Mersenne: Specific exponent reservation
Hi, Is it possible to reserve a specific exponent through Primenet? I would like to test an exponent that is not randomly assigned to me but don't want to loose the credit for it on my Primenet account. Thanks, Attila _ Unsubscribe list info -- http://www.scruz.net/~luke/signup.htm Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
Re: Mersenne: (2^2^n) + 1
www.mersenne.org/ecmg.htm for current ECM factoring limits on Fermat numbers, Oops, should read: ecmf.htm Ciao, Alex. _ Unsubscribe list info -- http://www.scruz.net/~luke/signup.htm Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
Re: Mersenne: (2^2^n) + 1
Nayan Hajratwala wrote: problem; I need to find the largest known prime of the form: (2^2^n)+1 congratulations by the way on finding the largest Mersenne prime!!! These are Fermat numbers, Fermat conjectured that all numbers of this form would be prime and proved it for F_0=3, F_1=5, F_2=17, F_3=257,F4=65537. However, not a single prime beyond those 5 has been found so far, all 4 n 31, and many 31, are proved composite. It is commonly believed that indeed there is not a single Fermat prime except the first five. Find another one and you'll be famous! See http://www.perfsci.com/prizes.html for a Fermat factoring contest, www.mersenne.org/ecmg.htm for current ECM factoring limits on Fermat numbers, and http://vamri.xray.ufl.edu/proths/fermat.html for overall status of Fermat numbers (prime, genuine composites, some factors known, completely factored) Ciao, Alex. _ Unsubscribe list info -- http://www.scruz.net/~luke/signup.htm Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
Re: Mersenne: p-1 and trial factoring
On 25 Feb 00, at 15:23, Reto Keiser wrote: > Why can't we do first first the factorization up to n-2 bits (1/4) of the > trial factoring time, then start the P-1 factoring up to 1/3 of the B1 > value, after this, we can complete the trial factoring process and at the > end we complete the P-1 (using the save file od intermediate file). (the > parameters can be optimized) This sounds fairly sensible. However, this entails splitting the factoring part into fairly small sub-assignments, which may cause unneccessary complications with the server. Also, trial factoring and P-1 are quite different from the point of view of system requirements - trial factoring uses very little memory (in practice it runs almost entirely in the L1 cache) whereas P-1 is actually more of a memory hog than LL testing. So I suspect we want some bias towards early trial factoring rather than P-1. > until now >210 factors are found for 10megadiginumbers and more than 280 > exponents were factored up to 68 bits. Some (about 7) 67 digit factors > were found but none with 68 bits. This is likely to be a statistical anomoly. A sample size of 7 is a bit small to condemn the data as biased. > A lot of factors of exponents between 1 and 100 were found using > the new P-1 method. Is there a database which contains which exponent were > tested using which B1 and maybe a database od the save files? Yes - I think we need this database - with or without savefiles, it's a waste of effort to inadvertently duplicate work done before. Since P-1 is deterministic (like trial factoring, but unlike Pollard's rho or ECM) you should get the same result every if you use the same limits on the same exponent. If anyone has any data to contribute, I'd be willing to assemble publish the database. I also have adequate storage space on my anon ftp server for save files. Regards Brian Beesley _ Unsubscribe list info -- http://www.scruz.net/~luke/signup.htm Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
Re: Mersenne: p-1 and trial factoring
Hi, At 03:23 PM 2/25/00 +0100, Reto Keiser wrote: parallel use of p-1 and trial factoring --- Why can't we do first first the factorization up to n-2 bits (1/4) of the trial factoring time, then start the P-1 factoring up to 1/3 of the B1 value, after this, we can complete the trial factoring process and at the end we complete the P-1 (using the save file od intermediate file). (the parameters can be optimized) I can't see any flaws in your reasoning, although it would be a bit unwieldy to implement. no 68 bit factors - until now 210 factors are found for 10megadiginumbers and more than 280 exponents were factored up to 68 bits. Some (about 7) 67 digit factors were found but none with 68 bits. My database has: 3321966173867482830512390441 3322338783006905661336745889 33221387123317319076102495049 33235409128314644111933147703 33238463131707491089550166169 33230671139408728702078150121 33224957193425473534465274127 That's 6 67-bit factors and 1 68-bit factor. Not the expected distribution, but nothing to be concerned about yet either. organization of p-1 factoring - A lot of factors of exponents between 1 and 100 were found using the new P-1 method. Is there a database which contains which exponent were tested using which B1 and maybe a database od the save files? All exponents from 2 to 11 were done with B1=1M and B2=40M Exponents from 11 to 60 (still in progress) were done with B1=100K and B2=4M. I still have the save files for exponents below 11. I think Alex has the save files for the larger exponents. However, it must be pointed out that at some point you are better off switching to ECM rather than expanding the P-1 bounds. I'm not sure what that point is. Regards, George _ Unsubscribe list info -- http://www.scruz.net/~luke/signup.htm Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
Mersenne: Perfect numbers
Can someone please outline a proof as to why (2^p-1)(2(p-1)) is a perfect number if 2^p-1 is prime? 2^6972593 - 1 is prime. e^(i*pi) + 1 = 0. This is the e-mail address of Simon Rubinstein-Salzedo. When you read this e-mail, Simon will probably be at a math contest. Don't forget to check Simon's website at http://www.albanyconsort.com/simon Thanks SJRS _ Unsubscribe list info -- http://www.scruz.net/~luke/signup.htm Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
RE: Mersenne: Perfect numbers
A number N is perfect if an only if sigma(N)=2N, where the sigma function is the sum of alldivisors of N, including 1 and N. The sigma function verify: i) sigma(p)=p+1, if p is prime ii) sigma(p^n)=1+p+p^2+...+p^n=(p^(n+1)-p)/(p-1), if p is prime iii) sigma(a·b)=sigma(a)·sigma(b), if gcd(a,b)=1 (it's a multiplicative function) Then, if N=2^(p-1)(2^p-1), with 2^p-1 prime 8a Mersenne prime), we have sigma(N)=sigma(2^(p-1)(2^p-1))=sigma(2^(p-1))sigma(2^p-1)=((2^p-1)/(2-1))(2^ p-1+1)= (2^p-1)2^p=2(2^(p-1))(2^p-1))=2N. There is a partial converse: If N is perfect AND EVEN, then N=2^(p-1)(2^p-1), with 2^p-1 prime. It is not proved the inexistence of perfect odd numbers, althought the minimum cote is very high. Un saludo, Ignacio Larrosa Cañestro A Coruña (España) [EMAIL PROTECTED] - Original Message - From: Simon J Rubinstein-Salzedo [EMAIL PROTECTED] To: [EMAIL PROTECTED] Sent: Saturday, February 26, 2000 12:27 AM Subject: Mersenne: Perfect numbers Can someone please outline a proof as to why (2^p-1)(2(p-1)) is a perfect number if 2^p-1 is prime? 2^6972593 - 1 is prime. e^(i*pi) + 1 = 0. This is the e-mail address of Simon Rubinstein-Salzedo. When you read this e-mail, Simon will probably be at a math contest. Don't forget to check Simon's website at http://www.albanyconsort.com/simon Thanks SJRS _ Unsubscribe list info -- http://www.scruz.net/~luke/signup.htm Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers _ Unsubscribe list info -- http://www.scruz.net/~luke/signup.htm Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
Mersenne: A couple quick questions
I just joined GIMPS (now 6% done testing a number with exponent just short of 10M if it makes a difference) and I have been looking into the theory behind Mersenne primes. Can anyone show me or at least point me to a webpage with the proof that the exponent of a Mersenne prime must be prime? How about a proof that the LL test works? I have had math through DiffEq I. It is intuitively obvious to me that every Mersenne number with even composite exponent will be found by the formula M(p) = 4M(p-2)+3. Since M(2)=3 is a multiple of 3, all these numbers will also be multiples, and therefore composite. However, I can't understand why this is true of numbers whose exponents have higher This is particularly easy to see when the numbers are written in binary. Since it is difficult to hand-compute the factors of Mersenne composites much above 1023, I cannot easily search for patterns in higher Mersenne numbers with composite exponent that has no factor of 2. A completely unrelated question: Why, on the PrimeNet stats summery page, are far more numbers listed as "finished LL" than as "available for doublecheck"? Does this simply mean that some numbers do not require doublechecking because they were turned in by a proven computer? Or have they been already double-checked and turned in since the page was last updated? __ Get Your Private, Free Email at http://www.hotmail.com _ Unsubscribe list info -- http://www.scruz.net/~luke/signup.htm Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
