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A291240 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S^3)^2.

%I A291240 #8 Sep 03 2017 21:40:09
%S A291240 0,0,2,0,6,3,12,18,24,63,66,173,222,438,722,1146,2142,3213,5958,9327,
%T A291240 16210,26898,44400,75875,123252,210240,344160,578052,958200,1588404,
%U A291240 2650008,4370292,7285684,12022704,19960488,33008505,54594504,90368550,149168350
%N A291240 p-INVERT of (0,1,0,1,0,1,...), where p(S) = (1 - S^3)^2.
%C A291240 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 A291240 See A291219 for a guide to related sequences.
%H A291240 Clark Kimberling, <a href="/A291240/b291240.txt">Table of n, a(n) for n = 0..1000</a>
%H A291240 <a href="/index/Rec#order_12">Index entries for linear recurrences with constant coefficients</a>, signature (0,6,2,-15,-6,19,6,-15,-2,6,0,-1)
%F A291240 G.f.: -((x^2 (-2 + 6 x^2 + x^3 - 6 x^4 + 2 x^6))/((-1 + x + x^2)^2 (1 + x - x^2 - x^3 + x^4)^2)).
%F A291240 a(n) = 6*a(n-2) + 2*a(n-3) - 15*a(n-4) - 6*a(n-5) + 19*a(n-6) + 6*a(n-7) - 15*a(n-8) - 2*a(n-9) + 6*a(n-10) - a(n-12) for n >= 13.
%t A291240 z = 60; s = x/(1 - x^2); p = (1 - s^3)^2;
%t A291240 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000035 *)
%t A291240 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291240 *)
%t A291240 LinearRecurrence[{0, 6, 2, -15, -6, 19, 6, -15, -2, 6, 0, -1}, {0, 0, 2, 0, 6, 3, 12, 18, 24, 63, 66, 173}, 40] (* _Vincenzo Librandi_, Aug 29 2017 *)
%Y A291240 Cf. A000035, A291219.
%K A291240 nonn,easy
%O A291240 0,3
%A A291240 _Clark Kimberling_, Aug 28 2017