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A291018 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - 6 S + S^2.

%I A291018 #4 Aug 23 2017 23:37:20
%S A291018 6,41,280,1912,13056,89152,608768,4156928,28385280,193826816,
%T A291018 1323532288,9037643776,61712891904,421401985024,2877512744960,
%U A291018 19648886079488,134170986676224,916176804773888,6256046544781312,42718957920059392,291703291002224640
%N A291018 p-INVERT of (1,1,1,1,1,...), where p(S) = 1 - 6 S + S^2.
%C A291018 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 A291018 See A291000 for a guide to related sequences.
%H A291018 Clark Kimberling, <a href="/A291018/b291018.txt">Table of n, a(n) for n = 0..1000</a>
%H A291018 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (8, -8)
%F A291018 G.f.: (6 - 7 x)/(1 - 8 x + 8 x^2).
%F A291018 a(n) = 8*a(n-1) - 8*a(n-2) n >= 3.
%t A291018 z = 60; s = x/(1 - x); p = 1 - 6 s + s^2;
%t A291018 Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000012 *)
%t A291018 Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291018 *)
%Y A291018 Cf. A000012, A289780, A291000.
%K A291018 nonn,easy
%O A291018 0,1
%A A291018 _Clark Kimberling_, Aug 23 2017