Cont Fract Where n Divides Each Numerator

[Hans Havermann]

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?


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.
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