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A292482 p-INVERT of the odd positive integers, where p(S) = (1 - S)^2.

%I A292482 #6 Oct 03 2017 08:48:51
%S A292482 2,9,32,112,384,1296,4320,14256,46656,151632,489888,1574640,5038848,
%T A292482 16061328,51018336,161558064,510183360,1607077584,5050815264,
%U A292482 15841193328,49589822592,154968195600,483500770272,1506290861232,4686238234944,14560811658576
%N A292482 p-INVERT of the odd positive integers, where p(S) = (1 - S)^2.
%C A292482 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 A292482 See A292480 for a guide to related sequences.
%H A292482 Clark Kimberling, <a href="/A292482/b292482.txt">Table of n, a(n) for n = 0..1000</a>
%H A292482 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (6, -9)
%F A292482 G.f.: ((1 + x) (2 - 5 x + x^2))/(-1 + 3 x)^2.
%F A292482 a(n) = 6*a(n-1) - 9*a(n-2) for n >= 3.
%F A292482 a(n) = 16*3^(n-3)*(4 + n) for n>1. - _Colin Barker_, Oct 03 2017
%t A292482 z = 60; s = x (x + 1)/(1 - x)^2; p = 1 - s^3;
%t A292482 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
%t A292482 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292482 *)
%Y A292482 Cf. A005408, A292480.
%K A292482 nonn,easy
%O A292482 0,1
%A A292482 _Clark Kimberling_, Oct 02 2017