A292399 - OEIS

A292399

p-INVERT of (1,2,3,5,8,...) (distinct Fibonacci numbers), where p(S) = (1 - S)^2.

2, 7, 22, 69, 212, 644, 1936, 5772, 17088, 50288, 147232, 429136, 1245888, 3604544, 10396160, 29900992, 85784064, 245548800, 701402624, 1999734016, 5691409408, 16172221440, 45885403136, 130011401216, 367902195712, 1039836672000, 2935713865728, 8279592292352

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,1

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).

LINKS

Clark Kimberling, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (4,0,-8,-4)

FORMULA

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

a(n) = 4*a(n-1) - 8*a(n-3) - 4*a(n-4) for n >= 5.

a(n) = Sum_{k=0..n+1} (k+1) * A155112(n+1,k). - Alois P. Heinz, Sep 29 2022

MATHEMATICA

z = 60; s = x (x + 1)/(1 - x - x^2); p = (1 - s)^2;

Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000045 *)

Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A292399 *)

CROSSREFS