A292491 p-INVERT of the odd positive integers, where p(S) = 1 + S - 2 S^2.

-1, 0, 8, 16, 24, 96, 360, 1008, 2808, 8640, 26568, 79056, 235224, 707616, 2128680, 6380208, 19123128, 57386880, 172213128, 516586896, 1549603224, 4648967136, 13947373800, 41841649008, 125523529848, 376572006720, 1129720271688, 3389156563536, 10167456936024

Offset: 0

Author

Clark Kimberling, Oct 03 2017

Keywords

• easy
• sign

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 (3, -3, 9)

Crossrefs

Cf. A005408, A292480.

Programs

Mathematica

z = 60; s = x (x + 1)/(1 - x)^2; p = 1 + s - 2 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292491 *)

PARI

x='x+O('x^99); Vec(((1+x)*(-1+4*x+x^2))/((1-3*x)*(1+3*x^2))) \\ Altug Alkan, Oct 03 2017

Formula

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

a(n) = 3*a(n-1) - 3*a(n-2) + 9*a(n-3) for n >= 5.