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A291023 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - 3 S^2 + 2 S^3.

%I A291023 #6 Aug 24 2017 12:58:08
%S A291023 0,3,4,12,24,56,120,264,568,1224,2616,5576,11832,25032,52792,111048,
%T A291023 233016,487880,1019448,2126280,4427320,9204168,19107384,39612872,
%U A291023 82021944,169636296,350457400,723284424,1491308088,3072094664,6323146296,13004206536,26724240952
%N A291023 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - 3 S^2 + 2 S^3.
%C A291023 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 A291023 See A291000 for a guide to related sequences.
%H A291023 Clark Kimberling, <a href="/A291023/b291023.txt">Table of n, a(n) for n = 0..1000</a>
%H A291023 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3, 0, -4)
%F A291023 G.f.: (3 x - 5 x^2)/((1 + x) (-1 + 2 x)^2).
%F A291023 a(n) = 3*a(n-1) - 4*a(n-3) for n >= 4.
%F A291023 a(n) = (16*((-1)^(1+n) + 2^n) + 3*2^n*n) / 18. - _Colin Barker_, Aug 24 2017
%t A291023 z = 60; s = x/(1 - x); p = 1 - 3 s^2 + 2 s^3;
%t A291023 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
%t A291023 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291023 *)
%o A291023 (PARI) concat(0, Vec(x*(3 - 5*x) / ((1 + x)*(1 - 2*x)^2) + O(x^40))) \\ _Colin Barker_, Aug 24 2017
%Y A291023 Cf. A000012, A289780, A291000.
%K A291023 nonn,easy
%O A291023 0,2
%A A291023 _Clark Kimberling_, Aug 24 2017