[seqfan] Re: nth divisor of a number

[Charles Greathouse]

> Why is it clearly enough?

For n to be the 5th-smallest divisor of the number, it must divide the
number.  This happens with probability 1/n, so n is the 5th-smallest
divisor with probability < 1/12 for n > 12.

Charles Greathouse
Analyst/Programmer
Case Western Reserve University

On Fri, Mar 30, 2012 at 3:09 PM,  <israel at math.ubc.ca> wrote:
> 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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