%N a(1) = 1 and a(n) is the least integer such that the continued fraction for 1/a(1) + 1/a(2) +...+ 1/a(n) contains exactly n elements.
%C Conjecture: limsup n-->infty a(n)=infinity. More precisely we claim that log(a(n))/log(n) is bounded and doesn't converge to zero (see related link). Does a(n)=2 infinitely many times or does it exist M>=2 such that a(n)<=M infinitely many times?
%H Benoit Cloitre, <a href="/A201267/a201267.png">Plot of log(a(n))/log(n) (black dot) vs arithmetic mean of log(a(n))/log(n) (red line)</a>
%e 1/a(1)+1/a(2)+1/a(3)+1/a(4)=1+1/2+1/3+1/11=127/66 and the continued fraction expansion is [1, 1, 12, 5] containing 4 elements. Next k=2 is the smallest integer >=1 such that the continued fraction expansion of 127/66+1/k which is [2, 2, 2, 1, 4] contains 5 elements, thus a(5)=2.
%o (PARI) v=;for(m=2,100,k=1;while(abs(length(contfrac(1/k+sum(i=1,length(v),1/v[i])))-m)>0,k++);v=concat(v,[k]));a(n)=v[n];
%Y Cf. A071012 (version with increasing terms).
%A _Benoit Cloitre_, Jan 09 2013