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A289795 p-INVERT of (3n), where p(S) = 1 - S - S^2.

%I A289795 #11 Aug 14 2017 14:17:08
%S A289795 3,24,162,1083,7260,48681,326406,2188536,14674041,98388840,659693103,
%T A289795 4423214952,29657473194,198852130383,1333295304660,8939689838877,
%U A289795 59940250397646,401896898269128,2694702070258437,18067865859946320,121144292846335179,812267469938047224
%N A289795 p-INVERT of (3n), where p(S) = 1 - S - S^2.
%C A289795 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 A289795 See A289780 for a guide to related sequences.
%H A289795 Clark Kimberling, <a href="/A289795/b289795.txt">Table of n, a(n) for n = 0..1000</a>
%H A289795 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (7, -3, 7, -1)
%F A289795 G.f.: 3 (1 + x + x^2)/(1 - 7 x + 3 x^2 - 7 x^3 + x^4).
%F A289795 a(n) = 7*a(n-1) - 3*a(n-2) + 7*a(n-3) - a(n-4).
%t A289795 z = 60; s = 3*x/(1 - x)^2; p = 1 - s - s^2;
%t A289795 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A008585 *)
%t A289795 u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289795 *)
%t A289795 u/3  (* A289796 *)
%Y A289795 Cf. A008585, A289780, A289796.
%K A289795 nonn,easy
%O A289795 0,1
%A A289795 _Clark Kimberling_, Aug 12 2017