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

0,8

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

Links

Alois P. Heinz, Rows n = 0..140, flattened

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.

A. Hamdi and J. 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.

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 n<k. - Wolfdieter 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)

Example

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)

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 n<k or n<0 or k<0 then
        0;
    elif n=0 and k= 0 then
        1;
    else
        procname(n-1, k-1)+(n-1)^2*procname(n-2, k) ;
    end if;
end proc: # R. 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 *)

Crossrefs

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

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

Sequence in context: A318299 A164652 A127557 * A133843 A215083 A221308

Adjacent sequences: A060521 A060522 A060523 * A060525 A060526 A060527

Keyword

easy,nonn,tabl

Author

Vladeta Jovovic, Apr 01 2001

Status

approved