touchard polynomials, stirling numbers and random permutations · touchard polynomials, stirling...
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![Page 1: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/1.jpg)
Touchard Polynomials, Stirling Numbers andRandom Permutations
Ross Pinsky
Department of Mathematics, Technion32000 Haifa, ISRAEL
September 3, 2018
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 2: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/2.jpg)
Notation:
[n] = {1, 2, · · · , n}
Sn = permutations of [n]
|Sn| = n!
S6 3 σ1 =
(1 2 3 4 5 62 4 6 1 5 3
)= (124)(36)(5),
S6 3 σ2 =
(1 2 3 4 5 63 4 6 5 1 2
)= (136245) = (624513)
σ2 is a 6-cycle in S6. There are 5! different 6-cycles in S6.
There are (n − 1)! different n-cycles in Sn.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 3: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/3.jpg)
Notation:
[n] = {1, 2, · · · , n}
Sn = permutations of [n]
|Sn| = n!
S6 3 σ1 =
(1 2 3 4 5 62 4 6 1 5 3
)= (124)(36)(5),
S6 3 σ2 =
(1 2 3 4 5 63 4 6 5 1 2
)= (136245) = (624513)
σ2 is a 6-cycle in S6. There are 5! different 6-cycles in S6.
There are (n − 1)! different n-cycles in Sn.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 4: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/4.jpg)
Notation:
[n] = {1, 2, · · · , n}
Sn = permutations of [n]
|Sn| = n!
S6 3 σ1 =
(1 2 3 4 5 62 4 6 1 5 3
)= (124)(36)(5),
S6 3 σ2 =
(1 2 3 4 5 63 4 6 5 1 2
)= (136245) = (624513)
σ2 is a 6-cycle in S6. There are 5! different 6-cycles in S6.
There are (n − 1)! different n-cycles in Sn.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 5: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/5.jpg)
Notation:
[n] = {1, 2, · · · , n}
Sn = permutations of [n]
|Sn| = n!
S6 3 σ1 =
(1 2 3 4 5 62 4 6 1 5 3
)= (124)(36)(5),
S6 3 σ2 =
(1 2 3 4 5 63 4 6 5 1 2
)= (136245) = (624513)
σ2 is a 6-cycle in S6. There are 5! different 6-cycles in S6.
There are (n − 1)! different n-cycles in Sn.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 6: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/6.jpg)
Notation:
[n] = {1, 2, · · · , n}
Sn = permutations of [n]
|Sn| = n!
S6 3 σ1 =
(1 2 3 4 5 62 4 6 1 5 3
)= (124)(36)(5),
S6 3 σ2 =
(1 2 3 4 5 63 4 6 5 1 2
)= (136245) = (624513)
σ2 is a 6-cycle in S6. There are 5! different 6-cycles in S6.
There are (n − 1)! different n-cycles in Sn.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 7: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/7.jpg)
Notation:
[n] = {1, 2, · · · , n}
Sn = permutations of [n]
|Sn| = n!
S6 3 σ1 =
(1 2 3 4 5 62 4 6 1 5 3
)= (124)(36)(5),
S6 3 σ2 =
(1 2 3 4 5 63 4 6 5 1 2
)= (136245) = (624513)
σ2 is a 6-cycle in S6. There are 5! different 6-cycles in S6.
There are (n − 1)! different n-cycles in Sn.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 8: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/8.jpg)
Rising Factorials
x (n) := x(x +1) · · · (x +n−1)
(≡
n∑k=1
ankxk), x ∈ R, n ≥ 1. (1)
Unsigned Stirling Numbers of the First Kind:|s(n, k)| :=the number of permutations in Sn with exactly k cycles, 1 ≤ k ≤ nRecurrence relation
|s(n, n)| = 1, |s(n, 1)| = (n − 1)!, n ≥ 1;
|s(n + 1, k)| = n|s(n, k)|+ |s(n, k − 1)|, 2 ≤ k ≤ n.(2)
It is easy to check that the coefficients {ank} in (1) also satisfy(2); thus, ank = |s(n, k)|. Therefore
x (n) =n∑
k=1
|s(n, k)|xk . (3)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 9: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/9.jpg)
Rising Factorials
x (n) := x(x +1) · · · (x +n−1)(≡
n∑k=1
ankxk), x ∈ R, n ≥ 1. (1)
Unsigned Stirling Numbers of the First Kind:|s(n, k)| :=the number of permutations in Sn with exactly k cycles, 1 ≤ k ≤ nRecurrence relation
|s(n, n)| = 1, |s(n, 1)| = (n − 1)!, n ≥ 1;
|s(n + 1, k)| = n|s(n, k)|+ |s(n, k − 1)|, 2 ≤ k ≤ n.(2)
It is easy to check that the coefficients {ank} in (1) also satisfy(2); thus, ank = |s(n, k)|. Therefore
x (n) =n∑
k=1
|s(n, k)|xk . (3)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 10: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/10.jpg)
Rising Factorials
x (n) := x(x +1) · · · (x +n−1)(≡
n∑k=1
ankxk), x ∈ R, n ≥ 1. (1)
Unsigned Stirling Numbers of the First Kind:|s(n, k)| :=the number of permutations in Sn with exactly k cycles, 1 ≤ k ≤ n
Recurrence relation
|s(n, n)| = 1, |s(n, 1)| = (n − 1)!, n ≥ 1;
|s(n + 1, k)| = n|s(n, k)|+ |s(n, k − 1)|, 2 ≤ k ≤ n.(2)
It is easy to check that the coefficients {ank} in (1) also satisfy(2); thus, ank = |s(n, k)|. Therefore
x (n) =n∑
k=1
|s(n, k)|xk . (3)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 11: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/11.jpg)
Rising Factorials
x (n) := x(x +1) · · · (x +n−1)(≡
n∑k=1
ankxk), x ∈ R, n ≥ 1. (1)
Unsigned Stirling Numbers of the First Kind:|s(n, k)| :=the number of permutations in Sn with exactly k cycles, 1 ≤ k ≤ nRecurrence relation
|s(n, n)| = 1, |s(n, 1)| = (n − 1)!, n ≥ 1;
|s(n + 1, k)| = n|s(n, k)|+ |s(n, k − 1)|, 2 ≤ k ≤ n.(2)
It is easy to check that the coefficients {ank} in (1) also satisfy(2); thus, ank = |s(n, k)|. Therefore
x (n) =n∑
k=1
|s(n, k)|xk . (3)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 12: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/12.jpg)
Rising Factorials
x (n) := x(x +1) · · · (x +n−1)(≡
n∑k=1
ankxk), x ∈ R, n ≥ 1. (1)
Unsigned Stirling Numbers of the First Kind:|s(n, k)| :=the number of permutations in Sn with exactly k cycles, 1 ≤ k ≤ nRecurrence relation
|s(n, n)| = 1, |s(n, 1)| = (n − 1)!, n ≥ 1;
|s(n + 1, k)| = n|s(n, k)|+ |s(n, k − 1)|, 2 ≤ k ≤ n.(2)
It is easy to check that the coefficients {ank} in (1) also satisfy(2); thus, ank = |s(n, k)|.
Therefore
x (n) =n∑
k=1
|s(n, k)|xk . (3)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 13: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/13.jpg)
Rising Factorials
x (n) := x(x +1) · · · (x +n−1)(≡
n∑k=1
ankxk), x ∈ R, n ≥ 1. (1)
Unsigned Stirling Numbers of the First Kind:|s(n, k)| :=the number of permutations in Sn with exactly k cycles, 1 ≤ k ≤ nRecurrence relation
|s(n, n)| = 1, |s(n, 1)| = (n − 1)!, n ≥ 1;
|s(n + 1, k)| = n|s(n, k)|+ |s(n, k − 1)|, 2 ≤ k ≤ n.(2)
It is easy to check that the coefficients {ank} in (1) also satisfy(2); thus, ank = |s(n, k)|. Therefore
x (n) =n∑
k=1
|s(n, k)|xk . (3)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 14: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/14.jpg)
Falling Factorials
(x)n := x(x − 1) · · · (x − n + 1), x ∈ R, n ≥ 1. (4)
Substituting −x for x in (4), we have
(−x)n = −x(−x−1) · · · (−x−n+1) = (−1)nx(x+1) · · · (x+n−1)= (−1)nx (n).Substituting above −x for x , we have
(x)n = (−1)n(−x)(n)from (3)
= (−1)nn∑
k=1
|s(n, k)|(−x)k =
n∑k=1
(−1)n−k |s(n, k)|xk .(5)
Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called theStirling Numbers of the First Kind. From (5) we have
(x)n =n∑
k=1
s(n, k)xk . (6)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 15: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/15.jpg)
Falling Factorials
(x)n := x(x − 1) · · · (x − n + 1), x ∈ R, n ≥ 1. (4)
Substituting −x for x in (4), we have
(−x)n = −x(−x−1) · · · (−x−n+1) =
(−1)nx(x+1) · · · (x+n−1)= (−1)nx (n).Substituting above −x for x , we have
(x)n = (−1)n(−x)(n)from (3)
= (−1)nn∑
k=1
|s(n, k)|(−x)k =
n∑k=1
(−1)n−k |s(n, k)|xk .(5)
Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called theStirling Numbers of the First Kind. From (5) we have
(x)n =n∑
k=1
s(n, k)xk . (6)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 16: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/16.jpg)
Falling Factorials
(x)n := x(x − 1) · · · (x − n + 1), x ∈ R, n ≥ 1. (4)
Substituting −x for x in (4), we have
(−x)n = −x(−x−1) · · · (−x−n+1) = (−1)nx(x+1) · · · (x+n−1)
= (−1)nx (n).Substituting above −x for x , we have
(x)n = (−1)n(−x)(n)from (3)
= (−1)nn∑
k=1
|s(n, k)|(−x)k =
n∑k=1
(−1)n−k |s(n, k)|xk .(5)
Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called theStirling Numbers of the First Kind. From (5) we have
(x)n =n∑
k=1
s(n, k)xk . (6)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 17: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/17.jpg)
Falling Factorials
(x)n := x(x − 1) · · · (x − n + 1), x ∈ R, n ≥ 1. (4)
Substituting −x for x in (4), we have
(−x)n = −x(−x−1) · · · (−x−n+1) = (−1)nx(x+1) · · · (x+n−1)= (−1)nx (n).
Substituting above −x for x , we have
(x)n = (−1)n(−x)(n)from (3)
= (−1)nn∑
k=1
|s(n, k)|(−x)k =
n∑k=1
(−1)n−k |s(n, k)|xk .(5)
Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called theStirling Numbers of the First Kind. From (5) we have
(x)n =n∑
k=1
s(n, k)xk . (6)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 18: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/18.jpg)
Falling Factorials
(x)n := x(x − 1) · · · (x − n + 1), x ∈ R, n ≥ 1. (4)
Substituting −x for x in (4), we have
(−x)n = −x(−x−1) · · · (−x−n+1) = (−1)nx(x+1) · · · (x+n−1)= (−1)nx (n).Substituting above −x for x , we have
(x)n = (−1)n(−x)(n)from (3)
= (−1)nn∑
k=1
|s(n, k)|(−x)k
=
n∑k=1
(−1)n−k |s(n, k)|xk .(5)
Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called theStirling Numbers of the First Kind. From (5) we have
(x)n =n∑
k=1
s(n, k)xk . (6)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 19: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/19.jpg)
Falling Factorials
(x)n := x(x − 1) · · · (x − n + 1), x ∈ R, n ≥ 1. (4)
Substituting −x for x in (4), we have
(−x)n = −x(−x−1) · · · (−x−n+1) = (−1)nx(x+1) · · · (x+n−1)= (−1)nx (n).Substituting above −x for x , we have
(x)n = (−1)n(−x)(n)from (3)
= (−1)nn∑
k=1
|s(n, k)|(−x)k =
n∑k=1
(−1)n−k |s(n, k)|xk .(5)
Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called theStirling Numbers of the First Kind. From (5) we have
(x)n =n∑
k=1
s(n, k)xk . (6)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 20: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/20.jpg)
Falling Factorials
(x)n := x(x − 1) · · · (x − n + 1), x ∈ R, n ≥ 1. (4)
Substituting −x for x in (4), we have
(−x)n = −x(−x−1) · · · (−x−n+1) = (−1)nx(x+1) · · · (x+n−1)= (−1)nx (n).Substituting above −x for x , we have
(x)n = (−1)n(−x)(n)from (3)
= (−1)nn∑
k=1
|s(n, k)|(−x)k =
n∑k=1
(−1)n−k |s(n, k)|xk .(5)
Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called theStirling Numbers of the First Kind.
From (5) we have
(x)n =n∑
k=1
s(n, k)xk . (6)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 21: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/21.jpg)
Falling Factorials
(x)n := x(x − 1) · · · (x − n + 1), x ∈ R, n ≥ 1. (4)
Substituting −x for x in (4), we have
(−x)n = −x(−x−1) · · · (−x−n+1) = (−1)nx(x+1) · · · (x+n−1)= (−1)nx (n).Substituting above −x for x , we have
(x)n = (−1)n(−x)(n)from (3)
= (−1)nn∑
k=1
|s(n, k)|(−x)k =
n∑k=1
(−1)n−k |s(n, k)|xk .(5)
Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called theStirling Numbers of the First Kind. From (5) we have
(x)n =n∑
k=1
s(n, k)xk . (6)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 22: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/22.jpg)
Stirling Numbers of the Second Kind
S(n, k) := number of ways to partition the set [n] into k nonemptysets.
Example: S(4, 3) = 5:{1, 2}, {3}, {4}{1, 3}, {2}, {4}{1, 4}, {2}, {3}{2, 3}, {1}, {4}{2, 4}, {1}, {3}
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 23: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/23.jpg)
Stirling Numbers of the Second Kind
S(n, k) := number of ways to partition the set [n] into k nonemptysets.Example: S(4, 3) = 5:
{1, 2}, {3}, {4}{1, 3}, {2}, {4}{1, 4}, {2}, {3}{2, 3}, {1}, {4}{2, 4}, {1}, {3}
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 24: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/24.jpg)
Stirling Numbers of the Second Kind
S(n, k) := number of ways to partition the set [n] into k nonemptysets.Example: S(4, 3) = 5:{1, 2}, {3}, {4}{1, 3}, {2}, {4}{1, 4}, {2}, {3}{2, 3}, {1}, {4}{2, 4}, {1}, {3}
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 25: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/25.jpg)
Stirling Numbers of the Second Kind
S(n, k) := number of ways to partition the set [n] into k nonemptysets.
Theorem.
xn =n∑
k=1
S(n, k)(x)k , (7)
Proof. Consider functions f : [n]→ X , where |X | = x ∈ N.How many such functions are there?The answer by direct count: xn.
An alternative indirect count:For k = 1, 2, · · · , n, let ck denote the number of such functionswith |Im(f )| = k . (If x < n, then ck = 0 for x < k ≤ n.)So the answer by this indirect count is
∑nk=1 ck .
Thus, xn =∑n
k=1 ck . To complete the proof we now show thatck = S(n, k)(x)k .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 26: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/26.jpg)
Stirling Numbers of the Second Kind
S(n, k) := number of ways to partition the set [n] into k nonemptysets.Theorem.
xn =n∑
k=1
S(n, k)(x)k , (7)
Proof. Consider functions f : [n]→ X , where |X | = x ∈ N.How many such functions are there?The answer by direct count: xn.
An alternative indirect count:For k = 1, 2, · · · , n, let ck denote the number of such functionswith |Im(f )| = k . (If x < n, then ck = 0 for x < k ≤ n.)So the answer by this indirect count is
∑nk=1 ck .
Thus, xn =∑n
k=1 ck . To complete the proof we now show thatck = S(n, k)(x)k .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Stirling Numbers of the Second Kind
S(n, k) := number of ways to partition the set [n] into k nonemptysets.Theorem.
xn =n∑
k=1
S(n, k)(x)k , (7)
Proof. Consider functions f : [n]→ X , where |X | = x ∈ N.
How many such functions are there?The answer by direct count: xn.
An alternative indirect count:For k = 1, 2, · · · , n, let ck denote the number of such functionswith |Im(f )| = k . (If x < n, then ck = 0 for x < k ≤ n.)So the answer by this indirect count is
∑nk=1 ck .
Thus, xn =∑n
k=1 ck . To complete the proof we now show thatck = S(n, k)(x)k .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 28: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/28.jpg)
Stirling Numbers of the Second Kind
S(n, k) := number of ways to partition the set [n] into k nonemptysets.Theorem.
xn =n∑
k=1
S(n, k)(x)k , (7)
Proof. Consider functions f : [n]→ X , where |X | = x ∈ N.How many such functions are there?
The answer by direct count: xn.
An alternative indirect count:For k = 1, 2, · · · , n, let ck denote the number of such functionswith |Im(f )| = k . (If x < n, then ck = 0 for x < k ≤ n.)So the answer by this indirect count is
∑nk=1 ck .
Thus, xn =∑n
k=1 ck . To complete the proof we now show thatck = S(n, k)(x)k .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 29: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/29.jpg)
Stirling Numbers of the Second Kind
S(n, k) := number of ways to partition the set [n] into k nonemptysets.Theorem.
xn =n∑
k=1
S(n, k)(x)k , (7)
Proof. Consider functions f : [n]→ X , where |X | = x ∈ N.How many such functions are there?The answer by direct count: xn.
An alternative indirect count:For k = 1, 2, · · · , n, let ck denote the number of such functionswith |Im(f )| = k . (If x < n, then ck = 0 for x < k ≤ n.)So the answer by this indirect count is
∑nk=1 ck .
Thus, xn =∑n
k=1 ck . To complete the proof we now show thatck = S(n, k)(x)k .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 30: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/30.jpg)
Stirling Numbers of the Second Kind
S(n, k) := number of ways to partition the set [n] into k nonemptysets.Theorem.
xn =n∑
k=1
S(n, k)(x)k , (7)
Proof. Consider functions f : [n]→ X , where |X | = x ∈ N.How many such functions are there?The answer by direct count: xn.
An alternative indirect count:
For k = 1, 2, · · · , n, let ck denote the number of such functionswith |Im(f )| = k . (If x < n, then ck = 0 for x < k ≤ n.)So the answer by this indirect count is
∑nk=1 ck .
Thus, xn =∑n
k=1 ck . To complete the proof we now show thatck = S(n, k)(x)k .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 31: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/31.jpg)
Stirling Numbers of the Second Kind
S(n, k) := number of ways to partition the set [n] into k nonemptysets.Theorem.
xn =n∑
k=1
S(n, k)(x)k , (7)
Proof. Consider functions f : [n]→ X , where |X | = x ∈ N.How many such functions are there?The answer by direct count: xn.
An alternative indirect count:For k = 1, 2, · · · , n, let ck denote the number of such functionswith |Im(f )| = k . (If x < n, then ck = 0 for x < k ≤ n.)
So the answer by this indirect count is∑n
k=1 ck .Thus, xn =
∑nk=1 ck . To complete the proof we now show that
ck = S(n, k)(x)k .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 32: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/32.jpg)
Stirling Numbers of the Second Kind
S(n, k) := number of ways to partition the set [n] into k nonemptysets.Theorem.
xn =n∑
k=1
S(n, k)(x)k , (7)
Proof. Consider functions f : [n]→ X , where |X | = x ∈ N.How many such functions are there?The answer by direct count: xn.
An alternative indirect count:For k = 1, 2, · · · , n, let ck denote the number of such functionswith |Im(f )| = k . (If x < n, then ck = 0 for x < k ≤ n.)So the answer by this indirect count is
∑nk=1 ck .
Thus, xn =∑n
k=1 ck . To complete the proof we now show thatck = S(n, k)(x)k .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 33: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/33.jpg)
Stirling Numbers of the Second Kind
S(n, k) := number of ways to partition the set [n] into k nonemptysets.Theorem.
xn =n∑
k=1
S(n, k)(x)k , (7)
Proof. Consider functions f : [n]→ X , where |X | = x ∈ N.How many such functions are there?The answer by direct count: xn.
An alternative indirect count:For k = 1, 2, · · · , n, let ck denote the number of such functionswith |Im(f )| = k . (If x < n, then ck = 0 for x < k ≤ n.)So the answer by this indirect count is
∑nk=1 ck .
Thus, xn =∑n
k=1 ck .
To complete the proof we now show thatck = S(n, k)(x)k .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 34: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/34.jpg)
Stirling Numbers of the Second Kind
S(n, k) := number of ways to partition the set [n] into k nonemptysets.Theorem.
xn =n∑
k=1
S(n, k)(x)k , (7)
Proof. Consider functions f : [n]→ X , where |X | = x ∈ N.How many such functions are there?The answer by direct count: xn.
An alternative indirect count:For k = 1, 2, · · · , n, let ck denote the number of such functionswith |Im(f )| = k . (If x < n, then ck = 0 for x < k ≤ n.)So the answer by this indirect count is
∑nk=1 ck .
Thus, xn =∑n
k=1 ck . To complete the proof we now show thatck = S(n, k)(x)k .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 35: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/35.jpg)
Consider functions f : [n]→ X , where |X | = x ∈ N. Let ckdenote the number of such functions with |Im(f )| = k.Proof that ck = S(n, k)(x)k :
For f with |Im(f )| = k, let Im(f ) = {x1, · · · , xk}. Then
f −1({x1}), f −1({x2}), · · · , f −1({xk}) is a partition of [n] into knon-empty sets, which we will denote by {B1, · · · ,Bk}.
Now:1. There are S(n, k) ways to choose the sets {B1, · · · ,Bk};
2. There are(xk
)was to choose the {x1, · · · , xk};
3. There are k! ways to make the correspondence betweenf −1({x1}), f −1({x2}), · · · , f −1({xk}) and {B1, · · · ,Bk}.
Thus ck = S(n, k)×(xk
)× k! = S(n, k) x!
(x−k)! = S(n, k)(x)k . �
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Consider functions f : [n]→ X , where |X | = x ∈ N. Let ckdenote the number of such functions with |Im(f )| = k.Proof that ck = S(n, k)(x)k :
For f with |Im(f )| = k, let Im(f ) = {x1, · · · , xk}.
Then
f −1({x1}), f −1({x2}), · · · , f −1({xk}) is a partition of [n] into knon-empty sets, which we will denote by {B1, · · · ,Bk}.
Now:1. There are S(n, k) ways to choose the sets {B1, · · · ,Bk};
2. There are(xk
)was to choose the {x1, · · · , xk};
3. There are k! ways to make the correspondence betweenf −1({x1}), f −1({x2}), · · · , f −1({xk}) and {B1, · · · ,Bk}.
Thus ck = S(n, k)×(xk
)× k! = S(n, k) x!
(x−k)! = S(n, k)(x)k . �
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Consider functions f : [n]→ X , where |X | = x ∈ N. Let ckdenote the number of such functions with |Im(f )| = k.Proof that ck = S(n, k)(x)k :
For f with |Im(f )| = k, let Im(f ) = {x1, · · · , xk}. Then
f −1({x1}), f −1({x2}), · · · , f −1({xk}) is a partition of [n] into knon-empty sets, which we will denote by {B1, · · · ,Bk}.
Now:1. There are S(n, k) ways to choose the sets {B1, · · · ,Bk};
2. There are(xk
)was to choose the {x1, · · · , xk};
3. There are k! ways to make the correspondence betweenf −1({x1}), f −1({x2}), · · · , f −1({xk}) and {B1, · · · ,Bk}.
Thus ck = S(n, k)×(xk
)× k! = S(n, k) x!
(x−k)! = S(n, k)(x)k . �
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 38: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/38.jpg)
Consider functions f : [n]→ X , where |X | = x ∈ N. Let ckdenote the number of such functions with |Im(f )| = k.Proof that ck = S(n, k)(x)k :
For f with |Im(f )| = k, let Im(f ) = {x1, · · · , xk}. Then
f −1({x1}), f −1({x2}), · · · , f −1({xk}) is a partition of [n] into knon-empty sets, which we will denote by {B1, · · · ,Bk}.
Now:1. There are S(n, k) ways to choose the sets {B1, · · · ,Bk};
2. There are(xk
)was to choose the {x1, · · · , xk};
3. There are k! ways to make the correspondence betweenf −1({x1}), f −1({x2}), · · · , f −1({xk}) and {B1, · · · ,Bk}.
Thus ck = S(n, k)×(xk
)× k! = S(n, k) x!
(x−k)! = S(n, k)(x)k . �
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Consider functions f : [n]→ X , where |X | = x ∈ N. Let ckdenote the number of such functions with |Im(f )| = k.Proof that ck = S(n, k)(x)k :
For f with |Im(f )| = k, let Im(f ) = {x1, · · · , xk}. Then
f −1({x1}), f −1({x2}), · · · , f −1({xk}) is a partition of [n] into knon-empty sets, which we will denote by {B1, · · · ,Bk}.
Now:1. There are S(n, k) ways to choose the sets {B1, · · · ,Bk};
2. There are(xk
)was to choose the {x1, · · · , xk};
3. There are k! ways to make the correspondence betweenf −1({x1}), f −1({x2}), · · · , f −1({xk}) and {B1, · · · ,Bk}.
Thus ck = S(n, k)×(xk
)× k! = S(n, k) x!
(x−k)! = S(n, k)(x)k . �
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 40: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/40.jpg)
Consider functions f : [n]→ X , where |X | = x ∈ N. Let ckdenote the number of such functions with |Im(f )| = k.Proof that ck = S(n, k)(x)k :
For f with |Im(f )| = k, let Im(f ) = {x1, · · · , xk}. Then
f −1({x1}), f −1({x2}), · · · , f −1({xk}) is a partition of [n] into knon-empty sets, which we will denote by {B1, · · · ,Bk}.
Now:1. There are S(n, k) ways to choose the sets {B1, · · · ,Bk};
2. There are(xk
)was to choose the {x1, · · · , xk};
3. There are k! ways to make the correspondence betweenf −1({x1}), f −1({x2}), · · · , f −1({xk}) and {B1, · · · ,Bk}.
Thus ck = S(n, k)×(xk
)× k! = S(n, k) x!
(x−k)! = S(n, k)(x)k . �
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 41: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/41.jpg)
Consider functions f : [n]→ X , where |X | = x ∈ N. Let ckdenote the number of such functions with |Im(f )| = k.Proof that ck = S(n, k)(x)k :
For f with |Im(f )| = k, let Im(f ) = {x1, · · · , xk}. Then
f −1({x1}), f −1({x2}), · · · , f −1({xk}) is a partition of [n] into knon-empty sets, which we will denote by {B1, · · · ,Bk}.
Now:1. There are S(n, k) ways to choose the sets {B1, · · · ,Bk};
2. There are(xk
)was to choose the {x1, · · · , xk};
3. There are k! ways to make the correspondence betweenf −1({x1}), f −1({x2}), · · · , f −1({xk}) and {B1, · · · ,Bk}.
Thus ck = S(n, k)×(xk
)× k!
= S(n, k) x!(x−k)! = S(n, k)(x)k . �
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 42: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/42.jpg)
Consider functions f : [n]→ X , where |X | = x ∈ N. Let ckdenote the number of such functions with |Im(f )| = k.Proof that ck = S(n, k)(x)k :
For f with |Im(f )| = k, let Im(f ) = {x1, · · · , xk}. Then
f −1({x1}), f −1({x2}), · · · , f −1({xk}) is a partition of [n] into knon-empty sets, which we will denote by {B1, · · · ,Bk}.
Now:1. There are S(n, k) ways to choose the sets {B1, · · · ,Bk};
2. There are(xk
)was to choose the {x1, · · · , xk};
3. There are k! ways to make the correspondence betweenf −1({x1}), f −1({x2}), · · · , f −1({xk}) and {B1, · · · ,Bk}.
Thus ck = S(n, k)×(xk
)× k! = S(n, k) x!
(x−k)!
= S(n, k)(x)k . �
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 43: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/43.jpg)
Consider functions f : [n]→ X , where |X | = x ∈ N. Let ckdenote the number of such functions with |Im(f )| = k.Proof that ck = S(n, k)(x)k :
For f with |Im(f )| = k, let Im(f ) = {x1, · · · , xk}. Then
f −1({x1}), f −1({x2}), · · · , f −1({xk}) is a partition of [n] into knon-empty sets, which we will denote by {B1, · · · ,Bk}.
Now:1. There are S(n, k) ways to choose the sets {B1, · · · ,Bk};
2. There are(xk
)was to choose the {x1, · · · , xk};
3. There are k! ways to make the correspondence betweenf −1({x1}), f −1({x2}), · · · , f −1({xk}) and {B1, · · · ,Bk}.
Thus ck = S(n, k)×(xk
)× k! = S(n, k) x!
(x−k)! = S(n, k)(x)k . �
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 44: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/44.jpg)
xn =∑n
k=1 S(n, k)(x)k , n ≥ 1
(x)n =∑n
k=1 s(n, k)xk , n ≥ 1
S(n, k) (Stirling number of the second kind) is the number ofways to partition [n] into k nonempty subsets
|s(n, k)| (unsigned Stirling number of the first kind) is thenumber of permutations of [n] which have k cycles
s(n, k) = (−1)n−k |s(n, k)| (Stirling number of the first kind)
Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 bydefining S(n, k) = s(n, k) = 0, for k > n.
Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).Corollary. Ss = sS = Id.
That is,∑∞
j=1 S(n, j)s(j ,m) =
{1, if n = m;
0, if n 6= m.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 45: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/45.jpg)
xn =∑n
k=1 S(n, k)(x)k , n ≥ 1
(x)n =∑n
k=1 s(n, k)xk , n ≥ 1
S(n, k) (Stirling number of the second kind) is the number ofways to partition [n] into k nonempty subsets
|s(n, k)| (unsigned Stirling number of the first kind) is thenumber of permutations of [n] which have k cycles
s(n, k) = (−1)n−k |s(n, k)| (Stirling number of the first kind)
Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 bydefining S(n, k) = s(n, k) = 0, for k > n.
Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).Corollary. Ss = sS = Id.
That is,∑∞
j=1 S(n, j)s(j ,m) =
{1, if n = m;
0, if n 6= m.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 46: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/46.jpg)
xn =∑n
k=1 S(n, k)(x)k , n ≥ 1
(x)n =∑n
k=1 s(n, k)xk , n ≥ 1
S(n, k) (Stirling number of the second kind) is the number ofways to partition [n] into k nonempty subsets
|s(n, k)| (unsigned Stirling number of the first kind) is thenumber of permutations of [n] which have k cycles
s(n, k) = (−1)n−k |s(n, k)| (Stirling number of the first kind)
Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 bydefining S(n, k) = s(n, k) = 0, for k > n.
Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).Corollary. Ss = sS = Id.
That is,∑∞
j=1 S(n, j)s(j ,m) =
{1, if n = m;
0, if n 6= m.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
![Page 47: Touchard Polynomials, Stirling Numbers and Random Permutations · Touchard Polynomials, Stirling Numbers and Random Permutations Ross Pinsky Department of Mathematics, Technion 32000](https://reader033.vdocuments.pub/reader033/viewer/2022042621/5f663395daa7ba38e43a8b55/html5/thumbnails/47.jpg)
xn =∑n
k=1 S(n, k)(x)k , n ≥ 1
(x)n =∑n
k=1 s(n, k)xk , n ≥ 1
S(n, k) (Stirling number of the second kind) is the number ofways to partition [n] into k nonempty subsets
|s(n, k)| (unsigned Stirling number of the first kind) is thenumber of permutations of [n] which have k cycles
s(n, k) = (−1)n−k |s(n, k)| (Stirling number of the first kind)
Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 bydefining S(n, k) = s(n, k) = 0, for k > n.
Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).
Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).Corollary. Ss = sS = Id.
That is,∑∞
j=1 S(n, j)s(j ,m) =
{1, if n = m;
0, if n 6= m.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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xn =∑n
k=1 S(n, k)(x)k , n ≥ 1
(x)n =∑n
k=1 s(n, k)xk , n ≥ 1
S(n, k) (Stirling number of the second kind) is the number ofways to partition [n] into k nonempty subsets
|s(n, k)| (unsigned Stirling number of the first kind) is thenumber of permutations of [n] which have k cycles
s(n, k) = (−1)n−k |s(n, k)| (Stirling number of the first kind)
Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 bydefining S(n, k) = s(n, k) = 0, for k > n.
Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).
Corollary. Ss = sS = Id.
That is,∑∞
j=1 S(n, j)s(j ,m) =
{1, if n = m;
0, if n 6= m.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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xn =∑n
k=1 S(n, k)(x)k , n ≥ 1
(x)n =∑n
k=1 s(n, k)xk , n ≥ 1
S(n, k) (Stirling number of the second kind) is the number ofways to partition [n] into k nonempty subsets
|s(n, k)| (unsigned Stirling number of the first kind) is thenumber of permutations of [n] which have k cycles
s(n, k) = (−1)n−k |s(n, k)| (Stirling number of the first kind)
Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 bydefining S(n, k) = s(n, k) = 0, for k > n.
Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).Corollary. Ss = sS = Id.
That is,∑∞
j=1 S(n, j)s(j ,m) =
{1, if n = m;
0, if n 6= m.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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xn =∑n
k=1 S(n, k)(x)k , n ≥ 1
(x)n =∑n
k=1 s(n, k)xk , n ≥ 1
S(n, k) (Stirling number of the second kind) is the number ofways to partition [n] into k nonempty subsets
|s(n, k)| (unsigned Stirling number of the first kind) is thenumber of permutations of [n] which have k cycles
s(n, k) = (−1)n−k |s(n, k)| (Stirling number of the first kind)
Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 bydefining S(n, k) = s(n, k) = 0, for k > n.
Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).Corollary. Ss = sS = Id.
That is,∑∞
j=1 S(n, j)s(j ,m) =
{1, if n = m;
0, if n 6= m.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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xn =∑n
k=1 S(n, k)(x)k , n ≥ 1 (*)(x)n =
∑nk=1 s(n, k)xk , n ≥ 1 (**)
Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 bydefining S(n, k) = s(n, k) = 0, for k > n.Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).Corollary. Ss = sS = Id.Proof.
Consider the real vector space V of polynomials of theform
∑nk=1 ckx
k , n ≥ 1, ck ∈ R. .
Of course,B1 := {x , x2, x3, · · · } is a basis for V .
By (*), B2 := {(x)1, (x)2, (x)3, · · · } is also a basis for V .
By (*) and (**), the matrices S and s transform between thetwo basis, and thus Ss = sS = I . �
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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xn =∑n
k=1 S(n, k)(x)k , n ≥ 1 (*)(x)n =
∑nk=1 s(n, k)xk , n ≥ 1 (**)
Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 bydefining S(n, k) = s(n, k) = 0, for k > n.Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).Corollary. Ss = sS = Id.Proof. Consider the real vector space V of polynomials of theform
∑nk=1 ckx
k , n ≥ 1, ck ∈ R.
.
Of course,B1 := {x , x2, x3, · · · } is a basis for V .
By (*), B2 := {(x)1, (x)2, (x)3, · · · } is also a basis for V .
By (*) and (**), the matrices S and s transform between thetwo basis, and thus Ss = sS = I . �
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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xn =∑n
k=1 S(n, k)(x)k , n ≥ 1 (*)(x)n =
∑nk=1 s(n, k)xk , n ≥ 1 (**)
Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 bydefining S(n, k) = s(n, k) = 0, for k > n.Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).Corollary. Ss = sS = Id.Proof. Consider the real vector space V of polynomials of theform
∑nk=1 ckx
k , n ≥ 1, ck ∈ R. .
Of course,B1 := {x , x2, x3, · · · } is a basis for V .
By (*), B2 := {(x)1, (x)2, (x)3, · · · } is also a basis for V .
By (*) and (**), the matrices S and s transform between thetwo basis, and thus Ss = sS = I . �
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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xn =∑n
k=1 S(n, k)(x)k , n ≥ 1 (*)(x)n =
∑nk=1 s(n, k)xk , n ≥ 1 (**)
Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 bydefining S(n, k) = s(n, k) = 0, for k > n.Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).Corollary. Ss = sS = Id.Proof. Consider the real vector space V of polynomials of theform
∑nk=1 ckx
k , n ≥ 1, ck ∈ R. .
Of course,B1 := {x , x2, x3, · · · } is a basis for V .
By (*), B2 := {(x)1, (x)2, (x)3, · · · } is also a basis for V .
By (*) and (**), the matrices S and s transform between thetwo basis, and thus Ss = sS = I . �
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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xn =∑n
k=1 S(n, k)(x)k , n ≥ 1 (*)(x)n =
∑nk=1 s(n, k)xk , n ≥ 1 (**)
Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 bydefining S(n, k) = s(n, k) = 0, for k > n.Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).Corollary. Ss = sS = Id.Proof. Consider the real vector space V of polynomials of theform
∑nk=1 ckx
k , n ≥ 1, ck ∈ R. .
Of course,B1 := {x , x2, x3, · · · } is a basis for V .
By (*), B2 := {(x)1, (x)2, (x)3, · · · } is also a basis for V .
By (*) and (**), the matrices S and s transform between thetwo basis, and thus Ss = sS = I . �
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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S(n, k) (Stirling number of the second kind) is the number ofways to partition [n] into k nonempty subsets
|s(n, k)| (unsigned Stirling number of the first kind) is thenumber of permutations of [n] which have k cycles
s(n, k) = (−1)n−k |s(n, k)| (Stirling number of the first kind)
Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 bydefining S(n, k) = s(n, k) = 0, for k > n.
Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).Corollary. Ss = sS = Id
That is,∑∞
j=1 S(n, j)s(j ,m) =
{1, if n = m;
0, if n 6= m.
Example: n = 8,m = 4:0 =
∑∞j=1 S(8, j)s(j , 4) =
S(8, 4)−S(8, 5)|s(5, 4)|+S(8, 6)|s(6, 4)|−S(8, 7)|s(7, 4)|+|s(8, 4)|.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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S(n, k) (Stirling number of the second kind) is the number ofways to partition [n] into k nonempty subsets
|s(n, k)| (unsigned Stirling number of the first kind) is thenumber of permutations of [n] which have k cycles
s(n, k) = (−1)n−k |s(n, k)| (Stirling number of the first kind)
Extend the definitions of S(n, k) and s(n, k) to all n, k ≥ 1 bydefining S(n, k) = s(n, k) = 0, for k > n.
Let S = {Snk}∞n,k=1 denote the ∞×∞ matrix with entriesSnk = S(n, k).Let s = {snk}∞n,k=1 denote the ∞×∞ matrix with entriessnk = s(n, k).Corollary. Ss = sS = Id
That is,∑∞
j=1 S(n, j)s(j ,m) =
{1, if n = m;
0, if n 6= m.Example: n = 8,m = 4:0 =
∑∞j=1 S(8, j)s(j , 4) =
S(8, 4)−S(8, 5)|s(5, 4)|+S(8, 6)|s(6, 4)|−S(8, 7)|s(7, 4)|+|s(8, 4)|.Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Recalling the Poisson distribution
Let X be a random variable with the distribution Poiss(λ), λ > 0:
P(X = j) = e−λ λj
j! , j = 0, 1, 2, · · · .
nth moment: µn;λ := EX n
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Recalling the Poisson distribution
Let X be a random variable with the distribution Poiss(λ), λ > 0:
P(X = j) = e−λ λj
j! , j = 0, 1, 2, · · · .
nth moment: µn;λ := EX n
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Let X be a random variable with the distribution Poiss(λ), λ > 0:
P(X = j) = e−λ λj
j! , j = 1, 2, · · · .nth moment: µn;λ := EX n
µn;λ =∞∑j=0
(e−λ
λj
j!
)jn
(7)=∞∑j=0
(e−λ
λj
j!
) n∑k=1
S(n, k)(j)k
=
=∞∑j=0
(e−λ
λj
j!
)min(j ,n)∑k=1
S(n, k)(j)k(because (j)k = 0, k > j)
= e−λn∑
k=1
S(n, k)∞∑j=k
λj
j!(j)k .
But∑∞
j=kλj
j! (j)k =∑∞
j=kλj
(j−k)! = λk∑∞
j=kλj−k
(j−k)! = λk∑∞
l=0λl
l! =
λkeλ. Thus,
µn;λ =n∑
k=1
S(n, k)λk .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Let X be a random variable with the distribution Poiss(λ), λ > 0:
P(X = j) = e−λ λj
j! , j = 1, 2, · · · .nth moment: µn;λ := EX n
µn;λ =∞∑j=0
(e−λ
λj
j!
)jn
(7)=∞∑j=0
(e−λ
λj
j!
) n∑k=1
S(n, k)(j)k =
=∞∑j=0
(e−λ
λj
j!
)min(j ,n)∑k=1
S(n, k)(j)k
(because (j)k = 0, k > j)
= e−λn∑
k=1
S(n, k)∞∑j=k
λj
j!(j)k .
But∑∞
j=kλj
j! (j)k =∑∞
j=kλj
(j−k)! = λk∑∞
j=kλj−k
(j−k)! = λk∑∞
l=0λl
l! =
λkeλ. Thus,
µn;λ =n∑
k=1
S(n, k)λk .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Let X be a random variable with the distribution Poiss(λ), λ > 0:
P(X = j) = e−λ λj
j! , j = 1, 2, · · · .nth moment: µn;λ := EX n
µn;λ =∞∑j=0
(e−λ
λj
j!
)jn
(7)=∞∑j=0
(e−λ
λj
j!
) n∑k=1
S(n, k)(j)k =
=∞∑j=0
(e−λ
λj
j!
)min(j ,n)∑k=1
S(n, k)(j)k(because (j)k = 0, k > j)
= e−λn∑
k=1
S(n, k)∞∑j=k
λj
j!(j)k .
But∑∞
j=kλj
j! (j)k =∑∞
j=kλj
(j−k)! = λk∑∞
j=kλj−k
(j−k)! = λk∑∞
l=0λl
l! =
λkeλ. Thus,
µn;λ =n∑
k=1
S(n, k)λk .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Let X be a random variable with the distribution Poiss(λ), λ > 0:
P(X = j) = e−λ λj
j! , j = 1, 2, · · · .nth moment: µn;λ := EX n
µn;λ =∞∑j=0
(e−λ
λj
j!
)jn
(7)=∞∑j=0
(e−λ
λj
j!
) n∑k=1
S(n, k)(j)k =
=∞∑j=0
(e−λ
λj
j!
)min(j ,n)∑k=1
S(n, k)(j)k(because (j)k = 0, k > j)
= e−λn∑
k=1
S(n, k)∞∑j=k
λj
j!(j)k .
But∑∞
j=kλj
j! (j)k =∑∞
j=kλj
(j−k)! = λk∑∞
j=kλj−k
(j−k)! = λk∑∞
l=0λl
l! =
λkeλ. Thus,
µn;λ =n∑
k=1
S(n, k)λk .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Let X be a random variable with the distribution Poiss(λ), λ > 0:
P(X = j) = e−λ λj
j! , j = 1, 2, · · · .nth moment: µn;λ := EX n
µn;λ =∞∑j=0
(e−λ
λj
j!
)jn
(7)=∞∑j=0
(e−λ
λj
j!
) n∑k=1
S(n, k)(j)k =
=∞∑j=0
(e−λ
λj
j!
)min(j ,n)∑k=1
S(n, k)(j)k(because (j)k = 0, k > j)
= e−λn∑
k=1
S(n, k)∞∑j=k
λj
j!(j)k .
But∑∞
j=kλj
j! (j)k =∑∞
j=kλj
(j−k)!
= λk∑∞
j=kλj−k
(j−k)! = λk∑∞
l=0λl
l! =
λkeλ. Thus,
µn;λ =n∑
k=1
S(n, k)λk .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Let X be a random variable with the distribution Poiss(λ), λ > 0:
P(X = j) = e−λ λj
j! , j = 1, 2, · · · .nth moment: µn;λ := EX n
µn;λ =∞∑j=0
(e−λ
λj
j!
)jn
(7)=∞∑j=0
(e−λ
λj
j!
) n∑k=1
S(n, k)(j)k =
=∞∑j=0
(e−λ
λj
j!
)min(j ,n)∑k=1
S(n, k)(j)k(because (j)k = 0, k > j)
= e−λn∑
k=1
S(n, k)∞∑j=k
λj
j!(j)k .
But∑∞
j=kλj
j! (j)k =∑∞
j=kλj
(j−k)! = λk∑∞
j=kλj−k
(j−k)!
= λk∑∞
l=0λl
l! =
λkeλ. Thus,
µn;λ =n∑
k=1
S(n, k)λk .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Let X be a random variable with the distribution Poiss(λ), λ > 0:
P(X = j) = e−λ λj
j! , j = 1, 2, · · · .nth moment: µn;λ := EX n
µn;λ =∞∑j=0
(e−λ
λj
j!
)jn
(7)=∞∑j=0
(e−λ
λj
j!
) n∑k=1
S(n, k)(j)k =
=∞∑j=0
(e−λ
λj
j!
)min(j ,n)∑k=1
S(n, k)(j)k(because (j)k = 0, k > j)
= e−λn∑
k=1
S(n, k)∞∑j=k
λj
j!(j)k .
But∑∞
j=kλj
j! (j)k =∑∞
j=kλj
(j−k)! = λk∑∞
j=kλj−k
(j−k)! = λk∑∞
l=0λl
l!
=
λkeλ. Thus,
µn;λ =n∑
k=1
S(n, k)λk .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Let X be a random variable with the distribution Poiss(λ), λ > 0:
P(X = j) = e−λ λj
j! , j = 1, 2, · · · .nth moment: µn;λ := EX n
µn;λ =∞∑j=0
(e−λ
λj
j!
)jn
(7)=∞∑j=0
(e−λ
λj
j!
) n∑k=1
S(n, k)(j)k =
=∞∑j=0
(e−λ
λj
j!
)min(j ,n)∑k=1
S(n, k)(j)k(because (j)k = 0, k > j)
= e−λn∑
k=1
S(n, k)∞∑j=k
λj
j!(j)k .
But∑∞
j=kλj
j! (j)k =∑∞
j=kλj
(j−k)! = λk∑∞
j=kλj−k
(j−k)! = λk∑∞
l=0λl
l! =
λkeλ.
Thus,
µn;λ =n∑
k=1
S(n, k)λk .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Let X be a random variable with the distribution Poiss(λ), λ > 0:
P(X = j) = e−λ λj
j! , j = 1, 2, · · · .nth moment: µn;λ := EX n
µn;λ =∞∑j=0
(e−λ
λj
j!
)jn
(7)=∞∑j=0
(e−λ
λj
j!
) n∑k=1
S(n, k)(j)k =
=∞∑j=0
(e−λ
λj
j!
)min(j ,n)∑k=1
S(n, k)(j)k(because (j)k = 0, k > j)
= e−λn∑
k=1
S(n, k)∞∑j=k
λj
j!(j)k .
But∑∞
j=kλj
j! (j)k =∑∞
j=kλj
(j−k)! = λk∑∞
j=kλj−k
(j−k)! = λk∑∞
l=0λl
l! =
λkeλ. Thus,
µn;λ =n∑
k=1
S(n, k)λk .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Let X be a random variable with the distribution Poiss(λ), λ > 0:
P(X = j) = e−λ λj
j! , j = 1, 2, · · · .nth moment: µn;λ := EX n =
∑∞j=0
(e−λ λj
j!
)jn
µn;λ =∑n
k=1 S(n, k)λk
Touchard Polynomial: Tn(x) :=∑n
k=1 S(n, k)xk
µn;λ = Tn(λ)(Compare Tn with (6).)
∞∑j=0
(e−λ
λj
j!
)jn = EX n = µn;λ = Tn(λ) :=
n∑k=1
S(n, k)λk
When λ = 1:
1
e
∞∑j=0
jn
j!= EX n = µn;1 = Tn(1) :=
n∑k=1
S(n, k) := Bn,
where Bn is the nth Bell number:Bn = the total number of partitions of [n].
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Let X be a random variable with the distribution Poiss(λ), λ > 0:
P(X = j) = e−λ λj
j! , j = 1, 2, · · · .nth moment: µn;λ := EX n =
∑∞j=0
(e−λ λj
j!
)jn
µn;λ =∑n
k=1 S(n, k)λk
Touchard Polynomial: Tn(x) :=∑n
k=1 S(n, k)xk
µn;λ = Tn(λ)(Compare Tn with (6).)
∞∑j=0
(e−λ
λj
j!
)jn = EX n = µn;λ = Tn(λ) :=
n∑k=1
S(n, k)λk
When λ = 1:
1
e
∞∑j=0
jn
j!= EX n = µn;1 = Tn(1) :=
n∑k=1
S(n, k) := Bn,
where Bn is the nth Bell number:Bn = the total number of partitions of [n].
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Let X be a random variable with the distribution Poiss(λ), λ > 0:
P(X = j) = e−λ λj
j! , j = 1, 2, · · · .nth moment: µn;λ := EX n =
∑∞j=0
(e−λ λj
j!
)jn
µn;λ =∑n
k=1 S(n, k)λk
Touchard Polynomial: Tn(x) :=∑n
k=1 S(n, k)xk
µn;λ = Tn(λ)
(Compare Tn with (6).)
∞∑j=0
(e−λ
λj
j!
)jn = EX n = µn;λ = Tn(λ) :=
n∑k=1
S(n, k)λk
When λ = 1:
1
e
∞∑j=0
jn
j!= EX n = µn;1 = Tn(1) :=
n∑k=1
S(n, k) := Bn,
where Bn is the nth Bell number:Bn = the total number of partitions of [n].
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Let X be a random variable with the distribution Poiss(λ), λ > 0:
P(X = j) = e−λ λj
j! , j = 1, 2, · · · .nth moment: µn;λ := EX n =
∑∞j=0
(e−λ λj
j!
)jn
µn;λ =∑n
k=1 S(n, k)λk
Touchard Polynomial: Tn(x) :=∑n
k=1 S(n, k)xk
µn;λ = Tn(λ)(Compare Tn with (6).)
∞∑j=0
(e−λ
λj
j!
)jn = EX n = µn;λ = Tn(λ) :=
n∑k=1
S(n, k)λk
When λ = 1:
1
e
∞∑j=0
jn
j!= EX n = µn;1 = Tn(1) :=
n∑k=1
S(n, k) := Bn,
where Bn is the nth Bell number:Bn = the total number of partitions of [n].
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Let X be a random variable with the distribution Poiss(λ), λ > 0:
P(X = j) = e−λ λj
j! , j = 1, 2, · · · .nth moment: µn;λ := EX n =
∑∞j=0
(e−λ λj
j!
)jn
µn;λ =∑n
k=1 S(n, k)λk
Touchard Polynomial: Tn(x) :=∑n
k=1 S(n, k)xk
µn;λ = Tn(λ)(Compare Tn with (6).)
∞∑j=0
(e−λ
λj
j!
)jn = EX n = µn;λ = Tn(λ) :=
n∑k=1
S(n, k)λk
When λ = 1:
1
e
∞∑j=0
jn
j!= EX n = µn;1 = Tn(1) :=
n∑k=1
S(n, k) := Bn,
where Bn is the nth Bell number:Bn = the total number of partitions of [n].
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Let X be a random variable with the distribution Poiss(λ), λ > 0:
P(X = j) = e−λ λj
j! , j = 1, 2, · · · .nth moment: µn;λ := EX n =
∑∞j=0
(e−λ λj
j!
)jn
µn;λ =∑n
k=1 S(n, k)λk
Touchard Polynomial: Tn(x) :=∑n
k=1 S(n, k)xk
µn;λ = Tn(λ)(Compare Tn with (6).)
∞∑j=0
(e−λ
λj
j!
)jn = EX n = µn;λ = Tn(λ) :=
n∑k=1
S(n, k)λk
When λ = 1:
1
e
∞∑j=0
jn
j!= EX n = µn;1 = Tn(1) :=
n∑k=1
S(n, k)
:= Bn,
where Bn is the nth Bell number:Bn = the total number of partitions of [n].
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Let X be a random variable with the distribution Poiss(λ), λ > 0:
P(X = j) = e−λ λj
j! , j = 1, 2, · · · .nth moment: µn;λ := EX n =
∑∞j=0
(e−λ λj
j!
)jn
µn;λ =∑n
k=1 S(n, k)λk
Touchard Polynomial: Tn(x) :=∑n
k=1 S(n, k)xk
µn;λ = Tn(λ)(Compare Tn with (6).)
∞∑j=0
(e−λ
λj
j!
)jn = EX n = µn;λ = Tn(λ) :=
n∑k=1
S(n, k)λk
When λ = 1:
1
e
∞∑j=0
jn
j!= EX n = µn;1 = Tn(1) :=
n∑k=1
S(n, k) := Bn,
where Bn is the nth Bell number:Bn = the total number of partitions of [n].
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Let X be a random variable with the distribution Poiss(λ), λ > 0:
P(X = j) = e−λ λj
j! , j = 1, 2, · · · .nth moment: µn;λ := EX n
µn;λ =∑n
k=1 S(n, k)λk
Touchard Polynomial: Tn(x) :=∑n
k=1 S(n, k)xk
µn;λ = Tn(λ)
∞∑j=0
(e−λ
λj
j!
)jn = EX n = µn;λ = Tn(λ) :=
n∑k=1
S(n, k)λk
When λ = 1:
1
e
∞∑j=0
jn
j!= EX n = µn;λ = Tn(1) :=
n∑k=1
S(n, k) := Bn, (8)
where Bn is the nth Bell number:Bn = the total number of partitions of [n].(8) is known as Dobinski’s formula.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Random Permutations
Uniform probability measure on Sn: Pn(σ) := 1n! , σ ∈ Sn
We now define a family of non-uniform probability measures on Sn:
Let kn(σ) = the number of cycles in σ.
For θ > 0, define the following probability measure on Sn:
P(θ)n (σ) = θkn(σ)
Nn,θ, where Nn,θ =
∑σ∈Sn θ
kn(σ) is the normalizingconstant.
We have Nn,θ =∑
σ∈Sn θkn(σ) =
∑nk=1 θ
k |s(n, k)| (3)= θ(n).
Thus, P(θ)n (σ) = θkn(σ)
θ(n), σ ∈ Sn.
θ > 1: the measure favors permutations with many cyclesθ ∈ (0, 1): the measure favors permutations with few cyclesθ = 1: uniform measure
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Random Permutations
Uniform probability measure on Sn: Pn(σ) := 1n! , σ ∈ Sn
We now define a family of non-uniform probability measures on Sn:
Let kn(σ) = the number of cycles in σ.
For θ > 0, define the following probability measure on Sn:
P(θ)n (σ) = θkn(σ)
Nn,θ, where Nn,θ =
∑σ∈Sn θ
kn(σ) is the normalizingconstant.
We have Nn,θ =∑
σ∈Sn θkn(σ) =
∑nk=1 θ
k |s(n, k)| (3)= θ(n).
Thus, P(θ)n (σ) = θkn(σ)
θ(n), σ ∈ Sn.
θ > 1: the measure favors permutations with many cyclesθ ∈ (0, 1): the measure favors permutations with few cyclesθ = 1: uniform measure
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Random Permutations
Uniform probability measure on Sn: Pn(σ) := 1n! , σ ∈ Sn
We now define a family of non-uniform probability measures on Sn:
Let kn(σ) = the number of cycles in σ.
For θ > 0, define the following probability measure on Sn:
P(θ)n (σ) = θkn(σ)
Nn,θ, where Nn,θ =
∑σ∈Sn θ
kn(σ) is the normalizingconstant.
We have Nn,θ =∑
σ∈Sn θkn(σ) =
∑nk=1 θ
k |s(n, k)| (3)= θ(n).
Thus, P(θ)n (σ) = θkn(σ)
θ(n), σ ∈ Sn.
θ > 1: the measure favors permutations with many cyclesθ ∈ (0, 1): the measure favors permutations with few cyclesθ = 1: uniform measure
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Random Permutations
Uniform probability measure on Sn: Pn(σ) := 1n! , σ ∈ Sn
We now define a family of non-uniform probability measures on Sn:
Let kn(σ) = the number of cycles in σ.
For θ > 0, define the following probability measure on Sn:
P(θ)n (σ) = θkn(σ)
Nn,θ, where Nn,θ =
∑σ∈Sn θ
kn(σ) is the normalizingconstant.
We have Nn,θ =∑
σ∈Sn θkn(σ) =
∑nk=1 θ
k |s(n, k)| (3)= θ(n).
Thus, P(θ)n (σ) = θkn(σ)
θ(n), σ ∈ Sn.
θ > 1: the measure favors permutations with many cyclesθ ∈ (0, 1): the measure favors permutations with few cyclesθ = 1: uniform measure
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Random Permutations
Uniform probability measure on Sn: Pn(σ) := 1n! , σ ∈ Sn
We now define a family of non-uniform probability measures on Sn:
Let kn(σ) = the number of cycles in σ.
For θ > 0, define the following probability measure on Sn:
P(θ)n (σ) = θkn(σ)
Nn,θ, where Nn,θ =
∑σ∈Sn θ
kn(σ) is the normalizingconstant.
We have Nn,θ =∑
σ∈Sn θkn(σ) =
∑nk=1 θ
k |s(n, k)| (3)= θ(n).
Thus, P(θ)n (σ) = θkn(σ)
θ(n), σ ∈ Sn.
θ > 1: the measure favors permutations with many cyclesθ ∈ (0, 1): the measure favors permutations with few cyclesθ = 1: uniform measure
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Random Permutations
Uniform probability measure on Sn: Pn(σ) := 1n! , σ ∈ Sn
We now define a family of non-uniform probability measures on Sn:
Let kn(σ) = the number of cycles in σ.
For θ > 0, define the following probability measure on Sn:
P(θ)n (σ) = θkn(σ)
Nn,θ, where Nn,θ =
∑σ∈Sn θ
kn(σ) is the normalizingconstant.
We have Nn,θ =∑
σ∈Sn θkn(σ) =
∑nk=1 θ
k |s(n, k)|
(3)= θ(n).
Thus, P(θ)n (σ) = θkn(σ)
θ(n), σ ∈ Sn.
θ > 1: the measure favors permutations with many cyclesθ ∈ (0, 1): the measure favors permutations with few cyclesθ = 1: uniform measure
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Random Permutations
Uniform probability measure on Sn: Pn(σ) := 1n! , σ ∈ Sn
We now define a family of non-uniform probability measures on Sn:
Let kn(σ) = the number of cycles in σ.
For θ > 0, define the following probability measure on Sn:
P(θ)n (σ) = θkn(σ)
Nn,θ, where Nn,θ =
∑σ∈Sn θ
kn(σ) is the normalizingconstant.
We have Nn,θ =∑
σ∈Sn θkn(σ) =
∑nk=1 θ
k |s(n, k)| (3)= θ(n).
Thus, P(θ)n (σ) = θkn(σ)
θ(n), σ ∈ Sn.
θ > 1: the measure favors permutations with many cyclesθ ∈ (0, 1): the measure favors permutations with few cyclesθ = 1: uniform measure
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Random Permutations
Uniform probability measure on Sn: Pn(σ) := 1n! , σ ∈ Sn
We now define a family of non-uniform probability measures on Sn:
Let kn(σ) = the number of cycles in σ.
For θ > 0, define the following probability measure on Sn:
P(θ)n (σ) = θkn(σ)
Nn,θ, where Nn,θ =
∑σ∈Sn θ
kn(σ) is the normalizingconstant.
We have Nn,θ =∑
σ∈Sn θkn(σ) =
∑nk=1 θ
k |s(n, k)| (3)= θ(n).
Thus, P(θ)n (σ) = θkn(σ)
θ(n), σ ∈ Sn.
θ > 1: the measure favors permutations with many cyclesθ ∈ (0, 1): the measure favors permutations with few cyclesθ = 1: uniform measure
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Random Permutations
Uniform probability measure on Sn: Pn(σ) := 1n! , σ ∈ Sn
We now define a family of non-uniform probability measures on Sn:
Let kn(σ) = the number of cycles in σ.
For θ > 0, define the following probability measure on Sn:
P(θ)n (σ) = θkn(σ)
Nn,θ, where Nn,θ =
∑σ∈Sn θ
kn(σ) is the normalizingconstant.
We have Nn,θ =∑
σ∈Sn θkn(σ) =
∑nk=1 θ
k |s(n, k)| (3)= θ(n).
Thus, P(θ)n (σ) = θkn(σ)
θ(n), σ ∈ Sn.
θ > 1: the measure favors permutations with many cyclesθ ∈ (0, 1): the measure favors permutations with few cyclesθ = 1: uniform measure
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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P(θ)n (σ) = θkn(σ)
θ(n), σ ∈ Sn.
θ > 1: the measure favors permutations with many cyclesθ ∈ (0, 1): the measure favors permutations with few cyclesθ = 1: uniform measure
Let C(n)j (σ) = the number of j-cycles in σ ∈ Sn, j = 1, · · · , n.
Then for each θ > 0, we can think of {C (n)j }nj=1 as random
variables on the probability space (Sn,P(θ)n ).
Theorem. For any j, the distribution of C(n)j under P
(θ)n converges
weakly to the distribution Poiss( θj ) as n→∞. That is,
limn→∞ P(θ)n (C
(n)j = m) = e−
θj( θj)m
m! , for m = 0, 1, · · · .
Remark. C(n)1 (σ) is the number of fixed points of σ:
C(n)1 = |{k ∈ [n] : σk = k}|. So under P
(θ)n , the distribution of
the number of fixed points in a permutation in Sn converges asn→∞ to the Poisson distribution with parameter θ. Inparticular, under the uniform measure (θ = 1), it converges tothe Poisson distribution with parameter 1.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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P(θ)n (σ) = θkn(σ)
θ(n), σ ∈ Sn.
θ > 1: the measure favors permutations with many cyclesθ ∈ (0, 1): the measure favors permutations with few cyclesθ = 1: uniform measureLet C
(n)j (σ) = the number of j-cycles in σ ∈ Sn, j = 1, · · · , n.
Then for each θ > 0, we can think of {C (n)j }nj=1 as random
variables on the probability space (Sn,P(θ)n ).
Theorem. For any j, the distribution of C(n)j under P
(θ)n converges
weakly to the distribution Poiss( θj ) as n→∞. That is,
limn→∞ P(θ)n (C
(n)j = m) = e−
θj( θj)m
m! , for m = 0, 1, · · · .
Remark. C(n)1 (σ) is the number of fixed points of σ:
C(n)1 = |{k ∈ [n] : σk = k}|. So under P
(θ)n , the distribution of
the number of fixed points in a permutation in Sn converges asn→∞ to the Poisson distribution with parameter θ. Inparticular, under the uniform measure (θ = 1), it converges tothe Poisson distribution with parameter 1.
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P(θ)n (σ) = θkn(σ)
θ(n), σ ∈ Sn.
θ > 1: the measure favors permutations with many cyclesθ ∈ (0, 1): the measure favors permutations with few cyclesθ = 1: uniform measureLet C
(n)j (σ) = the number of j-cycles in σ ∈ Sn, j = 1, · · · , n.
Then for each θ > 0, we can think of {C (n)j }nj=1 as random
variables on the probability space (Sn,P(θ)n ).
Theorem. For any j, the distribution of C(n)j under P
(θ)n converges
weakly to the distribution Poiss( θj ) as n→∞. That is,
limn→∞ P(θ)n (C
(n)j = m) = e−
θj( θj)m
m! , for m = 0, 1, · · · .
Remark. C(n)1 (σ) is the number of fixed points of σ:
C(n)1 = |{k ∈ [n] : σk = k}|. So under P
(θ)n , the distribution of
the number of fixed points in a permutation in Sn converges asn→∞ to the Poisson distribution with parameter θ. Inparticular, under the uniform measure (θ = 1), it converges tothe Poisson distribution with parameter 1.
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P(θ)n (σ) = θkn(σ)
θ(n), σ ∈ Sn.
θ > 1: the measure favors permutations with many cyclesθ ∈ (0, 1): the measure favors permutations with few cyclesθ = 1: uniform measureLet C
(n)j (σ) = the number of j-cycles in σ ∈ Sn, j = 1, · · · , n.
Then for each θ > 0, we can think of {C (n)j }nj=1 as random
variables on the probability space (Sn,P(θ)n ).
Theorem. For any j, the distribution of C(n)j under P
(θ)n converges
weakly to the distribution Poiss( θj ) as n→∞.
That is,
limn→∞ P(θ)n (C
(n)j = m) = e−
θj( θj)m
m! , for m = 0, 1, · · · .
Remark. C(n)1 (σ) is the number of fixed points of σ:
C(n)1 = |{k ∈ [n] : σk = k}|. So under P
(θ)n , the distribution of
the number of fixed points in a permutation in Sn converges asn→∞ to the Poisson distribution with parameter θ. Inparticular, under the uniform measure (θ = 1), it converges tothe Poisson distribution with parameter 1.
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P(θ)n (σ) = θkn(σ)
θ(n), σ ∈ Sn.
θ > 1: the measure favors permutations with many cyclesθ ∈ (0, 1): the measure favors permutations with few cyclesθ = 1: uniform measureLet C
(n)j (σ) = the number of j-cycles in σ ∈ Sn, j = 1, · · · , n.
Then for each θ > 0, we can think of {C (n)j }nj=1 as random
variables on the probability space (Sn,P(θ)n ).
Theorem. For any j, the distribution of C(n)j under P
(θ)n converges
weakly to the distribution Poiss( θj ) as n→∞. That is,
limn→∞ P(θ)n (C
(n)j = m) = e−
θj( θj)m
m! , for m = 0, 1, · · · .
Remark. C(n)1 (σ) is the number of fixed points of σ:
C(n)1 = |{k ∈ [n] : σk = k}|. So under P
(θ)n , the distribution of
the number of fixed points in a permutation in Sn converges asn→∞ to the Poisson distribution with parameter θ. Inparticular, under the uniform measure (θ = 1), it converges tothe Poisson distribution with parameter 1.
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P(θ)n (σ) = θkn(σ)
θ(n), σ ∈ Sn.
θ > 1: the measure favors permutations with many cyclesθ ∈ (0, 1): the measure favors permutations with few cyclesθ = 1: uniform measureLet C
(n)j (σ) = the number of j-cycles in σ ∈ Sn, j = 1, · · · , n.
Then for each θ > 0, we can think of {C (n)j }nj=1 as random
variables on the probability space (Sn,P(θ)n ).
Theorem. For any j, the distribution of C(n)j under P
(θ)n converges
weakly to the distribution Poiss( θj ) as n→∞. That is,
limn→∞ P(θ)n (C
(n)j = m) = e−
θj( θj)m
m! , for m = 0, 1, · · · .
Remark. C(n)1 (σ) is the number of fixed points of σ:
C(n)1 = |{k ∈ [n] : σk = k}|. So under P
(θ)n , the distribution of
the number of fixed points in a permutation in Sn converges asn→∞ to the Poisson distribution with parameter θ. Inparticular, under the uniform measure (θ = 1), it converges tothe Poisson distribution with parameter 1.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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Theorem. For any j, the distribution of C(n)j under P
(θ)n converges
weakly to the distribution Poiss( θj ) as n→∞. That is,
limn→∞ P(θ)n (C
(n)j = m) = e−
θm
( θj)m
m! , for m = 0, 1, · · · .
Proof for θ = 1 (the same method works with virtually no extrawork for all θ).
Method of moments: It is enough to show that for all m, the
mth moment of C(n)j converges to the mth moment of the
distribution Poiss(1j ).
That is, it is enough to show that
limn→∞
E(1)n
(C
(n)j
)m= µm; 1
j= Tm
(1
j
).
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Theorem. For any j, the distribution of C(n)j under P
(θ)n converges
weakly to the distribution Poiss( θj ) as n→∞. That is,
limn→∞ P(θ)n (C
(n)j = m) = e−
θm
( θj)m
m! , for m = 0, 1, · · · .
Proof for θ = 1 (the same method works with virtually no extrawork for all θ).
Method of moments: It is enough to show that for all m, the
mth moment of C(n)j converges to the mth moment of the
distribution Poiss(1j ).
That is, it is enough to show that
limn→∞
E(1)n
(C
(n)j
)m= µm; 1
j= Tm
(1
j
).
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Theorem. For any j, the distribution of C(n)j under P
(θ)n converges
weakly to the distribution Poiss( θj ) as n→∞. That is,
limn→∞ P(θ)n (C
(n)j = m) = e−
θm
( θj)m
m! , for m = 0, 1, · · · .
Proof for θ = 1 (the same method works with virtually no extrawork for all θ).
Method of moments: It is enough to show that for all m, the
mth moment of C(n)j converges to the mth moment of the
distribution Poiss(1j ).
That is, it is enough to show that
limn→∞
E(1)n
(C
(n)j
)m= µm; 1
j= Tm
(1
j
).
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Proof that limn→∞ E(1)n
(C
(n)j
)m= µm; 1
j= Tm
(1j
).
We will show that
E(1)n
(C
(n)j
)m= µm; 1
j= Tm
(1
j
), for n ≥ mj .
For D ⊂ [n] with |D| = j , define
1D(σ) =
{1, if σ has a cycle consisting of the elements of D;
0, otherwise.
Then C(n)j (σ) =
∑D⊂[n]:|D|=j 1D(σ) and(
C(n)j (σ)
)m=( ∑
D1⊂[n]:|D1|=j
1D1(σ))· · ·( ∑
Dm⊂[n]:|Dm|=j
1Dm(σ))
=
∑(D1,D2,··· ,Dm
):|Dl |=···=|Dm|=j
m∏l=1
1Dl(σ).
SoE(1)n
(C
(n)j
)m=∑(
D1,D2,··· ,Dm
):|Dl |=···=|Dm|=j
P(1)n
(∏ml=1 1Dl
= 1).
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Proof that limn→∞ E(1)n
(C
(n)j
)m= µm; 1
j= Tm
(1j
).
We will show that
E(1)n
(C
(n)j
)m= µm; 1
j= Tm
(1
j
), for n ≥ mj .
For D ⊂ [n] with |D| = j , define
1D(σ) =
{1, if σ has a cycle consisting of the elements of D;
0, otherwise.
Then C(n)j (σ) =
∑D⊂[n]:|D|=j 1D(σ) and(
C(n)j (σ)
)m=( ∑
D1⊂[n]:|D1|=j
1D1(σ))· · ·( ∑
Dm⊂[n]:|Dm|=j
1Dm(σ))
=
∑(D1,D2,··· ,Dm
):|Dl |=···=|Dm|=j
m∏l=1
1Dl(σ).
SoE(1)n
(C
(n)j
)m=∑(
D1,D2,··· ,Dm
):|Dl |=···=|Dm|=j
P(1)n
(∏ml=1 1Dl
= 1).
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Proof that limn→∞ E(1)n
(C
(n)j
)m= µm; 1
j= Tm
(1j
).
We will show that
E(1)n
(C
(n)j
)m= µm; 1
j= Tm
(1
j
), for n ≥ mj .
For D ⊂ [n] with |D| = j , define
1D(σ) =
{1, if σ has a cycle consisting of the elements of D;
0, otherwise.
Then C(n)j (σ) =
∑D⊂[n]:|D|=j 1D(σ) and(
C(n)j (σ)
)m=( ∑
D1⊂[n]:|D1|=j
1D1(σ))· · ·( ∑
Dm⊂[n]:|Dm|=j
1Dm(σ))
=
∑(D1,D2,··· ,Dm
):|Dl |=···=|Dm|=j
m∏l=1
1Dl(σ).
SoE(1)n
(C
(n)j
)m=∑(
D1,D2,··· ,Dm
):|Dl |=···=|Dm|=j
P(1)n
(∏ml=1 1Dl
= 1).
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Proof that limn→∞ E(1)n
(C
(n)j
)m= µm; 1
j= Tm
(1j
).
We will show that
E(1)n
(C
(n)j
)m= µm; 1
j= Tm
(1
j
), for n ≥ mj .
For D ⊂ [n] with |D| = j , define
1D(σ) =
{1, if σ has a cycle consisting of the elements of D;
0, otherwise.
Then C(n)j (σ) =
∑D⊂[n]:|D|=j 1D(σ)
and(C
(n)j (σ)
)m=( ∑
D1⊂[n]:|D1|=j
1D1(σ))· · ·( ∑
Dm⊂[n]:|Dm|=j
1Dm(σ))
=
∑(D1,D2,··· ,Dm
):|Dl |=···=|Dm|=j
m∏l=1
1Dl(σ).
SoE(1)n
(C
(n)j
)m=∑(
D1,D2,··· ,Dm
):|Dl |=···=|Dm|=j
P(1)n
(∏ml=1 1Dl
= 1).
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Proof that limn→∞ E(1)n
(C
(n)j
)m= µm; 1
j= Tm
(1j
).
We will show that
E(1)n
(C
(n)j
)m= µm; 1
j= Tm
(1
j
), for n ≥ mj .
For D ⊂ [n] with |D| = j , define
1D(σ) =
{1, if σ has a cycle consisting of the elements of D;
0, otherwise.
Then C(n)j (σ) =
∑D⊂[n]:|D|=j 1D(σ) and(
C(n)j (σ)
)m=( ∑
D1⊂[n]:|D1|=j
1D1(σ))· · ·( ∑
Dm⊂[n]:|Dm|=j
1Dm(σ))
=
∑(D1,D2,··· ,Dm
):|Dl |=···=|Dm|=j
m∏l=1
1Dl(σ).
SoE(1)n
(C
(n)j
)m=∑(
D1,D2,··· ,Dm
):|Dl |=···=|Dm|=j
P(1)n
(∏ml=1 1Dl
= 1).
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Proof that limn→∞ E(1)n
(C
(n)j
)m= µm; 1
j= Tm
(1j
).
We will show that
E(1)n
(C
(n)j
)m= µm; 1
j= Tm
(1
j
), for n ≥ mj .
For D ⊂ [n] with |D| = j , define
1D(σ) =
{1, if σ has a cycle consisting of the elements of D;
0, otherwise.
Then C(n)j (σ) =
∑D⊂[n]:|D|=j 1D(σ) and(
C(n)j (σ)
)m=( ∑
D1⊂[n]:|D1|=j
1D1(σ))· · ·( ∑
Dm⊂[n]:|Dm|=j
1Dm(σ))
=
∑(D1,D2,··· ,Dm
):|Dl |=···=|Dm|=j
m∏l=1
1Dl(σ).
SoE(1)n
(C
(n)j
)m=∑(
D1,D2,··· ,Dm
):|Dl |=···=|Dm|=j
P(1)n
(∏ml=1 1Dl
= 1).
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E(1)n
(C
(n)j
)m=
∑(D1,D2,··· ,Dm
):|Dl |=···=|Dm|=j
P(1)n
( m∏l=1
1Dl= 1).
Now∏m
l=1 1Dl6≡ 0, (that is, there exists some σ ∈ Dn such that∏m
l=1 1Dl(σ) 6= 0),
if and only if for each pair i1, i2 ∈ [n], either Di1 = Di2 orDi1 ∩ Di2 = ∅.
That is, if and only if there exist a k ∈ [m] and disjoint sets{Ai}ki=1 such that {Dl}ml=1 = {Ai}ki=1.
In this case,∏m
l=1 1Dl=∏k
i=1 1Ai, and
P(1)n
(∏ml=1 1Dl
= 1)
= P(1)n
(∏ki=1 1Ai
= 1)
=
((j−1)!
)k(n−kj)!
n! .
(Here we use the assumption that n ≥ mj , which insures thatn ≥ kj .)
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E(1)n
(C
(n)j
)m=
∑(D1,D2,··· ,Dm
):|Dl |=···=|Dm|=j
P(1)n
( m∏l=1
1Dl= 1).
Now∏m
l=1 1Dl6≡ 0,
(that is, there exists some σ ∈ Dn such that∏ml=1 1Dl
(σ) 6= 0),if and only if for each pair i1, i2 ∈ [n], either Di1 = Di2 orDi1 ∩ Di2 = ∅.
That is, if and only if there exist a k ∈ [m] and disjoint sets{Ai}ki=1 such that {Dl}ml=1 = {Ai}ki=1.
In this case,∏m
l=1 1Dl=∏k
i=1 1Ai, and
P(1)n
(∏ml=1 1Dl
= 1)
= P(1)n
(∏ki=1 1Ai
= 1)
=
((j−1)!
)k(n−kj)!
n! .
(Here we use the assumption that n ≥ mj , which insures thatn ≥ kj .)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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E(1)n
(C
(n)j
)m=
∑(D1,D2,··· ,Dm
):|Dl |=···=|Dm|=j
P(1)n
( m∏l=1
1Dl= 1).
Now∏m
l=1 1Dl6≡ 0, (that is, there exists some σ ∈ Dn such that∏m
l=1 1Dl(σ) 6= 0),
if and only if for each pair i1, i2 ∈ [n], either Di1 = Di2 orDi1 ∩ Di2 = ∅.
That is, if and only if there exist a k ∈ [m] and disjoint sets{Ai}ki=1 such that {Dl}ml=1 = {Ai}ki=1.
In this case,∏m
l=1 1Dl=∏k
i=1 1Ai, and
P(1)n
(∏ml=1 1Dl
= 1)
= P(1)n
(∏ki=1 1Ai
= 1)
=
((j−1)!
)k(n−kj)!
n! .
(Here we use the assumption that n ≥ mj , which insures thatn ≥ kj .)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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E(1)n
(C
(n)j
)m=
∑(D1,D2,··· ,Dm
):|Dl |=···=|Dm|=j
P(1)n
( m∏l=1
1Dl= 1).
Now∏m
l=1 1Dl6≡ 0, (that is, there exists some σ ∈ Dn such that∏m
l=1 1Dl(σ) 6= 0),
if and only if for each pair i1, i2 ∈ [n], either Di1 = Di2 orDi1 ∩ Di2 = ∅.
That is, if and only if there exist a k ∈ [m] and disjoint sets{Ai}ki=1 such that {Dl}ml=1 = {Ai}ki=1.
In this case,∏m
l=1 1Dl=∏k
i=1 1Ai, and
P(1)n
(∏ml=1 1Dl
= 1)
= P(1)n
(∏ki=1 1Ai
= 1)
=
((j−1)!
)k(n−kj)!
n! .
(Here we use the assumption that n ≥ mj , which insures thatn ≥ kj .)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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E(1)n
(C
(n)j
)m=
∑(D1,D2,··· ,Dm
):|Dl |=···=|Dm|=j
P(1)n
( m∏l=1
1Dl= 1).
Now∏m
l=1 1Dl6≡ 0, (that is, there exists some σ ∈ Dn such that∏m
l=1 1Dl(σ) 6= 0),
if and only if for each pair i1, i2 ∈ [n], either Di1 = Di2 orDi1 ∩ Di2 = ∅.
That is, if and only if there exist a k ∈ [m] and disjoint sets{Ai}ki=1 such that {Dl}ml=1 = {Ai}ki=1.
In this case,∏m
l=1 1Dl=∏k
i=1 1Ai, and
P(1)n
(∏ml=1 1Dl
= 1)
= P(1)n
(∏ki=1 1Ai
= 1)
=
((j−1)!
)k(n−kj)!
n! .
(Here we use the assumption that n ≥ mj , which insures thatn ≥ kj .)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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E(1)n
(C
(n)j
)m=
∑(D1,D2,··· ,Dm
):|Dl |=···=|Dm|=j
P(1)n
( m∏l=1
1Dl= 1).
Now∏m
l=1 1Dl6≡ 0, (that is, there exists some σ ∈ Dn such that∏m
l=1 1Dl(σ) 6= 0),
if and only if for each pair i1, i2 ∈ [n], either Di1 = Di2 orDi1 ∩ Di2 = ∅.
That is, if and only if there exist a k ∈ [m] and disjoint sets{Ai}ki=1 such that {Dl}ml=1 = {Ai}ki=1.
In this case,∏m
l=1 1Dl=∏k
i=1 1Ai,
and
P(1)n
(∏ml=1 1Dl
= 1)
= P(1)n
(∏ki=1 1Ai
= 1)
=
((j−1)!
)k(n−kj)!
n! .
(Here we use the assumption that n ≥ mj , which insures thatn ≥ kj .)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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E(1)n
(C
(n)j
)m=
∑(D1,D2,··· ,Dm
):|Dl |=···=|Dm|=j
P(1)n
( m∏l=1
1Dl= 1).
Now∏m
l=1 1Dl6≡ 0, (that is, there exists some σ ∈ Dn such that∏m
l=1 1Dl(σ) 6= 0),
if and only if for each pair i1, i2 ∈ [n], either Di1 = Di2 orDi1 ∩ Di2 = ∅.
That is, if and only if there exist a k ∈ [m] and disjoint sets{Ai}ki=1 such that {Dl}ml=1 = {Ai}ki=1.
In this case,∏m
l=1 1Dl=∏k
i=1 1Ai, and
P(1)n
(∏ml=1 1Dl
= 1)
= P(1)n
(∏ki=1 1Ai
= 1)
=
((j−1)!
)k(n−kj)!
n! .
(Here we use the assumption that n ≥ mj , which insures thatn ≥ kj .)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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E(1)n
(C
(n)j
)m=
∑(D1,D2,··· ,Dm
):|Dl |=···=|Dm|=j
P(1)n
( m∏l=1
1Dl= 1).
Now∏m
l=1 1Dl6≡ 0, (that is, there exists some σ ∈ Dn such that∏m
l=1 1Dl(σ) 6= 0),
if and only if for each pair i1, i2 ∈ [n], either Di1 = Di2 orDi1 ∩ Di2 = ∅.
That is, if and only if there exist a k ∈ [m] and disjoint sets{Ai}ki=1 such that {Dl}ml=1 = {Ai}ki=1.
In this case,∏m
l=1 1Dl=∏k
i=1 1Ai, and
P(1)n
(∏ml=1 1Dl
= 1)
= P(1)n
(∏ki=1 1Ai
= 1)
=
((j−1)!
)k(n−kj)!
n! .
(Here we use the assumption that n ≥ mj , which insures thatn ≥ kj .)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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E(1)n
(C
(n)j
)m=∑(
D1,D2,··· ,Dm
):|Dl |=···=|Dm|=j
P(1)n
(∏ml=1 1Dl
= 1).
Now∏m
l=1 1Dl6≡ 0 if and only if there exist a k ∈ [m] and disjoint
sets {Ai}ki=1 such that {Dl}ml=1 = {Ai}ki=1. In this case,∏ml=1 1Dl
=∏k
i=1 1Ai, and
P(1)n
(∏ml=1 1Dl
= 1)
= P(1)n
(∏ki=1 1Ai
= 1)
=
((j−1)!
)k(n−kj)!
n! .
The number of ways to construct k disjoint (ordered) sets(A1, · · · ,Ak), each with j elements from [n], is(nj
)(n−jj
)· · ·(n−(m−1)j
j
)= n!
(j!)k (n−jk)! .
Given {Ai}ki=1, the number of ways to choose {Dl}ml=1 so that{Dl}ml=1 = {Ai}ki=1 is equal to S(m, k). So
E(1)n
(C
(n)j
)m=
m∑k=1
((j − 1)!
)k(n − kj)!
n!× n!
(j!)k(n − jk)!× S(m, k) =
m∑k=1
(1
j
)kS(m, k) = Tm(
1
j) = µm; 1
j.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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E(1)n
(C
(n)j
)m=∑(
D1,D2,··· ,Dm
):|Dl |=···=|Dm|=j
P(1)n
(∏ml=1 1Dl
= 1).
Now∏m
l=1 1Dl6≡ 0 if and only if there exist a k ∈ [m] and disjoint
sets {Ai}ki=1 such that {Dl}ml=1 = {Ai}ki=1. In this case,∏ml=1 1Dl
=∏k
i=1 1Ai, and
P(1)n
(∏ml=1 1Dl
= 1)
= P(1)n
(∏ki=1 1Ai
= 1)
=
((j−1)!
)k(n−kj)!
n! .The number of ways to construct k disjoint (ordered) sets(A1, · · · ,Ak), each with j elements from [n], is(nj
)(n−jj
)· · ·(n−(m−1)j
j
)= n!
(j!)k (n−jk)! .
Given {Ai}ki=1, the number of ways to choose {Dl}ml=1 so that{Dl}ml=1 = {Ai}ki=1 is equal to S(m, k). So
E(1)n
(C
(n)j
)m=
m∑k=1
((j − 1)!
)k(n − kj)!
n!× n!
(j!)k(n − jk)!× S(m, k) =
m∑k=1
(1
j
)kS(m, k) = Tm(
1
j) = µm; 1
j.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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E(1)n
(C
(n)j
)m=∑(
D1,D2,··· ,Dm
):|Dl |=···=|Dm|=j
P(1)n
(∏ml=1 1Dl
= 1).
Now∏m
l=1 1Dl6≡ 0 if and only if there exist a k ∈ [m] and disjoint
sets {Ai}ki=1 such that {Dl}ml=1 = {Ai}ki=1. In this case,∏ml=1 1Dl
=∏k
i=1 1Ai, and
P(1)n
(∏ml=1 1Dl
= 1)
= P(1)n
(∏ki=1 1Ai
= 1)
=
((j−1)!
)k(n−kj)!
n! .The number of ways to construct k disjoint (ordered) sets(A1, · · · ,Ak), each with j elements from [n], is(nj
)(n−jj
)· · ·(n−(m−1)j
j
)= n!
(j!)k (n−jk)! .
Given {Ai}ki=1, the number of ways to choose {Dl}ml=1 so that{Dl}ml=1 = {Ai}ki=1 is equal to S(m, k).
So
E(1)n
(C
(n)j
)m=
m∑k=1
((j − 1)!
)k(n − kj)!
n!× n!
(j!)k(n − jk)!× S(m, k) =
m∑k=1
(1
j
)kS(m, k) = Tm(
1
j) = µm; 1
j.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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E(1)n
(C
(n)j
)m=∑(
D1,D2,··· ,Dm
):|Dl |=···=|Dm|=j
P(1)n
(∏ml=1 1Dl
= 1).
Now∏m
l=1 1Dl6≡ 0 if and only if there exist a k ∈ [m] and disjoint
sets {Ai}ki=1 such that {Dl}ml=1 = {Ai}ki=1. In this case,∏ml=1 1Dl
=∏k
i=1 1Ai, and
P(1)n
(∏ml=1 1Dl
= 1)
= P(1)n
(∏ki=1 1Ai
= 1)
=
((j−1)!
)k(n−kj)!
n! .The number of ways to construct k disjoint (ordered) sets(A1, · · · ,Ak), each with j elements from [n], is(nj
)(n−jj
)· · ·(n−(m−1)j
j
)= n!
(j!)k (n−jk)! .
Given {Ai}ki=1, the number of ways to choose {Dl}ml=1 so that{Dl}ml=1 = {Ai}ki=1 is equal to S(m, k). So
E(1)n
(C
(n)j
)m=
m∑k=1
((j − 1)!
)k(n − kj)!
n!× n!
(j!)k(n − jk)!× S(m, k)
=
m∑k=1
(1
j
)kS(m, k) = Tm(
1
j) = µm; 1
j.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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E(1)n
(C
(n)j
)m=∑(
D1,D2,··· ,Dm
):|Dl |=···=|Dm|=j
P(1)n
(∏ml=1 1Dl
= 1).
Now∏m
l=1 1Dl6≡ 0 if and only if there exist a k ∈ [m] and disjoint
sets {Ai}ki=1 such that {Dl}ml=1 = {Ai}ki=1. In this case,∏ml=1 1Dl
=∏k
i=1 1Ai, and
P(1)n
(∏ml=1 1Dl
= 1)
= P(1)n
(∏ki=1 1Ai
= 1)
=
((j−1)!
)k(n−kj)!
n! .The number of ways to construct k disjoint (ordered) sets(A1, · · · ,Ak), each with j elements from [n], is(nj
)(n−jj
)· · ·(n−(m−1)j
j
)= n!
(j!)k (n−jk)! .
Given {Ai}ki=1, the number of ways to choose {Dl}ml=1 so that{Dl}ml=1 = {Ai}ki=1 is equal to S(m, k). So
E(1)n
(C
(n)j
)m=
m∑k=1
((j − 1)!
)k(n − kj)!
n!× n!
(j!)k(n − jk)!× S(m, k) =
m∑k=1
(1
j
)kS(m, k)
= Tm(1
j) = µm; 1
j.
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E(1)n
(C
(n)j
)m=∑(
D1,D2,··· ,Dm
):|Dl |=···=|Dm|=j
P(1)n
(∏ml=1 1Dl
= 1).
Now∏m
l=1 1Dl6≡ 0 if and only if there exist a k ∈ [m] and disjoint
sets {Ai}ki=1 such that {Dl}ml=1 = {Ai}ki=1. In this case,∏ml=1 1Dl
=∏k
i=1 1Ai, and
P(1)n
(∏ml=1 1Dl
= 1)
= P(1)n
(∏ki=1 1Ai
= 1)
=
((j−1)!
)k(n−kj)!
n! .The number of ways to construct k disjoint (ordered) sets(A1, · · · ,Ak), each with j elements from [n], is(nj
)(n−jj
)· · ·(n−(m−1)j
j
)= n!
(j!)k (n−jk)! .
Given {Ai}ki=1, the number of ways to choose {Dl}ml=1 so that{Dl}ml=1 = {Ai}ki=1 is equal to S(m, k). So
E(1)n
(C
(n)j
)m=
m∑k=1
((j − 1)!
)k(n − kj)!
n!× n!
(j!)k(n − jk)!× S(m, k) =
m∑k=1
(1
j
)kS(m, k) = Tm(
1
j)
= µm; 1j.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations
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E(1)n
(C
(n)j
)m=∑(
D1,D2,··· ,Dm
):|Dl |=···=|Dm|=j
P(1)n
(∏ml=1 1Dl
= 1).
Now∏m
l=1 1Dl6≡ 0 if and only if there exist a k ∈ [m] and disjoint
sets {Ai}ki=1 such that {Dl}ml=1 = {Ai}ki=1. In this case,∏ml=1 1Dl
=∏k
i=1 1Ai, and
P(1)n
(∏ml=1 1Dl
= 1)
= P(1)n
(∏ki=1 1Ai
= 1)
=
((j−1)!
)k(n−kj)!
n! .The number of ways to construct k disjoint (ordered) sets(A1, · · · ,Ak), each with j elements from [n], is(nj
)(n−jj
)· · ·(n−(m−1)j
j
)= n!
(j!)k (n−jk)! .
Given {Ai}ki=1, the number of ways to choose {Dl}ml=1 so that{Dl}ml=1 = {Ai}ki=1 is equal to S(m, k). So
E(1)n
(C
(n)j
)m=
m∑k=1
((j − 1)!
)k(n − kj)!
n!× n!
(j!)k(n − jk)!× S(m, k) =
m∑k=1
(1
j
)kS(m, k) = Tm(
1
j) = µm; 1
j.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations