A291142 - OEIS

A291142

a(n) = (1/4)*A291024(n).

0, 1, 2, 6, 16, 43, 114, 300, 784, 2037, 5266, 13554, 34752, 88799, 226210, 574680, 1456352, 3682409, 9292002, 23403102, 58842416, 147713043, 370262930, 926852868, 2317191024, 5786293597, 14433117938, 35964267594, 89528469088, 222666575815, 553319176770

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OFFSET

0,3

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 A291000 for a guide to related sequences.

LINKS

Clark Kimberling, Table of n, a(n) for n = 0..999

Index entries for linear recurrences with constant coefficients, signature (4, -2, -4, -1)

FORMULA

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

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

a(n) = (1/4)*A291024(n) for n >= 0.

a(n) = ((1+sqrt(2))^n*(3*sqrt(2) + 2*(-1+sqrt(2))*n) - (1-sqrt(2))^n*(3*sqrt(2) + 2*(1+sqrt(2))*n)) / 16. - Colin Barker, Aug 24 2017

MATHEMATICA

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

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