A156985 Triangle formed by coefficients of the expansion of p(x,n) = (1-x)^(2*n + 1)*Sum_{j >= 0} (1 +j +j^2)^n * x^j.

1, 1, 0, 1, 1, 4, 14, 4, 1, 1, 20, 175, 328, 175, 20, 1, 1, 72, 1708, 9784, 17190, 9784, 1708, 72, 1, 1, 232, 14189, 199616, 884498, 1431728, 884498, 199616, 14189, 232, 1, 1, 716, 108250, 3353948, 31986447, 115907544, 176287788, 115907544, 31986447, 3353948, 108250, 716, 1

Offset

0,6

Links

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

Formula

T(n, k) = coefficients of the expansion of p(x, n), where p(x,n) = (1-x)^(2*n + 1)*Sum_{j >= 0} (1 +j +j^2)^n * x^j.

Sum_{k=0..2*n} T(n, k) = A010050(n).

Example

Irregular triangle begins as:
  1;
  1,   0,     1;
  1,   4,    14,      4,      1;
  1,  20,   175,    328,    175,      20,      1;
  1,  72,  1708,   9784,  17190,    9784,   1708,     72,     1;
  1, 232, 14189, 199616, 884498, 1431728, 884498, 199616, 14189, 232, 1;

Mathematica

p[x_, n_] = (1-x)^(2*n+1)*Sum[(1+k+k^2)^n*x^k, {k, 0, Infinity}];
Table[CoefficientList[p[x, n], x], {n, 0, 10}]//Flatten

Prog

(Sage)
def T(n, k): return ( (1-x)^(2*n+1)*sum((j^2+j+1)^n*x^j for j in (0..2*n+1)) ).series(x, 2*n+2).list()[k]
flatten([1]+[[T(n, k) for k in (0..2*n)] for n in (1..12)]) # G. C. Greubel, Jan 07 2022

Crossrefs

Cf. A010050, A156896, A156890, A156901, A156918.

Sequence in context: A107775 A003117 A239465 * A138229 A131702 A276826

Adjacent sequences: A156982 A156983 A156984 * A156986 A156987 A156988

Keyword

nonn,tabf

Author

Roger L. Bagula, Feb 20 2009

Extensions

Edited by G. C. Greubel, Jan 07 2022

Status

approved