A289784 p-INVERT of the (4^n), where p(S) = 1 - S - S^2.

1, 6, 35, 201, 1144, 6477, 36557, 205950, 1158967, 6517653, 36638504, 205911129, 1157068585, 6501305814, 36527449211, 205222232433, 1152978556888, 6477584595765, 36391668781013, 204450911709582, 1148616498546991, 6452981164440861, 36253117007574920

Offset: 0

Author

Clark Kimberling, Aug 10 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).

Links

• Clark Kimberling, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (9, -19)

Crossrefs

Cf. A000302, A289780.

Programs

Mathematica

z = 60; s = x/(1 - 4*x); p = 1 - s - s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000302 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289784 *)

PARI

Vec(x*(1 - 3*x) / (1 - 9*x + 19*x^2) + O(x^30)) \\ Colin Barker, Aug 11 2017

Formula

G.f.: (1 - 3 x)/(1 - 9 x + 19 x^2).

a(n) = 9*a(n-1) - 19*a(n-2).

a(n) = (2^(-2-n)*((9-sqrt(5))^(n+1)*(-11+3*sqrt(5)) + (9+sqrt(5))^(n+1)*(11+3*sqrt(5)))) / (19*sqrt(5)). - Colin Barker, Aug 11 2017

a(n) = A081574(n+1)-3*A081574(n). - R. J. Mathar, Jul 08 2022