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%I #12 Aug 15 2017 19:30:23
%S 1,4,15,53,187,656,2301,8071,28308,99293,348275,1221603,4284864,
%T 15029495,52717114,184909361,648583888,2274958177,7979591823,
%U 27989035739,98173708464,344351878525,1207840857737,4236595263812,14860185689435,52123251095327
%N p-INVERT of A014217 (starting at n=1), where p(S) = 1 - S - S^2.
%C 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).
%C See A289780 for a guide to related sequences.
%F Conjectures from _Colin Barker_, Aug 15 2017: (Start)
%F 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).
%F 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.
%F (End)
%t z = 60; r = GoldenRatio; s = Sum[Floor[r^k] x^k, {k, 1, z}]; p = 1 - s - s^2;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A014217 shifted *)
%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x] , 1] (* A289927 *)
%Y Cf. A014217, A289780.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Aug 14 2017
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