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A291257 a(n) = (1/2)*A291228(n).

%I A291257 #4 Aug 25 2017 23:40:12
%S A291257 1,3,9,28,85,261,797,2440,7461,22827,69821,213588,653345,1998573,
%T A291257 6113529,18701072,57205769,174990195,535287793,1637423756,5008812525,
%U A291257 15321754293,46868623381,143369215128,438560602669,1341539064795,4103713486629,12553092811972
%N A291257 a(n) = (1/2)*A291228(n).
%C A291257 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 A291257 See A291219 for a guide to related sequences.
%H A291257 Clark Kimberling, <a href="/A291257/b291257.txt">Table of n, a(n) for n = 0..1000</a>
%H A291257 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2, 4, -2, -1)
%F A291257 G.f.: -((2 (-1 - x + x^2))/(1 - 2 x - 4 x^2 + 2 x^3 + x^4)).
%F A291257 a(n) = 2*a(n-1) + 4*a(n-2) - 2*a(n-3) - a(n-4) for n >= 5.
%F A291257 a(n) = (1/2)*A291228(n) for n >= 0.
%t A291257 z = 60; s = x/(1 - x^2); p = 1 - 2 s - 2 s^2;
%t A291257 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
%t A291257 u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291228 *)
%t A291257 u/2         (* A291257 *)
%Y A291257 Cf. A000035, A291219, A291228.
%K A291257 nonn,easy
%O A291257 0,2
%A A291257 _Clark Kimberling_, Aug 25 2017