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%I #2 Mar 30 2012 17:34:32
%S 2,1,1,17,-30,17,16,-10,-10,16,72,-176,256,-176,72,99,-57,78,78,-57,
%T 99,275,-282,1557,-1660,1557,-282,275,466,1180,2904,490,490,2904,1180,
%U 466,1058,5244,21704,4580,15468,4580,21704,5244,1058
%N Symmetrical triangle sequence from polynomials: q(x,n)=-((-1)^n*(Sum[(k + 1)^n*x^k/k^2, {k, 1, Infinity}] - PolyLog[2, x])*(x - 1)^(n - 1) + (-1)^n*n *(-1 + x)^(n - 1) Log[1 - x])/x; p(x,n)=q(x,n)+x^n*q(1/x,n).
%C Row sums are: 2*n!;
%C {2, 2, 4, 12, 48, 240, 1440, 10080, 80640, 725760,...}.
%C It was very difficult to separate out the polynomial from the Log and PolyLog terms.
%C This term:
%C (-1)^n*n *(-1 + x)^(n - 1) Log[1 - x];
%C is very strange.
%C General polynomials based on sums of the sort:
%C Sum[(k + 1)^n*x^k/k^m, {k, 1, Infinity}];m=Integer
%C that are Zeta[m] like probably exist.
%F q(x,n)=-((-1)^n*(Sum[(k + 1)^n*x^k/k^2, {k, 1, Infinity}] - PolyLog[2, x])*(x - 1)^(n - 1)
%F + (-1)^n*n *(-1 + x)^(n - 1) Log[1 - x])/x;
%F p(x,n)=q(x,n)+x^n*q(1/x,n);
%F t(n,m)=coefficients(p(x,n)).
%e {2},
%e {1, 1},
%e {17, -30, 17},
%e {16, -10, -10, 16},
%e {72, -176, 256, -176, 72},
%e {99, -57, 78, 78, -57, 99},
%e {275, -282, 1557, -1660, 1557, -282, 275},
%e {466, 1180, 2904, 490, 490, 2904, 1180, 466}, {1058, 5244, 21704, 4580, 15468, 4580, 21704, 5244, 1058}
%t Clear[p, x, n];
%t p[x_, n_] = -((-1)^n*(Sum[(k + 1)^n*x^k/k^2, {k, 1, Infinity}] - PolyLog[2, x])*(x - 1)^(n - 1) + (-1)^n*n *(-1 + x)^(n - 1) Log[1 - x])/x;
%t Table[FullSimplify[ExpandAll[p[x, n]]], {n, 1, 10}];
%t Table[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x] + Reverse[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x]], {n, 1, 10}];
%t Flatten[%]
%K sign,tabl,uned
%O 1,1
%A _Roger L. Bagula_, Jan 18 2009
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