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A291223 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S^3 - S^4.

%I A291223 #8 Aug 25 2017 06:23:02
%S A291223 0,0,1,1,3,5,8,17,25,52,83,159,271,497,868,1572,2762,4984,8784,15799,
%T A291223 27939,50089,88831,158880,282293,504179,896780,1600335,2848339,
%U A291223 5080363,9045953,16129172,28726972,51209648,91223508,162594868,289675121,516264093
%N A291223 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S^3 - S^4.
%C A291223 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 A291223 See A291219 for a guide to related sequences.
%H A291223 Clark Kimberling, <a href="/A291223/b291223.txt">Table of n, a(n) for n = 0..1000</a>
%H A291223 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (0, 4, 1, -5, -1, 4, 0, -1)
%F A291223 a(n) = 4*a(n-2) + a(n-3) - 5*a(n-4) - a(n-5) + 4*a(n-6) - a(n-8) for n >= 9.
%F A291223 G.f.: x^2*(1 + x - x^2) / (1 - 4*x^2 - x^3 + 5*x^4 + x^5 - 4*x^6 + x^8). - _Colin Barker_, Aug 25 2017
%t A291223 z = 60; s = x/(1 - x^2); p = 1 - s^3 - s^4;
%t A291223 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
%t A291223 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291223 *)
%o A291223 (PARI) concat(vector(2), Vec(x^2*(1 + x - x^2) / (1 - 4*x^2 - x^3 + 5*x^4 + x^5 - 4*x^6 + x^8) + O(x^50))) \\ _Colin Barker_, Aug 25 2017
%Y A291223 Cf. A000035, A291219.
%K A291223 nonn,easy
%O A291223 0,5
%A A291223 _Clark Kimberling_, Aug 24 2017