%I #7 Jan 08 2022 03:42:12

%S 1,1,-1,1,1,-6,7,-2,1,1,-23,46,-47,26,-3,1,1,-76,306,-536,459,-232,82,

%T -4,1,1,-237,1919,-5046,6965,-5995,3109,-958,247,-5,1,1,-722,11265,

%U -44634,91730,-113538,90417,-49398,17778,-3630,737,-6,1,1,-2179,62836,-381037,1099549,-1878718,2123525,-1658537,898985,-346886,93377,-13109,2200,-7,1

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

%C Row sums are one.

%H G. C. Greubel, <a href="/A156918/b156918.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} (2*j+1)^n*(-x + x^2)^j.

%F T(n, 1) = (-1)*A060188(n), for n >= 2. - _G. C. Greubel_, Jan 07 2022

%e Irregular triangle begins as:

%e 1;

%e 1, -1, 1;

%e 1, -6, 7, -2, 1;

%e 1, -23, 46, -47, 26, -3, 1;

%e 1, -76, 306, -536, 459, -232, 82, -4, 1;

%e 1, -237, 1919, -5046, 6965, -5995, 3109, -958, 247, -5, 1;

%e 1, -722, 11265, -44634, 91730, -113538, 90417, -49398, 17778, -3630, 737, -6, 1;

%t p[x_, n_] = (1+x-x^2)^(n+1)*Sum[(2*k+1)^n*(-x+x^2)^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-x^2)^(n+1)*sum((2*j+1)^n*(x^2-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. A060188, A156896, A156890, A156901.

%K sign,tabf

%O 0,6

%A _Roger L. Bagula_, Feb 18 2009

%E Edited by _G. C. Greubel_, Jan 07 2022