0,1

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 A291000 for a guide to related sequences.

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

Index entries for linear recurrences with constant coefficients, signature (6,-8).

a(n) = 6*a(n-1) - 8*a(n-2) for n >= 3.

From Colin Barker, Aug 23 2017: (Start)

G.f.: (4 - 7*x) / ((1 - 2*x)*(1 - 4*x)).

a(n) = 2^(n-1) * (9*2^n - 1). (End)

E.g.f.: (1/2)*(9*exp(4*x) - exp(2*x)). - Enrique Navarrete, Apr 03 2026

z = 60; s = x/(1-x); p = (1-s)(1-3s);

Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000012 *)

Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291009 *)

LinearRecurrence[{6, -8}, {4, 17}, 40] (* G. C. Greubel, Jun 04 2023 *)

(PARI) Vec((4-7*x)/((1-2*x)*(1-4*x)) + O(x^30)) \\ Colin Barker, Aug 23 2017

(Magma) [2^(n-1)*(9*2^n -1): n in [0..40]]; // G. C. Greubel, Jun 04 2023

(SageMath)

A291009=BinaryRecurrenceSequence(6, -8, 4, 17)

[A291009(n) for n in range(41)] # G. C. Greubel, Jun 04 2023

First differences of A010036.

Cf. A000012, A033453, A289780, A291000.

Sequence in context: A349695 A003127 A172447 * A017955 A017956 A136792

Adjacent sequences: A291006 A291007 A291008 * A291010 A291011 A291012

nonn,easy,changed

Clark Kimberling, Aug 23 2017

approved