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A291740 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = (1 - S)(1 - S^2).

%I A291740 #12 Oct 05 2017 15:31:14
%S A291740 1,2,3,7,9,18,25,47,65,118,165,290,408,702,992,1677,2379,3966,5643,
%T A291740 9300,13266,21654,30954,50116,71770,115388,165504,264475,379863,
%U A291740 603792,868267,1373621,1977413,3115222,4488843,7045205,10160427,15892794,22937999,35769390
%N A291740 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = (1 - S)(1 - S^2).
%C A291740 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 A291740 See A291728 for a guide to related sequences.
%H A291740 Clark Kimberling, <a href="/A291740/b291740.txt">Table of n, a(n) for n = 0..1000</a>
%H A291740 <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (1, 1, 0, 2, -3, 1, -3, 0, -1)
%F A291740 G.f.: -(((1 + x^2) (-1 - x + x^2 - x^3 + 2 x^4 + x^6))/((-1 + x + x^3)^2 (1 + x + x^3))).
%F A291740 a(n) = a(n-1) + a(n-2) + 2*a(n-4) - 3*a(n-5) + a(n-6) - 3*a(n-7) - a(n-9) for n >= 10.
%t A291740 z = 60; s = x + x^3; p = (1 - s) (1 - s^2);
%t A291740 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
%t A291740 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291740 *)
%o A291740 (PARI) x='x+O('x^99); Vec(((1+x^2)*(1+x-x^2+x^3-2*x^4-x^6))/((-1+x+x^3)^2*(1+x+x^3))) \\ _Altug Alkan_, Oct 04 2017
%Y A291740 Cf. A154272, A291728.
%K A291740 nonn,easy
%O A291740 0,2
%A A291740 _Clark Kimberling_, Sep 12 2017