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%I #3 Mar 30 2012 17:34:22
%S 1,0,2,0,0,4,-1,0,0,8,0,-4,0,0,16,0,0,-12,0,0,32,1,0,0,-32,0,0,64,0,6,
%T 0,0,-80,0,0,128,0,0,24,0,0,-192,0,0,256,-1,0,0,80,0,0,-448,0,0,512,0,
%U -8,0,0,240,0,0,-1024,0,0,1024
%N Triangular sequence from both a cubic expansion polynomial and a three deep polynomial recursion: Expansion polynomial: f(x,t)=1/(1 - 2*x*t + t^3); Recursion polynomials: p(x, n) = 2*x*p(x, n - 1) - p(x, n - 3);.
%C Row sums are A000071 (Fibonacci numbers A000045(n)-1).
%C This sequence was a designed experiment in Umbral Calculus
%C using a Weierstrass like cubic polynomial for the expansion base.
%C I was testing the recursion form in the polynomial as:
%C f(x,t)=1/(1-p(x,1)*t +t^m): m the depth of the recursion.
%C as a recursion of the form:
%C p(x,n)=p(x,1)*p(x,n-1)-p(x,n-m).
%F f(x,t)=1/(1 - 2*x*t + t^3); f(x,t)=Sum[q(x,n)*t^n,{n,1,Infinity}]; p(x,0)=1;p(x,1)=2*x;p(x,2)=4*x^2; p(x, n) = 2*x*p(x, n - 1) - p(x, n - 3);
%t (*expansion polynomial*) Clear[p, a] p[t_] = 1/(1 - 2*x*t + t^3) g = Table[ ExpandAll[SeriesCoefficient[ Series[p[t], {t, 0, 30}], n]], {n, 0, 10}]; a = Table[ CoefficientList[SeriesCoefficient[ Series[p[t], {t, 0, 30}], n], x], {n, 0, 10}]; Flatten[a] (* recursion polynomial*) Clear[p] p[x, 0] = 1; p[x, 1] = 2x; p[x, 2] = 4x^2; p[x_, n_] := p[x, n] = 2*x*p[x, n - 1] - p[x, n - 3]; Table[ExpandAll[p[x, n]], {n, 0, Length[g] - 1}]; Flatten[Table[CoefficientList[p[x, n], x], {n, 0, Length[g] - 1}]]
%Y Cf. A000071, A000045.
%K tabl,uned,sign
%O 1,3
%A _Roger L. Bagula_, Apr 12 2008