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A291417 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 4 S + 2 S^2.

%I A291417 #4 Sep 11 2017 12:06:20
%S A291417 4,18,76,322,1360,5744,24256,102428,432528,1826456,7712656,32568568,
%T A291417 137528704,580748416,2452351488,10355650832,43729255232,184657419808,
%U A291417 779760883392,3292730050592,13904353779456,58714516845824,247936332973056,1046971490364864
%N A291417 p-INVERT of (1,1,0,0,0,0,...), where p(S) = 1 - 4 S + 2 S^2.
%C A291417 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 A291417 See A291382 for a guide to related sequences.
%H A291417 Clark Kimberling, <a href="/A291417/b291417.txt">Table of n, a(n) for n = 0..1000</a>
%H A291417 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4, 2, -4, -2)
%F A291417 G.f.: -((2 (-1 + x) (1 + x) (2 + x))/(1 - 4 x - 2 x^2 + 4 x^3 + 2 x^4)).
%F A291417 a(n) = 4*a(n-1) + 2*a(n-2) - 4*a(n-3) - 2*a(n-4) for n >= 5.
%t A291417 z = 60; s = x + x^2; p = 1 - 4 s + 2 s^2;
%t A291417 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A019590 *)
%t A291417 u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291417 *)
%t A291417 u / 2  (*A291462)
%Y A291417 Cf. A019590, A291382, A291462.
%K A291417 nonn,easy
%O A291417 0,1
%A A291417 _Clark Kimberling_, Sep 07 2017