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A291229 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)(1 - 2 S).

%I A291229 #4 Aug 25 2017 23:39:57
%S A291229 3,7,18,45,111,272,663,1611,3906,9457,22875,55296,133611,322751,
%T A291229 779490,1882341,4545159,10974256,26496255,63970947,154444914,
%U A291229 372871721,900206067,2173312512,5246877459,12667142455,30581283762,73829906397,178241414367,430313249360
%N A291229 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S)(1 - 2 S).
%C A291229 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 A291229 See A291219 for a guide to related sequences.
%H A291229 Clark Kimberling, <a href="/A291229/b291229.txt">Table of n, a(n) for n = 0..1000</a>
%H A291229 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (3, 0, -3, -1)
%F A291229 G.f.: -((-3 + 2 x + 3 x^2)/((-1 + x + x^2) (-1 + 2 x + x^2))).
%F A291229 a(n) = 3*a(n-1) - 2*a(n-3) - a(n-4) for n >= 5.
%t A291229 z = 60; s = x/(1 - x^2); p = (1 - s)(1 - 2 s);
%t A291229 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
%t A291229 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291229 *)
%Y A291229 Cf. A000035, A291219.
%K A291229 nonn,easy
%O A291229 0,1
%A A291229 _Clark Kimberling_, Aug 25 2017