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A291239 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S^2) (1 - 2 S). 2
2, 5, 12, 31, 74, 184, 442, 1081, 2604, 6323, 15250, 36912, 89074, 215293, 519660, 1255223, 3030106, 7317032, 17664170, 42649553, 102963276, 248587051, 600137378, 1448890464, 3497918306, 8444802101, 20387522508, 49220043535, 118827609578, 286875776920 (list; graph; refs; listen; history; text; internal format)
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 (2, 4, -6, -4, 2, 1)

FORMULA

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

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

MATHEMATICA

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

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

Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291239 *)

CROSSREFS

Cf. A000035, A291219.

Sequence in context: A319557 A261937 A305311 * A193284 A238829 A125023

Adjacent sequences:  A291236 A291237 A291238 * A291240 A291241 A291242

KEYWORD

nonn,easy

AUTHOR

Clark Kimberling, Aug 28 2017

STATUS

approved

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Last modified June 19 00:37 EDT 2021. Contains 345125 sequences. (Running on oeis4.)