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a(n)=(a(n-1)^3*a(n-2)+1)/a(n-3) with a(0)=a(1)=a(2)=1.
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%I #23 Jul 04 2022 11:26:45

%S 1,1,1,2,9,1459,13975855106,442535332406378982945622818194705

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

%C This is the case a=1, b=3, 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 has 105 digits. - _Harvey P. Dale_, Jul 04 2022

%H Seiichi Manyama, <a href="/A208207/b208207.txt">Table of n, a(n) for n = 0..9</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 = -0.675130870566646070889621798150060480808032527677372732612153869841...

%F d2 = 0.4608111271891108834741240973014799919001128904578732982807715533323...

%F d3 = 3.2143197433775351874154977008485804889079196372194994343313823165091...

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

%F c1 = 0.8399660110229591295951614867364338523629139731316529610703364786466...

%F c2 = 0.5166029105674572719002224224720428001985297645051505025129589573676...

%F c3 = 1.0214282112585594227681235564690028577352359049566082298453239674712...

%F (End)

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

%t RecurrenceTable[{a[n] == (a[n - 1]^3*a[n - 2] + 1)/a[n - 3], a[0] == a[1] == a[2] == 1}, a, {n, 0, 8}] (* _Michael De Vlieger_, Mar 19 2017 *)

%t nxt[{a_, b_, c_}] := {b, c, (c^3 b + 1)/a}; NestList[nxt,{1,1,1},10][[All,1]] (* _Harvey P. Dale_, Jul 04 2022 *)

%Y Cf. A005246, A208206, A208208, A208213.

%K nonn

%O 0,4

%A _Matthew C. Russell_, Apr 23 2012