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A290895 p-INVERT of the positive integers, where p(S) = 1 - S^7.

%I A290895 #11 Aug 17 2017 16:04:07
%S A290895 0,0,0,0,0,0,1,14,105,560,2380,8568,27132,77521,203518,497826,1148126,
%T A290895 2527609,5401676,11508168,25437917,60978022,162008098,468103230,
%U A290895 1409724358,4259541790,12617126893,36241765553,100599743538,269998374114,702694008002
%N A290895 p-INVERT of the positive integers, where p(S) = 1 - S^7.
%C A290895 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 A290895 See A290890 for a guide to related sequences.
%H A290895 Clark Kimberling, <a href="/A290895/b290895.txt">Table of n, a(n) for n = 0..1000</a>
%H A290895 <a href="/index/Rec#order_14">Index entries for linear recurrences with constant coefficients</a>, signature (14, -91, 364, -1001, 2002, -3003, 3433, -3003, 2002, -1001, 364, -91, 14, -1)
%F A290895 a(n) = 14*a(n-1) - 91*a(n-2) + 364*a(n-3) - 1001*a(n-4) + 2002*a(n-5) - 3003*a(n-6) + 3433*a(n-7) - 3003*a(n-8) + 2002*a(n-9) - 1001*a(n-10) + 364*a(n-11) - 91*a(n-12) + 14*a(n-13) - a(n-14).
%F A290895 G.f.: x^6 / ((1 - 3*x + x^2)*(1 - 11*x + 57*x^2 - 182*x^3 + 398*x^4 - 626*x^5 + 727*x^6 - 626*x^7 + 398*x^8 - 182*x^9 + 57*x^10 - 11*x^11 + x^12)). - _Colin Barker_, Aug 16 2017
%t A290895 z = 60; s = x/(1 - x)^2; p = 1 - s^7;
%t A290895 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t A290895 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A290895 *)
%o A290895 (PARI) concat(vector(6), Vec(x^6 / ((1 - 3*x + x^2)*(1 - 11*x + 57*x^2 - 182*x^3 + 398*x^4 - 626*x^5 + 727*x^6 - 626*x^7 + 398*x^8 - 182*x^9 + 57*x^10 - 11*x^11 + x^12)) + O(x^50))) \\ _Colin Barker_, Aug 16 2017
%Y A290895 Cf. A000027, A290890.
%K A290895 nonn,easy
%O A290895 0,8
%A A290895 _Clark Kimberling_, Aug 15 2017