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%I #8 Mar 30 2012 18:39:17
%S 1,2,3,1,2,3,1,2,2,3,1,3,1,2,2,3,2,3,1,3,1,2,3,3,1,2,2,3,2,3,1,3,1,2,
%T 3,1,3,1,2,3,3,1,2,2,3,1,3,1,2,3,1,2,2,3,2,3,1,3,1,2,3,1,3,1,2,3,3,1,
%U 2,2,3,1,3,1,2,3,3,1,2,2,3,1,3,1,2,3,1,2,2
%N a(1) = 1, a(2) = 2, a(3) = 3; then a(a(1)+a(2)+..+a(k)) = a(k) and fill the "holes" using the rule : a(m) = 1 if a(m-1) = 3, a(m) = 2 if a(m-1) = 1, a(m) = 3 if a(m-1) = 2.
%F It seems that limit n ->infinity (1/n)*sum(k=1, n, a(k))= C >2 (C=2.07...)
%e a(1)+a(2)+a(3)=6 so a(6)=a(3)=3. Since a(3)=3 a(4)=1 therefore a(5)=2 and sequence begins : 1,2,3,1,2,3,
%Y Cf. A082847.
%K nonn
%O 1,2
%A _Benoit Cloitre_, Apr 14 2003
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