A291397 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S - S^3.

1, 2, 4, 10, 22, 46, 97, 209, 452, 972, 2084, 4472, 9608, 20645, 44345, 95238, 204552, 439366, 943734, 2027046, 4353861, 9351633, 20086392, 43143592, 92668072, 199041584, 427521184, 918272425, 1972356577, 4236422746, 9099408124, 19544609858, 41979848918

Offset: 0

Author

Clark Kimberling, Sep 06 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 (1, 1, 1, 3, 3, 1)

Crossrefs

Cf. A019590, A291382.

Programs

Mathematica

z = 60; s = x + x^2; p = 1 - s - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291397 *)

Formula

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

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