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a(n) = (a(n-1)^2*a(n-2)^2 + 1)/a(n-3) with a(0)=a(1)=a(2)=1.
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%I #37 May 16 2019 03:07:30

%S 1,1,1,2,5,101,127513,33172764857794,

%T 177153971843949087009428690473769185

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

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

%H Seiichi Manyama, <a href="/A208209/b208209.txt">Table of n, a(n) for n = 0..11</a>

%H Joshua Alman, Cesar Cuenca, and Jiaoyang Huang, <a href="https://doi.org/10.1007/s10801-015-0647-5">Laurent phenomenon sequences</a>, Journal of Algebraic Combinatorics 43(3) (2015), 589-633.

%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 = -1

%F d2 = (3-sqrt(5))/2 = 0.381966011250105151795413165634361882279690820194237...

%F d3 = (3+sqrt(5))/2 = 2.618033988749894848204586834365638117720309179805762...

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

%F c1 = 0.9084730936822995591913406002175634029260903950386034752117808169903...

%F c2 = 0.3198114201427769362008537317523839726550617444688426214134486371587...

%F c3 = 1.0375048945851318188473394167711806349224412339663566324740449820203...

%F (End)

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

%t a[0] = a[1] = a[2] = 1; a[n_] := a[n] = (a[n-1]^2*a[n-2]^2 + 1)/a[n-3];

%t Table[a[n], {n, 0, 11}] (* _Jean-François Alcover_, Nov 17 2017 *)

%Y Cf. A005246, A208202, A208206, A208210, A208213.

%K nonn

%O 0,4

%A _Matthew C. Russell_, Apr 23 2012