A291414 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 2 S + S^3.

2, 6, 15, 37, 89, 212, 500, 1173, 2741, 6388, 14860, 34524, 80138, 185904, 431075, 999280, 2315960, 5366755, 12435075, 28810731, 66748062, 154635086, 358234115, 829886167, 1922494024, 4453566092, 10316878400, 23899399028, 55363614076, 128251081293

Offset: 0

Author

Clark Kimberling, Sep 07 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 A291382 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 (2, 2, -1, -3, -3, -1)

Crossrefs

Cf. A019590, A291382.

Programs

Mathematica

z = 60; s = x + x^2; p = 1 - 2 s + s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291414 *)

Formula

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

a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3) - 3*a(n-4) - 3*a(n-5) - a(n-6) for n >= 7.