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A292401 p-INVERT of (1,0,2,0,2,0,2,0,2,0,...), where p(S) = (1 - S)^2.

%I A292401 #4 Sep 30 2017 16:05:31
%S A292401 2,3,8,17,34,71,144,289,578,1147,2264,4449,8706,16975,32992,63937,
%T A292401 123586,238323,458600,880753,1688482,3231639,6175728,11785313,
%U A292401 22460802,42754283,81290424,154396097,292953858,555334047,1051781312,1990373249,3763583618
%N A292401 p-INVERT of (1,0,2,0,2,0,2,0,2,0,...), where p(S) = (1 - S)^2.
%C A292401 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).
%H A292401 Clark Kimberling, <a href="/A292401/b292401.txt">Table of n, a(n) for n = 0..1000</a>
%H A292401 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (2, 1, 0, -3, -2, -1)
%F A292401 G.f.: -(((1 + x^2) (-2 + x + 2 x^2 + x^3))/(-1 + x + x^2 + x^3)^2).
%F A292401 a(n) = 2*a(n-1) + a(n-2) - 3*a(n-4) - 2*a(n-5) - s(n-6) for n >= 7.
%t A292401 z = 60; s = x (x^2 + 1)/(1 - x^2); p = (1 - s)^2;
%t A292401 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* abs. values of A176742 *)
%t A292401 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292401 *)
%Y A292401 Cf. A176742, A292400.
%K A292401 nonn,easy
%O A292401 0,1
%A A292401 _Clark Kimberling_, Sep 30 2017