Springboards to Mathematics: From Zombies to Zeckendorf

Transcription

1 Springboards to Mathematics: From Zombies to Zeckendorf Steven J. Miller

2 Zombies

3 General Advice: What are your tools and how can they be used?

4 Zombine Infection: Rules If share walls with 2 or more infected, become infected. Once infected, always infected.

5 Zombine Infection: Rules If share walls with 2 or more infected, become infected. Once infected, always infected.

6 Zombine Infection: Rules If share walls with 2 or more infected, become infected. Once infected, always infected.

7 Zombine Infection: Rules If share walls with 2 or more infected, become infected. Once infected, always infected.

8 Zombine Infection: Rules If share walls with 2 or more infected, become infected. Once infected, always infected.

9 Zombine Infection: Conquering The World Easiest initial state that ensures all eventually infected is...?

10 Zombine Infection: Conquering The World Easiest initial state that ensures all eventually infected is...?

11 Zombine Infection: Conquering The World Next simplest initial state ensuring all eventually infected...?

12 Zombine Infection: Conquering The World Next simplest initial state ensuring all eventually infected...?

13 Zombine Infection: Conquering The World Fewest number of initial infections needed to get all...?

14 Zombine Infection: Conquering The World Fewest number of initial infections needed to get all...?

15 Zombine Infection: Conquering The World Fewest number of initial infections needed to get all...?

16 Zombine Infection: Conquering The World Fewest number of initial infections needed to get all...?

17 Zombine Infection: Conquering The World Fewest number of initial infections needed to get all...?

18 Zombine Infection: Conquering The World Fewest number of initial infections needed to get all...?

19 Zombine Infection: Conquering The World Fewest number of initial infections needed to get all...?

20 Zombine Infection: Conquering The World Fewest number of initial infections needed to get all...?

21 Zombine Infection: Conquering The World Fewest number of initial infections needed to get all...?

22 Zombie Infection: Can n 1 infect all?

23 Zombie Infection: Can n 1 infect all? Perimeter of infection unchanged.

24 Zombie Infection: Can n 1 infect all? Perimeter of infection unchanged.

25 Zombie Infection: Can n 1 infect all? Perimeter of infection decreases by 2.

26 Zombie Infection: Can n 1 infect all? Perimeter of infection decreases by 4.

27 Zombie Infection: n 1 cannot infect all If n 1 infected, maximum perimeter is 4(n 1) = 4n 4.

28 Zombie Infection: n 1 cannot infect all If n 1 infected, maximum perimeter is 4(n 1) = 4n 4. Mono-variant: As time passes, perimeter of infection never increases.

29 Zombie Infection: n 1 cannot infect all If n 1 infected, maximum perimeter is 4(n 1) = 4n 4. Mono-variant: As time passes, perimeter of infection never increases. Perimeter of n n square is 4n, so at least 1 square safe!

30 Zeckendorf

31 Tiling the Plane with Squares Have n n square for each n, place one at a time so that shape formed is always connected and a rectangle.

32 Tiling the Plane with Squares Have n n square for each n, extra 1 1 square, place one at a time so that shape formed is always connected and a rectangle.

33 Tiling the Plane with Squares: 1 1, 1 1, 2 2, 3 3,....

34 Tiling the Plane with Squares: 1 1, 1 1, 2 2, 3 3,....

35 Tiling the Plane with Squares: 1 1, 1 1, 2 2, 3 3,....

36 Tiling the Plane with Squares: 1 1, 1 1, 2 2, 3 3,....

37 Tiling the Plane with Squares: 1 1, 1 1, 2 2, 3 3,.... 1, 1, 2, 3, 5,....

38 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 ).

39 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.

40 Previous Results Fibonacci Numbers: F n+1 = F n + F n 1 ; First few: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,....

41 Previous Results Fibonacci Numbers: F n+1 = F n + F n 1 ; First few: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,.... Zeckendorf s Theorem Every positive integer can be written uniquely as a sum of non-consecutive Fibonacci numbers.

42 Previous Results Fibonacci Numbers: F n+1 = F n + F n 1 ; First few: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,.... Zeckendorf s Theorem Every positive integer can be written uniquely as a sum of non-consecutive Fibonacci numbers. Example: 51 =?

43 Previous Results Fibonacci Numbers: F n+1 = F n + F n 1 ; First few: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,.... Zeckendorf s Theorem Every positive integer can be written uniquely as a sum of non-consecutive Fibonacci numbers. Example: 51 = = F

44 Previous Results Fibonacci Numbers: F n+1 = F n + F n 1 ; First few: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,.... Zeckendorf s Theorem Every positive integer can be written uniquely as a sum of non-consecutive Fibonacci numbers. Example: 51 = = F 8 + F

45 Previous Results Fibonacci Numbers: F n+1 = F n + F n 1 ; First few: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,.... Zeckendorf s Theorem Every positive integer can be written uniquely as a sum of non-consecutive Fibonacci numbers. Example: 51 = = F 8 + F 6 + F

46 Previous Results Fibonacci Numbers: F n+1 = F n + F n 1 ; First few: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,.... Zeckendorf s Theorem Every positive integer can be written uniquely as a sum of non-consecutive Fibonacci numbers. Example: 51 = = F 8 + F 6 + F 3 + F 1.

47 Previous Results Fibonacci Numbers: F n+1 = F n + F n 1 ; First few: 1, 2, 3, 5, 8, 13, 21, 34, 55, 89,... Zeckendorf s Theorem Every positive integer can be written uniquely as a sum of non-consecutive Fibonacci numbers. Example: 51 = = F 8 + F 6 + F 3 + F 1. Example: 83 = = F 9 + F 7 + F 4 + F 2.

48 Previous Results Fibonacci Numbers: F n+1 = F n + F n 1 ; Example: 51 = = F 8 + F 6 + F 3 + F 1. Example: 83 = = F 9 + F 7 + F 4 + F 2. Observe: 51 miles 82.1 kilometers.

49 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

50 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

51 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.

52 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.

53 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.

54 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):

55 55 Zombies Zeckendorf Gaussianity Kentucky and Quilts Future / Refs 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):

56 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):

57 57 Zombies Zeckendorf Gaussianity Kentucky and Quilts Future / Refs 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

58 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}.

59 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.

60 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.

61 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).

62 Gaussian Behavior

63 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.

64 (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.

65 (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).

66 (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

67 (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 )

68 (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)

69 (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).

70 (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.

71 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

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

73 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.

74 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,

75 75 Zombies Zeckendorf Gaussianity Kentucky and Quilts Future / Refs 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],

76 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,

77 77 Zombies Zeckendorf Gaussianity Kentucky and Quilts Future / Refs 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],

78 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,

79 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].

80 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.

81 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.

82 What s in a name?

83 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).

84 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).

85 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.

86 Fibonacci Spiral:

87 Fibonacci Spiral:

88 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.

89 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.

90 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

91 Future Work and References

92 Future Research Future Research Generalizing results beyond PLRS, signed decompositions, higher dimensions... Other systems such as f -Decompositions of Demontigny, Do, Miller and Varma.

93 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.