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

0,4

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).

Links

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

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).

Example

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;

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 *)

Prog

(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

Crossrefs

Cf. A060187.

Sequence in context: A246049 A316773 A006837 * A187558 A327547 A233293

Adjacent sequences: A158779 A158780 A158781 * A158783 A158784 A158785

Keyword

nonn,tabf

Author

Roger L. Bagula, Mar 26 2009

Extensions

Edited by G. C. Greubel, Mar 08 2022

Status

approved