A291395 - OEIS

A291395

p-INVERT of (1,1,0,0,0,0,...), where p(S) = (1 - 2 S)(1 - 3 S).

5, 24, 103, 425, 1704, 6715, 26153, 101052, 388303, 1486337, 5673840, 21616915, 82244873, 312603348, 1187325847, 4507385921, 17104894344, 64893555547, 246150297257, 933554883084, 3540272085535, 13424640644225, 50903370755040, 193007618806051, 731797403031305

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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 A291382 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 (5, -1, -12, -6)

FORMULA

G.f.: -(((1 + x) (-5 + 6 x + 6 x^2))/((-1 + 2 x + 2 x^2) (-1 + 3 x + 3 x^2))).

a(n) = 5*a(n-1) - a(n-2) - 12*a(n-3) - 6*a(n-4) for n >= 5.

MATHEMATICA

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

Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A019590 *)

u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291395 *)

CROSSREFS