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A084242
Least k, 1 <= k <= n, such that the number of elements of the continued fraction for n/k is maximum.
8
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, 18, 19, 18, 23, 19, 21, 22, 22, 23, 22, 25, 29, 23, 26, 25, 27, 26, 27, 29, 31, 30, 29, 28, 33, 33, 31, 34, 41, 32, 36, 33, 37, 33, 35, 37, 39, 47, 37, 41, 42, 40, 41, 41, 53, 45, 43
OFFSET
1,3
COMMENTS
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
LINKS
FORMULA
For k > 1, a(F(k)) = F(k-1) where F(k) denotes the k-th Fibonacci number.
Probably, lim_{n->oo} (1/n)*Sum_{k=1..n} a(k) = 1/phi = A094214.
PROG
(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)
CROSSREFS
KEYWORD
nonn
AUTHOR
Benoit Cloitre, Jun 21 2003
STATUS
approved