[seqfan] Re: nth divisor of a number

David Wilson davidwwilson at comcast.net
Sat Mar 31 20:00:00 CEST 2012


On 3/31/2012 1:58 AM, Benoît Jubin wrote:
> And these probabilities are obviously rational numbers, so two OEIS
> tables will be enough!
>
In this case, I almost think one tabular sequence would be enough, to wit

1
0 1
0 1 1
0 0 2 1
0 4 4 3 1
0 0 0 4 5 1

&c, with the understanding that the denominator of the nth row is LCM(1, 
.., n).



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