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A290921 p-INVERT of the positive integers, where p(S) = (1 - S)^6.

%I A290921 #13 Aug 24 2017 08:23:28
%S A290921 6,33,158,696,2886,11425,43590,161355,582340,2056818,7130388,24319054,
%T A290921 81757104,271353288,890327048,2891047695,9299683770,29658374355,
%U A290921 93843661530,294791108106,919849034686,2852495485953,8794877092878,26971256457596,82298545175130
%N A290921 p-INVERT of the positive integers, where p(S) = (1 - S)^6.
%C A290921 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 A290921 See A290890 for a guide to related sequences.
%H A290921 Clark Kimberling, <a href="/A290921/b290921.txt">Table of n, a(n) for n = 0..1000</a>
%H A290921 <a href="/index/Rec#order_12">Index entries for linear recurrences with constant coefficients</a>, signature (18, -141, 630, -1770, 3258, -3989, 3258, -1770, 630, -141, 18, -1)
%F A290921 a(n) = 18*a(n-1) - 141*a(n-2) + 630*a(n-3) - 1770*a(n-4) + 3258*a(n-5) - 3989*a(n-6) + 3258*a(n-7) - 1770*a(n-8) + 630*a(n-9) - 141*a(n-10) + 18*a(n-11) - a(n-12).
%F A290921 G.f.: (2 - x)*(1 - 2*x)*(1 - 5*x + 9*x^2 - 5*x^3 + x^4)*(3 - 15*x + 25*x^2 - 15*x^3 + 3*x^4) / (1 - 3*x + x^2)^6. - _Colin Barker_, Aug 24 2017
%t A290921 z = 60; s = x/(1 - x)^2; p = (1 - s)^6;
%t A290921 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t A290921 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290921 *)
%o A290921 (PARI) Vec((2 - x)*(1 - 2*x)*(1 - 5*x + 9*x^2 - 5*x^3 + x^4)*(3 - 15*x + 25*x^2 - 15*x^3 + 3*x^4) / (1 - 3*x + x^2)^6 + O(x^30)) \\ _Colin Barker_, Aug 24 2017
%Y A290921 Cf. A000027, A290890.
%K A290921 nonn,easy
%O A290921 0,1
%A A290921 _Clark Kimberling_, Aug 18 2017