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A290891 p-INVERT of the positive integers, where p(S) = 1 - S^3.

%I A290891 #6 Aug 17 2017 05:12:30
%S A290891 0,0,1,6,21,57,138,330,827,2175,5826,15519,40836,106584,277696,724968,
%T A290891 1897380,4972113,13029534,34125561,89336141,233831262,612074526,
%U A290891 1602358863,4195173507,10983645498,28756340047,75285234408,197097337248,516002648064
%N A290891 p-INVERT of the positive integers, where p(S) = 1 - S^3.
%C A290891 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 A290891 See A280890 for a guide to related sequences.
%H A290891 Clark Kimberling, <a href="/A290891/b290891.txt">Table of n, a(n) for n = 0..1000</a>
%H A290891 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6, -15, 21, -15, 6, -1)
%F A290891 G.f.: x^2/(1 - 6 x + 15 x^2 - 21 x^3 + 15 x^4 - 6 x^5 + x^6).
%F A290891 a(n) = 6*a(n-1) - 15*a(n-2) + 21*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6).
%t A290891 z = 60; s = x/(1 - x)^2; p = 1 - s^3;
%t A290891 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t A290891 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290891 *)
%Y A290891 Cf. A000027, A290890.
%K A290891 nonn,easy
%O A290891 0,4
%A A290891 _Clark Kimberling_, Aug 15 2017