%I #22 Sep 13 2022 12:56:17
%S 1,1,1,2,3,97,11786,33736797423001,79097781524295318019203322936641
%N a(n)=(a(n-1)*a(n-2)^5+1)/a(n-3) with a(0)=a(1)=a(2)=1.
%C This is the case a=5, b=1, 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(9)) has 96 digits. - _Harvey P. Dale_, Sep 13 2022
%H Seiichi Manyama, <a href="/A208205/b208205.txt">Table of n, a(n) for n = 0..11</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 = -1.903211925911553287485216224057094233775314050044332659604216582082...
%F d2 = 0.1939365664746304482560845569332033002552873106788960042162607290276...
%F d3 = 2.7092753594369228392291316671238909335200267393654366553879558530545...
%F are the roots of the equation d^3 + 1 = d^2 + 5*d and
%F c1 = 0.9741074409555962981370572554321352591111177638556227590517984272608...
%F c2 = 0.0499759123576461468686480770163694779918691526759585723897652462761...
%F c3 = 1.0272217210627198315132544386598971884129462517962425299212701250318...
%F (End)
%p a:=proc(n) if n<3 then return 1: fi: return (a(n-1)*a(n-2)^5+1)/a(n-3): end: seq(a(i),i=1..10);
%t RecurrenceTable[{a[n] == (a[n - 1] a[n - 2]^5 + 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*b^5+1)/a}; NestList[nxt,{1,1,1},10][[All,1]] (* _Harvey P. Dale_, Sep 13 2022 *)
%Y Cf. A005246, A208204.
%K nonn
%O 0,4
%A _Matthew C. Russell_, Apr 23 2012