0,2

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).

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

Index entries for linear recurrences with constant coefficients, signature (9, -19)

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

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

Cf. A000302, A289780.

Sequence in context: A370036 A079027 A081105 * A161727 A121838 A242629

Adjacent sequences: A289781 A289782 A289783 * A289785 A289786 A289787

nonn,easy

Clark Kimberling, Aug 10 2017

approved