A291023 - OEIS

A291023

p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - 3 S^2 + 2 S^3.

0, 3, 4, 12, 24, 56, 120, 264, 568, 1224, 2616, 5576, 11832, 25032, 52792, 111048, 233016, 487880, 1019448, 2126280, 4427320, 9204168, 19107384, 39612872, 82021944, 169636296, 350457400, 723284424, 1491308088, 3072094664, 6323146296, 13004206536, 26724240952

(list; graph; refs; listen; history; text; internal format)

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..1000

Index entries for linear recurrences with constant coefficients, signature (3, 0, -4)

FORMULA

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

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

a(n) = (16*((-1)^(1+n) + 2^n) + 3*2^n*n) / 18. - 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 *)

Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291023 *)