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%I #9 Feb 03 2013 08:24:17
%S 1,2,3,11,16,17,19,20,21,24,25,27,29,31,32,37,39,71,81,82,89,94,97,98,
%T 99,101,103,106,109,115,116,124,163,171,187,227,251,252,298,346,353,
%U 359,394,424,438,452,509,542,590,643,677,685,751,810,882,1063,1123
%N a(1)=1, a(n) is the smallest integer > a(n-1) such that the continued fraction for 1/a(1)+1/a(2)+....+1/a(n) contains strictly more elements than the continued fraction for 1/a(1)+1/a(2)+....+1/a(n-1).
%e Continued fraction for 1+1/2+1/3+1/11+1/16 is [1, 1, 74, 2, 3] which contains 5 elements, the continued fraction for 1+1/2+1/3+1/11+1/16+1/17 is [2, 21, 1, 17, 1, 1, 2, 4] which contains 8 elements, hence a(6)=17
%o (PARI) a(n)=if(n<0,0,s=a(n-1)+1; while(length(contfrac(1/s+sum(i=1,n-1,1/a(i))))<=length(contfrac(sum(i=1,n-1,1/a(i)))),s++); s)
%K nonn,cofr
%O 1,2
%A _Benoit Cloitre_, Nov 09 2002
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