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A208214 a(n)=(a(n-1)^3*a(n-2)^3+1)/a(n-3) with a(0)=a(1)=a(2)=1.

%I A208214 #22 Mar 19 2017 08:46:01
%S A208214 1,1,1,2,9,5833,72339160083737,
%T A208214 8347449602301100278574002746114271427525770715131218
%N A208214 a(n)=(a(n-1)^3*a(n-2)^3+1)/a(n-3) with a(0)=a(1)=a(2)=1.
%C A208214 This is the case a=3, b=3, y(0)=y(1)=y(2)=1 of the recurrence shown in the Example 3.2 of "The Laurent phenomenon" (see Link lines, p. 10).
%H A208214 Seiichi Manyama, <a href="/A208214/b208214.txt">Table of n, a(n) for n = 0..9</a>
%H A208214 Sergey Fomin and Andrei Zelevinsky, <a href="http://arxiv.org/abs/math/0104241">The Laurent phenomenon</a>, arXiv:math/0104241v1 [math.CO] (2001), Advances in Applied Mathematics 28 (2002), 119-144.
%F A208214 From _Vaclav Kotesovec_, May 20 2015: (Start)
%F A208214 a(n) ~ c1^(d1^n) * c2^(d2^n) * c3^(d3^n), where
%F A208214 d1 = -1
%F A208214 d2 = 2-sqrt(3) = 0.2679491924311227064725536584941276330571947461896193719...
%F A208214 d3 = 2+sqrt(3) = 3.7320508075688772935274463415058723669428052538103806280...
%F A208214 are the roots of the equation d^3 + 1 = 3*d^2 + 3*d and
%F A208214 c1 = 0.9085343342123995498629194372995408229585378171837724081842452659181...
%F A208214 c2 = 0.3811823487030541690662698257664022175009714305688428757048879374472...
%F A208214 c3 = 1.0119167333492916399265234093841995850496968884402785055210058839859...
%F A208214 (End)
%p A208214 a:=proc(n) if n<3 then return 1: fi: return (a(n-1)^3*a(n-2)^3+1)/a(n-3): end: seq(a(i),i=0..10);
%t A208214 RecurrenceTable[{a[n] == (a[n - 1]^3*a[n - 2]^3 + 1)/a[n - 3], a[0] == a[1] == a[2] == 1}, a, {n, 0, 7}] (* _Michael De Vlieger_, Mar 19 2017 *)
%Y A208214 Cf. A005246, A208210, A208213.
%K A208214 nonn
%O A208214 0,4
%A A208214 _Matthew C. Russell_, Apr 23 2012