1,12

Row sums are probably a repeating sequence:

{1, 0, -1, -1, 0, 1, 1, 0, -1, -1, 0}:

Joint work with Gary Adamson on the Catalan triangle suggested this sequence. Instead of the Boubaker/ Steinbach:

p(x, n) = x*p(x, n - 1) - p(x, n - 2)

I used:

p(x, n) = x^2*p(x, n - 1) - p(x, n - 2)

A general:

p(x, n) = x^n*p(x, n - 1) - p(x, n - 2); n->1,2,3...

p(x,0)=1;p(x,2)=x^2-1; p(x, n) = x^2*p(x, n - 1) - p(x, n - 2)

{1},

{-1, 0, 1},

{-1, 0, -1, 0, 1},

{1, 0, -2, 0, -1, 0, 1},

{1, 0, 2, 0, -3, 0, -1, 0, 1},

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

{-1, 0, -3, 0, 6, 0, 4, 0, -5, 0, -1, 0, 1},

{1, 0, -4, 0, -6, 0, 10, 0, 5, 0, -6, 0, -1, 0,1},

{1, 0, 4,0, -10, 0, -10, 0, 15, 0, 6, 0, -7, 0, -1, 0, 1},

{-1, 0, 5, 0, 10, 0, -20, 0, -15, 0, 21, 0, 7, 0, -8, 0, -1, 0, 1},

{-1, 0, -5, 0, 15, 0,20, 0, -35, 0, -21, 0, 28, 0, 8, 0, -9, 0, -1, 0, 1}

Clear[p, x, n] p[x, 0] = 1; p[x, 1] = x^2 - 1; p[x_, n_] := p[x, n] = x^2*p[x, n - 1] - p[x, n - 2]; Table[ExpandAll[p[x, n]], {n, 0, 10}]; a = Table[CoefficientList[p[x, n], x], {n, 0, 10}]; Flatten[a] Table[Apply[Plus, CoefficientList[p[x, n], x]], {n, 0, 10}];

Adjacent sequences: A136742 A136743 A136744 this_sequence A136746 A136747 A136748

Sequence in context: A033784 A082886 A097304 this_sequence A090465 A052344 A147768

uned,tabf,sign

Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Mar 19 2008