A291732 - OEIS

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A291732

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

4, 12, 36, 104, 288, 780, 2080, 5472, 14240, 36736, 94080, 239440, 606144, 1527360, 3833024, 9584768, 23890944, 59380160, 147207168, 364084224, 898569216, 2213388288, 5442392064, 13360097536, 32746992640, 80153705472, 195933828096, 478374127616

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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 (4, -4, 4, -8, 0, -4)

FORMULA

G.f.: -((4 (1 + x^2) (-1 + x + x^3))/(-1 + 2 x + 2 x^3)^2).

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

MATHEMATICA

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

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

u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291732 *)

u / 4 (*A291733)