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A290898 p-INVERT of the positive integers, where p(S) = 1 - S - S^4.

%I A290898 #10 Aug 18 2017 06:50:26
%S A290898 1,3,8,22,65,203,647,2053,6423,19811,60490,183750,557551,1693921,
%T A290898 5157224,15731043,48041589,146785994,448475954,1369853581,4182850121,
%U A290898 12769287055,38976737437,118967979141,363132913719,1108463577238,3383732698880,10329587789993
%N A290898 p-INVERT of the positive integers, where p(S) = 1 - S - S^4.
%C A290898 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 A290898 See A290890 for a guide to related sequences.
%H A290898 Clark Kimberling, <a href="/A290898/b290898.txt">Table of n, a(n) for n = 0..1000</a>
%H A290898 <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (9, -34, 71, -89, 71, -34, 9, -1)
%F A290898 a(n) = 9*a(n-1) - 34*a(n-2) + 71*a(n-3) - 89*a(n-4) + 71*a(n-5) - 34*a(n-6) + 9*a(n-7) - a(n-8).
%F A290898 G.f.: (1 - x + x^2)*(1 - 5*x + 9*x^2 - 5*x^3 + x^4) / (1 - 9*x + 34*x^2 - 71*x^3 + 89*x^4 - 71*x^5 + 34*x^6 - 9*x^7 + x^8). - _Colin Barker_, Aug 18 2017
%t A290898 z = 60; s = x/(1 - x)^2; p = 1 - s - s^4;
%t A290898 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t A290898 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290898 *)
%o A290898 (PARI) Vec((1 - x + x^2)*(1 - 5*x + 9*x^2 - 5*x^3 + x^4) / (1 - 9*x + 34*x^2 - 71*x^3 + 89*x^4 - 71*x^5 + 34*x^6 - 9*x^7 + x^8) + O(x^40)) \\ _Colin Barker_, Aug 18 2017
%Y A290898 Cf. A000027, A290890.
%K A290898 nonn,easy
%O A290898 0,2
%A A290898 _Clark Kimberling_, Aug 17 2017