A158782 Irregular triangle of coefficients of p(n, x) = (1 - x^2)^(n+1)*Sum_{j >= 0} (4*j+ 1)^n*x^(2*j), read by rows.

1, 1, 0, 3, 1, 0, 22, 0, 9, 1, 0, 121, 0, 235, 0, 27, 1, 0, 620, 0, 3446, 0, 1996, 0, 81, 1, 0, 3119, 0, 40314, 0, 63854, 0, 15349, 0, 243, 1, 0, 15618, 0, 422087, 0, 1434812, 0, 963327, 0, 112546, 0, 729, 1, 0, 78117, 0, 4157997, 0, 26672209, 0, 37898739, 0, 12960063, 0, 806047, 0, 2187

Offset: 0

Author

Roger L. Bagula, Mar 26 2009

Keywords

• nonn
• tabf

Comments

Define the series q(x, n) = (1 - x^2)^(n+1)*Sum_{j >= 1} (4*k+1)^n*x^(2*k) then the sum r(x, n) = p(x, n) + q(x, n) is symmetrical and gives r(x, n) =(x+1)^(2*n+1)*A060187(x, n).

Examples

			The irregular triangle begins as:
  1;
  1, 0,     3;
  1, 0,    22, 0,      9;
  1, 0,   121, 0,    235, 0,      27;
  1, 0,   620, 0,   3446, 0,    1996, 0,     81;
  1, 0,  3119, 0,  40314, 0,   63854, 0,  15349, 0,    243;
  1, 0, 15618, 0, 422087, 0, 1434812, 0, 963327, 0, 112546, 0, 729;
		

Links

• G. C. Greubel, Rows n = 0..50 of the irregular triangle, flattened

Crossrefs

Cf. A060187.

Programs

Mathematica

p[n_, x_]= (1-x^2)^(n+1)*Sum[(4*k+1)^n*x^(2*k), {k,0,Infinity}];
Table[FullSimplify[p[n,x]], {n,0,12}];
Table[CoefficientList[p[n,x], x], {n, 0, 12}]//Flatten (* modified by G. C. Greubel, Mar 08 2022 *)

Sage

def p(n,x): return (1-x^2)^(n+1)*sum( (4*j+1)^n*x^(2*j) for j in (0..n+1) )
def T(n,k): return ( p(n,x) ).series(x, 2*n+1).list()[k]
flatten([[T(n,k) for k in (0..2*n)] for n in (0..12)]) # G. C. Greubel, Mar 08 2022

Formula

T(n, k) = [x^k]( p(n, x) ), where p(n, x) = (1 - x^2)^(n+1)*Sum_{j >= 0} (4*j+ 1)^n*x^(2*j).

Extensions

Edited by G. C. Greubel, Mar 08 2022