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A292321 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = 1 - S^2 - S^3.

%I A292321 #4 Sep 23 2017 19:21:22
%S A292321 0,1,1,1,4,5,7,17,23,38,75,109,190,339,524,917,1563,2519,4360,7305,
%T A292321 12056,20621,34407,57452,97423,162672,272961,460454,770281,1294575,
%U A292321 2177777,3649129,6134192,10306017,17287962,29054244,48790024,81894794,137592982
%N A292321 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = 1 - S^2 - S^3.
%C A292321 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 A292321 See A291728 for a guide to related sequences.
%H A292321 Clark Kimberling, <a href="/A292321/b292321.txt">Table of n, a(n) for n = 0..1000</a>
%H A292321 <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (0, 1, 4, 0, -1, -3, 0, 0, 1)
%F A292321 G.f.: (x (-1 - x + x^3))/(-1 + x^2 + 4 x^3 - x^5 - 3 x^6 + x^9).
%F A292321 a(n) = a(n-2) + 4*a(n-3) - a(n-5) - 3*a(n-6)  + a(n-9) for n >= 10.
%t A292321 z = 60; s = x/(x - x^3); p = 1 - s^2 - s^3;
%t A292321 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A079978  *)
%t A292321 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292321 *)
%Y A292321 Cf. A079978, A292320.
%K A292321 nonn,easy
%O A292321 0,5
%A A292321 _Clark Kimberling_, Sep 15 2017