A291265 - OEIS

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A291265

a(n) = (1/3)*A291232(n).

2, 9, 38, 153, 596, 2268, 8480, 31275, 114086, 412443, 1479926, 5276664, 18711758, 66041901, 232129190, 812934621, 2837740232, 9877082004, 34288573484, 118752490863, 410394698534, 1415492232255, 4873386985130, 16750755602928, 57487476629594, 197013756414033

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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).

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 (6, -7, -6, -1)

FORMULA

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

a(n) = 6*a(n-1) - 7*a(n-2) -6*a(n-3) - a(n-4) for n >= 5.

a(n) = (((3-sqrt(13))/2)^n*(-3+sqrt(13))*(-39+17*sqrt(13)-39*n) + 2^(-n)*(3+sqrt(13))^(1+n)*(39+17*sqrt(13)+39*n)) / 338. - Colin Barker, Aug 26 2017

MATHEMATICA

z = 60; s = x/(1 - x^2); p = (1 - 3 s)^2;

Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000035 *)

u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291232 *)