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A292491 p-INVERT of the odd positive integers, where p(S) = 1 + S - 2 S^2.

%I A292491 #8 Oct 03 2017 20:53:51
%S A292491 -1,0,8,16,24,96,360,1008,2808,8640,26568,79056,235224,707616,2128680,
%T A292491 6380208,19123128,57386880,172213128,516586896,1549603224,4648967136,
%U A292491 13947373800,41841649008,125523529848,376572006720,1129720271688,3389156563536,10167456936024
%N A292491 p-INVERT of the odd positive integers, where p(S) = 1 + S - 2 S^2.
%C A292491 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 A292491 See A292480 for a guide to related sequences.
%H A292491 Clark Kimberling, <a href="/A292491/b292491.txt">Table of n, a(n) for n = 0..1000</a>
%H A292491 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3, -3, 9)
%F A292491 G.f.: -(((1 + x) (-1 + 4 x + x^2))/((-1 + 3 x) (1 + 3 x^2))).
%F A292491 a(n) = 3*a(n-1) - 3*a(n-2) + 9*a(n-3)  for n >= 5.
%t A292491 z = 60; s = x (x + 1)/(1 - x)^2; p = 1 + s - 2 s^2;
%t A292491 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
%t A292491 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292491 *)
%o A292491 (PARI) x='x+O('x^99); Vec(((1+x)*(-1+4*x+x^2))/((1-3*x)*(1+3*x^2))) \\ _Altug Alkan_, Oct 03 2017
%Y A292491 Cf. A005408, A292480.
%K A292491 easy,sign
%O A292491 0,3
%A A292491 _Clark Kimberling_, Oct 03 2017