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A290616 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = 1 - 2 S - S^2.

%I A290616 #4 Sep 14 2017 18:14:31
%S A290616 2,5,12,31,80,205,526,1350,3464,8889,22810,58532,150198,385420,989018,
%T A290616 2537899,6512450,16711463,42882940,110041025,282373998,724594076,
%U A290616 1859365870,4771280299,12243483684,31417750230,80620439004,206878440932,530866488090
%N A290616 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = 1 - 2 S - S^2.
%C A290616 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 A290616 See A291728 for a guide to related sequences.
%H A290616 Clark Kimberling, <a href="/A290616/b290616.txt">Table of n, a(n) for n = 0..1000</a>
%H A290616 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (2, 1, 2, -2, 0, -1)
%F A290616 G.f.: -((-2 - x + 2 x^3)/(1 - 2 x - x^2 - 2 x^3 + 2 x^4 + x^6)).
%F A290616 a(n) = 2*a(n-1) + a(n-2) + 2*a(n-3) - 2*a(n-4) - a(n-6) for n >= 7.
%t A290616 z = 60; s = x/(x - x^3); p = 1 - 2 s - s^2;
%t A290616 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A079978  *)
%t A290616 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290616 *)
%Y A290616 Cf. A079978, A289918, A291035.
%K A290616 nonn,easy
%O A290616 0,1
%A A290616 _Clark Kimberling_, Sep 14 2017