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A291019 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S - S^2 - S^3 + S^4.

%I A291019 #4 Aug 23 2017 23:37:28
%S A291019 1,3,9,25,68,185,504,1373,3739,10180,27714,75445,205376,559064,
%T A291019 1521840,4142609,11276581,30695881,83556891,227449066,619135745,
%U A291019 1685339900,4587637263,12487934387,33993205996,92532358762,251880840375,685640764594,1866371634554
%N A291019 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S - S^2 - S^3 + S^4.
%C A291019 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 A291019 See A291000 for a guide to related sequences.
%H A291019 Clark Kimberling, <a href="/A291019/b291019.txt">Table of n, a(n) for n = 0..1000</a>
%H A291019 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (5, -8, 6, -3)
%F A291019 G.f.: (1 - 2 x + 2 x^2 - 2 x^3)/(1 - 5 x + 8 x^2 - 6 x^3 + 3 x^4).
%F A291019 a(n) = 5*a(n-1) - 8*a(n-2) + 6*a(n-3) - 3*a(n-4) for n >= 5.
%t A291019 z = 60; s = x/(1 - x); p = 1 - s - s^2 - s^3 + s^4;
%t A291019 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
%t A291019 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291019 *)
%Y A291019 Cf. A000012, A289780, A291000.
%K A291019 nonn,easy
%O A291019 0,2
%A A291019 _Clark Kimberling_, Aug 23 2017