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A291222 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S^2 - S^3.

%I A291222 #8 Aug 25 2017 06:22:52
%S A291222 0,1,1,3,5,9,19,30,66,106,223,379,753,1345,2565,4723,8816,16456,30480,
%T A291222 57093,105677,197751,366697,684765,1272311,2371846,4412898,8218386,
%U A291222 15300891,28483823,53042669,98734485,183863833,342263703,637320032,1186464528,2209131168
%N A291222 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S^2 - S^3.
%C A291222 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 A291222 See A291219 for a guide to related sequences.
%H A291222 Clark Kimberling, <a href="/A291222/b291222.txt">Table of n, a(n) for n = 0..1000</a>
%H A291222 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0, 4, 1, -4, 0, 1)
%F A291222 a(n) = 4*a(n-2) + a(n-3) - 4*a(n-4) + a(n-6) for n >= 7.
%F A291222 G.f.: x*(1 + x - x^2) / (1 - 4*x^2 - x^3 + 4*x^4 - x^6). - _Colin Barker_, Aug 25 2017
%t A291222 z = 60; s = x/(1 - x^2); p = 1 - s^2 - s^3;
%t A291222 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
%t A291222 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291222 *)
%o A291222 (PARI) concat(0, Vec(x*(1 + x - x^2) / (1 - 4*x^2 - x^3 + 4*x^4 - x^6) + O(x^40))) \\ _Colin Barker_, Aug 25 2017
%Y A291222 Cf. A000035, A291219.
%K A291222 nonn,easy
%O A291222 0,4
%A A291222 _Clark Kimberling_, Aug 24 2017