[seqfan] Re: nth divisor of a number
Benoît Jubin
benoit.jubin at gmail.com
Sat Mar 31 07:58:36 CEST 2012
And these probabilities are obviously rational numbers, so two OEIS
tables will be enough!
In addition to the three identities you wrote, we have
p(n,n) = 1/(lcm{1,...,n}) = 1/A003418(n)
and
p(n,d) = 0 if n < tau(d) = A000005(d)
and
p(tau(d),d) = 1/d * product( (q-1)/q , q prime and there is an a with
q^a<d and q^a does not divide d )
and in particular, if r is prime, then
p(2,r) = 1/r * product( ( (q-1)/q , q prime, q<r ).
Forgetting p(1,1) = 1, the table begins:
1/2,
1/6, 1/6,
0, 1/6, 1/12,
1/15, 1/15, 1/20, 1/60,
0, 0, 1/15, 1/12, 1/60,
4/105, ...
Benoît
On Fri, Mar 30, 2012 at 4:33 PM, David Wilson <davidwwilson at comcast.net> wrote:
> 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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