Cont Fract Where n Divides Each Numerator

[Leroy Quet]

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