0,3

Row sums are all zero.

Based on Identity:

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

broken into parts of the (1-x)^2

and the count up polynomials:

q(x,n)= Sum[(k + 1)*(x)^k, {k, 0, n - 1}].

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

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

{0},

{1, -2, 1},

{1, 0, -3, 2},

{1, 0, 0, -4, 3},

{1, 0, 0, 0, -5, 4},

{1, 0, 0, 0, 0, -6, 5},

{1, 0, 0, 0, 0, 0, -7, 6},

{1, 0, 0, 0, 0, 0, 0, -8, 7},

{1, 0, 0, 0, 0, 0, 0, 0, -9, 8},

{1, 0, 0, 0, 0, 0, 0, 0, 0, -10, 9},

{1, 0, 0, 0, 0, 0, 0, 0, 0, 0, -11, 10}

Clear[p];

p[x_, n_] := (1 - x)^2*Sum[(k + 1)*x^k, {k, 0, n - 1}];

Table[FullSimplify[ExpandAll[p[x, n]]], {n, 0, 10}];

Table[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x], {n, 0, 10}];

Flatten[%]

Sequence in context: A029352 A055168 A085144 this_sequence A097230 A144789 A087117

Adjacent sequences: A156575 A156576 A156577 this_sequence A156579 A156580 A156581

sign,tabl,uned

Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Feb 10 2009