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A291252 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S^3)^3.

%I A291252 #6 Sep 03 2017 21:42:06
%S A291252 0,0,3,0,9,6,18,36,40,126,135,351,513,936,1755,2682,5373,8260,15525,
%T A291252 25731,44511,78030,129564,229617,381438,664038,1121144,1910790,
%U A291252 3263796,5500110,9404820,15824790,26910426,45388638,76700664,129564945,218084256,368095230
%N A291252 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S^3)^3.
%C A291252 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 A291252 See A291219 for a guide to related sequences.
%H A291252 Clark Kimberling, <a href="/A291252/b291252.txt">Table of n, a(n) for n = 0..1000</a>
%H A291252 <a href="/index/Rec#order_18">Index entries for linear recurrences with constant coefficients</a>, signature (0, 9, 3, -36, -18, 81, 45, -117, -59, 117, 45, -81, -18, 36, 3, -9, 0, 1)
%F A291252 G.f.: -((x^2 (3 - 18 x^2 - 3 x^3 + 45 x^4 + 9 x^5 - 59 x^6 - 9 x^7 + 45 x^8 + 3 x^9 - 18 x^10 + 3 x^12))/((-1 + x + x^2)^3 (1 + x - x^2 - x^3 + x^4)^3)).
%F A291252 a(n) = 9*a(n-2) + 3*a(n-3) - 36*a(n-4) - 18*a(n-5) + 81*a(n-6) + 45*a(n-7) - 117*a(n-8) - 59*a(n-9) + 117*a(n-10) + 45*a(n-11) - 81*a(n-12) - 18*a(n-13) + 36*a(n-14) + 3*a(n-15) - 9*a(n-16) + a(n-18)  for n >= 19.
%t A291252 z = 60; s = x/(1 - x^2); p = (1 - s^3)^3;
%t A291252 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
%t A291252 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291252 *)
%Y A291252 Cf. A000035, A291219.
%K A291252 nonn,easy
%O A291252 0,3
%A A291252 _Clark Kimberling_, Aug 29 2017