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A291405 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 2 S^2 - 2 S^4.

%I A291405 #4 Sep 07 2017 21:30:21
%S A291405 0,2,4,8,24,52,120,290,672,1592,3760,8824,20800,48976,115296,271588,
%T A291405 639488,1505816,3546032,8350064,19662944,46302800,109033952,256754760,
%U A291405 604609280,1423740736,3352643712,7894846528,18590881280,43778039424,103089066752
%N A291405 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 2 S^2 - 2 S^4.
%C A291405 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 A291405 See A291382 for a guide to related sequences.
%H A291405 Clark Kimberling, <a href="/A291405/b291405.txt">Table of n, a(n) for n = 0..1000</a>
%H A291405 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (0, 2, 4, 4, 8, 12, 8, 2)
%F A291405 G.f.: -((2 x (1 + x)^2 (1 + x^2 + 2 x^3 + x^4))/(-1 + 2 x^2 + 4 x^3 + 4 x^4 + 8 x^5 + 12 x^6 + 8 x^7 + 2 x^8)).
%F A291405 a(n) = 2*a(n-2) + 4*a(n-3) + 4*a(n-4) + 5*a(n-5) + 12*a(n-6) + 8*a(n-7) + 2*a(n-8) for n >= 9.
%t A291405 z = 60; s = x + x^2; p = 1 - 2 s^2 - 2 s^4;
%t A291405 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
%t A291405 u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291405 *)
%t A291405 u / 2 (* A291406 *)
%Y A291405 Cf. A019590, A291382, A291406.
%K A291405 nonn,easy
%O A291405 0,2
%A A291405 _Clark Kimberling_, Sep 07 2017