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Topic: 2^57,885,161 -1
Replies: 10   Last Post: Feb 9, 2013 1:28 PM

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gus gassmann

Posts: 60
Registered: 7/26/12
Re: 2^57,885,161 -1
Posted: Feb 8, 2013 9:01 AM
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On 08/02/2013 8:45 AM, David Bernier wrote:
> On 02/07/2013 03:56 PM, Pubkeybreaker wrote:
>> On Feb 7, 2:00 pm, Transfer Principle<david.l.wal...@lausd.net>
>> wrote:

>>> On Feb 7, 7:17 am, Frederick Williams<freddywilli...@btinternet.com>
>>> wrote:
>>>

>>>> Sam Wormley wrote:
>>>>> Largest Prime Number Discovered [to date]
>>>>>> http://www.scientificamerican.com/article.cfm?id=largest-prime-number...
>>>>>>
>>>>>> The number ? 2 raised to the 57,885,161 power minus 1 ? was
>>>>>> discovered by University of Central Missouri mathematician Curtis
>>>>>> Cooper as part of a giant network of volunteer computers devoted to
>>>>>> finding primes, similar to projects like SETI@Home, which downloads
>>>>>> and analyzes radio telescope data in the Search for Extraterrestrial
>>>>>> Intelligence (SETI). The network, called the Great Internet Mersenne
>>>>>> Prime Search (GIMPS) harnesses about 360,000 processors operating at
>>>>>> 150 trillion calculations per second. This is the third prime number
>>>>>> discovered by Cooper.

>>>> By Cooper or by GIMPS?
>>>
>>> By Cooper. GIMPS itself has discovered 14 primes.
>>>
>>> http://en.wikipedia.org/wiki/Great_Internet_Mersenne_Prime_Search

>>
>> Cooper found the specific prime, BUT it was a GIMPS
>> *** group effort** that sifted through many many thousands of
>> candidates
>> and eliminated them as possibilities.

>
> Yes, I agree Dr Silverman.
>
> This 48th known Mersenne prime has been added to Chris Caldwell's
> Prime Pages:
>
> http://primes.utm.edu/notes/faq/NextMersenne.html


Thanks for supplying this link. I think the analysis there is quite
interesting. One question not addressed is a probabilistic analysis of
whether the new prime is indeed the 48th or there is another prime in
there someplace. It seems this analysis could be attempted based on the
data points currently available.

> There's the heuristic that if y_n = log_2(log_2(M_n)), where
> M_n is the n'th Mersenne prime, then the y_n resemble the
> arrival times in a Poisson process.
>
> For the known y_n, n=1 to 48, I get:
>
> 0.664
> 1.489
> 2.308
> 2.805
> 3.700
> 4.087
> 4.247
> 4.954
> 5.930
> 6.475
> 6.741
> 6.988
> 9.025
> 9.245
> 10.320
> 11.105
> 11.155
> 11.651
> 12.054
> 12.110
> 13.242
> 13.279
> 13.452
> 14.283
> 14.405
> 14.502
> 15.441
> 16.396
> 16.753
> 17.010
> 17.721
> 19.529
> 19.713
> 20.262
> 20.415
> 21.505
> 21.526
> 22.733
> 23.682
> 24.323
> 24.518
> 24.630
> 24.857
> 24.957
> 25.147
> 25.345
> 25.361
> 25.786 (48 values).
>
> Chris Caldwell states that this gives:
> "a correlation coefficient R^2 = 0.9919".
>
> I get that this is the "Coefficient of determination" R^2:
> http://en.wikipedia.org/wiki/Coefficient_of_determination
>
> the square of Pearson's r:
> http://en.wikipedia.org/wiki/Pearson_product-moment_correlation_coefficient#For_a_sample
>
>
> These commands show my computation of R^2:
>
> ? ybar = (sum(X=1,48,y[X]))/48.0000000000000
> %94 = 14.538509898083550641780687795589356676
>
> ? A = sum(X=1,48, (x[X] - xbar)*(y[X]-ybar))
> %95 = 5088.1949729587201638754833533863947090
>
> ? B = sum(X=1,48, (x[X]-xbar)^2)
> %96 = 9212.0000000000000000000000000000000000
>
> ? C = sum(X=1,48, (y[X]-ybar)^2)
> %97 = 2833.5016529801736003925746709139375204
>
> ? B=sqrt(B)
> %98 = 95.979164405614617749102200287187885447
>
> ? C = sqrt(C)
> %99 = 53.230645806529283553703786993969466052
>
> ? r = A/(B*C)
> %100 = 0.99592135484160784915459792443098549831
>
> ? R2 = r*r
> %101 = 0.99185934502954377404245361220842196837
>
> ? R2
> %102 = 0.99185934502954377404245361220842196837
>
> Let's compare: Caldwell obtains R^2 = 0.9919 (Ok).
>
> ===
>
> Now, Poisson processes are memoryless
> http://en.wikipedia.org/wiki/Poisson_process
>
> So, if Y_i for i=1 to 48 are the arrival times (of the event number
> i) in a Poisson process, a sample,
>
> and X_i = i for i = 1 to 48, what is the distribution of
> Z= R^2, the square of Pearson's r ?
>
> And in particular what are the chances that Z >= 0.9919 ?
> (the R^2 for the Mersennes).
>
> David Bernier
>
>





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