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A290899 p-INVERT of the positive integers, where p(S) = 1 - S^2 - S^4.

%I A290899 #10 Aug 18 2017 18:49:08
%S A290899 0,1,4,12,36,110,332,983,2876,8380,24428,71357,208868,612178,1795228,
%T A290899 5264684,15436060,45248195,132616392,388652536,1138993032,3338020181,
%U A290899 9782903524,28671786116,84032220964,246284956558,721820483900,2115530739035,6200240318564
%N A290899 p-INVERT of the positive integers, where p(S) = 1 - S^2 - S^4.
%C A290899 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 A290899 See A290890 for a guide to related sequences.
%H A290899 Clark Kimberling, <a href="/A290899/b290899.txt">Table of n, a(n) for n = 0..1000</a>
%H A290899 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (8, -27, 52, -63, 52, -27, 8, -1)
%F A290899 a(n) = 8*a(n-1) - 27*a(n-2) + 52*a(n-3) - 63*a(n-4) + 52*a(n-5) - 27*a(n-6) + 8*a(n-7) - a(n-8).
%F A290899 G.f.: x*(1 - 4*x + 7*x^2 - 4*x^3 + x^4) / (1 - 8*x + 27*x^2 - 52*x^3 + 63*x^4 - 52*x^5 + 27*x^6 - 8*x^7 + x^8). - _Colin Barker_, Aug 18 2017
%t A290899 z = 60; s = x/(1 - x)^2; p = 1 - s^2 - s^4;
%t A290899 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t A290899 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290899 *)
%o A290899 (PARI) concat(0, Vec(x*(1 - 4*x + 7*x^2 - 4*x^3 + x^4) / (1 - 8*x + 27*x^2 - 52*x^3 + 63*x^4 - 52*x^5 + 27*x^6 - 8*x^7 + x^8) + O(x^40))) \\ _Colin Barker_, Aug 18 2017
%Y A290899 Cf. A000027, A290890.
%K A290899 nonn,easy
%O A290899 0,3
%A A290899 _Clark Kimberling_, Aug 17 2017