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A292487 p-INVERT of the odd positive integers, where p(S) = 1 - S - 4 S^2.

%I A292487 #6 Oct 03 2017 20:53:01
%S A292487 1,8,44,212,1020,4980,24348,118868,580156,2831924,13824092,67481876,
%T A292487 329408892,1607991540,7849328028,38316090836,187038012604,
%U A292487 913016364980,4456842098396,21755843899028,106200025265148,518409923170932,2530591191342108,12352949840710484
%N A292487 p-INVERT of the odd positive integers, where p(S) = 1 - S - 4 S^2.
%C A292487 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 A292487 See A292480 for a guide to related sequences.
%H A292487 Clark Kimberling, <a href="/A292487/b292487.txt">Table of n, a(n) for n = 0..1000</a>
%H A292487 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (5, -3, 11, 4)
%F A292487 G.f.: -(((1 + x) (1 + 2 x + 5 x^2))/(-1 + 5 x - 3 x^2 + 11 x^3 + 4 x^4)).
%F A292487 a(n) = 5*a(n-1) - 3*a(n-2) + 11*a(n-3) + 4*a(n-4)  for n >= 5.
%t A292487 z = 60; s = x (x + 1)/(1 - x)^2; p = 1 - s - 4 s^2;
%t A292487 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
%t A292487 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292487 *)
%o A292487 (PARI) x='x+O('x^99); Vec(((1+x)*(1+2*x+5*x^2))/(1-5*x+3*x^2-11*x^3-4*x^4)) \\ _Altug Alkan_, Oct 03 2017
%Y A292487 Cf. A005408, A292480.
%K A292487 nonn,easy
%O A292487 0,2
%A A292487 _Clark Kimberling_, Oct 03 2017