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 A201267 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. 1

%I

%S 1,2,3,11,2,16,4,7,4,12,5,2,41,3,11,13,3,4,22,19,2,12,27,29,9,18,8,39,

%T 94,14,13,35,101,44,122,36,2,4,60,11,7,129,4,25,18,27,19,77,62,35,14,

%U 229,74,7,29,4,32,88,132,30,2,154,511,71,59,9,127,2,47,20,34,54,22,34,57,56,68,16,45,298,57,169,13,134,45,39,120,77,109

%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=[1];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).

%K nonn

%O 1,2

%A _Benoit Cloitre_, Jan 09 2013

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Last modified August 15 16:02 EDT 2020. Contains 336505 sequences. (Running on oeis4.)