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%I #22 Apr 09 2016 15:58:52
%S 2,3,14,915,54718634,179054255520688074387,
%T 104910436622395373760090792140594526518830577847960206
%N Rank 2 cluster algebra recurrence with parameters b=c=3 and initial conditions a(0)=2 and a(1)=3.
%C This recurrence comes from taking a(0)=2, a(1)=3 with parameters b=c=3 in the well known (rank 2 cluster algebra) recurrence:
%C a(n) = (1+a(n-1)^b)/a(n-2) if n is even,
%C a(n) = (1+a(n-1)^c)/a(n-2) if n is odd.
%F a(n) = (1+a(n-1)^3)/a(n-2), n>=2, a(0)=2, a(1)=3.
%F a(n) ~ c^(((1+sqrt(5))/2)^(2*n)), where c = 1.46103748126346957816556... . - _Vaclav Kotesovec_, Apr 08 2016
%p # Here we set x=a(0) and y=a(1) to be appropriate initial conditions to produce an integer sequence.
%p # In general we do not obtain an integer sequence.
%p # The values x=2, y=3 are appropriate initial conditions when b=c=3.
%p a:=proc(n) local resp, j; option remember;
%p if n=0 then return x fi:
%p if n=1 then return y fi:
%p if type(n,even)=true then
%p resp:=simplify((1+a(n-1)^b)/a(n-2)):
%p else resp:=simplify((1+a(n-1)^c)/a(n-2)):
%p fi:
%p return simplify(resp)
%p end proc:
%t a[0] = 2; a[1] = 3; a[n_] := a[n] = (1 + a[n - 1]^3)/a[n - 2]; Array[a, 8, 0] (* _Michael De Vlieger_, Apr 08 2016 *)
%o (PARI) a(n) = if (n==0, 2, if (n==1, 3, if (n%2, (1+a(n-1)^3)/a(n-2), (1+a(n-1)^3)/a(n-2)))); \\ _Michel Marcus_, Apr 04 2016
%Y Cf. A271331.
%K nonn
%O 0,1
%A _Hector J. Blandin N._, Apr 04 2016
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