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%I #5 Mar 30 2012 18:39:16
%S 5,13,18,23,25,30,38,43,45,53,58,60,65,70,78,83,85,93,98,103,105,110,
%T 118,120,125,133,138,140,145,150,158,163,165,173,178,183,185,190,198,
%U 203,205,213,218,220,225,230,238,240,245,253,258,263,265,270,278,280
%N a(n)-th term of the continued fraction for sum(k>=0,1/2^(2^k)) is 2.
%F a(n)=5*n + 0 or + 3 and it appears that a(n)=5*n+3*A073089(n+2)
%e The continued fraction for sum(k>=0,1/2^(2^k)) is : 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1... hence a(1)=5
%Y Cf. A006466, A073089.
%K easy,nonn
%O 1,1
%A _Benoit Cloitre_, Apr 09 2003