A291024 - OEIS

A291024

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

0, 4, 8, 24, 64, 172, 456, 1200, 3136, 8148, 21064, 54216, 139008, 355196, 904840, 2298720, 5825408, 14729636, 37168008, 93612408, 235369664, 590852172, 1481051720, 3707411472, 9268764096, 23145174388, 57732471752, 143857070376, 358113876352, 890666303260

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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 A291000 for a guide to related sequences.

LINKS

Clark Kimberling, Table of n, a(n) for n = 0..999

Index entries for linear recurrences with constant coefficients, signature (4, -2, -4, -1)

FORMULA

G.f.: -((4 (-x + 2 x^2))/(-1 + 2 x + x^2)^2).

a(n) = 4*a(n-1) - 2 a(n-2) - 4*a(n-3) - a(n-4) for n >= 5.

a(n) = 4*A291142(n) for n >= 0.

a(n) = ((1+sqrt(2))^n*(3*sqrt(2) + 2*(-1+sqrt(2))*n) - (1-sqrt(2))^n*(3*sqrt(2) + 2*(1+sqrt(2))*n)) / 4. - Colin Barker, Aug 24 2017

MATHEMATICA

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

Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000012 *)