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A292486 p-INVERT of the odd positive integers, where p(S) = 1 - S - 3 S^2.

%I A292486 #8 Sep 29 2024 18:35:56
%S A292486 1,7,36,165,747,3420,15705,72063,330516,1515933,6953283,31893516,
%T A292486 146289393,671000247,3077745156,14117009877,64751939163,297004363452,
%U A292486 1362300384969,6248602953135,28661108314356,131462846314317,602994126047283,2765815028667756
%N A292486 p-INVERT of the odd positive integers, where p(S) = 1 - S - 3 S^2.
%C A292486 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 A292486 See A292480 for a guide to related sequences.
%H A292486 Clark Kimberling, <a href="/A292486/b292486.txt">Table of n, a(n) for n = 0..1000</a>
%H A292486 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (5, -4, 9, 3)
%F A292486 G.f.: -(((1 + x) (1 + x + 4 x^2))/(-1 + 5 x - 4 x^2 + 9 x^3 + 3 x^4)).
%F A292486 a(n) = 5*a(n-1) - 4*a(n-2) + 9*a(n-3) + 3*a(n-4)  for n >= 5.
%t A292486 z = 60; s = x (x + 1)/(1 - x)^2; p = 1 - s - 3 s^2;
%t A292486 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A005408 *)
%t A292486 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292486 *)
%t A292486 LinearRecurrence[{5,-4,9,3},{1,7,36,165},30] (* _Harvey P. Dale_, Sep 29 2024 *)
%o A292486 (PARI) x='x+O('x^99); Vec(((1+x)*(1+x+4*x^2))/(1-5*x+4*x^2-9*x^3-3*x^4)) \\ _Altug Alkan_, Oct 03 2017
%Y A292486 Cf. A005408, A292480.
%K A292486 nonn,easy
%O A292486 0,2
%A A292486 _Clark Kimberling_, Oct 03 2017