Continued Fractions Which Are Permutations

[Leroy Quet]

I just submitted the following.

>%S A000001 1,3,11,48
>%N A000001 Greatest numerator among the n! ratios equal to the continued 
>fractions which have the permutations of (1,2,3,...,n) for terms.
>%e A000001 a(4) = 48 because the continued fractions [4;2,1,3] (= 48/11) 
>and [3;1,2,4] (= 48/13) have the greatest numerators among continued 
>fraction which each have a permutation of (1,2,3,4) for terms.
>%O A000001 1
>%K A000001 ,more,nonn,


But with only 4 terms, the sequences of least numerators (1,3,9,37,...), 
greatest denominators (1,2,7,31,...), and least denominators 
(1,1,3,9,...) each bring up several hits.

Also the numerators (and probably the denominators as well) of the 
greatest ratios
(1,3,11/3,47/10,...) and of least ratios (1,3/2,9/7,38/31,...) among CFs 
which are permutations of (1,2,3,..,n) bring up several hits.

Could someone calculate more terms, or are these sequences already in the 
database under different names?

thanks,
Leroy Quet