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A136337 Triangular sequence from both a quartic expansion polynomial and a four deep polynomial recursion: Expansion polynomial: f(x,t)=1/(1 - 2*x*t + t^4); Recursion polynomials: p(x, n) = 2*x*p(x, n - 1) - p(x, n - 4);. 0
1, 0, 2, 0, 0, 4, 0, 0, 0, 8, -1, 0, 0, 0, 16, 0, -4, 0, 0, 0, 32, 0, 0, -12, 0, 0, 0, 64, 0, 0, 0, -32, 0, 0, 0, 128, 1, 0, 0, 0, -80, 0, 0, 0, 256, 0, 6, 0, 0, 0, -192, 0, 0, 0, 512, 0, 0, 24, 0, 0, 0, -448, 0, 0, 0, 1024 (list; table; graph; refs; listen; history; text; internal format)
OFFSET
1,3
COMMENTS
Row sums are A008937.
This sequence was a designed experiment in Umbral Calculus
using a quartic 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
FORMULA
f(x,t)=1/(1 - 2*x*t + t^4); f(x,t)=Sum[q(x,n)*t^n,{n,1,Infinity}]; p(x,-1)=0;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 - 4);
MATHEMATICA
(*expansion polynomial*) Clear[p, a] p[t_] = 1/(1 - 2*x*t + t^4) 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, -1]=0; 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 - 4]; Table[ExpandAll[p[x, n]], {n, 0, Length[g] - 1}]; Flatten[Table[CoefficientList[p[x, n], x], {n, 0, Length[g] - 1}]]
CROSSREFS
Cf. A008937.
Sequence in context: A321256 A130123 A319935 * A028601 A077958 A077959
KEYWORD
tabl,uned,sign
AUTHOR
Roger L. Bagula, Apr 12 2008
STATUS
approved

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Last modified May 5 18:56 EDT 2024. Contains 372277 sequences. (Running on oeis4.)