# [seqfan] Re: nth divisor of a number

israel at math.ubc.ca israel at math.ubc.ca
Fri Mar 30 09:02:17 CEST 2012

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.

The probability that the first five divisors are 1,2,4,5,8 is
(1/40)*(2/3)*(6/7)
The probability that the first five divisors are 1,2,4,7,8 is
(1/56)*(2/3)*(4/5)
These are the only ways the fifth divisor can be 8, so the total probability
that the fifth divisor is 8 is 1/42.

Similar calculations could be made for any of the other possibilities for
the fifth divisor.

Note that "almost every" integer has a fifth divisor, in fact for any N the
asymptotic density of n with Omega(n)<N is 0.

Robert Israel
University of British Columbia

On Mar 29 2012, Maximilian Hasler wrote:

>According to numerical data (which does not seem to
>vary when I change the interval from [1..1e6] to [1..1e7]),
>
>it seems that the 5th divisor is most likely to be 6
>(in about 20% of the cases, followed by 8 and 16 in 6% of the cases)
>
>and the 6th divisor is most likely to be 8 (in 9% of the cases)
>(closely followed in likelihood by 10, in  ~7.5% of the cases).
>
>Maximilian
>
> Counts for 5th divisor, per 1000: 0, 0, 0, 0, 42, 211, 36, 60, 52, 48,
> 35, 0, 29, 27, 34, 60, 20, 0, 18, 0, 19, 15, 15, 0, 19, 11, 7, 0, 12, 0,
> 11,... (i.e. 1,2,3,4 occur in 0% of the cases, 5 occurs in 4.2% of the
> cases, etc...)
>
>Counts for 6th divisor, per 1000:
>0, 0, 0, 0, 0, 49, 35, 91, 51, 77, 19, 51, 14, 55, 10, 12, 16, 44, 12,
>31, 12, 28, 9, 0, 10, 22, 11, 20, 7, 0, 6, 28, 10,...
>
>These percentages are of course approximate, calculated by neglecting
>cases where the 5th resp. 6th divisor is > 100.
>
>c=vector(100);for(n=1,1e7,5<#(d=divisors(n)) & d[6]<#c & c[d[6]]++)
>c*1e3\sum(i=1,#c,c[i])
>
>
>On Thu, Mar 29, 2012 at 10:52 PM, Marc LeBrun <mlb at well.com> wrote:
>>>="Alonso Del Arte" <alonso.delarte at gmail.com>
>>
>> Interesting puzzle!
>>
>> My first guess would be the n-th divisor of A061799(n), but...
>>
>>
>>
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