[seqfan] Re: nth divisor of a number
Charles Greathouse
charles.greathouse at case.edu
Fri Mar 30 09:17:58 CEST 2012
No, wait, it's clearly enough to check up to 12. In that case since
11 happens only 16/1155 of the time we're done -- 6 is the most
common.
Charles Greathouse
Analyst/Programmer
Case Western Reserve University
On Fri, Mar 30, 2012 at 3:05 AM, Charles Greathouse
<charles.greathouse at case.edu> wrote:
> 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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