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a(n)=(a(n-1)^4*a(n-2)+1)/a(n-3) with a(0)=a(1)=a(2)=1.
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%I #19 Mar 19 2017 08:46:11

%S 1,1,1,2,17,167043,6618080569762280805809

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

%C This is the case a=1, b=4, 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="/A208208/b208208.txt">Table of n, a(n) for n = 0..8</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 = -0.588363990685104156421284586508527584304318862407786509166141051262...

%F d2 = 0.4064206546327112651910488344937800073049991477253475806754539682375...

%F d3 = 4.1819433360523928912302357520147475769993197146824389284906870830246...

%F are the roots of the equation d^3 + 1 = 4*d^2 + d and

%F c1 = 0.8094826741348488413005600397911253102639462301397489110738060562305...

%F c2 = 0.5758908197062035276668941188013698534573120455706764136847247903030...

%F c3 = 1.0094396347013780675988108222508397688561313671701492219003321772184...

%F (End)

%p a:=proc(n) if n<3 then return 1: fi: return (a(n-1)^4*a(n-2)+1)/a(n-3): end: seq(a(i),i=0..10);

%t RecurrenceTable[{a[n] == (a[n - 1]^4*a[n - 2] + 1)/a[n - 3], a[0] == a[1] == a[2] == 1}, a, {n, 0, 7}] (* _Michael De Vlieger_, Mar 19 2017 *)

%Y Cf. A005246, A208207.

%K nonn

%O 0,4

%A _Matthew C. Russell_, Apr 23 2012