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A290893 p-INVERT of the positive integers, where p(S) = 1 - S^5.

%I A290893 #8 Aug 17 2017 15:52:09
%S A290893 0,0,0,0,1,10,55,220,715,2003,5025,11650,25850,57475,134883,345090,
%T A290893 952195,2722455,7765010,21615771,58293475,152593575,390679925,
%U A290893 988851150,2502813930,6394182650,16569837550,43533891575,115440190725,307108317769,815362167365
%N A290893 p-INVERT of the positive integers, where p(S) = 1 - S^5.
%C A290893 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 A290893 See A280890 for a guide to related sequences.
%H A290893 Clark Kimberling, <a href="/A290893/b290893.txt">Table of n, a(n) for n = 0..1000</a>
%H A290893 <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (10, -45, 120, -210, 253, -210, 120, -45, 10, -1)
%F A290893 a(n) = 10*a(n-1) - 45*a(n-2) + 120*a(n-3) - 210*a(n-4) + 253*a(n-5) - 210*a(n-6) + 120*a(n-7) - 45*a(n-8) + 10*a(n-9) - a(n-10).
%F A290893 G.f.: x^4 / ((1 - 3*x + x^2)*(1 - 7*x + 23*x^2 - 44*x^3 + 55*x^4 - 44*x^5 + 23*x^6 - 7*x^7 + x^8)). - _Colin Barker_, Aug 16 2017
%t A290893 z = 60; s = x/(1 - x)^2; p = 1 - s^5;
%t A290893 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t A290893 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290893 *)
%o A290893 (PARI) concat(vector(4), Vec(x^4 / ((1 - 3*x + x^2)*(1 - 7*x + 23*x^2 - 44*x^3 + 55*x^4 - 44*x^5 + 23*x^6 - 7*x^7 + x^8)) + O(x^50))) \\ _Colin Barker_, Aug 16 2017
%Y A290893 Cf. A000027, A290890.
%K A290893 nonn,easy
%O A290893 0,6
%A A290893 _Clark Kimberling_, Aug 15 2017