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A291243 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - 3 S + S^2.

%I A291243 #10 Aug 29 2017 03:34:09
%S A291243 3,8,24,71,210,621,1836,5428,16047,47440,140247,414612,1225716,
%T A291243 3623579,10712370,31668929,93622704,276776352,818232603,2418937120,
%U A291243 7151092203,21140739568,62498266944,184763326671,546214936050,1614772594421,4773744472356,14112597876668
%N A291243 p-INVERT of (0,1,0,1,0,1,...), where p(S) = 1 - 3 S + S^2.
%C A291243 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 A291243 See A291219 for a guide to related sequences.
%H A291243 Clark Kimberling, <a href="/A291243/b291243.txt">Table of n, a(n) for n = 0..1000</a>
%H A291243 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (3,1,-3,-1)
%F A291243 G.f.: (3 - x - 3*x^2)/(1 - 3*x - x^2 + 3*x^3 + x^4).
%F A291243 a(n) = 3*a(n-1) + a(n-2) - 3*a(n-3) - a(n-4) for n >= 5.
%t A291243 z = 60; s = x/(1 - x^2); p = 1 - 3 s - s^2;
%t A291243 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
%t A291243 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291243 *)
%Y A291243 Cf. A000035, A291219.
%K A291243 nonn,easy
%O A291243 0,1
%A A291243 _Clark Kimberling_, Aug 28 2017