A291036 - OEIS

A291036

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

2, 6, 16, 46, 132, 376, 1074, 3066, 8752, 24986, 71328, 203624, 581298, 1659462, 4737360, 13524006, 38607732, 110215648, 314638754, 898216794, 2564189568, 7320134930, 20897197344, 59656394448, 170304435554, 486177568038, 1387918211824, 3962167507006

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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 (2, 2, 2, -2, 0, -1)

FORMULA

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

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

MATHEMATICA

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

Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A079978 *)

u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291036 *)

u/2 (* A291037 *)