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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Cookie Monster Meets the FibonacciNumbers. Mmmmmm Theorems!
Steven J. Miller (MC96)http://www.williams.edu/Mathematics/sjmiller/public_html
Yale University, April 14, 2014
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Introduction
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
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.
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Pre-requisites: Probability Review
5 10
0.05
0.10
0.15
0.20
Let X be random variable with density p(x): p(x) 0; p(x)dx = 1; Prob (a X b) = ba p(x)dx .Mean: =
xp(x)dx .
Variance: 2 =(x )2p(x)dx .
Gaussian: Density (22)1/2 exp((x )2/22).4
Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Pre-requisites: Combinatorics Review
n!: number of ways to order n people, order matters.
n!k !(nk)! = nCk =
(nk): number of ways to choose k from n,
order doesnt matter.
Stirlings Formula: n! nnen2n.
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Previous Results
Fibonacci Numbers: Fn+1 = Fn + Fn1;
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Previous Results
Fibonacci Numbers: Fn+1 = Fn + Fn1;F1 = 1, F2 = 2, F3 = 3, F4 = 5, . . . .
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Previous Results
Fibonacci Numbers: Fn+1 = Fn + Fn1;F1 = 1, F2 = 2, F3 = 3, F4 = 5, . . . .Zeckendorfs TheoremEvery positive integer can be written uniquely as a sum ofnon-consecutive Fibonacci numbers.
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Previous Results
Fibonacci Numbers: Fn+1 = Fn + Fn1;F1 = 1, F2 = 2, F3 = 3, F4 = 5, . . . .Zeckendorfs TheoremEvery positive integer can be written uniquely as a sum ofnon-consecutive Fibonacci numbers.Example:2012 = 1597+ 377+ 34+ 3+ 1 = F16 + F13 + F8 + F3 + F1.
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Previous Results
Fibonacci Numbers: Fn+1 = Fn + Fn1;F1 = 1, F2 = 2, F3 = 3, F4 = 5, . . . .Zeckendorfs TheoremEvery positive integer can be written uniquely as a sum ofnon-consecutive Fibonacci numbers.Example:2012 = 1597+ 377+ 34+ 3+ 1 = F16 + F13 + F8 + F3 + F1.Lekkerkerkers Theorem (1952)The average number of summands in the Zeckendorfdecomposition for integers in [Fn,Fn+1) tends to n2+1 .276n,where = 1+
5
2 is the golden mean.
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Old Results
Central Limit Type TheoremAs n distribution of number of summands in Zeckendorfdecomposition for m [Fn,Fn+1) is Gaussian (normal).
500 520 540 560 580 600
0.005
0.010
0.015
0.020
0.025
0.030
Figure: Number of summands in [F2010,F2011); F2010 10420.11
Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
New Results: Bulk Gaps: m [Fn,Fn+1) and = 1+5
2
m =k(m)=nj=1
Fij , m;n(x) =1
k(m) 1k(m)j=2
(x (ij ij1)
).
Theorem (Zeckendorf Gap Distribution)Gap measures m;n converge almost surely to average gapmeasure where P(k) = 1/k for k 2.
5 10 15 20 25 30
0.1
0.2
0.3
0.4
5 10 15 20 25
0.5
1.0
1.5
2.0
Figure: Distribution of gaps in [F1000,F1001); F2010 10208.12
Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
New Results: Longest Gap
Theorem (Longest Gap)As n, the probability that m [Fn,Fn+1) has longest gapless than or equal to f (n) converges to
Prob (Ln(m) f (n)) eelog nf (n)/ log .
Immediate Corollary: If f (n) grows slower or faster thanlog n/ log, then Prob(Ln(m) f (n)) goes to 0 or 1,respectively.
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Preliminaries: The Cookie ProblemThe Cookie ProblemThe number of ways of dividing C identical cookies among Pdistinct people is
(C+P1P1
).
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Preliminaries: The Cookie ProblemThe Cookie ProblemThe number of ways of dividing C identical cookies among Pdistinct people is
(C+P1P1
).
Proof : Consider C + P 1 cookies in a line.Cookie Monster eats P 1 cookies: (C+P1P1 ) ways to do.Divides the cookies into P sets.
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Preliminaries: The Cookie ProblemThe Cookie ProblemThe number of ways of dividing C identical cookies among Pdistinct people is
(C+P1P1
).
Proof : Consider C + P 1 cookies in a line.Cookie Monster eats P 1 cookies: (C+P1P1 ) ways to do.Divides the cookies into P sets.Example: 8 cookies and 5 people (C = 8, P = 5):
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Preliminaries: The Cookie ProblemThe Cookie ProblemThe number of ways of dividing C identical cookies among Pdistinct people is
(C+P1P1
).
Proof : Consider C + P 1 cookies in a line.Cookie Monster eats P 1 cookies: (C+P1P1 ) ways to do.Divides the cookies into P sets.Example: 8 cookies and 5 people (C = 8, P = 5):
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Preliminaries: The Cookie ProblemThe Cookie ProblemThe number of ways of dividing C identical cookies among Pdistinct people is
(C+P1P1
).
Proof : Consider C + P 1 cookies in a line.Cookie Monster eats P 1 cookies: (C+P1P1 ) ways to do.Divides the cookies into P sets.Example: 8 cookies and 5 people (C = 8, P = 5):
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Preliminaries: The Cookie Problem: Reinterpretation
Reinterpreting the Cookie ProblemThe number of solutions to x1 + + xP = C with xi 0 is(C+P1
P1).
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Preliminaries: The Cookie Problem: Reinterpretation
Reinterpreting the Cookie ProblemThe number of solutions to x1 + + xP = C with xi 0 is(C+P1
P1).
Let pn,k = # {N [Fn,Fn+1): the Zeckendorf decomposition ofN has exactly k summands}.
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Preliminaries: The Cookie Problem: Reinterpretation
Reinterpreting the Cookie ProblemThe number of solutions to x1 + + xP = C with xi 0 is(C+P1
P1).
Let pn,k = # {N [Fn,Fn+1): the Zeckendorf decomposition ofN has exactly k summands}.For N [Fn,Fn+1), the largest summand is Fn.
N = Fi1 + Fi2 + + Fik1 + Fn,1 i1 < i2 < < ik1 < ik = n, ij ij1 2.
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Preliminaries: The Cookie Problem: Reinterpretation
Reinterpreting the Cookie ProblemThe number of solutions to x1 + + xP = C with xi 0 is(C+P1
P1).
Let pn,k = # {N [Fn,Fn+1): the Zeckendorf decomposition ofN has exactly k summands}.For N [Fn,Fn+1), the largest summand is Fn.
N = Fi1 + Fi2 + + Fik1 + Fn,1 i1 < i2 < < ik1 < ik = n, ij ij1 2.
d1 := i1 1, dj := ij ij1 2 (j > 1).d1 + d2 + + dk = n 2k + 1, dj 0.
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Preliminaries: The Cookie Problem: Reinterpretation
Reinterpreting the Cookie ProblemThe number of solutions to x1 + + xP = C with xi 0 is(C+P1
P1).
Let pn,k = # {N [Fn,Fn+1): the Zeckendorf decomposition ofN has exactly k summands}.For N [Fn,Fn+1), the largest summand is Fn.
N = Fi1 + Fi2 + + Fik1 + Fn,1 i1 < i2 < < ik1 < ik = n, ij ij1 2.
d1 := i1 1, dj := ij ij1 2 (j > 1).d1 + d2 + + dk = n 2k + 1, dj 0.
Cookie counting pn,k =(n2k+1 + k1
k1)=(nkk1).
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Gaussian Behavior
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Generalizing Lekkerkerker: Erdos-Kac type result
Theorem (KKMW 2010)As n, the distribution of the number of summands inZeckendorfs Theorem is a Gaussian.
Sketch of proof: Use Stirlings formula,
n! nnen2n
to approximates binomial coefficients, after a few pages ofalgebra find the probabilities are approximately Gaussian.
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
(Sketch of the) Proof of GaussianityThe probability density for the number of Fibonacci numbers that add up to an integer in [Fn , Fn+1) isfn(k) =
(n1kk
)/Fn1. Consider the density for the n + 1 case. Then we have, by Stirling
fn+1(k) =(n k
k
) 1Fn
=(n k)!
(n 2k)!k !1Fn
=12
(n k)nk+12
k(k+12 )(n 2k)n2k+
12
1Fn
plus a lower order correction term.Also we can write Fn = 15
n+1 = 5n for large n, where is the golden ratio (we are using relabeled
Fibonacci numbers where 1 = F1 occurs once to help dealing with uniqueness and F2 = 2). We can now split theterms that exponentially depend on n.
fn+1(k) =(
12
(n k)k(n 2k)
5
)(n
(n k)nk
kk (n 2k)n2k
).
Define
Nn =12
(n k)k(n 2k)
5
, Sn = n(n k)nk
kk (n 2k)n2k.
Thus, write the density function asfn+1(k) = NnSn
where Nn is the first term that is of order n1/2 and Sn is the second term with exponential dependence on n.26
Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
(Sketch of the) Proof of GaussianityModel the distribution as centered around the mean by the change of variable k = + x where and are themean and the standard deviation, and depend on n. The discrete weights of fn(k) will become continuous. Thisrequires us to use the change of variable formula to compensate for the change of scales:
fn(k)dk = fn( + x)dx.
Using the change of variable, we can write Nn as
Nn =12
n k
k(n 2k)5
=1
2n
1 k/n
(k/n)(1 2k/n)
5
=1
2n
1 ( + x)/n
(( + x)/n)(1 2( + x)/n)
5
=1
2n
1 C y
(C + y)(1 2C 2y)
5
where C = /n 1/( + 2) (note that 2 = + 1) and y = x/n. But for large n, the y term vanishes since n and thus y n1/2. Thus
Nn 1
2n
1 C
C(1 2C)
5
=1
2n
( + 1)( + 2)
5
=1
2n
5( + 2)
=
122
since 2 = n 5(+2) .27
Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
(Sketch of the) Proof of GaussianityFor the second term Sn , take the logarithm and once again change variables by k = + x,
log(Sn) = log(n
(n k)(nk)
kk (n 2k)(n2k)
)
= n log() + (n k) log(n k) (k) log(k) (n 2k) log(n 2k)
= n log() + (n ( + x)) log(n ( + x)) ( + x) log( + x) (n 2( + x)) log(n 2( + x))
= n log()
+ (n ( + x))(log(n ) + log
(1
xn
))
( + x)(log() + log
(1 +
x
))
(n 2( + x))(log(n 2) + log
(1
xn 2
))
= n log()
+ (n ( + x))(log( n 1)
+ log(1
xn
))
( + x) log(1 +
x
)
(n 2( + x))(log( n 2)
+ log(1
xn 2
)).
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
(Sketch of the) Proof of Gaussianity
Note that, since n/ = + 2 for large n, the constant terms vanish. We have log(Sn)
= n log() + (n k) log( n 1
) (n 2k) log
( n 2)
+ (n ( + x)) log(1
xn
)
( + x) log(1 +
x
) (n 2( + x)) log
(1
xn 2
)
= n log() + (n k) log ( + 1) (n 2k) log () + (n ( + x)) log(1
xn
)
( + x) log(1 +
x
) (n 2( + x)) log
(1
xn 2
)
= n( log() + log(2) log ()) + k(log(2) + 2 log()) + (n ( + x)) log
(1
xn
)
( + x) log(1 +
x
) (n 2( + x)) log
(1 2
xn 2
)
= (n ( + x)) log(1
xn
) ( + x) log
(1 +
x
)
(n 2( + x)) log(1 2
xn 2
).
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
(Sketch of the) Proof of GaussianityFinally, we expand the logarithms and collect powers of x/n.
log(Sn) = (n ( + x))(
xn
12
( xn
)2+ . . .
)
( + x)(x
12
( x
)2+ . . .
)
(n 2( + x))(2
xn 2
12
(2
xn 2
)2+ . . .
)
= (n ( + x)) x
n (+1)(+2)
12
xn (+1)(+2)
2 + . . .
( + x) xn
+2
12
xn
+2
2 + . . .
(n 2( + x)) 2x
n +2
12
2xn +2
2 + . . .
=xnn((1
1 + 2
)( + 2)( + 1)
1 + 2(1
2 + 2
) + 2
)
12
( xn
)2n(2
+ 2 + 1
+ + 2 + 1
+ 2( + 2) ( + 2) + 4 + 2
)
+O(n (x/n)3
)30
Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
(Sketch of the) Proof of Gaussianity
log(Sn) =xnn( + 1 + 2
+ 2 + 1
1 + 2
+ 2 + 2
)
12
( xn
)2n( + 2)
(
1 + 1
+ 1 +4
)
+O(n( x
n
)3)
= 12(x)2
n( + 2)
(3 + 4( + 1)
+ 1)
+ O(n( x
n
)3)
= 12(x)2
n( + 2)
(3 + 4 + 2 + 1
( + 1)
)+ O
(n( x
n
)3)
= 12x22
( 5( + 2)n
)+ O
(n (x/n)3
).
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
(Sketch of the) Proof of Gaussianity
But recall that2 =
n5( + 2)
.
Also, since n1/2, n(xn)3 n1/2. So for large n, the O (n ( xn )3
)term vanishes. Thus we are left
with
log Sn = 12x2
Sn = e12 x
2.
Hence, as n gets large, the density converges to the normal distribution:
fn(k)dk = NnSndk
=1
22
e12 x
2dx
=12
e12 x
2dx.
!
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Generalizations
Generalizing from Fibonacci numbers to linearly recursivesequences with arbitrary nonnegative coefficients.
Hn+1 = c1Hn + c2Hn1 + + cLHnL+1, n Lwith H1 = 1, Hn+1 = c1Hn+ c2Hn1+ + cnH1+ 1, n < L,coefficients ci 0; c1, cL > 0 if L 2; c1 > 1 if L = 1.
Zeckendorf: Every positive integer can be writtenuniquely asaiHi with natural constraints on the ai s(e.g. cannot use the recurrence relation to removeany summand).LekkerkerkerCentral Limit Type Theorem
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Generalizing Lekkerkerker
Generalized Lekkerkerkers TheoremThe average number of summands in the generalizedZeckendorf decomposition for integers in [Hn,Hn+1) tendsto Cn + d as n, where C > 0 and d are computableconstants determined by the ci s.
C = y(1)y(1) =
L1m=0(sm + sm+1 1)(sm+1 sm)ym(1)2L1m=0(m + 1)(sm+1 sm)ym(1) .
s0 = 0, sm = c1 + c2 + + cm.y(x) is the root of 1L1m=0sm+11j=sm x jym+1.y(1) is the root of 1 c1y c2y2 cLyL.
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Central Limit Type Theorem
Central Limit Type TheoremAs n, the distribution of the number of summands,i.e., a1 + a2 + + am in the generalized Zeckendorfdecompositionmi=1 aiHi for integers in [Hn,Hn+1) isGaussian.
1000 1050 1100 1150 1200
0.005
0.010
0.015
0.020
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Example: the Special Case of L = 1, c1 = 10Hn+1 = 10Hn, H1 = 1, Hn = 10n1.
Legal decomposition is decimal expansion: mi=1 aiHi :ai {0, 1, . . . , 9} (1 i < m), am {1, . . . , 9}.For N [Hn,Hn+1), m = n, i.e., first term isanHn = an10n1.Ai : the corresponding random variable of ai .The Ai s are independent.For large n, the contribution of An is immaterial.Ai (1 i < n) are identically distributed randomvariableswith mean 4.5 and variance 8.25.Central Limit Theorem: A2+A3+ +An Gaussianwith mean 4.5n +O(1)and variance 8.25n +O(1).
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Far-difference RepresentationTheorem (Alpert, 2009) (Analogue to Zeckendorf)Every integer can be written uniquely as a sum of theFns, such that every two terms of the same (opposite)sign differ in index by at least 4 (3).Example: 1900 = F17 F14 F10 + F6 + F2.K : # of positive terms, L: # of negative terms.Generalized Lekkerkerkers TheoremAs n, E [K ] and E [L] n/10.E [K ] E [L] = /2 .809.Central Limit Type TheoremAs n, K and L converges to a bivariate Gaussian.
corr(K , L) = (21 2)/(29+ 2) .551, =
5+12 .
K + L and K L are independent.37
Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Generating Function (Example: Binets Formula)Binets Formula
F 1 = F 2 = 1; F n = 15[(
1+5
2
)n(1+5
2
)n].
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Generating Function (Example: Binets Formula)Binets Formula
F 1 = F 2 = 1; F n = 15[(
1+5
2
)n(1+5
2
)n].
Recurrence relation: F n+1 = F n + F n1 (1)
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Generating Function (Example: Binets Formula)Binets Formula
F 1 = F 2 = 1; F n = 15[(
1+5
2
)n(1+5
2
)n].
Recurrence relation: F n+1 = F n + F n1 (1)Generating function: g(x) =n>0 F nxn.
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Generating Function (Example: Binets Formula)Binets Formula
F 1 = F 2 = 1; F n = 15[(
1+5
2
)n(1+5
2
)n].
Recurrence relation: F n+1 = F n + F n1 (1)Generating function: g(x) =n>0 F nxn.(1)
n2
F n+1xn+1 =n2
F nxn+1 +n2
F n1xn+1
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Generating Function (Example: Binets Formula)Binets Formula
F 1 = F 2 = 1; F n = 15[(
1+5
2
)n(1+5
2
)n].
Recurrence relation: F n+1 = F n + F n1 (1)Generating function: g(x) =n>0 F nxn.(1)
n2
F n+1xn+1 =n2
F nxn+1 +n2
F n1xn+1
n3
F nxn =n2
F nxn+1 +n1
F nxn+2
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Generating Function (Example: Binets Formula)Binets Formula
F 1 = F 2 = 1; F n = 15[(
1+5
2
)n(1+5
2
)n].
Recurrence relation: F n+1 = F n + F n1 (1)Generating function: g(x) =n>0 F nxn.(1)
n2
F n+1xn+1 =n2
F nxn+1 +n2
F n1xn+1
n3
F nxn =n2
F nxn+1 +n1
F nxn+2
n3
F nxn = xn2
F nxn + x2n1
F nxn
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Generating Function (Example: Binets Formula)Binets Formula
F 1 = F 2 = 1; F n = 15[(
1+5
2
)n(1+5
2
)n].
Recurrence relation: F n+1 = F n + F n1 (1)Generating function: g(x) =n>0 F nxn.(1)
n2
F n+1xn+1 =n2
F nxn+1 +n2
F n1xn+1
n3
F nxn =n2
F nxn+1 +n1
F nxn+2
n3
F nxn = xn2
F nxn + x2n1
F nxn
g(x) F 1x F 2x2 = x(g(x) F 1x) + x2g(x)
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Generating Function (Example: Binets Formula)Binets Formula
F 1 = F 2 = 1; F n = 15[(
1+5
2
)n(1+5
2
)n].
Recurrence relation: F n+1 = F n + F n1 (1)Generating function: g(x) =n>0 F nxn.(1)
n2
F n+1xn+1 =n2
F nxn+1 +n2
F n1xn+1
n3
F nxn =n2
F nxn+1 +n1
F nxn+2
n3
F nxn = xn2
F nxn + x2n1
F nxn
g(x) F 1x F 2x2 = x(g(x) F 1x) + x2g(x) g(x) = x/(1 x x2).
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Partial Fraction Expansion (Example: Binets Formula)
Generating function: g(x) =n>0 F nxn = x1xx2 .
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Partial Fraction Expansion (Example: Binets Formula)
Generating function: g(x) =n>0 F nxn = x1xx2 .Partial fraction expansion:
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Partial Fraction Expansion (Example: Binets Formula)
Generating function: g(x) =n>0 F nxn = x1xx2 .Partial fraction expansion:
g(x) = x1 x x2 =15
(1+5
2 x1 1+
5
2 x
1+52 x
1 1+5
2 x
).
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Partial Fraction Expansion (Example: Binets Formula)
Generating function: g(x) =n>0 F nxn = x1xx2 .Partial fraction expansion:
g(x) = x1 x x2 =15
(1+5
2 x1 1+
5
2 x
1+52 x
1 1+5
2 x
).
Coefficient of xn (power series expansion):F n = 15
[(1+5
2
)n (1+52 )n] - Binets Formula!(using geometric series: 11r = 1+ r + r2 + r3 + ).
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
Differentiating Identities and Method of MomentsDifferentiating identitiesExample: Given a random variable X such thatPr(X = 1) = 12 , Pr(X = 2) = 14 , Pr(X = 3) = 18, ....then whats the mean of X (i.e., E [X ])?Solution: Let f (x) = 12x + 14x2 + 18x3 + = 11x/2 1.
f (x) = 1 12 + 2 14x + 3 18x2 + .f (1) = 1 12 + 2 14 + 3 18 + = E [X ].
Method of moments: Random variables X1, X2, . . . .If th moments E [X n ] converges those of standardnormal then Xn converges to a Gaussian.Standard normal distribution:2mth moment: (2m 1)!! = (2m 1)(2m 3) 1,(2m 1)th moment: 0.
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Intro Gaussianity Gaps (Bulk) Longest Gap Future / Refs
New Approach: Case of Fibonacci Numberspn,k = # {N [Fn,Fn+1): the Zeckendorf decomposition ofN has exactly k summands}.Recurrence relation:N [Fn+1,Fn+2): N = Fn+1 + Ft + , t n 1.
pn+1,k+1 = pn1,k + pn2,k +
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New Approach: Case of Fibonacci Numberspn,k = # {N [Fn,Fn+1): the Zeckendorf decomposition ofN has exactly k summands}.Recurrence relation:N [Fn+1,Fn+2): N = Fn+1 + Ft + , t n 1.
pn+1,k+1 = pn1,k + pn2,k + pn,k+1 = pn2,k + pn3,k +
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New Approach: Case of Fibonacci Numberspn,k = # {N [Fn,Fn+1): the Zeckendorf decomposition ofN has exactly k summands}.Recurrence relation:N [Fn+1,Fn+2): N = Fn+1 + Ft + , t n 1.
pn+1,k+1 = pn1,k + pn2,k + pn,k+1 = pn2,k + pn3,k +
pn+1,k+1 = pn,k+1 + pn1,k .
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New Approach: Case of Fibonacci Numberspn,k = # {N [Fn,Fn+1): the Zeckendorf decomposition ofN has exactly k summands}.Recurrence relation:N [Fn+1,Fn+2): N = Fn+1 + Ft + , t n 1.
pn+1,k+1 = pn1,k + pn2,k + pn,k+1 = pn2,k + pn3,k +
pn+1,k+1 = pn,k+1 + pn1,k .Generating function: n,k>0 pn,kxkyn = y1yxy2 .Partial fraction expansion:
y1 y xy2 =
yy1(x) y2(x)
( 1y y1(x)
1y y2(x)
)where y1(x) and y2(x) are the roots of 1 y xy2 = 0.Coefficient of yn: g(x) =k>0 pn,kxk .
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New Approach: Case of Fibonacci Numbers (Continued)Kn: the corresponding random variable associated with k .g(x) =k>0 pn,kxk .Differentiating identities:g(1) =k>0 pn,k = Fn+1 Fn,g(x) =k>0 kpn,kxk1, g(1) = g(1)E [Kn],(xg(x)) =k>0 k2pn,kxk1,(xg(x)) |x=1 = g(1)E [K 2n ],(x (xg(x))
) |x=1 = g(1)E [K 3n ], ...Similar results hold for the centralized Kn:K n = Kn E [Kn].Method of moments (for normalized K n):E [(K n)2m]/(SD(K n))2m (2m 1)!!,E [(K n)2m1]/(SD(K n))2m1 0. Kn Gaussian.
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New Approach: General CaseLet pn,k = # {N [Hn,Hn+1): the generalized Zeckendorfdecomposition of N has exactly k summands}.
Recurrence relation:Fibonacci: pn+1,k+1 = pn,k+1 + pn,k .General: pn+1,k =
L1m=0
sm+11j=sm pnm,kj .
where s0 = 0, sm = c1 + c2 + + cm.Generating function:Fibonacci: y1yxy2 .General:
nL pn,kxkyn L1
m=0sm+11
j=sm x jym+1
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New Approach: General Case (Continued)
Partial fraction expansion:Fibonacci: yy1(x)y2(x)
(1
yy1(x) 1yy2(x)).
General: 1sL1
j=sL1 xjLi=1
B(x , y)(y yi(x))
j =i (yj(x) yi(x))
.
B(x , y) =nL
pn,kxkyn L1m=0
sm+11j=sm
x jym+1
n0 pn,kxk .Differentiating identitiesMethod of moments: implies Kn Gaussian.
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Gaps in the Bulk
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Distribution of Gaps
For Fi1 + Fi2 + + Fin , the gaps are the differencesin in1, in1 in2, . . . , i2 i1.
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Distribution of Gaps
For Fi1 + Fi2 + + Fin , the gaps are the differencesin in1, in1 in2, . . . , i2 i1.
Example: For F1 + F8 + F18, the gaps are 7 and 10.
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Distribution of Gaps
For Fi1 + Fi2 + + Fin , the gaps are the differencesin in1, in1 in2, . . . , i2 i1.
Example: For F1 + F8 + F18, the gaps are 7 and 10.
Let Pn(k) be the probability that a gap for a decompositionin [Fn,Fn+1) is of length k .
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Distribution of Gaps
For Fi1 + Fi2 + + Fin , the gaps are the differencesin in1, in1 in2, . . . , i2 i1.
Example: For F1 + F8 + F18, the gaps are 7 and 10.
Let Pn(k) be the probability that a gap for a decompositionin [Fn,Fn+1) is of length k .
What is P(k) = limnPn(k)?
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Distribution of Gaps
For Fi1 + Fi2 + + Fin , the gaps are the differencesin in1, in1 in2, . . . , i2 i1.
Example: For F1 + F8 + F18, the gaps are 7 and 10.
Let Pn(k) be the probability that a gap for a decompositionin [Fn,Fn+1) is of length k .
What is P(k) = limnPn(k)?
Can ask similar questions about binary or otherexpansions: 2012 = 210 + 29 + 28 + 27 + 26 + 24 + 23 + 22.
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Main Results
Theorem (Base B Gap Distribution)For base B decompositions, P(0) = (B1)(B2)B2 , and fork 1, P(k) = cBBk , with cB = (B1)(3B2)B2 .
Theorem (Zeckendorf Gap Distribution)For Zeckendorf decompositions, P(k) = (1)k for k 2,with = 1+
5
2 the golden mean.
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Main Results
TheoremLet Hn+1 = c1Hn + c2Hn1 + + cLHn+1L be a positivelinear recurrence of length L where ci 1 for all 1 i L.Then P(j) =1 ( a1CLek )(
n+21 n+11 + 211 + a11 3) j = 0
11 (1
CLek )(1(1 2a1) + a1) j = 1(1 1)2
(a1CLek
)j1 j 2
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Proof of Fibonacci Result
Lekkerkerker total number of gaps Fn1 n2+1 .
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Proof of Fibonacci Result
Lekkerkerker total number of gaps Fn1 n2+1 .
Let Xi ,j = #{m [Fn,Fn+1): decomposition of m includesFi , Fj , but not Fq for i < q < j}.
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Proof of Fibonacci Result
Lekkerkerker total number of gaps Fn1 n2+1 .
Let Xi ,j = #{m [Fn,Fn+1): decomposition of m includesFi , Fj , but not Fq for i < q < j}.
P(k) = limn
nki=1 Xi ,i+kFn1 n2+1
.
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Calculating Xi ,i+k
How many decompositions contain a gap from Fi to Fi+k?
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Calculating Xi ,i+k
How many decompositions contain a gap from Fi to Fi+k?
Number of choices is Fnk2iFi1:
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Calculating Xi ,i+k
How many decompositions contain a gap from Fi to Fi+k?
Number of choices is Fnk2iFi1:
For the indices less than i : Fi1 choices. Why? Have Fi ,dont have Fi1. Follows by Zeckendorf: like the interval[Fi ,Fi+1) as have Fi , number elements is Fi+1 Fi = Fi1.
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Calculating Xi ,i+k
How many decompositions contain a gap from Fi to Fi+k?
Number of choices is Fnk2iFi1:
For the indices less than i : Fi1 choices. Why? Have Fi ,dont have Fi1. Follows by Zeckendorf: like the interval[Fi ,Fi+1) as have Fi , number elements is Fi+1 Fi = Fi1.
For the indices greater than i + k : Fnki2 choices. Why?Have Fn, dont have Fi+k+1. Like Zeckendorf with potentialsummands Fi+k+2, . . . ,Fn. Shifting, like summandsF1, . . . ,Fnki1, giving Fnki2.
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Determining P(k)
nki=1
Xi ,i+k = Fnk1 +nk2i=1
Fi1Fnki2
nk3i=0 FiFnki3 is the xnk3 coefficient of (g(x))2,
where g(x) is the generating function of theFibonaccis.
Alternatively, use Binets formula and get sums ofgeometric series.
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Determining P(k)
nki=1
Xi ,i+k = Fnk1 +nk2i=1
Fi1Fnki2
nk3i=0 FiFnki3 is the xnk3 coefficient of (g(x))2,
where g(x) is the generating function of theFibonaccis.
Alternatively, use Binets formula and get sums ofgeometric series.
P(k) = C/k for a constant C, so P(k) = 1/k .74
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Proof sketch of almost sure convergence
m =k(m)j=1 Fij ,m;n(x) = 1k(m)1
k(m)j=2 (x (ij ij1)) .
m,n(t) = xtdm;n(x).
Show Em[m;n(t)] equals average gap moments, (t).
Show Em[(m;n(t) (t))2] and Em[(m;n(t) (t))4]tend to zero.
Key ideas: (1) Replace k(m) with average (Gaussianity);(2) use Xi ,i+g1,j ,j+g2.
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Longest Gap
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Longest Gap
For most recurrences, our central result isTheorem (Mean and Variance of Longest Gap)Let 1 be the largest eigenvalue of the recurrence, beEulers constant, and K a constant that is a polynomial in1. Then the mean and variance of the longeset gap are:
n =log (nK )log1
+
log 1 12 + o(1)
2n =2
6(log 1)2+ o(1).
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Strategy
Our argument follows three main steps:Find a rational generating function Sf (x) for thenumber of m (Hn,Hn+1] with longest gap less thanf .Obtain an approximate formula for the CDF of thelongest gap.Estimate the mean and variance using PartialSummation and the Euler Maclaurin Formula.
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Fibonacci case
For the fibonacci numbers, our generating function is
Sf (x) =x
1 x x2 + xf .
From this we obtain
Theorem (Longest Gap Asymptotic CDF)As n, the probability that m [Fn,Fn+1) has longestgap less than or equal to f (n) converges to
Prob (Ln(m) f (n)) eelog nf (n)/ log.
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Generating Function: I
For k fixed the number of m [Fn,Fn+1) with k summandsand longest gap less than f equals the coefficient of xnfor in the expression
11 x
f (n)2
j=2xjk1
.
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Generating Function: II
Why the nth coefficient of 11x(f (n)1
j=2 xj)k1
?
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Generating Function: II
Why the nth coefficient of 11x(f (n)1
j=2 xj)k1
?Let m = Fn + Fng1 + Fng1g2 + + Fng1gn1. Thegaps uniquely identify m because of ZeckendorfsTheorem! And we have the following:
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Generating Function: II
Why the nth coefficient of 11x(f (n)1
j=2 xj)k1
?Let m = Fn + Fng1 + Fng1g2 + + Fng1gn1. Thegaps uniquely identify m because of ZeckendorfsTheorem! And we have the following:
The sum of the gaps of x is n.
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Generating Function: II
Why the nth coefficient of 11x(f (n)1
j=2 xj)k1
?Let m = Fn + Fng1 + Fng1g2 + + Fng1gn1. Thegaps uniquely identify m because of ZeckendorfsTheorem! And we have the following:
The sum of the gaps of x is n.Each gap is 2.
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Generating Function: II
Why the nth coefficient of 11x(f (n)1
j=2 xj)k1
?Let m = Fn + Fng1 + Fng1g2 + + Fng1gn1. Thegaps uniquely identify m because of ZeckendorfsTheorem! And we have the following:
The sum of the gaps of x is n.Each gap is 2.Each gap is < f .
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Generating Function: III
If we sum over k we get the total number ofm [Fn,Fn+1) with longest gap < f . Its the nth coefficientof
F (x) = 11 xk=1
(x2 xf21 x
)k1=
x1 x x2 + xf .
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Obtaining the CDF
We analyze asymptotic behavior of the coefficients of
Sf (x) =x
1 x x2 + xfas n, f vary.
Use a partial fraction decomposition.Problem: What happens to the roots of1 x x2 + xf as f varies?Solution: 1 x x2 + xf has a unique smallest rootf which converges to 1/ for large f .The contribution of f dominates, allowing us toobtain an approximate CDF .
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Numerical Results
Convergence to mean is at best approximately n forsome small > 0. Computing numerics is difficult:
Fn+1 = Fn + Fn1: Sampling 100 numbers from [Fn,Fn+1)with n = 1, 000, 000.
Mean predicted : 28.73 vs. observed: 28.51Variance predicted : 2.67 vs. observed: 2.44
an+1 = 2an + 4an1: Sampling 100 numbers from[an, an+1) with n = 51, 200.
Mean predicted : 9.95 vs. observed: 9.91Variance predicted : 1.09 vs. observed: 1.22
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Numerical Results pt 2
Fn+1 = Fn + Fn1: Sampling 20 numbers from [Fn,Fn+1)with n = 10, 000, 000.
Mean predicted : 33.52 vs. observed: 33.60Variance predicted : 2.67 vs. observed: 2.33
an+1 = 2an + 4an1: Sampling 100 numbers from[an, an+1) with n = 102, 400.
Mean predicted : 10.54 vs. observed: 10.45Variance predicted : 1.09 vs. observed: 1.10
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Future Workand References
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Future Research
Future Research
Generalizing results to all PLRS and signeddecompositions.
Other systems such as f-Decompositions ofDemontigny, Do, Miller and Varma.
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References
ReferencesBeckwith, Bower, Gaudet, Insoft, Li, Miller and Tosteson: The AverageGap Distribution for Generalized Zeckendorf Decompositions. TheFibonacci Quarterly 51 (2013), 1327.
http://arxiv.org/abs/1208.5820.
Bower, Insoft, Li, Miller and Tosteson: Distribution of gaps ingeneralized Zeckendorf decompositions, preprint 2014.
http://arxiv.org/abs/1402.3912.
Kologlu, Kopp, Miller and Wang: On the number of summands inZeckendorf decompositions, Fibonacci Quarterly 49 (2011), no. 2,116130.
http://arxiv.org/pdf/1008.3204.
Miller and Wang: Gaussian Behavior in Generalized ZeckendorfDecompositions, to appear in the conference proceedings of the 2011Combinatorial and Additive Number Theory Conference.
http://arxiv.org/pdf/1107.2718.pdf.92
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