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Search: id:A136334
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| A136334 |
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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);. |
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+0
1
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| 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
(list; table; graph; listen)
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OFFSET
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1,3
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COMMENT
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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).
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FORMULA
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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);
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MATHEMATICA
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(*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}]]
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CROSSREFS
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Cf. A000071, A000045.
Sequence in context: A140668 A071390 A061669 this_sequence A155039 A106235 A118965
Adjacent sequences: A136331 A136332 A136333 this_sequence A136335 A136336 A136337
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KEYWORD
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tabl,uned,sign
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AUTHOR
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Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Apr 12 2008
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