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A291220 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^4.

%I A291220 #8 Aug 25 2017 06:22:29
%S A291220 1,1,2,4,7,15,27,55,101,199,370,718,1347,2595,4898,9397,17803,34066,
%T A291220 64682,123561,234917,448289,852979,1626689,3096695,5903316,11241426,
%U A291220 21424775,40805833,77759648,148118585,282229961,537636210,1024373916,1951472023,3718072991
%N A291220 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - S - S^4.
%C A291220 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 A291220 See A291219 for a guide to related sequences.
%H A291220 Clark Kimberling, <a href="/A291220/b291220.txt">Table of n, a(n) for n = 0..1000</a>
%H A291220 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (1, 4, -3, -5, 3, 4, -1, -1)
%F A291220 a(n) = a(n-1) + 4*a(n-2) - 3*a(n-3) - 5*a(n-4) + 3*a(n-5) + 4*a(n-6) - a(n-7) - a(n-8) for n >= 9.
%F A291220 G.f.: (1 + x - x^2)*(1 - x - x^2 + x^3 + x^4) / ((1 - x - 2*x^2 + x^4)*(1 - 2*x^2 + x^3 + x^4)). - _Colin Barker_, Aug 25 2017
%t A291220 z = 60; s = x/(1 - x^2); p = 1 - s - s^4;
%t A291220 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
%t A291220 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291220 *)
%o A291220 (PARI) Vec((1 + x - x^2)*(1 - x - x^2 + x^3 + x^4) / ((1 - x - 2*x^2 + x^4)*(1 - 2*x^2 + x^3 + x^4)) + O(x^30)) \\ _Colin Barker_, Aug 25 2017
%Y A291220 Cf. A000035, A291219.
%K A291220 nonn,easy
%O A291220 0,3
%A A291220 _Clark Kimberling_, Aug 24 2017