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Search: id:A142071
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| A142071 |
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A triangle sequence of coefficients of an infinite sum polynomial: p(x,n)=Sum[k^n*(x/(1 + x))^k, {k, 0, Infinity}]=PolyLog[ -n,x/(1+x)]. |
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1
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| 0, 1, 0, 1, 1, 0, 1, 3, 2, 0, 1, 7, 12, 6, 0, 1, 15, 50, 60, 24, 0, 1, 31, 180, 390, 360, 120, 0, 1, 63, 602, 2100, 3360, 2520, 720, 0, 1, 127, 1932, 10206, 25200, 31920, 20160, 5040, 0, 1, 255, 6050, 46620, 166824, 317520, 332640, 181440, 40320, 0, 1, 511, 18660
(list; graph; listen)
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OFFSET
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1,8
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COMMENT
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Row sums are:
{1, 2, 6, 26, 150, 1082, 9366, 94586, 1091670, 14174522, 204495126}.
If you divide the PolyLog function by x you get A028246.
I got this polynomial sequence looking for a Eulerian number like infinite sum polynomial.
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FORMULA
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p(x,n)=Sum[k^n*(x/(1 + x))^k, {k, 0, Infinity}]=PolyLog[ -n,x/(1+x)]; t(n,m)=Coefficients(p(x,n)).
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EXAMPLE
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{0, 1},
{0, 1, 1},
{0, 1, 3, 2},
{0, 1, 7, 12, 6},
{0, 1, 15, 50, 60, 24},
{0, 1, 31, 180, 390, 360, 120},
{0, 1, 63, 602, 2100, 3360, 2520, 720},
{0, 1, 127, 1932, 10206, 25200, 31920, 20160, 5040},
{0, 1, 255, 6050, 46620, 166824, 317520, 332640, 181440, 40320},
{0, 1, 511, 18660, 204630, 1020600, 2739240, 4233600, 3780000, 1814400, 362880}, {0, 1, 1023, 57002, 874500, 5921520, 21538440, 46070640, 59875200, 46569600, 19958400, 3628800}
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MATHEMATICA
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p[x_, n_] = Sum[k^n*(x/(1 + x))^k, {k, 0, Infinity}]; Table[FullSimplify[ExpandAll[p[x, n]]], {n, 0, 10}]; Table[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x], {n, 0, 10}]; Flatten[%]
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CROSSREFS
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Cf. A028246 .
Sequence in context: A139144 A081576 A054654 this_sequence A118972 A145878 A112606
Adjacent sequences: A142068 A142069 A142070 this_sequence A142072 A142073 A142074
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KEYWORD
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nonn,uned
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AUTHOR
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Roger L. Bagula and Gary W. Adamson (rlbagulatftn(AT)yahoo.com), Sep 15 2008
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