[seqfan] Re: nth divisor of a number
davidwwilson at comcast.net
Sat Mar 31 01:33:18 CEST 2012
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.
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