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A290920 p-INVERT of the positive integers, where p(S) = (1 - S)^5.

%I A290920 #13 Aug 24 2017 08:23:21
%S A290920 5,25,110,450,1746,6505,23465,82435,283270,955258,3169520,10368490,
%T A290920 33497790,107028120,338582738,1061557195,3301399385,10191612315,
%U A290920 31250047480,95226980516,288523285450,869559080385,2607834545025,7785230674580,23142279699355
%N A290920 p-INVERT of the positive integers, where p(S) = (1 - S)^5.
%C A290920 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 A290920 See A290890 for a guide to related sequences.
%H A290920 Clark Kimberling, <a href="/A290920/b290920.txt">Table of n, a(n) for n = 0..1000</a>
%H A290920 <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (15, -95, 330, -685, 873, -685, 330, -95, 15, -1)
%F A290920 a(n) = 15*a(n-1) - 95*a(n-2) + 330*a(n-3) - 685*a(n-4) + 873*a(n-5) - 695*a(n-6) + 330*a(n-7) - 95*a(n-8) + 15*a(n-9) - a(n-10).
%F A290920 G.f.: (5 - 50*x + 210*x^2 - 475*x^3 + 621*x^4 - 475*x^5 + 210*x^6 - 50*x^7 + 5*x^8) / (1 - 3*x + x^2)^5. - _Colin Barker_, Aug 24 2017
%t A290920 z = 60; s = x/(1 - x)^2; p = (1 - s)^5;
%t A290920 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t A290920 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290920 *)
%o A290920 (PARI) Vec((5 - 50*x + 210*x^2 - 475*x^3 + 621*x^4 - 475*x^5 + 210*x^6 - 50*x^7 + 5*x^8) / (1 - 3*x + x^2)^5 + O(x^30)) \\ _Colin Barker_, Aug 24 2017
%Y A290920 Cf. A000027, A290890.
%K A290920 nonn,easy
%O A290920 0,1
%A A290920 _Clark Kimberling_, Aug 18 2017