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%I #2 Mar 30 2012 17:26:14
%S 1,1,2,3,2,1,1,2,3,2,1,1,2,3,2,1,1,2,3,2,1,1
%N For n>2, a(n) > 0 is such that a(n-1)^2+4*a(n-2)*a(n) is a minimal square, with a(1)=1, a(2)=1.
%C The sequence depends on seed terms a(1) and a(2); if a(1)=1, a(3)=a(2)+1. All(?) sequences end with cycle={1,2,3,2,1} (or {2,4,6,4,2}, which essentially the same cycle) of length=5.
%Y Cf. A105736 - A105746.
%K nonn
%O 1,3
%A _Zak Seidov_, Apr 18 2005
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