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%I #28 Jul 31 2023 23:27:30
%S 1,1,2,3,3,4,4,5,5,6,7,7,8,9,11,9,10,11,11,11,13,13,14,13,14,15,17,17,
%T 18,19,18,23,19,21,22,22,23,22,25,29,23,26,25,27,26,27,29,31,30,29,28,
%U 33,33,31,34,41,32,36,33,37,33,35,37,39,47,37,41,42,40,41,41,53,45,43
%N Least k, 1 <= k <= n, such that the number of elements of the continued fraction for n/k is maximum.
%C Also, for n > 1, the smallest number k such that the Euclidean algorithm for (n,k) requires the maximum number of steps, A034883(n). - _T. D. Noe_, Mar 24 2011
%H John Metcalf, <a href="/A084242/b084242.txt">Table of n, a(n) for n = 1..10000</a>
%F For k > 1, a(F(k)) = F(k-1) where F(k) denotes the k-th Fibonacci number.
%F Probably, lim_{n->oo} (1/n)*Sum_{k=1..n} a(k) = 1/phi = A094214.
%o (PARI) a(n)=if(n<0,0,s=1; while(abs(vecmax(vector(n,i,length(contfrac(n/i))))-length(contfrac(n/s)))>0,s++); s)
%Y Cf. A000045, A071677, A094214.
%K nonn
%O 1,3
%A _Benoit Cloitre_, Jun 21 2003
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