%I #4 Oct 01 2012 11:24:16
%S 1,1,1,0,-1,1,0,-1,1,0,-2,0,1,1,0,-2,0,1,1,0,-3,0,3,0,-1,1,0,-3,0,3,0,
%T -1,1,0,-4,0,6,0,-4,0,1,1,0,-4,0,6,0,-4,0,1,1,0,-5,0,10,0,-10,0,5,0,-1
%N A golden mean based polynomials set that behaves like an even powered Pascal triangle: p(x,n) = (x - phi)^floor(n/2)*(x + phi)^floor(n/2).
%F p(x,n)=(x - GoldenRatio)^Floor[n/2]*(x + GoldenRatio)^Floor[n/2]; t(n,m)=GoldenRatio^((-m + 1))*Reverse(Coefficients(p(x,n))).
%e {1}, {1}, {1, 0, -1}, {1, 0, -1}, {1, 0, -2, 0, 1}, {1, 0, -2, 0, 1}, {1, 0, -3, 0, 3, 0, -1}, {1, 0, -3, 0, 3, 0, -1}, {1, 0, -4, 0, 6, 0, -4, 0, 1}, {1, 0, -4, 0, 6, 0, -4, 0, 1}, {1, 0, -5, 0, 10, 0, -10, 0, 5, 0, -1}
%t Clear[p, x, n]; p[x_, n_] = (x - GoldenRatio)^Floor[n/2]*(x + GoldenRatio)^Floor[n/2] a = Table[Reverse[CoefficientList[FullSimplify[ExpandAll[p[x, n]]], x]], {n, 0, 10}]; Flatten[%]; b = Table[a[[n]][[m]]*GoldenRatio^((-m + 1)), {n, 1, Length[a]}, {m, 1, Length[a[[n]]]}]; Flatten[b]
%K tabf,sign,more,uned
%O 0,11
%A _Roger L. Bagula_, Nov 05 2008