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A291265 a(n) = (1/3)*A291232(n).

%I A291265 #8 Aug 26 2017 21:37:12
%S A291265 2,9,38,153,596,2268,8480,31275,114086,412443,1479926,5276664,
%T A291265 18711758,66041901,232129190,812934621,2837740232,9877082004,
%U A291265 34288573484,118752490863,410394698534,1415492232255,4873386985130,16750755602928,57487476629594,197013756414033
%N A291265 a(n) = (1/3)*A291232(n).
%C A291265 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 A291265 See A291219 for a guide to related sequences.
%H A291265 Clark Kimberling, <a href="/A291265/b291265.txt">Table of n, a(n) for n = 0..1000</a>
%H A291265 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (6, -7, -6, -1)
%F A291265 G.f.: (2 - 3 x - 2 x^2)/(-1 + 3 x + x^2)^2.
%F A291265 a(n) = 6*a(n-1) - 7*a(n-2) -6*a(n-3) - a(n-4) for n >= 5.
%F A291265 a(n) = (((3-sqrt(13))/2)^n*(-3+sqrt(13))*(-39+17*sqrt(13)-39*n) + 2^(-n)*(3+sqrt(13))^(1+n)*(39+17*sqrt(13)+39*n)) / 338. - _Colin Barker_, Aug 26 2017
%t A291265 z = 60; s = x/(1 - x^2); p = (1 - 3 s)^2;
%t A291265 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
%t A291265 u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291232 *)
%t A291265 u/3   (* A291265 *)
%o A291265 (PARI) Vec((2 + x)*(1 - 2*x) / (1 - 3*x - x^2)^2 + O(x^30)) \\ _Colin Barker_, Aug 26 2017
%Y A291265 Cf. A000035, A291232, A291219.
%K A291265 nonn,easy
%O A291265 0,1
%A A291265 _Clark Kimberling_, Aug 26 2017