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A290918 p-INVERT of the positive integers, where p(S) = (1 - S)^3.

%I A290918 #13 Aug 24 2017 12:57:54
%S A290918 3,12,43,147,486,1566,4944,15351,47009,142278,426315,1266300,3732705,
%T A290918 10928910,31806583,92069229,265215756,760621914,2172669846,6183333681,
%U A290918 17538237677,49590486888,139817553417,393157465848,1102792703055,3086146454592,8617872504643
%N A290918 p-INVERT of the positive integers, where p(S) = (1 - S)^3.
%C A290918 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 A290918 See A290890 for a guide to related sequences.
%H A290918 Clark Kimberling, <a href="/A290918/b290918.txt">Table of n, a(n) for n = 0..1000</a>
%H A290918 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (9, -30, 45, -30, 9, -1)
%F A290918 G.f.: (3 - 15 x + 25 x^2 - 15 x^3 + 3 x^4)/(1 - 3 x + x^2)^3.
%F A290918 a(n) = 9*a(n-1) - 30*a(n-2) + 45*a(n-3) - 30*a(n-4) + 9*a(n-5) - a(n-6).
%F A290918 (a(n)) is the p-INVERT of (1,1,1,1,1...) using p(S) = (1 - S - S^2)^3.
%t A290918 z = 60; s = x/(1 - x)^2; p = (1 - s)^3;
%t A290918 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000027 *)
%t A290918 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A290918 *)
%Y A290918 Cf. A000027, A290890.
%K A290918 nonn,easy
%O A290918 0,1
%A A290918 _Clark Kimberling_, Aug 18 2017