A292482 p-INVERT of the odd positive integers, where p(S) = (1 - S)^2.

2, 9, 32, 112, 384, 1296, 4320, 14256, 46656, 151632, 489888, 1574640, 5038848, 16061328, 51018336, 161558064, 510183360, 1607077584, 5050815264, 15841193328, 49589822592, 154968195600, 483500770272, 1506290861232, 4686238234944, 14560811658576

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, -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]  (* A292482 *)

Formula

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

a(n) = 6*a(n-1) - 9*a(n-2) for n >= 3.

a(n) = 16*3^(n-3)*(4 + n) for n>1. - Colin Barker, Oct 03 2017