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A290894 p-INVERT of the positive integers, where p(S) = 1 - S^6.

%I A290894 #9 Aug 17 2017 15:56:57
%S A290894 0,0,0,0,0,1,12,78,364,1365,4368,12377,31848,75882,170560,370266,
%T A290894 803712,1827099,4531980,12346791,35783396,105681186,308229948,
%U A290894 873545479,2392395276,6336768804,16309261148,41095234896,102361858716,254804224832,640481466012
%N A290894 p-INVERT of the positive integers, where p(S) = 1 - S^6.
%C A290894 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 A290894 See A290890 for a guide to related sequences.
%H A290894 Clark Kimberling, <a href="/A290894/b290894.txt">Table of n, a(n) for n = 0..1000</a>
%H A290894 <a href="/index/Rec#order_12">Index entries for linear recurrences with constant coefficients</a>, signature (12, -66, 220, -495, 792, -923, 792, -495, 220, -66, 12, -1)
%F A290894 a(n) = 12*a(n-1) - 66*a(n-2) + 220*a(n-3) - 495*a(n-4) + 792*a(n-5) - 923*a(n-6) + 792*a(n-7) - 495*a(n-8) + 22*a(n-9) - 66*a(n-10) + 12*a(n-11) - a(n-12).
%F A290894 G.f.: x^5 / ((1 - 3*x + x^2)*(1 - x + x^2)*(1 - 5*x + 9*x^2 - 5*x^3 + x^4)*(1 - 3*x + 5*x^2 - 3*x^3 + x^4)). - _Colin Barker_, Aug 16 2017
%t A290894 z = 60; s = x/(1 - x)^2; p = 1 - s^6;
%t A290894 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t A290894 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290894 *)
%o A290894 (PARI) concat(vector(5), Vec(x^5 / ((1 - 3*x + x^2)*(1 - x + x^2)*(1 - 5*x + 9*x^2 - 5*x^3 + x^4)*(1 - 3*x + 5*x^2 - 3*x^3 + x^4)) + O(x^50))) \\ _Colin Barker_, Aug 16 2017
%Y A290894 Cf. A000027, A290890.
%K A290894 nonn,easy
%O A290894 0,7
%A A290894 _Clark Kimberling_, Aug 15 2017