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A154991
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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).
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2, 1, 1, 17, -30, 17, 16, -10, -10, 16, 72, -176, 256, -176, 72, 99, -57, 78, 78, -57, 99, 275, -282, 1557, -1660, 1557, -282, 275, 466, 1180, 2904, 490, 490, 2904, 1180, 466, 1058, 5244, 21704, 4580, 15468, 4580, 21704, 5244, 1058
(list; table; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| Row sums are: 2*n!;
{2, 2, 4, 12, 48, 240, 1440, 10080, 80640, 725760,...}.
It was very difficult to separate out the polynomial from the Log and PolyLog terms.
This term:
(-1)^n*n *(-1 + x)^(n - 1) Log[1 - x];
is very strange.
General polynomials based on sums of the sort:
Sum[(k + 1)^n*x^k/k^m, {k, 1, Infinity}];m=Integer
that are Zeta[m] like probably exist.
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FORMULA
| 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);
t(n,m)=coefficients(p(x,n)).
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EXAMPLE
| {2},
{1, 1},
{17, -30, 17},
{16, -10, -10, 16},
{72, -176, 256, -176, 72},
{99, -57, 78, 78, -57, 99},
{275, -282, 1557, -1660, 1557, -282, 275},
{466, 1180, 2904, 490, 490, 2904, 1180, 466}, {1058, 5244, 21704, 4580, 15468, 4580, 21704, 5244, 1058}
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MATHEMATICA
| Clear[p, x, n];
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;
Table[FullSimplify[ExpandAll[p[x, n]]], {n, 1, 10}];
Table[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x] + Reverse[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x]], {n, 1, 10}];
Flatten[%]
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CROSSREFS
| Sequence in context: A156697 A173504 A174918 * A090163 A179071 A124001
Adjacent sequences: A154988 A154989 A154990 * A154992 A154993 A154994
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KEYWORD
| sign,tabl,uned
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AUTHOR
| Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Jan 18 2009
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