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A291020 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S - S^2 - S^3 - S^4 + S^5.

%I A291020 #4 Aug 23 2017 23:37:35
%S A291020 1,3,9,27,79,228,656,1889,5445,15701,45275,130544,376388,1085199,
%T A291020 3128841,9021083,26009635,74991112,216214692,623391005,1797363157,
%U A291020 5182163781,14941232871,43078615236,124204414928,358106605227,1032494220505,2976890957419
%N A291020 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - S - S^2 - S^3 - S^4 + S^5.
%C A291020 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 A291020 See A291000 for a guide to related sequences.
%H A291020 Clark Kimberling, <a href="/A291020/b291020.txt">Table of n, a(n) for n = 0..1000</a>
%H A291020 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (6, -13, 14, -7)
%F A291020 G.f.: -((-1 + 3 x - 4 x^2 + 2 x^3 + x^4)/(1 - 6 x + 13 x^2 - 14 x^3 + 7 x^4) ).
%F A291020 a(n) = 6*a(n-1) - 13*a(n-2) + 14*a(n-3) - 7*a(n-4) for n >= 6.
%t A291020 z = 60; s = x/(1 - x); p = 1 - s - s^2 - s^3 - s^4 + s^5;
%t A291020 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
%t A291020 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291020 *)
%Y A291020 Cf. A000012, A289780, A291000.
%K A291020 nonn,easy
%O A291020 0,2
%A A291020 _Clark Kimberling_, Aug 23 2017