# [seqfan] Re: nth divisor of a number

Charles Greathouse charles.greathouse at case.edu
Fri Mar 30 09:05:58 CEST 2012

Right -- I was composing an email as I received yours detailing that
process.  Basically, the counts near 10^8 are higher because of the
number of times that there are not enough divisors or too-large
divisors.

6 will occur 1/12 of the time.

5 happens 1/60 of the time.

7 happens 1/70 of the time.

8 happens 1/42 of the time.

9 happens 13/630.

10 happens 2/105.

1, 2, 3, 4, 12, 18, 20, ... never happen.

I did these by hand, but a computer could do them faster I'm sure.  In
any case the numbers are too small to give a definitive proof that 6
is the most common, but it's pretty clear that it's so.

Charles Greathouse
Analyst/Programmer
Case Western Reserve University

On Fri, Mar 30, 2012 at 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.
>
> 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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