A289789 p-INVERT of A016777, where p(S) = 1 - S - S^2.

1, 6, 26, 111, 460, 1905, 7910, 32880, 136675, 568050, 2360825, 9811650, 40777750, 169474875, 704348000, 2927312625, 12166086250, 50562982500, 210142784375, 873366003750, 3629761440625, 15085506018750, 62696266831250, 260569441284375, 1082942209562500

Offset: 0

Author

Clark Kimberling, Aug 11 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 A289780 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 (5, -5, 5, 5)

Crossrefs

Cf. A016777, A289780.

Programs

Mathematica

z = 60; s = x (1 + 2*x)/(1 - x)^2; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A016777 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289789 *)

Formula

G.f.: (-1 - x - x^2 - 6 x^3)/(-1 + 5 x - 5 x^2 + 5 x^3 + 5 x^4).

a(n) = 5*a(n-1) - 5*a(n-2) + 5*a(n-3) + 5*a(n-4).