[seqfan] Re: nth divisor of a number

[israel at math.ubc.ca]

Why is it clearly enough?

Robert Israel
University of British Columbia

On Mar 30 2012, Charles Greathouse wrote:

>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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