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a(n)=(a(n-1)^2*a(n-2)+1)/a(n-3) with a(0)=a(1)=a(2)=1.
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%I #20 Mar 19 2017 08:46:39

%S 1,1,1,2,5,51,6503,431347892,23724602128927104843,

%T 37334625705205335653803036700733450756576803

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

%C This is the case a=1, b=2, 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).

%C The next term (a(10)) has 98 digits. - _Harvey P. Dale_, Oct 04 2014

%H Seiichi Manyama, <a href="/A208206/b208206.txt">Table of n, a(n) for n = 0..12</a>

%H 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 From _Vaclav Kotesovec_, May 20 2015: (Start)

%F a(n) ~ c1^(d1^n) * c2^(d2^n) * c3^(d3^n), where

%F d1 = -0.80193773580483825247220463901489010233183832426371430010712484639886484...

%F d2 = 0.554958132087371191422194871006410481067288862470910089376025968205157535...

%F d3 = 2.246979603717467061050009768008479621264549461792804210731098878193707304...

%F are the roots of the equation d^3 + 1 = 2*d^2 + d and

%F c1 = 0.874335057499939749225491691816700793966151250175012051621456437468590379...

%F c2 = 0.402356411273897640287204171338236092104516307383060911032953286637247174...

%F c3 = 1.071117422488325114038954501945557033632156032599675833309484054582086570...

%F (End)

%p a:=proc(n) if n<3 then return 1: fi: return (a(n-1)^2*a(n-2)+1)/a(n-3): end: seq(a(i),i=0..10);

%t RecurrenceTable[{a[0]==a[1]==a[2]==1,a[n]==(a[n-1]^2 a[n-2]+1)/a[n-3]},a,{n,10}] (* _Harvey P. Dale_, Oct 04 2014 *)

%Y Cf. A005246, A208202, A208207, A208209.

%K nonn

%O 0,4

%A _Matthew C. Russell_, Apr 23 2012