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A291000
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p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S - S^2 - S^3.
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57
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1, 3, 9, 26, 74, 210, 596, 1692, 4804, 13640, 38728, 109960, 312208, 886448, 2516880, 7146144, 20289952, 57608992, 163568448, 464417728, 1318615104, 3743926400, 10630080640, 30181847168, 85694918912, 243312448256, 690833811712, 1961475291648, 5569190816256
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OFFSET
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0,2
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COMMENTS
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Suppose s = (c(0), c(1), c(2),...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).
In the following guide to p-INVERT sequences using s = (1,1,1,1,1,...) = A000012, in some cases t(1,1,1,1,1,...) is a shifted version of the cited sequence:
p(S) t(1,1,1,1,1,...)
(1 - S)(1 - 2 S)(1 - 3 S)(1 - 4 S) A291003
1 + S - S^2 A000045 (Fibonacci numbers starting with -1)
1 - S - S^2 - S^3 - S^4 - S^5 A291007
1 - S - S^2 - S^3 - S^4 + S^5 A291020
1 - S - S^2 - S^3 + S^4 + S^5 A291021
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LINKS
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FORMULA
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G.f.: (-1 + x - x^2)/(-1 + 4 x - 4 x^2 + 2 x^3).
a(n) = 4*a(n-1) - 4*a(n-2) + 2*a(n-3) for n >= 4.
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MATHEMATICA
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z = 60; s = x/(1 - x); p = 1 - s - s^2 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291000 *)
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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