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A208221 a(n)=(a(n-1)^2*a(n-3)^2+a(n-2))/a(n-4) with a(0)=a(1)=a(2)=a(3)=1.

%I A208221 #16 Dec 17 2017 14:14:15
%S A208221 1,1,1,1,2,5,27,2921,106653026,1658455747832683945,
%T A208221 869174798276372512100586962107665002957113
%N A208221 a(n)=(a(n-1)^2*a(n-3)^2+a(n-2))/a(n-4) with a(0)=a(1)=a(2)=a(3)=1.
%C A208221 This is the case a=2, b=1, c=2, y(0)=y(1)=y(2)=y(3)=1 of the recurrence shown in the Example 3.3 of "The Laurent phenomenon" (see Link lines, p. 10).
%C A208221 The next term (a(11)) has 97 digits. - _Harvey P. Dale_, Dec 17 2017
%H A208221 Seiichi Manyama, <a href="/A208221/b208221.txt">Table of n, a(n) for n = 0..13</a>
%H A208221 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.
%p A208221 y:=proc(n) if n<4 then return 1: fi: return (y(n-1)^2*y(n-3)^2+y(n-2))/y(n-4): end:
%p A208221 seq(y(n),n=0..11);
%t A208221 a[n_] := a[n] = If[n <= 3, 1, (a[n-1]^2*a[n-3]^2 + a[n-2])/a[n-4]];
%t A208221 Table[a[n], {n, 0, 13}] (* _Jean-François Alcover_, Nov 24 2017 *)
%t A208221 RecurrenceTable[{a[0]==a[1]==a[2]==a[3]==1,a[n]==(a[n-1]^2 a[n-3]^2+ a[n-2])/ a[n-4]},a,{n,12}] (* _Harvey P. Dale_, Dec 17 2017 *)
%Y A208221 Cf. A048736, A208218, A208220, A208222, A208224.
%K A208221 nonn
%O A208221 0,5
%A A208221 _Matthew C. Russell_, Apr 25 2012