From Fibonacci Numbers to Central Limit Type Theorems

From Fibonacci Numbers to Central Limit Type Theorems Murat Koloğlu, Gene Kopp, Steven J. Miller and Yinghui Wang Williams College, October 1st, 010

Introduction

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

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

5 Intro Lekkerkerker Erdos-Kac Generalizations Method Open Questions Appendix Previous Results Fibonacci Numbers: F n+1 = F n + F n 1 ; F 1 = 1, F =, F 3 = 3, F 4 = 5,... Zeckendorf s Theorem Every positive integer can be written uniquely as a sum of non-consecutive Fibonacci numbers.

Previous Results Fibonacci Numbers: F n+1 = F n + F n 1 ; F 1 = 1, F =, 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: 010 = 1597+ 377+34+ = F 16 + F 13 + F 8 + F.

7 Intro Lekkerkerker Erdos-Kac Generalizations Method Open Questions Appendix Previous Results Fibonacci Numbers: F n+1 = F n + F n 1 ; F 1 = 1, F =, 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: 010 = 1597+ 377+34+ = F 16 + F 13 + F 8 + F. Lekkerkerker s Theorem (195) The average number of summands in the Zeckendorf n decomposition for integers in [F n, F n+1 ) tends to φ +1.76n, where φ = 1+ 5 is the golden mean.

New 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). 0.00 0.015 0.010 0.005 1000 1050 1100 1150 100 Figure: Number of summands in [F 010, F 011 )

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

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 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.

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

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 i + +F ik 1 + F n, 1 i 1 < i < < i k 1 < i k = n, i j i j 1.

Lekkerkerker s Theorem (note slightly different notation)

Preliminaries E(n) := n 1 k=0 ( n 1 k k k ).

Preliminaries E(n) := n 1 k=0 ( n 1 k k k ). Average number of summands in [F n, F n+1 ) is E(n) F n 1 + 1.

Preliminaries E(n) := n 1 k=0 ( n 1 k k k ). Average number of summands in [F n, F n+1 ) is E(n) F n 1 + 1. Recurrence Relation for E(n) E(n)+E(n ) = (n )F n 3.

Recurrence Relation Recurrence Relation for E(n) E(n)+E(n ) = (n )F n 3. Proof by algebra (details in appendix): E(n) = n 1 k=0 = (n ) ( ) n 1 k k k n 3 l=0 ( n 3 l = (n )F n 3 E(n ). l ) n 3 l=0 ( ) n 3 l l l

Solving Recurrence Relation Formula for E(n) (i.e., Lekkerkerker s Theorem) E(n) = nf n 1 φ + 1 + O(F n ). = n 3 ( 1) l (E(n l)+e(n (l+1))) l=0 n 3 ( 1) l (n l)f n 3 l. l=0 Result follows from Binet s formula, the geometric series formula, and differentiating identities: m j=0 jx j = x (m+1)xm (x 1) (x m+1 1). Details in appendix. (x 1)

An Erdos-Kac Type Theorem (note slightly different notation)

Generalizing Lekkerkerker Theorem (KKMW 010) As n, the distribution of the number of summands in Zeckendorf s Theorem is a Gaussian. 0.00 0.015 0.010 0.005 1000 1050 1100 1150 100 Figure: Number of summands in [F 010, F 011 )

Generalizing Lekkerkerker: Erdos-Kac type result Theorem (KKMW 010) As n, the distribution of the number of summands in Zeckendorf s Theorem is a Gaussian. Numerics: At F 100,000 : Ratio of m th moment σ m to (m 1)!!σ m is between.999955 and 1 for m 10. Sketch of proof: Use Stirling s formula, n! n n e n πn to approximates binomial coefficients, after a few pages of algebra find the probabilities are approximately Gaussian.

(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 k)!k! F n 1 (n k) (n k+ 1) π k (k+ 1) (n k) (n k+ 1) F n 1 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 = ). We can now split the terms that exponentially depend on n. Define f n+1 (k) = ( 1 π (n k) k(n k) )( 5 φ n (n k) (n k) ) φ k k (n k) (n k) N n = 1 (n k) 5 (n k) π k(n k) φ, Sn = (n k) φ n k k (n k) (n k). Thus, write the density function as f n+1 (k) = N ns n where N n is the first term that is of order n 1/ and S n is the second term with exponential dependence on n.

(Sketch of the) Proof of Gaussianity (cont) 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 π k(n k) 5 1 1 k/n πn 1 πn 1 πn (k/n)(1 k/n) 5 φ 1 (μ + σx)/n 5 ((μ + σx)/n)(1 (μ + σx)/n) φ 1 C y 5 (C + y)(1 C y) φ where C = μ/n 1/(φ + ) (note that φ = φ + 1) and y = σx/n. But for large n, the y term vanishes since σ n and thus y n 1/. Thus N n 1 1 C 5 πn C(1 C) φ = 1 (φ + 1)(φ + ) 5 πn φ φ = 1 5(φ + ) 1 = πn φ πσ since σ φ = n 5(φ+).

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

(Sketch of the) Proof of Gaussianity (cont) Note that, since n/μ = φ + 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 k) log ) (n (μ + xσ)) log ) μ ( 1 xσ n μ = n log(φ) + (n k) log(φ + 1) (n k) log(φ) + (n (μ + xσ)) log ( + (n (μ + xσ)) log 1 xσ n μ ) ( 1 xσ ) n μ ( (μ + xσ) log 1 + xσ ) ( (n (μ + xσ)) log 1 xσ ) μ n μ ( = n( log(φ) + log φ ) log(φ)) + k(log(φ ) + log(φ)) + (n (μ + xσ)) log ( (μ + xσ) log 1 + xσ ) ( (n (μ + xσ)) log μ ( = (n (μ + xσ)) log 1 xσ n μ ( (n (μ + xσ)) log xσ 1 n μ ) ( (μ + xσ) log 1 + xσ μ ). ) xσ 1 n μ ) ( 1 xσ ) n μ )

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

(Sketch of the) Proof of Gaussianity (cont) = ( xσ n n φ + 1 ) φ + φ + φ + 1 1 + φ φ + φ + φ 1 ( ) xσ n(φ + )( 1 n φ + 1 + 1 + 4 ) φ ( ( ) ) xσ 3 +O n n = 1 (xσ) ( ) ( ( ) ) 3φ + 4 xσ 3 (φ + ) n φ(φ + 1) + 1 + O n n = 1 (xσ) ( ) ( ( ) ) 3φ + 4 + φ + 1 xσ 3 (φ + ) + O n n φ(φ + 1) n = 1 ( ) x σ 5(φ + ) + O (n(xσ/n) 3) φn

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

Generalizations

Generalizations Generalizing from Fibonacci numbers to linearly recursive sequences with arbitrary nonnegative coefficients. H n+1 = c 1 H n + c H n 1 + +c L H n L+1, n L. with H 1 = 1, H n+1 = c 1 H n + c H n 1 + +c n H 1 + 1, n < L, coefficients c i 0; c 1, c L > 0 if L ; 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

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. C = y (1) L 1 y(1) = m=0 (s m + s m+1 1)(s m+1 s m )y m (1) L 1 m=0 (m+1)(s. m+1 s m )y m (1) s 0 = 0, s m = c 1 + c + +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 y c L y L.

Central Limit Type Theorem Central Limit Type Theorem As n, the distribution of the number of summands, i.e., a 1 + a + +a m in the generalized Zeckendorf decomposition m i=1 a ih i for integers in [H n, H n+1 ) is Gaussian. 0.00 0.015 0.010 0.005 1000 1050 1100 1150 100

Example: the Special Case of L = 1 H n+1 = c 1 H n, H 1 = 1. H n = c n 1 1. Legal decomposition m i=1 a ih i : a i {0, 1,..., c 1 1} (1 i < m), a m {1,..., c 1 1}, equivalent to the c 1 -base expansion. For N [H n, H n+1 ), m = n, i.e., the first term is a n H n. 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 (c 1 1)/ and variance (c 1 1)/1. Central Limit Theorem: A + A 3 + +A n Gaussian with mean n(c 1 1)/+O(1) and variance n(c 1 1)/1+O(1).

Far-difference Representation Theorem (Alpert, 009) (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. 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/.809.

Method of General Proof

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

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

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

55 Intro Lekkerkerker Erdos-Kac Generalizations Method Open Questions Appendix Preliminaries Generating Function (Example: Binet s Formula) Binet s Formula F 1 = F = 1; F n = 1 5 [ ( 1+ 5 ) n ( ) ] n 1+ 5. Recurrence relation: F n+1 = F n + F n 1 (1) Generating function: g(x) = n>0 F nx n. (1) n F n+1 x n+1 = n F n x n+1 + n F n 1 x n+1 F n x n = F n x n+1 + F n x n+ n 3 n n 1 F n x n = x F n x n + x F n x n n 3 n n 1

57 Intro Lekkerkerker Erdos-Kac Generalizations Method Open Questions Appendix Preliminaries Generating Function (Example: Binet s Formula) Binet s Formula F 1 = F = 1; F n = 1 5 [ ( 1+ 5 ) n ( ) ] n 1+ 5. Recurrence relation: F n+1 = F n + F n 1 (1) Generating function: g(x) = n>0 F nx n. (1) n F n+1 x n+1 = n F n x n+1 + n F n 1 x n+1 F n x n = F n x n+1 + F n x n+ n 3 n n 1 F n x n = x F n x n + x F n x n n 3 n n 1 g(x) F 1 x F x = x(g(x) F 1 x)+x g(x) g(x) = x/(1 x x ).

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

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

Preliminaries Partial Fraction Expansion (Example: Binet s Formula) Generating function: g(x) = n>0 F nx n = Partial fraction expansion: ( 1 1 x x = 1 1 5 x 1+ 5. x 1 x x. 1 x 1 5 )

Preliminaries Partial Fraction Expansion (Example: Binet s Formula) Generating function: g(x) = n>0 F nx n = Partial fraction expansion: ( 1 1 x x = 1 5. g(x) = x 1 x x 1 x 1+ 5 ( = 1 5 x x 1+ 5 x 1 x x. 1 x 1 5 ) x x 1 5 )

Preliminaries Partial Fraction Expansion (Example: Binet s Formula) Generating function: g(x) = n>0 F nx n = Partial fraction expansion: ( 1 1 x x = 1 5. g(x) = x 1 x x 1 x 1+ 5 ( = 1 5 = ( 1 1+ 5 x x 1+ 5 5 x x 1 x x. 1 x 1 5 1 1+ 5 x ) x x 1 5 ) 1+ 5 x 1 1+ 5 x ).

Preliminaries Differentiating Identities and Method of Moments Differentiating identities

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

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

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+ ): N = F n+1 + F t +, t n 1. p n+1,k+1 = p n 1,k + p n,k +

75 Intro Lekkerkerker Erdos-Kac Generalizations Method Open Questions Appendix 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+ ): N = F n+1 + F t +, t n 1. p n+1,k+1 = p n 1,k + p n,k + = p n,k + p n 3,k + p n,k+1 p n+1,k+1 = p n,k+1 + p n 1,k. Generating function: p n,kx k y n y = n,k>0 1 y xy. Partial fraction expansion: ( ) y 1 y xy = y 1 y 1 (x) y (x) y y 1 (x) 1 y y (x), where y1 (x) and y (x) are the roots of 1 y xy = 0. Coefficient of y n : g(x) = n,k>0 p n,kx k.

New Approach: Case of Fibonacci Numbers (Continued) K n : the corresponding random variable associated with k. g(x) = n,k>0 p n,kx k. Differentiating identities: g(1) = n,k>0 p n,k = F n+1 F n,

77 Intro Lekkerkerker Erdos-Kac Generalizations Method Open Questions Appendix New Approach: Case of Fibonacci Numbers (Continued) K n : the corresponding random variable associated with k. g(x) = n,k>0 p n,kx k. Differentiating identities: g(1) = n,k>0 p n,k = F n+1 F n, g (x) = n,k>0 kp n,kx k 1, g (1) = g(1)e[k n ],

New Approach: Case of Fibonacci Numbers (Continued) K n : the corresponding random variable associated with k. g(x) = n,k>0 p n,kx k. Differentiating identities: g(1) = n,k>0 p n,k = F n+1 F n, g (x) = n,k>0 kp n,kx k 1, g (1) = g(1)e[k n ], (xg (x)) = n,k>0 k p n,k x k 1, (xg (x)) x=1 = g(1)e[k 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 ].

New Approach: General Case Let p n,k = # {N [H n, H n+1 ): the generalized Zeckendorf decomposition of N has exactly k summands}.

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:

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.

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 + +c m. Generating function: Fibonacci: y 1 y xy.

New Approach: General Case (Continued) Partial fraction expansion:

New Approach: General Case (Continued) Partial fraction expansion: ( y Fibonacci: y 1 (x) y (x) 1 y y 1 (x) 1 y y (x) ).

New Approach: General Case (Continued) Partial fraction expansion: ( y Fibonacci: y 1 (x) y (x) General: 1 sl 1 j=s L 1 x j i=1 B(x, y) = n L p n,k x k y n 1 y y 1 (x) 1 y y (x) ). L B(x, y) (y y i (x)) ( j =i yj (x) y i (x) ). L 1 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. Coefficient of y n : g(x) = n,k>0 p n,kx k. Differentiating identities Method of moments K n Gaussian. p n,k x k y n,

Future Research

Further Research 1 Are there similar results for linearly recursive sequences with arbitrary integer coefficients (i.e. negative coefficients are allowed in the defining relation)? Lekkerkerker s theorem, and the Gaussian extension, are for the behavior in intervals [F n, F n+1 ). Do the limits exist if we consider other intervals, say [F n + g 1 (F n ), F n + g (F n )) for some functions g 1 and g? If yes, what must be true about the growth rates of g 1 and g? 3 For the generalized recurrence relations, what happens if instead of looking at n i=1 a i we study n i=1 min(1, a i)? In other words, we only care about how many distinct H i s occur in the decomposition. 4 What can we say about the distribution of the largest gap between summands in the Zeckendorf decomposition? Appropriately normalized, how does the distribution of gaps between the summands behave?

Appendix: Details of Computations

Needed Binomial Identity Binomial identity involving Fibonacci Numbers Let F m denote the m th Fibonacci number, with F 1 = 1, F =, F 3 = 3, F 4 = 5 and so on. Then n 1 ( ) n 1 k = F n 1. k k=0 Proof by induction: The base case is trivially ( verified. ) Assume our claim holds for n and show that it holds for n + 1. We may extend the sum to n 1, as n 1 k = 0 whenever k > n 1. Using the standard identity that k ( ( ) ( m m m + 1 ) + l l + 1 ( and the convention that m l) = 0 if l is a negative integer, we find = l + 1 ), n ( ) n k k=0 k = = = n [( n 1 k k 1 k=0 n ( n 1 k ) + ( )] n 1 k k ) n ( ) n 1 k + k 1 k k=1 k=0 n ( ) n (k 1) n ( ) n 1 k + = F n + F n 1 k 1 k k=1 k=0 by the inductive assumption; noting F n + F n 1 = F n completes the proof.

Derivation of Recurrence Relation for E(n) E(n) = = = = = n 1 ( ) n 1 k k k k=0 n 1 (n 1 k)! k k!(n 1 k)! k=1 n 1 (n k)! (n 1 k) (k 1)!(n 1 k)! k=1 n 1 (n 3 (k 1)! (n (k 1)) (k 1)!(n 3 (k 1))! k=1 n 3 ( ) n 3 l (n l) l l=0 n 3 ( n 3 l = (n ) l l=0 = (n )F n 3 E(n ), ) n 3 l=0 ( ) n 3 l l l which proves the claim (note we used the binomial identity to replace the sum of binomial coefficients with a Fibonacci number).

Formula for E(n) Formula for E(n) E(n) = nf n 1 φ + 1 + O(F n ). Proof: The proof follows from using telescoping sums to get an expression for E(n), which is then evaluated by inputting Binet s formula and differentiating identities. Explicitly, consider n 3 ( 1) l (E(n l) + E(n (l + 1))) = l=0 = = n 3 l=0 ( 1) l (n 3 l)f n 3 l + n 3 l=0 n 3 ( 1) l (n 3 l)f n 3 l + O(F n ); l=0 ( 1) l (n l)f n 3 l n 3 ( 1) l (l)f n 3 l l=0 while we could evaluate the last sum exactly, trivially estimating it suffices to obtain the main term (as we have a sum of every other Fibonacci number, the sum is at most the next Fibonacci number after the largest one in our sum).

Formula for E(n) (continued) We now use Binet s formula to convert the sum into a geometric series. Lettingφ = 1+ 5 be the golden mean, we have F n = φ φ n 1 φ (1 φ) n 5 5 (our constants are because our counting has F 1 = 1, F = and so on). As 1 φ < 1, the error from dropping the (1 φ) n term is O( l n n) = O(n ) = o(f n ), and may thus safely be absorbed in our error term. We thus find E(n) = = n 3 φ (n 3 l)( 1) l φ n 3 l + O(F n ) 5 l=0 n 3 n 3 φ n 5 (n 3) ( φ ) l l( φ ) l + O(F n ). l=0 l=0

Formula for E(n) (continued) We use the geometric series formula to evaluate the first term. We drop the upper boundary term of ( φ 1 ) n 3, as this term is negligible since φ > 1. We may also move the 3 from the n 3 into the error term, and are left with E(n) = = φ n 5 n 3 n 1 + φ l( φ ) l + O(F n ) l=0 φ n [ ( n )] n 3 S, φ + O(F 5 1 + φ n ), where S(m, x) = m jx j. j=0 There is a simple formula for S(m, x). As m j=0 x j = xm+1 1, x 1 applying the operator x d dx gives S(m, x) = m j=0 jx j = x (m + 1)xm (x 1) (x m+1 1) = mx m+ (m + 1)x m+1 + x. (x 1) (x 1)

Formula for E(n) (continued) Taking x = φ, we see that the contribution from this piece may safely be absorbed into the error term O(F n ), leaving us with E(n) = nφ n 5(1 + φ ) + O(F n ) = nφ n 5(φ + 1) + O(F n ). Noting that for large n we have F n 1 = φn 5 + O(1), we finally obtain E(n) = nf n 1 φ + 1 + O(F n ).