A292329 p-INVERT of the Fibonacci sequence (A000045), where p(S) = 1 - S^3.

0, 0, 1, 3, 9, 23, 57, 138, 332, 798, 1920, 4626, 11157, 26925, 64997, 156921, 378861, 914692, 2208324, 5331444, 12871324, 31074180, 75019701, 181113471, 437246349, 1055605659, 2548456957, 6152518758, 14853493752, 35859505946, 86572506132, 209004519918

Offset: 0

Author

Clark Kimberling, Sep 15 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).

Links

• Clark Kimberling, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (3, 0, -4, 0, 3, 1)

Crossrefs

Cf. A000045, A292328.

Programs

Mathematica

z = 60; s = x/(1 - x - x^2); p = 1 - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000045 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292329 *)

Formula

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

a(n) = 3*a(n-1) - 4*a(n-3) + 3*a(n-5) + a(n-6) for n >= 7.