A289927 - OEIS

A289927

p-INVERT of A014217 (starting at n=1), where p(S) = 1 - S - S^2.

1, 4, 15, 53, 187, 656, 2301, 8071, 28308, 99293, 348275, 1221603, 4284864, 15029495, 52717114, 184909361, 648583888, 2274958177, 7979591823, 27989035739, 98173708464, 344351878525, 1207840857737, 4236595263812, 14860185689435, 52123251095327

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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

Table of n, a(n) for n=0..25.

FORMULA

Conjectures from Colin Barker, Aug 15 2017: (Start)

G.f.: (1 - x^2 + x^3)*(1 + x - x^3) / (1 - 3*x - 4*x^2 + 7*x^3 + 5*x^4 - 7*x^5 - 4*x^6 + 3*x^7 + x^8).

a(n) = 3*a(n-1) + 4*a(n-2) - 7*a(n-3) - 5*a(n-4) + 7*a(n-5) + 4*a(n-6) - 3*a(n-7) - a(n-8) for n>7.

(End)

MATHEMATICA

z = 60; r = GoldenRatio; s = Sum[Floor[r^k] x^k, {k, 1, z}]; p = 1 - s - s^2;

Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A014217 shifted *)

Drop[CoefficientList[Series[1/p, {x, 0, z}], x] , 1] (* A289927 *)