Download Distribution of Summands in Generalized

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History
f -Decompositions
3-Bin Decompositions
Future Directions
Distribution of Summands in Generalized
Zeckendorf Decompositions
Archit Kulkarni, Carnegie Mellon University
Umang Varma, Kalamazoo College
[email protected], [email protected]
(with Philippe Demontigny, Thao Do, and David Moon)
Advisor: Steven J Miller
SMALL REU, Williams College
Young Mathematicians Conference, 2013
The Ohio State University
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History
f -Decompositions
3-Bin Decompositions
Future Directions
Summary
Zeckendorf’s Theorem
Our research is inspired by an elegant theorem of Zeckendorf.
Theorem
Write the Fibonacci numbers as F1 = 1, F2 = 2, Fn = Fn−1 + Fn−2
for n > 2. All natural numbers can be uniquely written as a sum of
non-consecutive Fibonacci numbers.
Proof of existence uses the greedy algorithm and uniqueness
follows by induction.
Example
2013 = 1597 + 377 + 34 + 5 = F16 + F13 + F8 + F4
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History
f -Decompositions
3-Bin Decompositions
Future Directions
Summary
Previous research
The number of summands in the Zeckendorf decomposition of
integers in [Fn , Fn+1 ) converges to the Gaussian distribution
as n → ∞.
This result extends to non-negative recurrence relations of the
form
Hn+1 = c1 Hn + c2 Hn−1 + · · · + cL Hn−L+1 ,
where c1 , c2 , . . . , cL are nonnegative integers and L, c1 , cL
are positive.
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History
f -Decompositions
3-Bin Decompositions
Future Directions
Summary
Going the other way
Previous work: linear recurrence sequence → notion of legality.
Our work: notion of legality → linear recurrence sequence.
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History
f -Decompositions
3-Bin Decompositions
Future Directions
Definition and Examples
f -Decompositions
We focused on constructing sequences from notions of legal
decomposition.
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History
f -Decompositions
3-Bin Decompositions
Future Directions
Definition and Examples
f -Decompositions
We focused on constructing sequences from notions of legal
decomposition.
Many notions of “legal” decompositions can be encoded as
f -decompositions.
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History
f -Decompositions
3-Bin Decompositions
Future Directions
Definition and Examples
f -Decompositions
We focused on constructing sequences from notions of legal
decomposition.
Many notions of “legal” decompositions can be encoded as
f -decompositions.
Definition
Let f : N0 → N0 . A sum x = ki=0 ani of terms of {an } is an
f -decomposition of x using {an } if for every ani , the previous f (ni )
terms are not in the sum.
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History
f -Decompositions
3-Bin Decompositions
Future Directions
Definition and Examples
Definition
Let f : N0 → N0 . A sum x = ki=0 ani of terms of {an } is an
f -decomposition of x using {an } if for every ani , the previous f (ni )
terms are not in the sum.
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Example (Base b representation)
Base b representation can be interpreted as f -decompositions. For
example, consider base 5:
{an } = 1, 2, 3, 4, 5, 10, 15, 20, 25, 50, 75, 100, . . .
Here, an = 5an−4 .
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History
f -Decompositions
3-Bin Decompositions
Future Directions
Definition and Examples
Definition
Let f : N0 → N0 . A sum x = ki=0 ani of terms of {an } is an
f -decomposition of x using {an } if for every ani , the previous f (ni )
terms are not in the sum.
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Example (Zeckendorf Decompositions)
Consecutive terms may not be chosen: this is equivalent to saying
f (n) = 1 for all n ∈ N.
We construct a sequence by appending the smallest number that
cannot be decomposed using previous terms.
{Fn } =
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History
f -Decompositions
3-Bin Decompositions
Future Directions
Definition and Examples
Definition
Let f : N0 → N0 . A sum x = ki=0 ani of terms of {an } is an
f -decomposition of x using {an } if for every ani , the previous f (ni )
terms are not in the sum.
P
Example (Zeckendorf Decompositions)
Consecutive terms may not be chosen: this is equivalent to saying
f (n) = 1 for all n ∈ N.
We construct a sequence by appending the smallest number that
cannot be decomposed using previous terms.
{Fn } = 1,
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History
f -Decompositions
3-Bin Decompositions
Future Directions
Definition and Examples
Definition
Let f : N0 → N0 . A sum x = ki=0 ani of terms of {an } is an
f -decomposition of x using {an } if for every ani , the previous f (ni )
terms are not in the sum.
P
Example (Zeckendorf Decompositions)
Consecutive terms may not be chosen: this is equivalent to saying
f (n) = 1 for all n ∈ N.
We construct a sequence by appending the smallest number that
cannot be decomposed using previous terms.
{Fn } = 1, 2,
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History
f -Decompositions
3-Bin Decompositions
Future Directions
Definition and Examples
Definition
Let f : N0 → N0 . A sum x = ki=0 ani of terms of {an } is an
f -decomposition of x using {an } if for every ani , the previous f (ni )
terms are not in the sum.
P
Example (Zeckendorf Decompositions)
Consecutive terms may not be chosen: this is equivalent to saying
f (n) = 1 for all n ∈ N.
We construct a sequence by appending the smallest number that
cannot be decomposed using previous terms.
{Fn } = 1, 2, 3,
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History
f -Decompositions
3-Bin Decompositions
Future Directions
Definition and Examples
Definition
Let f : N0 → N0 . A sum x = ki=0 ani of terms of {an } is an
f -decomposition of x using {an } if for every ani , the previous f (ni )
terms are not in the sum.
P
Example (Zeckendorf Decompositions)
Consecutive terms may not be chosen: this is equivalent to saying
f (n) = 1 for all n ∈ N.
We construct a sequence by appending the smallest number that
cannot be decomposed using previous terms.
{Fn } = 1, 2, 3, 5,
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History
f -Decompositions
3-Bin Decompositions
Future Directions
Definition and Examples
Definition
Let f : N0 → N0 . A sum x = ki=0 ani of terms of {an } is an
f -decomposition of x using {an } if for every ani , the previous f (ni )
terms are not in the sum.
P
Example (Zeckendorf Decompositions)
Consecutive terms may not be chosen: this is equivalent to saying
f (n) = 1 for all n ∈ N.
We construct a sequence by appending the smallest number that
cannot be decomposed using previous terms.
{Fn } = 1, 2, 3, 5, 8, 13, 21, 34, . . .
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History
f -Decompositions
3-Bin Decompositions
Future Directions
Analogue to Zeckendorf’s Theorem
Theorem
For any f : N0 → N0 , there exists a sequence of natural numbers
{an }∞
n=0 such that every positive integer has a unique
f -decomposition using {an }.
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f -Decompositions
3-Bin Decompositions
Future Directions
Analogue to Zeckendorf’s Theorem
Theorem
For any f : N0 → N0 , there exists a sequence of natural numbers
{an }∞
n=0 such that every positive integer has a unique
f -decomposition using {an }.
Construction of {an }.
Let a0 = 1 and an = an−1 + an−1−f (n−1) (assume an = 1
when n < 0). The proof of existence is by induction.
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History
f -Decompositions
3-Bin Decompositions
Future Directions
Analogue to Zeckendorf’s Theorem
Theorem
For any f : N0 → N0 , there exists a sequence of natural numbers
{an }∞
n=0 such that every positive integer has a unique
f -decomposition using {an }.
Construction of {an }.
Let a0 = 1 and an = an−1 + an−1−f (n−1) (assume an = 1
when n < 0). The proof of existence is by induction.
Consider any x ∈ [am , am+1 ) = [am , am + am−f (m) ). Clearly,
x − am ∈ [0, am−f (m) ).
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History
f -Decompositions
3-Bin Decompositions
Future Directions
Analogue to Zeckendorf’s Theorem
Theorem
For any f : N0 → N0 , there exists a sequence of natural numbers
{an }∞
n=0 such that every positive integer has a unique
f -decomposition using {an }.
Construction of {an }.
Let a0 = 1 and an = an−1 + an−1−f (n−1) (assume an = 1
when n < 0). The proof of existence is by induction.
Consider any x ∈ [am , am+1 ) = [am , am + am−f (m) ). Clearly,
x − am ∈ [0, am−f (m) ).
By the induction hypothesis, x − am has an f -decomposition
m−f (m)−1
in {an }n=0
.
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f -Decompositions
3-Bin Decompositions
Future Directions
Linear Recurrence
Theorem
If f : N0 → N0 is periodic, then the corresponding {an } satisfies a
linear recurrence relation.
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History
f -Decompositions
3-Bin Decompositions
Future Directions
Linear Recurrence
Theorem
If f : N0 → N0 is periodic, then the corresponding {an } satisfies a
linear recurrence relation.
Using linear algebra, we can show that the subsequence
{ai,n } = ai , ai+b , ai+2b , ai+3b , . . . satisfies some linear
recurrence relation for all i ∈ {0, 1, 2, . . . , b − 1}.
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History
f -Decompositions
3-Bin Decompositions
Future Directions
Linear Recurrence
Theorem
If f : N0 → N0 is periodic, then the corresponding {an } satisfies a
linear recurrence relation.
Using linear algebra, we can show that the subsequence
{ai,n } = ai , ai+b , ai+2b , ai+3b , . . . satisfies some linear
recurrence relation for all i ∈ {0, 1, 2, . . . , b − 1}.
By finding a common recurrence relation we can show that
the interleaved sequence satisfies a linearly recurrent sequence.
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f -Decompositions
3-Bin Decompositions
Future Directions
Linear Recurrence
We can combine two linear recurrence relations by finding the
recurrence relation corresponding to the product of their respective
characteristic polynomials.
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History
f -Decompositions
3-Bin Decompositions
Future Directions
Linear Recurrence
We can combine two linear recurrence relations by finding the
recurrence relation corresponding to the product of their respective
characteristic polynomials.
Example
{Fn } = 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, . . .
satisfies Fn = Fn−1 + Fn−2 .
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History
f -Decompositions
3-Bin Decompositions
Future Directions
Linear Recurrence
We can combine two linear recurrence relations by finding the
recurrence relation corresponding to the product of their respective
characteristic polynomials.
Example
{Fn } = 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, . . .
satisfies Fn = Fn−1 + Fn−2 .
{an } = 1, 4, 15, 56, 209, 780, 2911, 10864, 40545, . . .
satisfies an = 4an−1 − an−2 .
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History
f -Decompositions
3-Bin Decompositions
Future Directions
Linear Recurrence
We can combine two linear recurrence relations by finding the
recurrence relation corresponding to the product of their respective
characteristic polynomials.
Example
{Fn } = 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, . . .
satisfies Fn = Fn−1 + Fn−2 .
{an } = 1, 4, 15, 56, 209, 780, 2911, 10864, 40545, . . .
satisfies an = 4an−1 − an−2 .
pF (x )pa (x ) = (x 2 −x −1)(x 2 −4x +1) = x 4 −5x 3 +4x 2 +3x −1.
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History
f -Decompositions
3-Bin Decompositions
Future Directions
Linear Recurrence
We can combine two linear recurrence relations by finding the
recurrence relation corresponding to the product of their respective
characteristic polynomials.
Example
{Fn } = 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, . . .
satisfies Fn = Fn−1 + Fn−2 .
{an } = 1, 4, 15, 56, 209, 780, 2911, 10864, 40545, . . .
satisfies an = 4an−1 − an−2 .
pF (x )pa (x ) = (x 2 −x −1)(x 2 −4x +1) = x 4 −5x 3 +4x 2 +3x −1.
Both sequences satisfy sn = 5sn−1 − 4sn−2 − 3sn−3 + sn−4 .
(sn − sn−1 − sn−2 ) − 4(sn−1 − sn−2 − sn−3 )
+ (sn−2 − sn−3 − sn−4 ) = 0
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f -Decompositions
3-Bin Decompositions
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Bins
Motivation
Previous research only handles linear recurrence relations with
nonnegative coefficients.
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f -Decompositions
3-Bin Decompositions
Future Directions
Bins
Motivation
Previous research only handles linear recurrence relations with
nonnegative coefficients.
Question
Do notions of legal decomposition exist for other sequences?
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History
f -Decompositions
3-Bin Decompositions
Future Directions
Bins
Motivation
Previous research only handles linear recurrence relations with
nonnegative coefficients.
Question
Do notions of legal decomposition exist for other sequences?
Answer
Yes! Some functions f yield sequences {an } that can only be
described by linear recurrence relations with at least one negative
coefficient.
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History
f -Decompositions
3-Bin Decompositions
Future Directions
Bins
Definition of 3-Bin Decomposition
Consider the periodic function {f (n)} = 1, 1, 2, 1, 1, 2, . . . .
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History
f -Decompositions
3-Bin Decompositions
Future Directions
Bins
Definition of 3-Bin Decomposition
Consider the periodic function {f (n)} = 1, 1, 2, 1, 1, 2, . . . .
We have “bins” of width 3, and a legal decomposition contains no
consecutive terms and at most one term from each bin.
{an } = 1, 2, 3, 4, 7, 11, 15, 26, 41, 56, 97, 153, 209, 362, 571, . . .
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History
f -Decompositions
3-Bin Decompositions
Future Directions
Bins
Definition of 3-Bin Decomposition
Consider the periodic function {f (n)} = 1, 1, 2, 1, 1, 2, . . . .
We have “bins” of width 3, and a legal decomposition contains no
consecutive terms and at most one term from each bin.
{an } = 1, 2, 3, 4, 7, 11, 15, 26, 41, 56, 97, 153, 209, 362, 571, . . .
We can show that no linear recurrence relation describing this
sequence has all nonnegative coefficients, and that the shortest
such relation is an = 4an−3 − an−6 . Thus, these decompositions are
out of the scope of previous work.
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History
f -Decompositions
3-Bin Decompositions
Future Directions
Bins
Gaussian Behavior of the Number of Summands
Question
Consider the number of summands for an integer randomly chosen
from the interval [0, a3n ). As n → ∞, what can we say about the
distribution?
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History
f -Decompositions
3-Bin Decompositions
Future Directions
Bins
Gaussian Behavior of the Number of Summands
Question
Consider the number of summands for an integer randomly chosen
from the interval [0, a3n ). As n → ∞, what can we say about the
distribution?
For positive linear recurrence sequences, previous work has shown
that the distribution approaches a Gaussian. Can we extend this to
our new sequences?
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f -Decompositions
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Bins
Gaussian Behavior of the Number of Summands
pn,k
150 000
100 000
50 000
2
4
6
First 10 bins
8
10
k
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3-Bin Decompositions
Future Directions
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Bins
Gaussian Behavior of the Number of Summands
pn,k
6 ´ 1010
5 ´ 1010
4 ´ 1010
3 ´ 1010
2 ´ 1010
1 ´ 1010
5
10
First 20 bins
15
20
k
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3-Bin Decompositions
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Bins
Gaussian Behavior of the Number of Summands
pn,k
1 ´ 1022
8 ´ 1021
6 ´ 1021
4 ´ 1021
2 ´ 1021
10
20
First 40 bins
30
40
k
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f -Decompositions
3-Bin Decompositions
Future Directions
Refs
Bins
Gaussian Behavior of the Number of Summands
pn,k
6 ´ 1044
5 ´ 1044
4 ´ 1044
3 ´ 1044
2 ´ 1044
1 ´ 1044
20
40
First 80 bins
60
80
k
History
f -Decompositions
3-Bin Decompositions
Future Directions
Proof of Gaussianity
Number of summands
Let pn,k be the number of integers whose decomposition uses k
summands all from the first n bins.
Claim
pn,k = pn−1,k + 3pn−1,k−1 − pn−2,k−2
We count the number of ways to choose the summands.
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f -Decompositions
3-Bin Decompositions
Future Directions
Proof of Gaussianity
Number of summands
Let pn,k be the number of integers whose decomposition uses k
summands all from the first n bins.
Claim
pn,k = pn−1,k + 3pn−1,k−1 − pn−2,k−2
We count the number of ways to choose the summands.
Either no summand is chosen from the first bin, or...
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History
f -Decompositions
3-Bin Decompositions
Future Directions
Proof of Gaussianity
Number of summands
Let pn,k be the number of integers whose decomposition uses k
summands all from the first n bins.
Claim
pn,k = pn−1,k + 3pn−1,k−1 − pn−2,k−2
We count the number of ways to choose the summands.
Either no summand is chosen from the first bin, or...
One of the three summands is chosen from the first bin.
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History
f -Decompositions
3-Bin Decompositions
Future Directions
Proof of Gaussianity
Number of summands
Let pn,k be the number of integers whose decomposition uses k
summands all from the first n bins.
Claim
pn,k = pn−1,k + 3pn−1,k−1 − pn−2,k−2
We count the number of ways to choose the summands.
Either no summand is chosen from the first bin, or...
One of the three summands is chosen from the first bin.
Can’t choose both the last element of the first bin and the
first element of the second bin.
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f -Decompositions
3-Bin Decompositions
Future Directions
Proof of Gaussianity
Generating Function
Begin with
pn,k = pn−1,k + 3pn−1,k−1 − pn−2,k−2
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f -Decompositions
3-Bin Decompositions
Future Directions
Proof of Gaussianity
Generating Function
Begin with
pn,k = pn−1,k + 3pn−1,k−1 − pn−2,k−2
Let F (x , y ) =
P
n,k≥0 pn,k x
ny k .
We can write
F (x , y ) = xF (x , y ) + 3xyF (x , y ) − x 2 y 2 F (x , y ) + 1
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f -Decompositions
3-Bin Decompositions
Future Directions
Proof of Gaussianity
Generating Function
Begin with
pn,k = pn−1,k + 3pn−1,k−1 − pn−2,k−2
Let F (x , y ) =
P
n,k≥0 pn,k x
ny k .
We can write
F (x , y ) = xF (x , y ) + 3xyF (x , y ) − x 2 y 2 F (x , y ) + 1
Simplifying gives us
F (x , y ) =
1
1 − x − 3xy + x 2 y 2
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f -Decompositions
3-Bin Decompositions
Future Directions
Proof of Gaussianity
Generating Function
Begin with
pn,k = pn−1,k + 3pn−1,k−1 − pn−2,k−2
Let F (x , y ) =
P
n,k≥0 pn,k x
ny k .
We can write
F (x , y ) = xF (x , y ) + 3xyF (x , y ) − x 2 y 2 F (x , y ) + 1
Simplifying gives us
F (x , y ) =
1
1 − x − 3xy + x 2 y 2
Define gn (y ) = k≥0 pn,k y k to isolate x n terms. Using partial
fractions and Taylor series,
P
√
gn
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(3y +1+
(y )=
)
5y 2 +6y +1
n+1
2n+1
√
(
√
− 3y +1−
5y 2 +6y +1
)
5y 2 +6y +1
n+1
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f -Decompositions
3-Bin Decompositions
Future Directions
Proof of Gaussianity
Mean and Variance
We can use gn (y ) = k≥0 pn,k y k to compute mean and
variance of the random variable Xn , the number of summands
for an integer randomly chosen from [0, a3n ).
P
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History
f -Decompositions
3-Bin Decompositions
Future Directions
Proof of Gaussianity
Mean and Variance
We can use gn (y ) = k≥0 pn,k y k to compute mean and
variance of the random variable Xn , the number of summands
for an integer randomly chosen from [0, a3n ).
P
The mean is
√
8 + 3 12
gn0 (1)
=√ µn =
√ n + O(1)
gn (1)
12 4 + 12
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f -Decompositions
3-Bin Decompositions
Future Directions
Proof of Gaussianity
Mean and Variance
We can use gn (y ) = k≥0 pn,k y k to compute mean and
variance of the random variable Xn , the number of summands
for an integer randomly chosen from [0, a3n ).
P
The mean is
√
8 + 3 12
gn0 (1)
=√ µn =
√ n + O(1)
gn (1)
12 4 + 12
The variance is
σn2
=
d
dy
√
[ygn0 (y )]
g(1)
y =1
2
−µ =
3
n + O(1)
9
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f -Decompositions
3-Bin Decompositions
Future Directions
Proof of Gaussianity
Moment Generating Function
If we normalize Xn to Yn = (Xn − µn )/σn , the moment
generating function of Yn is
X pn,k e t
tYn
P
MYn (t) = E(e ) =
k≥0
50
(k−µn )
σn
k≥0 pn,k
=
gn (e t/σn )e −tµn /σn
gn (1)
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f -Decompositions
3-Bin Decompositions
Future Directions
Proof of Gaussianity
Moment Generating Function
If we normalize Xn to Yn = (Xn − µn )/σn , the moment
generating function of Yn is
X pn,k e t
tYn
P
MYn (t) = E(e ) =
k≥0
(k−µn )
σn
k≥0 pn,k
=
gn (e t/σn )e −tµn /σn
gn (1)
After multiple Taylor Series expansions, we get
log(MYn (t)) =
as n → ∞.
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t2
+ o(1)
2
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f -Decompositions
3-Bin Decompositions
Future Directions
Proof of Gaussianity
Moment Generating Function
If we normalize Xn to Yn = (Xn − µn )/σn , the moment
generating function of Yn is
X pn,k e t
tYn
P
MYn (t) = E(e ) =
k≥0
(k−µn )
σn
k≥0 pn,k
=
gn (e t/σn )e −tµn /σn
gn (1)
After multiple Taylor Series expansions, we get
log(MYn (t)) =
t2
+ o(1)
2
as n → ∞.
Hence, the distribution of Yn converges to the standard
normal distribution.
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Future Directions
Proof of Gaussianity
Generalizations
This method can easily be generalized for large classes of
f -decompositions.
We have proved Gaussian behavior in the following cases:
every bin is of width b for any b ≥ 3
bins have alternating widths x and y
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3-Bin Decompositions
Future Directions
Future Directions
Does Gaussian behavior hold for all f -decompositions if f is
periodic?
Can we find interesting notions of legal decomposition that
are more general than f -decompositions?
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f -Decompositions
3-Bin Decompositions
Future Directions
Thanks
We would like to thank
Our advisor, Prof. Steven Miller.
Our co-authors Philippe Demontigny, Thao Do, and David
Moon.
Everyone involved with the SMALL REU at Williams College.
Williams College and the NSF (Grant DMS0850577) for
funding our research.
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3-Bin Decompositions
Future Directions
References
Kologlu, Kopp, Miller and Wang: Gaussianity for Fibonacci case,
Fibonacci Quarterly 49 (2011), no. 2, 116–130.
S. J. Miller and Y. Wang, From Fibonacci numbers to Central Limit
Type Theorems, Journal of Combinatorial Theory, Series A 119
(2012), no. 7, 1398–1413.
E. Zeckendorf, Représentation des nombres naturels par une somme
des nombres de Fibonacci ou de nombres de Lucas, Bulletin de la
Société Royale des Sciences de Liège 41 (1972), pages 179–182.
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