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A291732 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = (1 - 2 S)^2.

%I A291732 #4 Sep 11 2017 20:04:49
%S A291732 4,12,36,104,288,780,2080,5472,14240,36736,94080,239440,606144,
%T A291732 1527360,3833024,9584768,23890944,59380160,147207168,364084224,
%U A291732 898569216,2213388288,5442392064,13360097536,32746992640,80153705472,195933828096,478374127616
%N A291732 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = (1 - 2 S)^2.
%C A291732 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 A291732 See A291728 for a guide to related sequences.
%H A291732 Clark Kimberling, <a href="/A291732/b291732.txt">Table of n, a(n) for n = 0..1000</a>
%H A291732 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (4, -4, 4, -8, 0, -4)
%F A291732 G.f.: -((4 (1 + x^2) (-1 + x + x^3))/(-1 + 2 x + 2 x^3)^2).
%F A291732 a(n) = 4*a(n-1) - 4*a(n-2) + 4*a(n-3) - 8*a(n-4) - 4*a(n-6) for n >= 7.
%t A291732 z = 60; s = x + x^3; p = (1 - 2 s)^2;
%t A291732 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
%t A291732 u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291732 *)
%t A291732 u / 4  (*A291733)
%Y A291732 Cf. A154272, A291728, A291733.
%K A291732 nonn,easy
%O A291732 0,1
%A A291732 _Clark Kimberling_, Sep 11 2017