A291400 - OEIS

A291400

p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S^2 - S^4.

0, 1, 2, 3, 8, 15, 26, 52, 100, 193, 378, 726, 1396, 2699, 5210, 10065, 19444, 37528, 72448, 139890, 270104, 521547, 1007026, 1944336, 3754132, 7248558, 13995676, 27023186, 52176848, 100743849, 194517966, 375578833, 725174524, 1400180233, 2703493026

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OFFSET

0,3

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 (0, 1, 2, 2, 4, 6, 4, 1)

FORMULA

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

a(n) = a(n-2) + 2*a(n-3) + 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^2 - s^4;

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

u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291400 *)

CROSSREFS