Cont Fract Where n Divides Each Numerator

[Leroy Quet]

I wrote:
>I just submitted this sequence:
>
>>%S A000001 1,1,1,2,4,2
>>%N A000001 a(1) = 1. a(n) is the lowest positive integer such that the 
>>continued fraction [a(1),a(2),a(3),...,a(n)] has a numerator divisible by n.
>>%e A000001 a(5) = 4 because 4 is the lowest positive integer m such that 
>>the continued fraction [1,1,1,2,m] has a numerator divisible by 5.
>>1 + 1/(1 + 1/(1 + 1/(2 + 1/4))) = 35/22, and 35 is divisible by 5.
>>%O A000001 1
>>%K A000001 ,cofr,more,nonn,
>
>Could someone please calculate and submit more terms?
>
>Maybe someone could also submit the numerator and denominator sequences 
>of the convergents (and maybe also submit the sequence which is the nth 
>numerator divided by n).
>1/1, 2/1, 3/2, 8/5, 35/22,...
>
>Numerator divided by n:
>1,1,1,2,7,...
>
>thanks,
>Leroy Quet

Hans Havermann then wrote:
>I think this one might be finite. The numerator of [1, 1, 1, 2, 4, 2,  
>7, 2, 7, 10, 5, 10, 4, 2, 7, 2, 9, x] is (713003632 + 6751100673 x)  
>and this does not appear to be divisible by 18.

I just had an (unoriginal) idea for 4 sequences based on the above.

1) If there is no numerator divisible by a particular n, simply skip to 
the next higher n which works.

2) Another sequence would be the list of n's skipped.

3) Also, we can have the sequence where the numerator must be divisible 
by m(n), where m(n) is the lowest positive integer not occurring earlier 
in {m(n)} and which divides the numerator of the nth convergent. (So with 
this sequence, we may come back to a numerator divisible by 18.)

4) Is the sequence {m(n)} a permutation of the positive integers? Or are 
some positive integers missing from it?

thanks,
Leroy Quet