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A291723 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - S^3.

%I A291723 #7 Jun 07 2022 11:06:15
%S A291723 0,0,1,0,3,1,3,6,2,15,9,21,36,27,85,72,141,222,231,513,540,945,1422,
%T A291723 1741,3222,3876,6337,9339,12447,20809,27135,42546,62195,86709,136866,
%U A291723 187278,286113,417303,595852,910431,1281810,1926984,2810883,4064571,6097464
%N A291723 p-INVERT of (1,0,1,0,0,0,0,...), where p(S) = 1 - S^3.
%C A291723 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 A291723 See A291728 for a guide to related sequences.
%H A291723 Clark Kimberling, <a href="/A291723/b291723.txt">Table of n, a(n) for n = 0..1000</a>
%H A291723 <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (0, 0, 1, 0, 3, 0, 3, 0, 1)
%F A291723 G.f.: -((x^2 (1 + x^2)^3)/((1 - x + x^2) (-1 + x + x^3) (1 + 2 x + 2 x^2 + x^3 + x^4))).
%F A291723 a(n) = a(n-3) + 3*a(n-5) + 3*a(n-7) + a(n-9) for n >= 10.
%t A291723 z = 60; s = x + x^3; p = 1 - s^3;
%t A291723 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A154272 *)
%t A291723 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291723 *)
%t A291723 LinearRecurrence[{0,0,1,0,3,0,3,0,1},{0,0,1,0,3,1,3,6,2},60] (* _Harvey P. Dale_, Jun 07 2022 *)
%Y A291723 Cf. A154272, A291728.
%K A291723 nonn,easy
%O A291723 0,5
%A A291723 _Clark Kimberling_, Sep 08 2017