A291227 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - 2*S^2.

1, 3, 6, 17, 37, 96, 221, 551, 1302, 3189, 7625, 18528, 44537, 107835, 259830, 628105, 1515053, 3659808, 8832085, 21328159, 51481638, 124302381, 300068689, 724468416, 1748959153, 4222461747, 10193761254, 24610180673, 59413804789, 143438304480, 346289581709

Offset: 0

Author

Clark Kimberling, Aug 25 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 A291219 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 (1,4,-1,-1).

Crossrefs

Cf. A000035, A291219.

Programs

Mathematica

z = 60; s = x/(1 - x^2); p = 1 - s - 2 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291227 *)

Formula

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

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

a(n-1) = (4*A000129(n) + (-1)^n*A000045(n))/3 for n >= 1. - Greg Dresden, Jan 01 2021