A289790 p-INVERT of A016789, where p(S) = 1 - S - S^2.

2, 13, 72, 385, 2056, 11000, 58872, 315065, 1686086, 9023167, 48287964, 258415702, 1382925814, 7400803253, 39605804028, 211952630117, 1134276112400, 6070140759292, 32484690838716, 173843602765153, 930333564584074, 4978731041147699, 26643951936925764

Offset: 0

Author

Clark Kimberling, Aug 12 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 (6, -5, 8, 1)

Crossrefs

Cf. A016789, A289780.

Programs

Mathematica

z = 60; s = x (2 + x)/(1 - x)^2; p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A016789 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289790 *)

Formula

G.f.: (-2 - x - 4 x^2 - 2 x^3)/(-1 + 6 x - 5 x^2 + 8 x^3 + x^4).

a(n) = 6*a(n-1) - 5*a(n-2) + 8*a(n-3) + a(n-4).