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%I #15 Jan 07 2022 02:50:47
%S 1,1,1,-1,1,1,-4,5,-2,1,1,-11,22,-23,14,-3,1,1,-26,92,-158,145,-82,32,
%T -4,1,1,-57,359,-906,1265,-1135,649,-238,67,-5,1,1,-120,1311,-4798,
%U 9630,-12132,10163,-5970,2406,-620,135,-6,1,1,-247,4540,-24205,66769,-113626,131045,-106889,62261,-26426,8033,-1517,268,-7,1
%N Triangle formed by coefficients of the expansion of p(x, n), where p(x,n) = (1+x-x^2)^(n+1)*Sum_{j >= 0} (j+1)^n*(-x + x^2)^j.
%C Row sums are equal to 1.
%H G. C. Greubel, <a href="/A156890/b156890.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-x^2)^(n + 1)*Sum_{j >= 0} (j+1)^n*(-x + x^2)^j.
%F T(n, 1) = (-1)*A000295(n) for n >= 2. - _G. C. Greubel_, Jan 06 2022
%e Irregular triangle begins as:
%e 1;
%e 1;
%e 1, -1, 1;
%e 1, -4, 5, -2, 1;
%e 1, -11, 22, -23, 14, -3, 1;
%e 1, -26, 92, -158, 145, -82, 32, -4, 1;
%e 1, -57, 359, -906, 1265, -1135, 649, -238, 67, -5, 1;
%e 1, -120, 1311, -4798, 9630, -12132, 10163, -5970, 2406, -620, 135, -6, 1;
%t p[x_, n_]:= ((1+x-x^2)^(n+1))*Sum[(j+1)^n*(-x+x^2)^j, {j,0,Infinity}];
%t Table[CoefficientList[p[x, n], x], {n, 0, 10}]//Flatten
%o (Sage)
%o def T(n,k): return ( (1+x-x^2)^(n+1)*sum((j+1)^n*(x^2-x)^j for j in (0..2*n+1)) ).series(x, 2*n+3).list()[k]
%o [1]+flatten([[T(n,k) for k in (0..2*n-2)] for n in (0..12)]) # _G. C. Greubel_, Jan 06 2022
%Y Cf. A000295, A156896, A156901, A156918.
%K tabf,sign
%O 0,7
%A _Roger L. Bagula_, Feb 17 2009
%E Edited by _G. C. Greubel_, Jan 06 2022
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