A291224 - OEIS

A291224

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

4, 10, 24, 55, 120, 254, 524, 1059, 2104, 4120, 7968, 15244, 28888, 54284, 101240, 187537, 345268, 632122, 1151408, 2087485, 3768280, 6775322, 12136940, 21666712, 38555100, 68401582, 121011800, 213521067, 375813760, 659910710, 1156204452, 2021495767

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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 A291219 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 (4, -2, -8, 5, 8, -2, -4, -1)

FORMULA

a(n) = 4*a(n-1) - 2*a(n-2) - 8*a(n-3) + 5*a(n-4) + 8*a(n-5) - 2*a(n-6) - 4*a(n-7) - a(n-8) for n >= 9.

G.f.: (2 - x - 2*x^2)*(2 - 2*x - 3*x^2 + 2*x^3 + 2*x^4) / (1 - x - x^2)^4. - Colin Barker, Aug 25 2017

MATHEMATICA

z = 60; s = x/(1 - x^2); p = (1 - s)^4;

Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000035 *)

Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291224 *)

PROG