Touchard Polynomials, Stirling Numbers and Random Permutations
Ross Pinsky
Department of Mathematics, Technion 32000 Haifa, ISRAEL
September 3, 2018
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations |Sn| = n! 1 2 3 4 5 6 S 3 σ = = (124)(36)(5), 6 1 2 4 6 1 5 3 1 2 3 4 5 6 S 3 σ = = (136245) = (624513) 6 2 3 4 6 5 1 2
σ2 is a 6-cycle in S6. There are 5! different 6-cycles in S6.
There are (n − 1)! different n-cycles in Sn.
Notation: [n] = {1, 2, ··· , n}
Sn = permutations of [n]
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations 1 2 3 4 5 6 S 3 σ = = (124)(36)(5), 6 1 2 4 6 1 5 3 1 2 3 4 5 6 S 3 σ = = (136245) = (624513) 6 2 3 4 6 5 1 2
σ2 is a 6-cycle in S6. There are 5! different 6-cycles in S6.
There are (n − 1)! different n-cycles in Sn.
Notation: [n] = {1, 2, ··· , n}
Sn = permutations of [n]
|Sn| = n!
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations 1 2 3 4 5 6 S 3 σ = = (136245) = (624513) 6 2 3 4 6 5 1 2
σ2 is a 6-cycle in S6. There are 5! different 6-cycles in S6.
There are (n − 1)! different n-cycles in Sn.
Notation: [n] = {1, 2, ··· , n}
Sn = permutations of [n]
|Sn| = n! 1 2 3 4 5 6 S 3 σ = = (124)(36)(5), 6 1 2 4 6 1 5 3
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations σ2 is a 6-cycle in S6. There are 5! different 6-cycles in S6.
There are (n − 1)! different n-cycles in Sn.
Notation: [n] = {1, 2, ··· , n}
Sn = permutations of [n]
|Sn| = n! 1 2 3 4 5 6 S 3 σ = = (124)(36)(5), 6 1 2 4 6 1 5 3 1 2 3 4 5 6 S 3 σ = = (136245) = (624513) 6 2 3 4 6 5 1 2
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations There are (n − 1)! different n-cycles in Sn.
Notation: [n] = {1, 2, ··· , n}
Sn = permutations of [n]
|Sn| = n! 1 2 3 4 5 6 S 3 σ = = (124)(36)(5), 6 1 2 4 6 1 5 3 1 2 3 4 5 6 S 3 σ = = (136245) = (624513) 6 2 3 4 6 5 1 2
σ2 is a 6-cycle in S6. There are 5! different 6-cycles in S6.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Notation: [n] = {1, 2, ··· , n}
Sn = permutations of [n]
|Sn| = n! 1 2 3 4 5 6 S 3 σ = = (124)(36)(5), 6 1 2 4 6 1 5 3 1 2 3 4 5 6 S 3 σ = = (136245) = (624513) 6 2 3 4 6 5 1 2
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 n X k ≡ ank x , x ∈ R, n ≥ 1. (1) k=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; (2) |s(n + 1, k)| = n|s(n, k)| + |s(n, k − 1)|, 2 ≤ k ≤ n.
It is easy to check that the coefficients {ank } in (1) also satisfy (2); thus, ank = |s(n, k)|. Therefore n X x(n) = |s(n, k)|xk . (3) k=1
Rising Factorials x(n) := x(x +1) ··· (x +n−1)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations 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; (2) |s(n + 1, k)| = n|s(n, k)| + |s(n, k − 1)|, 2 ≤ k ≤ n.
It is easy to check that the coefficients {ank } in (1) also satisfy (2); thus, ank = |s(n, k)|. Therefore n X x(n) = |s(n, k)|xk . (3) k=1
Rising Factorials
n (n) X k x := x(x +1) ··· (x +n−1) ≡ ank x , x ∈ R, n ≥ 1. (1) k=1
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Recurrence relation |s(n, n)| = 1, |s(n, 1)| = (n − 1)!, n ≥ 1; (2) |s(n + 1, k)| = n|s(n, k)| + |s(n, k − 1)|, 2 ≤ k ≤ n.
It is easy to check that the coefficients {ank } in (1) also satisfy (2); thus, ank = |s(n, k)|. Therefore n X x(n) = |s(n, k)|xk . (3) k=1
Rising Factorials
n (n) X k x := x(x +1) ··· (x +n−1) ≡ ank x , x ∈ R, n ≥ 1. (1) k=1 Unsigned Stirling Numbers of the First Kind: |s(n, k)| := the number of permutations in Sn with exactly k cycles, 1 ≤ k ≤ n
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations It is easy to check that the coefficients {ank } in (1) also satisfy (2); thus, ank = |s(n, k)|. Therefore n X x(n) = |s(n, k)|xk . (3) k=1
Rising Factorials
n (n) X k x := x(x +1) ··· (x +n−1) ≡ ank x , x ∈ R, n ≥ 1. (1) k=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; (2) |s(n + 1, k)| = n|s(n, k)| + |s(n, k − 1)|, 2 ≤ k ≤ n.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Therefore
n X x(n) = |s(n, k)|xk . (3) k=1
Rising Factorials
n (n) X k x := x(x +1) ··· (x +n−1) ≡ ank x , x ∈ R, n ≥ 1. (1) k=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; (2) |s(n + 1, k)| = n|s(n, k)| + |s(n, k − 1)|, 2 ≤ k ≤ n.
It is easy to check that the coefficients {ank } in (1) also satisfy (2); thus, ank = |s(n, k)|.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Rising Factorials
n (n) X k x := x(x +1) ··· (x +n−1) ≡ ank x , x ∈ R, n ≥ 1. (1) k=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; (2) |s(n + 1, k)| = n|s(n, k)| + |s(n, k − 1)|, 2 ≤ k ≤ n.
It is easy to check that the coefficients {ank } in (1) also satisfy (2); thus, ank = |s(n, k)|. Therefore n X x(n) = |s(n, k)|xk . (3) k=1
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Substituting −x for x in (4), we have n (−x)n = −x(−x −1) ··· (−x −n+1) = (−1) x(x +1) ··· (x +n−1) = (−1)nx(n). Substituting above −x for x, we have n n (n) from (3) n X k (x)n = (−1) (−x) = (−1) |s(n, k)|(−x) = k=1 n (5) X (−1)n−k |s(n, k)|xk . k=1 Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called the Stirling Numbers of the First Kind. From (5) we have n X k (x)n = s(n, k)x . (6) k=1
Falling Factorials
(x)n := x(x − 1) ··· (x − n + 1), x ∈ R, n ≥ 1. (4)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations (−1)nx(x +1) ··· (x +n−1) = (−1)nx(n). Substituting above −x for x, we have n n (n) from (3) n X k (x)n = (−1) (−x) = (−1) |s(n, k)|(−x) = k=1 n (5) X (−1)n−k |s(n, k)|xk . k=1 Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called the Stirling Numbers of the First Kind. From (5) we have n X k (x)n = s(n, k)x . (6) k=1
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) =
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations = (−1)nx(n). Substituting above −x for x, we have n n (n) from (3) n X k (x)n = (−1) (−x) = (−1) |s(n, k)|(−x) = k=1 n (5) X (−1)n−k |s(n, k)|xk . k=1 Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called the Stirling Numbers of the First Kind. From (5) we have n X k (x)n = s(n, k)x . (6) k=1
Falling Factorials
(x)n := x(x − 1) ··· (x − n + 1), x ∈ R, n ≥ 1. (4) Substituting −x for x in (4), we have n (−x)n = −x(−x −1) ··· (−x −n+1) = (−1) x(x +1) ··· (x +n−1)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Substituting above −x for x, we have n n (n) from (3) n X k (x)n = (−1) (−x) = (−1) |s(n, k)|(−x) = k=1 n (5) X (−1)n−k |s(n, k)|xk . k=1 Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called the Stirling Numbers of the First Kind. From (5) we have n X k (x)n = s(n, k)x . (6) k=1
Falling Factorials
(x)n := x(x − 1) ··· (x − n + 1), x ∈ R, n ≥ 1. (4) Substituting −x for x in (4), we have n (−x)n = −x(−x −1) ··· (−x −n+1) = (−1) x(x +1) ··· (x +n−1) = (−1)nx(n).
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations =
n (5) X (−1)n−k |s(n, k)|xk . k=1 Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called the Stirling Numbers of the First Kind. From (5) we have n X k (x)n = s(n, k)x . (6) k=1
Falling Factorials
(x)n := x(x − 1) ··· (x − n + 1), x ∈ R, n ≥ 1. (4) Substituting −x for x in (4), we have n (−x)n = −x(−x −1) ··· (−x −n+1) = (−1) x(x +1) ··· (x +n−1) = (−1)nx(n). Substituting above −x for x, we have n n (n) from (3) n X k (x)n = (−1) (−x) = (−1) |s(n, k)|(−x) k=1
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called the Stirling Numbers of the First Kind. From (5) we have n X k (x)n = s(n, k)x . (6) k=1
Falling Factorials
(x)n := x(x − 1) ··· (x − n + 1), x ∈ R, n ≥ 1. (4) Substituting −x for x in (4), we have n (−x)n = −x(−x −1) ··· (−x −n+1) = (−1) x(x +1) ··· (x +n−1) = (−1)nx(n). Substituting above −x for x, we have n n (n) from (3) n X k (x)n = (−1) (−x) = (−1) |s(n, k)|(−x) = k=1 n (5) X (−1)n−k |s(n, k)|xk . k=1
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations From (5) we have n X k (x)n = s(n, k)x . (6) k=1
Falling Factorials
(x)n := x(x − 1) ··· (x − n + 1), x ∈ R, n ≥ 1. (4) Substituting −x for x in (4), we have n (−x)n = −x(−x −1) ··· (−x −n+1) = (−1) x(x +1) ··· (x +n−1) = (−1)nx(n). Substituting above −x for x, we have n n (n) from (3) n X k (x)n = (−1) (−x) = (−1) |s(n, k)|(−x) = k=1 n (5) X (−1)n−k |s(n, k)|xk . k=1 Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called the Stirling Numbers of the First Kind.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Falling Factorials
(x)n := x(x − 1) ··· (x − n + 1), x ∈ R, n ≥ 1. (4) Substituting −x for x in (4), we have n (−x)n = −x(−x −1) ··· (−x −n+1) = (−1) x(x +1) ··· (x +n−1) = (−1)nx(n). Substituting above −x for x, we have n n (n) from (3) n X k (x)n = (−1) (−x) = (−1) |s(n, k)|(−x) = k=1 n (5) X (−1)n−k |s(n, k)|xk . k=1 Define s(n, k) := (−1)n−k |s(n, k)|. The s(n, k) are called the Stirling Numbers of the First Kind. From (5) we have n X k (x)n = s(n, k)x . (6) k=1
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations 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}
Stirling Numbers of the Second Kind S(n, k) := number of ways to partition the set [n] into k nonempty sets.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations {1, 2}, {3}, {4} {1, 3}, {2}, {4} {1, 4}, {2}, {3} {2, 3}, {1}, {4} {2, 4}, {1}, {3}
Stirling Numbers of the Second Kind S(n, k) := number of ways to partition the set [n] into k nonempty sets. Example: S(4, 3) = 5:
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Stirling Numbers of the Second Kind S(n, k) := number of ways to partition the set [n] into k nonempty sets. 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 Theorem.
n n X x = S(n, k)(x)k , (7) k=1
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 functions with |Im(f )| = k. (If x < n, then ck = 0 for x < k ≤ n.) Pn So the answer by this indirect count is k=1 ck . n Pn Thus, x = k=1 ck . To complete the proof we now show that ck = S(n, k)(x)k .
Stirling Numbers of the Second Kind S(n, k) := number of ways to partition the set [n] into k nonempty sets.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations 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 functions with |Im(f )| = k. (If x < n, then ck = 0 for x < k ≤ n.) Pn So the answer by this indirect count is k=1 ck . n Pn Thus, x = k=1 ck . To complete the proof we now show that ck = S(n, k)(x)k .
Stirling Numbers of the Second Kind S(n, k) := number of ways to partition the set [n] into k nonempty sets. Theorem.
n n X x = S(n, k)(x)k , (7) k=1
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations 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 functions with |Im(f )| = k. (If x < n, then ck = 0 for x < k ≤ n.) Pn So the answer by this indirect count is k=1 ck . n Pn Thus, x = k=1 ck . To complete the proof we now show that ck = S(n, k)(x)k .
Stirling Numbers of the Second Kind S(n, k) := number of ways to partition the set [n] into k nonempty sets. Theorem.
n n X x = S(n, k)(x)k , (7) k=1
Proof. Consider functions f :[n] → X , where |X | = x ∈ N.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations The answer by direct count: xn. An alternative indirect count: For k = 1, 2, ··· , n, let ck denote the number of such functions with |Im(f )| = k. (If x < n, then ck = 0 for x < k ≤ n.) Pn So the answer by this indirect count is k=1 ck . n Pn Thus, x = k=1 ck . To complete the proof we now show that ck = S(n, k)(x)k .
Stirling Numbers of the Second Kind S(n, k) := number of ways to partition the set [n] into k nonempty sets. Theorem.
n n X x = S(n, k)(x)k , (7) k=1
Proof. Consider functions f :[n] → X , where |X | = x ∈ N. How many such functions are there?
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations An alternative indirect count: For k = 1, 2, ··· , n, let ck denote the number of such functions with |Im(f )| = k. (If x < n, then ck = 0 for x < k ≤ n.) Pn So the answer by this indirect count is k=1 ck . n Pn Thus, x = k=1 ck . To complete the proof we now show that ck = S(n, k)(x)k .
Stirling Numbers of the Second Kind S(n, k) := number of ways to partition the set [n] into k nonempty sets. Theorem.
n n X x = S(n, k)(x)k , (7) k=1
Proof. Consider functions f :[n] → X , where |X | = x ∈ N. How many such functions are there? The answer by direct count: xn.
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations For k = 1, 2, ··· , n, let ck denote the number of such functions with |Im(f )| = k. (If x < n, then ck = 0 for x < k ≤ n.) Pn So the answer by this indirect count is k=1 ck . n Pn Thus, x = k=1 ck . To complete the proof we now show that ck = S(n, k)(x)k .
Stirling Numbers of the Second Kind S(n, k) := number of ways to partition the set [n] into k nonempty sets. Theorem.
n n X x = S(n, k)(x)k , (7) k=1
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:
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Pn So the answer by this indirect count is k=1 ck . n Pn Thus, x = k=1 ck . To complete the proof we now show that ck = S(n, k)(x)k .
Stirling Numbers of the Second Kind S(n, k) := number of ways to partition the set [n] into k nonempty sets. Theorem.
n n X x = S(n, k)(x)k , (7) k=1
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 functions with |Im(f )| = k. (If x < n, then ck = 0 for x < k ≤ n.)
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations n Pn Thus, x = k=1 ck . To complete the proof we now show that ck = S(n, k)(x)k .
Stirling Numbers of the Second Kind S(n, k) := number of ways to partition the set [n] into k nonempty sets. Theorem.
n n X x = S(n, k)(x)k , (7) k=1
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 functions with |Im(f )| = k. (If x < n, then ck = 0 for x < k ≤ n.) Pn So the answer by this indirect count is k=1 ck .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations To complete the proof we now show that ck = S(n, k)(x)k .
Stirling Numbers of the Second Kind S(n, k) := number of ways to partition the set [n] into k nonempty sets. Theorem.
n n X x = S(n, k)(x)k , (7) k=1
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 functions with |Im(f )| = k. (If x < n, then ck = 0 for x < k ≤ n.) Pn So the answer by this indirect count is k=1 ck . n Pn Thus, x = k=1 ck .
Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Stirling Numbers of the Second Kind S(n, k) := number of ways to partition the set [n] into k nonempty sets. Theorem.
n n X x = S(n, k)(x)k , (7) k=1
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 functions with |Im(f )| = k. (If x < n, then ck = 0 for x < k ≤ n.) Pn So the answer by this indirect count is k=1 ck . n Pn Thus, x = k=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 For f with |Im(f )| = k, let Im(f ) = {x1, ··· , xk }. Then −1 −1 −1 f ({x1}), f ({x2}), ··· , f ({xk }) is a partition of [n] into k non-empty sets, which we will denote by {B1, ··· , Bk }. Now: 1. There are S(n, k) ways to choose the sets {B1, ··· , Bk }; x 2. There are k was to choose the {x1, ··· , xk }; 3. There are k! ways to make the correspondence between −1 −1 −1 f ({x1}), f ({x2}), ··· , f ({xk }) and {B1, ··· , Bk }.