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 jSnj = 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] = f1; 2; ··· ; ng 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] = f1; 2; ··· ; ng Sn = permutations of [n] jSnj = 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] = f1; 2; ··· ; ng Sn = permutations of [n] jSnj = 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] = f1; 2; ··· ; ng Sn = permutations of [n] jSnj = 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] = f1; 2; ··· ; ng Sn = permutations of [n] jSnj = 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] = f1; 2; ··· ; ng Sn = permutations of [n] jSnj = 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 2 R; n ≥ 1: (1) k=1 Unsigned Stirling Numbers of the First Kind: js(n; k)j := the number of permutations in Sn with exactly k cycles; 1 ≤ k ≤ n Recurrence relation js(n; n)j = 1; js(n; 1)j = (n − 1)!; n ≥ 1; (2) js(n + 1; k)j = njs(n; k)j + js(n; k − 1)j; 2 ≤ k ≤ n: It is easy to check that the coefficients fank g in (1) also satisfy (2); thus, ank = js(n; k)j. Therefore n X x(n) = js(n; k)jxk : (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: js(n; k)j := the number of permutations in Sn with exactly k cycles; 1 ≤ k ≤ n Recurrence relation js(n; n)j = 1; js(n; 1)j = (n − 1)!; n ≥ 1; (2) js(n + 1; k)j = njs(n; k)j + js(n; k − 1)j; 2 ≤ k ≤ n: It is easy to check that the coefficients fank g in (1) also satisfy (2); thus, ank = js(n; k)j. Therefore n X x(n) = js(n; k)jxk : (3) k=1 Rising Factorials n (n) X k x := x(x +1) ··· (x +n−1) ≡ ank x ; x 2 R; n ≥ 1: (1) k=1 Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Recurrence relation js(n; n)j = 1; js(n; 1)j = (n − 1)!; n ≥ 1; (2) js(n + 1; k)j = njs(n; k)j + js(n; k − 1)j; 2 ≤ k ≤ n: It is easy to check that the coefficients fank g in (1) also satisfy (2); thus, ank = js(n; k)j. Therefore n X x(n) = js(n; k)jxk : (3) k=1 Rising Factorials n (n) X k x := x(x +1) ··· (x +n−1) ≡ ank x ; x 2 R; n ≥ 1: (1) k=1 Unsigned Stirling Numbers of the First Kind: js(n; k)j := 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 fank g in (1) also satisfy (2); thus, ank = js(n; k)j. Therefore n X x(n) = js(n; k)jxk : (3) k=1 Rising Factorials n (n) X k x := x(x +1) ··· (x +n−1) ≡ ank x ; x 2 R; n ≥ 1: (1) k=1 Unsigned Stirling Numbers of the First Kind: js(n; k)j := the number of permutations in Sn with exactly k cycles; 1 ≤ k ≤ n Recurrence relation js(n; n)j = 1; js(n; 1)j = (n − 1)!; n ≥ 1; (2) js(n + 1; k)j = njs(n; k)j + js(n; k − 1)j; 2 ≤ k ≤ n: Ross Pinsky Touchard Polynomials, Stirling Numbers and Random Permutations Therefore n X x(n) = js(n; k)jxk : (3) k=1 Rising Factorials n (n) X k x := x(x +1) ··· (x +n−1) ≡ ank x ; x 2 R; n ≥ 1: (1) k=1 Unsigned Stirling Numbers of the First Kind: js(n; k)j := the number of permutations in Sn with exactly k cycles; 1 ≤ k ≤ n Recurrence relation js(n; n)j = 1; js(n; 1)j = (n − 1)!; n ≥ 1; (2) js(n + 1; k)j = njs(n; k)j + js(n; k − 1)j; 2 ≤ k ≤ n: It is easy to check that the coefficients fank g in (1) also satisfy (2); thus, ank = js(n; k)j. 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 2 R; n ≥ 1: (1) k=1 Unsigned Stirling Numbers of the First Kind: js(n; k)j := the number of permutations in Sn with exactly k cycles; 1 ≤ k ≤ n Recurrence relation js(n; n)j = 1; js(n; 1)j = (n − 1)!; n ≥ 1; (2) js(n + 1; k)j = njs(n; k)j + js(n; k − 1)j; 2 ≤ k ≤ n: It is easy to check that the coefficients fank g in (1) also satisfy (2); thus, ank = js(n; k)j. Therefore n X x(n) = js(n; k)jxk : (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) js(n; k)j(−x) = k=1 n (5) X (−1)n−k js(n; k)jxk : k=1 Define s(n; k) := (−1)n−k js(n; k)j. 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 2 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) js(n; k)j(−x) = k=1 n (5) X (−1)n−k js(n; k)jxk : k=1 Define s(n; k) := (−1)n−k js(n; k)j. 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 2 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) js(n; k)j(−x) = k=1 n (5) X (−1)n−k js(n; k)jxk : k=1 Define s(n; k) := (−1)n−k js(n; k)j. 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 2 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) js(n; k)j(−x) = k=1 n (5) X (−1)n−k js(n; k)jxk : k=1 Define s(n; k) := (−1)n−k js(n; k)j.