A291245 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - 5 S + S^2.

5, 24, 120, 599, 2990, 14925, 74500, 371876, 1856265, 9265776, 46251265, 230868900, 1152410620, 5752399899, 28713814350, 143328549649, 715442152480, 3571217840400, 17826174791885, 88981552487776, 444162405876285, 2217091490069376, 11066885918992400

Offset: 0

Author

Clark Kimberling, Aug 28 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 (5,1,-5,-1)

Crossrefs

Cf. A000035, A291219.

Programs

Mathematica

z = 60; s = x/(1 - x^2); p = 1 - 5 s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291245 *)

Formula

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

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