Mersenne: Specific exponent reservation

2000-02-25 Thread Attila Megyeri
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 _

Re: Mersenne: (2^2^n) + 1

2000-02-25 Thread Alexander Kruppa
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 --

Re: Mersenne: (2^2^n) + 1

2000-02-25 Thread Alexander Kruppa
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

Re: Mersenne: p-1 and trial factoring

2000-02-25 Thread Brian J. Beesley
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

Re: Mersenne: p-1 and trial factoring

2000-02-25 Thread George Woltman
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,

Mersenne: Perfect numbers

2000-02-25 Thread Simon J Rubinstein-Salzedo
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

RE: Mersenne: Perfect numbers

2000-02-25 Thread Ignacio Larrosa Cañestro
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

Mersenne: A couple quick questions

2000-02-25 Thread Nathan Russell
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