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A291404 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S^2 - 2 S^4.

%I A291404 #4 Sep 07 2017 21:29:56
%S A291404 0,1,2,4,12,23,42,89,188,404,856,1763,3652,7641,16030,33612,70252,
%T A291404 146623,306334,640637,1340024,2802056,5857264,12243403,25596040,
%U A291404 53515853,111889138,233922392,489039852,1022399607,2137493106,4468804953,9342779572,19532522316
%N A291404 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - S^2 - 2 S^4.
%C A291404 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 A291404 See A291382 for a guide to related sequences.
%H A291404 Clark Kimberling, <a href="/A291404/b291404.txt">Table of n, a(n) for n = 0..1000</a>
%H A291404 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (0, 1, 2, 3, 8, 12, 8, 2)
%F A291404 G.f.: -((x (1 + x)^2 (1 + 2 x^2 + 4 x^3 + 2 x^4))/((1 + x^2 + 2 x^3 + x^4) (-1 + 2 x^2 + 4 x^3 + 2 x^4))).
%F A291404 a(n) = a(n-2) + 2*a(n-3) + 3*a(n-4) + 8*a(n-5) + 12*a(n-6) + 8*a(n-7) + 2*a(n-8) for n >= 9.
%t A291404 z = 60; s = x + x^2; p = 1 - s^2 - 2 s^4;
%t A291404 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
%t A291404 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291404 *)
%Y A291404 Cf. A019590, A291382.
%K A291404 nonn,easy
%O A291404 0,3
%A A291404 _Clark Kimberling_, Sep 07 2017