A289843 - OEIS

A289843

p-INVERT of (1,0,2,0,3,0,4,0,5,...) (A027656), where p(S) = 1 - S - S^2.

1, 2, 5, 13, 29, 73, 168, 410, 962, 2317, 5483, 13131, 31193, 74509, 177311, 423025, 1007505, 2402354, 5723761, 13644587, 32514730, 77501115, 184698088, 440216833, 1049148789, 2500520812, 5959478837, 14203542282, 33851496564, 80679640434, 192285583548

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OFFSET

0,2

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 A289780 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 (1, 5, -2, -6, 1, 4, 0, -1)

FORMULA

G.f.: (1 + x - 2 x^2 + x^4)/(1 - x - 5 x^2 + 2 x^3 + 6 x^4 - x^5 - 4 x^6 + x^8).

a(n) = a(n-1) + 5*a(n-2) - 2*a(n-3) - 6*a(n-4) + a(n-5) + 4*a(n-6) - a(n-8).

MATHEMATICA

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

Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A027656 *)

Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289843 *)

CROSSREFS