A060524 Triangle read by rows: T(n,k) = number of degree-n permutations with k odd cycles, k=0..n, n >= 0.

1, 0, 1, 1, 0, 1, 0, 5, 0, 1, 9, 0, 14, 0, 1, 0, 89, 0, 30, 0, 1, 225, 0, 439, 0, 55, 0, 1, 0, 3429, 0, 1519, 0, 91, 0, 1, 11025, 0, 24940, 0, 4214, 0, 140, 0, 1, 0, 230481, 0, 122156, 0, 10038, 0, 204, 0, 1, 893025, 0, 2250621, 0, 463490, 0, 21378, 0, 285, 0, 1, 0

Offset: 0

Author

Vladeta Jovovic, Apr 01 2001

Keywords

• easy
• nonn
• tabl

Comments

The row polynomials t(n,x):=Sum_{k=0..n} T(n,k)*x^k satisfy the recurrence relation t(n,x) = x*t(n-1,x) + ((n-1)^2)*t(n-2,x); t(-1,x)=0, t(0,x)=1. - Wolfdieter Lang, see above.

This is an example of a Sheffer triangle (coefficient triangle for Sheffer polynomials). In the umbral calculus (see the Roman reference given under A048854) s(n,x) := Sum_{k=0..n} T(n,k)*x^k would be called Sheffer polynomials for (1/cosh(t),tanh(t)), which translates to the e.g.f. for column number k>=0 given by (1/sqrt(1-x^2))*((arctanh(x))^k)/k!. The e.g.f. given below is rewritten in this Sheffer context as (1/sqrt(1-x^2))*exp(y*log(sqrt((1+x)/(1-x))))= (1/sqrt(1-x^2))*exp(y*arctanh(x)). The rows of the Jabotinsky type triangle |A049218| provide the coefficients of the associated polynomials. - Wolfdieter Lang, Feb 24 2005

The solution of the differential-difference relation f(n+1,x)= (d/dx)f(n,x) + (n^2)*f(n-1,x), n >= 1, with inputs f(0,x) and f(1,x) = (d/dx)f(0,x) is f(n,x) = t(n,d_x)*f(0,x), with the differential operator d_x:=d/dx and the row polynomials t(n,x) defined above. This problem appears in a computation of thermo field dynamics where f(0,x)=1/cosh(x). See the triangle A060081. - Wolfdieter Lang, Feb 24 2005

The inverse of the Sheffer matrix T with elements T(n,k) is the Sheffer matrix A060081. - Wolfdieter Lang, Jul 22 2005

T(n,k)=0 if n-k= 1(mod 2), else T(n,k) = sum of M2(n,p), p from {1,...,A000041(n)} restricted to partitions with exactly k odd parts and any nonnegative number of even parts. For the M2-multinomial numbers in A-St order see A036039(n,p). - Wolfdieter Lang, Aug 07 2007

Examples

			Triangle begins:
  [1],
  [0, 1],
  [1, 0, 1],
  [0, 5, 0, 1],
  [9, 0, 14, 0, 1],
  [0, 89, 0, 30, 0, 1],
  [225, 0, 439, 0, 55, 0, 1],
  [0, 3429, 0, 1519, 0, 91, 0, 1],
  [11025, 0, 24940, 0, 4214, 0, 140, 0, 1],
  [0, 230481, 0, 122156, 0, 10038, 0, 204, 0, 1],
  [893025, 0, 2250621, 0, 463490, 0, 21378, 0, 285, 0, 1],
  [0, 23941125, 0, 14466221, 0, 1467290, 0, 41778, 0, 385, 0, 1],
  ...
Signed version begins:
  [1],
  [0, 1],
  [-1, 0, 1],
  [0, -5, 0, 1],
  [9, 0, -14, 0, 1],
  [0, 89, 0, -30, 0, 1],
  [-225, 0, 439, 0, -55, 0, 1],
  [0, -3429, 0, 1519, 0, -91, 0, 1],
  ...
From _Peter Bala_, Feb 23 2024: (Start)
Maple can verify the following series for Pi:
Row 1 polynomial R(1, x) = x:
Pi = 3 + 4*Sum_{n >= 1} (-1)^(n+1)/((2*n + 1)*R(1, 2*n)*R(1, 2*n+2)).
Row 3 polynomial R(3, x) = 5*x + x^3:
(3/2)^2 * Pi = 7 + 4*(3^4)*Sum_{n >= 1} (-1)^(n+1)/((2*n + 1)*R(3, 2*n)*R(3, 2*n+2)).
Row 5 polynomial R(5, x) = 89*x + 30*x^3 + x^5:
((3*5)/(2*4))^2 * Pi = 11 + 4*(3*5)^4*Sum_{n >= 1} (-1)^(n+1)/((2*n + 1)*R(5, 2*n)*R(5, 2*n+2)). (End)
		

Links

• Alois P. Heinz, Rows n = 0..140, flattened
• Marcelo Aguiar, Sarah Brauner, and Vic Reiner, Configuration spaces and peak representations, Sém. Lotharingien Comb., Proc. 36th Conf. Formal Power Series Alg. Comb. (2024) Vol. 91B, Art. No. 13. See p. 12.
• Peter Bala, Meixner polynomials and Brouncker's continued fraction for Pi
• Paul L. Butzer and Tom H. Koornwinder, Josef Meixner: His life and his orthogonal polynomials, Indagationes Mathematicae, Volume 30, Issue 1, January 2019, Pages 250-264.
• Adel Hamdi and Jiang Zeng, Orthogonal polynomials and operator orderings, J. Math. Phys., 51:043506, (2010); arXiv:1006.0808 [math.CO], 2010.
• Eric Weisstein's World of Mathematics, Meixner polynomial of the first kind.

Crossrefs

Cf. A060338, A060523, A094368, A028353 (col 1), A103916 (col 2), A103917 (col 3), A103918 (col 4).

Cf. A111594 (associated Sheffer polynomials), A142979, A142983.

Programs

Maple

with(combinat):
b:= proc(n, i) option remember; expand(`if`(n=0, 1, `if`(i<1, 0,
      add(multinomial(n, n-i*j, i$j)*(i-1)!^j/j!*b(n-i*j, i-1)*
      `if`(irem(i, 2)=1, x^j, 1), j=0..n/i))))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n$2)):
seq(T(n), n=0..12);  # Alois P. Heinz, Mar 09 2015
# alternative
A060524 := proc(n,k)
    option remember;
    if nR. J. Mathar, Jul 06 2023

Mathematica

nn = 6; Range[0, nn]! CoefficientList[
   Series[(1 - x^2)^(-1/2) ((1 + x)/(1 - x))^(y/2), {x, 0, nn}], {x, y}] // Grid  (* Geoffrey Critzer, Aug 28 2012 *)

Formula

E.g.f.: (1+x)^((y-1)/2)/(1-x)^((y+1)/2).

T(n, k) = T(n-1, k-1) + ((n-1)^2)*T(n-2, k); T(-1, k):=0, T(n, -1):=0, T(0, 0)=1, T(n, k)=0 if nWolfdieter Lang, see above.

The Meixner polynomials defined by S_0(x)=1, S_1(x) = x; S_{n+1}(x) = x*S_n(x) - n^2*S_{n-1}(x) give a signed version of this triangle (cf. A060338). - N. J. A. Sloane, May 30 2013

From Peter Bala, Apr 10 2024: (Start)

The n-th row polynomial R(n, x) satisfies

(4*n + 2)*R(n, x) = (x + 1)*R(n, x+2) - (x - 1)*R(n, x-2).

Series for Pi involving the row polynomials R(n, x): for n >= 0 there holds

((2*n + 1)!!/(2^n*n!))^2 * Pi = (4*n + 3) + 4*((2*n + 1)!!^4) * Sum_{k >= 1} (-1)^(k+1)/((2*k + 1)*R(2*n+1, 2*k)*R(2*n+1, 2*k+2)). Cf. A142979 and A142983.

R(2*n, 0) = A001147(n)^2 = A001818(n); R(2*n+1, 0) = 0.

R(n, 1) = n! = A000142(n).

R(2*n, 2) = (4*n + 1)*A001147(n)^2 = (4*n + 1)*((2*n)!/(2^n*n!))^2;

R(2*n+1, 2) = 2*A001447(n+1)^2 = 2*(2*n + 1)!^2/(n!^2*4^n).

R(n, 3) = (2*n + 1)*n! = A007680(n). (End)