[seqfan] Re: nth divisor of a number

[David Wilson]

So presumably there is an algorithm to compute p(n, d), the probability 
that the nth divisor of an integer is d.

Given that almost all integers have n divisors, it would follow that 
SUM(d = 1..inf, p(n, d)) = 1.

Since d divides 1/d integers, I would suppose that SUM(n = 1..inf, p(n, 
d)) = 1/d.

Clearly p(n, d) = 0 for n > d.

We would be interested in the d which maximizes p(n, d) (the d which 
occurs most often as an nth divisor). Perhaps we could find an upper 
bound f(d) of p(n, d) which would make this computation tractable.

On 3/30/2012 3:02 AM, israel at math.ubc.ca wrote:
> The probability that the first five divisors of a "randomly chosen" n 
> are 1,2,3,4,6 (i.e. that $n$ is divisible by 12 but not by 5) is 
> (1/12)*(4/5) = 1/15. The probability that the first five divisors are 
> 1,2,3,5,6 (i.e. that $n$ is divisible by 30 but not by 4) is 1/60. 
> Those are the only ways the fifth divisor can be 6, so the total 
> probability that the fifth divisor is 6 is 1/15 + 1/60 = 1/12.
>