%I
%S 2,9,32,112,384,1296,4320,14256,46656,151632,489888,1574640,5038848,
%T 16061328,51018336,161558064,510183360,1607077584,5050815264,
%U 15841193328,49589822592,154968195600,483500770272,1506290861232,4686238234944,14560811658576
%N pINVERT of the odd positive integers, where p(S) = (1  S)^2.
%C 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 pINVERT 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 pINVERT is a generalization of the "INVERT" transform (e.g., A033453).
%C See A292480 for a guide to related sequences.
%H Clark Kimberling, <a href="/A292482/b292482.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (6, 9)
%F G.f.: ((1 + x) (2  5 x + x^2))/(1 + 3 x)^2.
%F a(n) = 6*a(n1)  9*a(n2) for n >= 3.
%F a(n) = 16*3^(n3)*(4 + n) for n>1.  _Colin Barker_, Oct 03 2017
%t z = 60; s = x (x + 1)/(1  x)^2; p = 1  s^3;
%t Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
%t Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A292482 *)
%Y Cf. A005408, A292480.
%K nonn,easy
%O 0,1
%A _Clark Kimberling_, Oct 02 2017
