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A292493 p-INVERT of the odd positive integers, where p(S) = 1 + S - 3 S^2.

%I A292493 #6 Oct 05 2017 09:50:42
%S A292493 -1,1,12,25,61,266,963,3053,10220,35413,120345,405682,1376119,4676201,
%T A292493 15859212,53768225,182400581,618792826,2098887003,7119249973,
%U A292493 24149097580,81915342653,277858469505,942504046562,3197013067439,10844389616401,36784545696012
%N A292493 p-INVERT of the odd positive integers, where p(S) = 1 + S - 3 S^2.
%C A292493 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 A292493 See A292480 for a guide to related sequences.
%H A292493 Clark Kimberling, <a href="/A292493/b292493.txt">Table of n, a(n) for n = 0..1000</a>
%H A292493 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (3, -2, 11, 1)
%F A292493 G.f.: -(((1 + x) (-1 + 5 x + 2 x^2))/(-1 + 3 x - 2 x^2 + 11 x^3 + x^4)).
%F A292493 a(n) = 3*a(n-1) - 2*a(n-2) + 11*a(n-3) + a(n-4) for n >= 5.
%t A292493 z = 60; s = x (x + 1)/(1 - x)^2; p = 1 + s + 3 s^2;
%t A292493 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
%t A292493 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292493 *)
%o A292493 (PARI) x='x+O('x^99); Vec(((1+x)*(1-5*x-2*x^2))/(-1+3*x-2*x^2+11*x^3+x^4)) \\ _Altug Alkan_, Oct 05 2017
%Y A292493 Cf. A005408, A292480.
%K A292493 easy,sign
%O A292493 0,3
%A A292493 _Clark Kimberling_, Oct 04 2017