A291725 - OEIS

A291725

p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = (1 - S)^2.

2, 3, 6, 11, 18, 30, 50, 81, 130, 208, 330, 520, 816, 1275, 1984, 3077, 4758, 7337, 11286, 17322, 26532, 40563, 61908, 94336, 143540, 218112, 331008, 501749, 759726, 1149159, 1736534, 2621751, 3954826, 5960902, 8977686, 13511461, 20320854, 30542064, 45875998

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OFFSET

0,1

COMMENTS

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).

See A291728 for a guide to related sequences.

LINKS

Clark Kimberling, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (2, -1, 2, -2, 0, -1)

FORMULA

G.f.: -(((-1 + x) (1 + x^2) (2 + x + x^2))/(-1 + x + x^3)^2).

a(n) = 2*a(n-1) - a(n-2) + 2*a(n-3) - 2*a(n-4) - a(n-6) for n >= 7.

MATHEMATICA

z = 60; s = x + x^3; p = (1 - s)^2;

Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A154272 *)

Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291725 *)

LinearRecurrence[{2, -1, 2, -2, 0, -1}, {2, 3, 6, 11, 18, 30}, 40] (* Vincenzo Librandi, Sep 10 2017 *)