A176669 A triangle of polynomial coefficients:p(x,n)=Sum[(2*k - 1)^n*Binomial[x, k], {k, 0, Infinity}]/2^x

1, -1, 1, 1, -1, 1, -1, 0, 0, 1, 1, 0, -3, 2, 1, -1, 5, -5, -5, 5, 1, 1, -5, 30, -35, 0, 9, 1, -1, -56, 91, 35, -105, 21, 14, 1, 1, 56, -518, 812, -245, -196, 70, 20, 1, -1, 1329, -2610, -42, 2793, -1449, -210, 162, 27, 1, 1, -1329, 15075, -28110, 14385, 4053, -4515, 90

Offset

0,13

Comments

Row sums are:

{1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1,...}.

Links

Table of n, a(n) for n=0..62.

Formula

p(x,n)=Sum[(2*k - 1)^n*Binomial[x, k], {k, 0, Infinity}]/2^x;

t(n,m)=coefficients(p(x,n))

Example

{1},
{-1, 1},
{1, -1, 1},
{-1, 0, 0, 1},
{1, 0, -3, 2, 1},
{-1, 5, -5, -5, 5, 1},
{1, -5, 30, -35, 0, 9, 1},
{-1, -56, 91, 35, -105, 21, 14, 1},
{1, 56, -518, 812, -245, -196, 70, 20, 1},
{-1, 1329, -2610, -42, 2793, -1449, -210, 162, 27, 1},
{1, -1329, 15075, -28110, 14385, 4053, -4515, 90, 315, 35, 1}

Mathematica

Clear[p, x, n]
p[x_, n_] = Sum[(2*k - 1)^n*Binomial[x, k], {k, 0, Infinity}]/2^x;
Table[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x], {n, 0, 10}];
Flatten[%]

Crossrefs

Sequence in context: A269233 A123396 A225800 * A327615 A058280 A113185

Adjacent sequences: A176666 A176667 A176668 * A176670 A176671 A176672

Keyword

sign,tabl,uned

Author

Roger L. Bagula, Apr 23 2010

Status

approved