A291401 - OEIS

A291401

p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S - S^4.

1, 2, 3, 6, 14, 32, 67, 134, 266, 538, 1110, 2304, 4760, 9770, 19991, 40931, 83976, 172519, 354452, 727830, 1493768, 3065341, 6291208, 12914136, 26511196, 54423052, 111715200, 229312168, 470697488, 966192481, 1983312305, 4071174986, 8356928055, 17154242334

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OFFSET

0,2

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 A291382 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 (1, 1, 0, 1, 4, 6, 4, 1)

FORMULA

G.f.: -(((1 + x) (1 + x + x^2) (1 - x + 2 x^3 + x^4))/(-1 + x + x^2 + x^4 + 4 x^5 + 6 x^6 + 4 x^7 + x^8)).

a(n) = a(n-1) + a(n-2) + a(n-4) + 4*a(n-5) + 6*a(n-6) + 4*a(n-7) + a(n-8) for n >= 9.

MATHEMATICA

z = 60; s = x + x^2; p = 1 - s - s^4;

Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A019590 *)

u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291401 *)

CROSSREFS