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A208226 a(n)=(a(n-1)*a(n-3)^4+a(n-2))/a(n-4) with a(0)=a(1)=a(2)=a(3)=1.

%I A208226 #14 Dec 29 2018 15:52:30
%S A208226 1,1,1,1,2,3,5,83,3364,700861,6652337263549,
%T A208226 10264082055393717193904815,
%U A208226 736193034562641516492404723890409674438627151,2057106833431631102316572923185391939849261245309254135929044995902093016346478213863681606
%N A208226 a(n)=(a(n-1)*a(n-3)^4+a(n-2))/a(n-4) with a(0)=a(1)=a(2)=a(3)=1.
%C A208226 This is the case a=4, b=1, c=1, 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).
%H A208226 Seiichi Manyama, <a href="/A208226/b208226.txt">Table of n, a(n) for n = 0..16</a>
%H A208226 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 A208226 y:=proc(n) if n<4 then return 1: fi: return (y(n-1)*y(n-3)^4+y(n-2))/y(n-4): end:
%p A208226 seq(y(n),n=0..13);
%t A208226 a[n_]:=If[n<4,1, (a[n - 1] *a[n- 3]^4 + a[n - 2])/a[n - 4]]; Table[a[n], {n, 0, 12}] (* _Indranil Ghosh_, Mar 19 2017 *)
%t A208226 RecurrenceTable[{a[0]==a[1]==a[2]==a[3]==1,a[n]==(a[n-1]a[n-3]^4+ a[n-2])/ a[n-4]},a,{n,14}] (* _Harvey P. Dale_, Dec 29 2018 *)
%Y A208226 Cf. A048736, A208223, A208227.
%K A208226 nonn
%O A208226 0,5
%A A208226 _Matthew C. Russell_, Apr 25 2012
%E A208226 One more term from _Harvey P. Dale_, Dec 29 2018