%I #7 Jan 08 2022 03:42:35
%S 1,1,0,1,1,4,14,4,1,1,20,175,328,175,20,1,1,72,1708,9784,17190,9784,
%T 1708,72,1,1,232,14189,199616,884498,1431728,884498,199616,14189,232,
%U 1,1,716,108250,3353948,31986447,115907544,176287788,115907544,31986447,3353948,108250,716,1
%N 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.
%H G. C. Greubel, <a href="/A156985/b156985.txt">Rows n = 0..50 of the irregular triangle, flattened</a>
%F 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.
%F Sum_{k=0..2*n} T(n, k) = A010050(n).
%e Irregular triangle begins as:
%e 1;
%e 1, 0, 1;
%e 1, 4, 14, 4, 1;
%e 1, 20, 175, 328, 175, 20, 1;
%e 1, 72, 1708, 9784, 17190, 9784, 1708, 72, 1;
%e 1, 232, 14189, 199616, 884498, 1431728, 884498, 199616, 14189, 232, 1;
%t p[x_, n_] = (1-x)^(2*n+1)*Sum[(1+k+k^2)^n*x^k, {k, 0, Infinity}];
%t Table[CoefficientList[p[x, n], x], {n, 0, 10}]//Flatten
%o (Sage)
%o 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]
%o flatten([1]+[[T(n, k) for k in (0..2*n)] for n in (1..12)]) # _G. C. Greubel_, Jan 07 2022
%Y Cf. A010050, A156896, A156890, A156901, A156918.
%K nonn,tabf
%O 0,6
%A _Roger L. Bagula_, Feb 20 2009
%E Edited by _G. C. Greubel_, Jan 07 2022