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A291038 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = (1 - 2 S)^2.

%I A291038 #4 Sep 14 2017 18:14:58
%S A291038 4,12,32,84,216,544,1348,3300,8000,19236,45936,109056,257604,605820,
%T A291038 1419232,3313396,7711944,17900320,41445764,95746260,220735616,
%U A291038 507934276,1166792160,2676017408,6128381316,14015556588,32012831648,73033858964,166434905016
%N A291038 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = (1 - 2 S)^2.
%C A291038 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 A291038 See A291728 for a guide to related sequences.
%H A291038 Clark Kimberling, <a href="/A291038/b291038.txt">Table of n, a(n) for n = 0..1000</a>
%H A291038 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (4, -4, 2, -4, 0, -1)
%F A291038 G.f.: -((4 (-1 + x + x^3))/(-1 + 2 x + x^3)^2).
%F A291038 a(n) = 4*a(n-1) - 4*a(n-2) + 2*a(n-3) - 4*a(n-4) - a(n-6) for n >= 7.
%t A291038 z = 60; s = x/(x - x^3); p = (1 - 2 s)^2;
%t A291038 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A079978  *)
%t A291038 u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291038 *)
%t A291038 u/4 (* A291039 *)
%Y A291038 Cf. A079978, A290616, A291039.
%K A291038 nonn,easy
%O A291038 0,1
%A A291038 _Clark Kimberling_, Sep 14 2017