A289804 p-INVERT of the even bisection (A001519) of the Fibonacci numbers, where p(S) = 1 - S - S^2.

1, 3, 9, 29, 96, 321, 1077, 3617, 12149, 40802, 137009, 459991, 1544169, 5183201, 17396800, 58387097, 195950657, 657602545, 2206838633, 7405775266, 24852220929, 83398067755, 279861976377, 939138581941, 3151475258656, 10575403936625, 35487807890381

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 (7, -15, 9, 1)

Crossrefs

Cf. A001519, A289780, A289803.

Programs

Mathematica

z = 60; s = x (1 - 2*x)/(1 - 3*x + x^2); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A001519 shifted *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289804 *)

Formula

G.f.: (-1 + 4 x - 3 x^2 - 2 x^3)/(-1 + 7 x - 15 x^2 + 9 x^3 + x^4).

a(n) = 7*a(n-1) - 15*a(n-2) + 9*a(n-3) + a(n-4).