many decompositions contain a gap from F i to F i+k?

70 Calculating X i,i+k How many decompositions contain a gap from F i to F i+k? Number of choices is F n k 2 i F i 1 :

71 Calculating X i,i+k How many decompositions contain a gap from F i to F i+k? Number of choices is F n k 2 i F i 1 : For the indices less than i: F i 1 choices. Why? Have F i, don t have F i 1. Follows by Zeckendorf: like the interval [F i, F i+1 ) as have F i, number elements is F i+1 F i = F i 1.

72 Calculating X i,i+k How many decompositions contain a gap from F i to F i+k? Number of choices is F n k 2 i F i 1 : For the indices less than i: F i 1 choices. Why? Have F i, don t have F i 1. Follows by Zeckendorf: like the interval [F i, F i+1 ) as have F i, number elements is F i+1 F i = F i 1. For the indices greater than i + k: F n k i 2 choices. Why? Have F n, don t have F i+k+1. Like Zeckendorf with potential summands F i+k+2,...,f n. Shifting, like summands F 1,...,F n k i 1, giving F n k i 2.

73 Determining P(k) n k X i,i+k = F n k 1 + i=1 n k 2 i=1 F i 1 F n k i 2 n k 3 i=0 F i F n k i 3 is the x n k 3 coefficient of (g(x)) 2, where g(x) is the generating function of the Fibonaccis. Alternatively, use Binet s formula and get sums of geometric series.

74 Determining P(k) n k X i,i+k = F n k 1 + i=1 n k 2 i=1 F i 1 F n k i 2 n k 3 i=0 F i F n k i 3 is the x n k 3 coefficient of (g(x)) 2, where g(x) is the generating function of the Fibonaccis. Alternatively, use Binet s formula and get sums of geometric series. P(k) = C/φ k for a constant C, so P(k) = 1/φ k.

75 75 Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs 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.

76 Longest Gap

77 77 Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs Longest Gap For most recurrences, our central result is Theorem (Mean and Variance of Longest Gap) Let λ 1 be the largest eigenvalue of the recurrence, γ be Euler s constant, and K a constant that is a polynomial in λ 1. Then the mean and variance of the longeset gap are: µ n = log(nk) logλ 1 σ 2 n = π 2 6(logλ 1 ) 2 + o(1). + γ logλ o(1)

78 Strategy Our argument follows three main steps: Find a rational generating function S f (x) for the number of m (H n, H n+1 ] with longest gap less than f. Obtain an approximate formula for the CDF of the longest gap. Estimate the mean and variance using Partial Summation and the Euler Maclaurin Formula.

79 Fibonacci case For the fibonacci numbers, our generating function is From this we obtain S f (x) = x 1 x x 2 + x f. Theorem (Longest Gap Asymptotic CDF) As n, the probability that m [F n, F n+1 ) has longest gap less than or equal to f(n) converges to Prob(L n (m) f(n)) e elog n f(n)/ logφ.

80 Generating Function: I For k fixed the number of m [F n, F n+1 ) with k summands and longest gap less than f equals the coefficient of x n for in the expression 1 1 x f(n) 2 x j j=2 k 1.

81 Generating Function: II Why the n th coefficient of ( f(n) 1 ) k x j=2 x j?

82 Generating Function: II Why the n th coefficient of ( f(n) 1 ) k x j=2 x j? Let m = F n + F n g1 + F n g1 g 2 + +F n g1 g n 1. The gaps uniquely identify m because of Zeckendorf s Theorem! And we have the following:

83 Generating Function: II Why the n th coefficient of ( f(n) 1 ) k x j=2 x j? Let m = F n + F n g1 + F n g1 g 2 + +F n g1 g n 1. The gaps uniquely identify m because of Zeckendorf s Theorem! And we have the following: The sum of the gaps of x is n.

84 Generating Function: II Why the n th coefficient of ( f(n) 1 ) k x j=2 x j? Let m = F n + F n g1 + F n g1 g 2 + +F n g1 g n 1. The gaps uniquely identify m because of Zeckendorf s Theorem! And we have the following: The sum of the gaps of x is n. Each gap is 2.

85 Generating Function: II Why the n th coefficient of ( f(n) 1 ) k x j=2 x j? Let m = F n + F n g1 + F n g1 g 2 + +F n g1 g n 1. The gaps uniquely identify m because of Zeckendorf s Theorem! And we have the following: The sum of the gaps of x is n. Each gap is 2. Each gap is < f.

86 Generating Function: III If we sum over k we get the total number of m [F n, F n+1 ) with longest gap < f. It s the n th coefficient of F(x) = 1 1 x ( ) x 2 x f 2 k 1 = 1 x k=1 x 1 x x 2 + x f.

87 Obtaining the CDF We analyze asymptotic behavior of the coefficients of as n, f vary. S f (x) = x 1 x x 2 + x f Use a partial fraction decomposition. Problem: What happens to the roots of 1 x x 2 + x f as f varies? Solution: 1 x x 2 + x f has a unique smallest root α f which converges to 1/φ for large f. The contribution of α f dominates, allowing us to obtain an approximate CDF.

88 Numerical Results Convergence to mean is at best approximately n δ for some small δ > 0. Computing numerics is difficult: F n+1 = F n + F n 1 : Sampling 100 numbers from [F n, F n+1 ) with n = 1, 000, 000. Mean predicted : vs. observed: Variance predicted : 2.67 vs. observed: 2.44 a n+1 = 2a n + 4a n 1 : Sampling 100 numbers from [a n, a n+1 ) with n = 51, 200. Mean predicted : 9.95 vs. observed: 9.91 Variance predicted : 1.09 vs. observed: 1.22

89 Numerical Results pt 2 F n+1 = F n + F n 1 : Sampling 20 numbers from [F n, F n+1 ) with n = 10, 000, 000. Mean predicted : vs. observed: Variance predicted : 2.67 vs. observed: 2.33 a n+1 = 2a n + 4a n 1 : Sampling 100 numbers from [a n, a n+1 ) with n = 102, 400. Mean predicted : vs. observed: Variance predicted : 1.09 vs. observed: 1.10

90 Future Work and References

91 Future Research Future Research Generalizing results to all PLRS and signed decompositions. Other systems such as f-decompositions of Demontigny, Do, Miller and Varma.

92 References References Beckwith, Bower, Gaudet, Insoft, Li, Miller and Tosteson: The Average Gap Distribution for Generalized Zeckendorf Decompositions. The Fibonacci Quarterly 51 (2013), Bower, Insoft, Li, Miller and Tosteson: Distribution of gaps in generalized Zeckendorf decompositions, preprint Kologlu, Kopp, Miller and Wang: On the number of summands in Zeckendorf decompositions, Fibonacci Quarterly 49 (2011), no. 2, Miller and Wang: Gaussian Behavior in Generalized Zeckendorf Decompositions, to appear in the conference proceedings of the 2011 Combinatorial and Additive Number Theory Conference.