A136334 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);.

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, 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, -8, 0, 0, 240, 0, 0, -1024, 0, 0, 1024

Offset

1,3

Comments

Row sums are A000071 (Fibonacci numbers A000045(n)-1).

This sequence was a designed experiment in Umbral Calculus

using a Weierstrass like cubic polynomial for the expansion base.

I was testing the recursion form in the polynomial as:

f(x,t)=1/(1-p(x,1)*t +t^m): m the depth of the recursion.

as a recursion of the form:

p(x,n)=p(x,1)*p(x,n-1)-p(x,n-m).

Links

Table of n, a(n) for n=1..66.

Formula

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);

Mathematica

(*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}]]

Crossrefs

Cf. A000071, A000045.

Sequence in context: A354352 A273346 A369731 * A155039 A238858 A106235

Adjacent sequences: A136331 A136332 A136333 * A136335 A136336 A136337

Keyword

tabl,uned,sign

Author

Roger L. Bagula, Apr 12 2008

Status

approved