A291235 - OEIS

A291235

p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^2 - S^3 - S^4 - S^5.

1, 2, 5, 12, 29, 69, 166, 394, 944, 2245, 5365, 12781, 30506, 72734, 173520, 413838, 987130, 2354465, 5615889, 13395047, 31949764, 76206828, 181768094, 433554067, 1034112065, 2466566144, 5883251633, 14032736684, 33470882601, 79834762768, 190421890053

(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 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 (1, 6, -3, -12, 5, 12, -3, -6, 1, 1)

FORMULA

G.f.: -((1 + x - 3 x^2 - 2 x^3 + 5 x^4 + 2 x^5 - 3 x^6 - x^7 + x^8)/(-1 + x + 6 x^2 - 3 x^3 - 12 x^4 + 5 x^5 + 12 x^6 - 3 x^7 - 6 x^8 + x^9 + x^10))

a(n) = a(n-1) + 6*a(n-2) - 3*a(n-3) - 12*a(n-4) + 5*a(n-5) + 12*a(n-6) - 3*a(n-7) - 6*a(n-8) + a(n-9) + a(n-10) for n >= 11.

MATHEMATICA

z = 60; s = x/(1 - x^2); p = 1 - s - s^2 - s^3 - s^4 - s^5;

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

Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291235 *)

CROSSREFS