A292481 p-INVERT of the odd positive integers, where p(S) = 1 - S^3.

0, 0, 1, 9, 42, 139, 381, 984, 2685, 8061, 25434, 79695, 242577, 721584, 2131785, 6333633, 18984618, 57194883, 172319157, 517851144, 1552599333, 4651054101, 13939132698, 41810229351, 125475990057, 376585031520, 1129975049169, 3389800055481, 10168040440746

Offset: 0

Author

Clark Kimberling, Oct 02 2017

Keywords

• nonn
• easy

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 A292480 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 (6, -15, 21, -12, 9)

Crossrefs

Cf. A005408, A292480.

Programs

Mathematica

z = 60; s = x (x + 1)/(1 - x)^2; p = 1 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292481 *)
LinearRecurrence[{6,-15,21,-12,9},{0,0,1,9,42,139},30] (* Harvey P. Dale, Jun 06 2024 *)

Formula

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

a(n) = 6*a(n-1) - 25*a(n-2) + 21*a(n-3) - 12*a(n-4) + 9*a(n-5) for n >= 6.