Mersenne: question
Hi, I wonder if any one can help me on the following: First consider the set of all mersenne numbers 2^n - 1, then we know that an infinite number of these are NOT prime, e.g., the set 2^n - 1 with n itself NOT prime. Now lets only focus on the set 2^p - 1 with p prime, i.e., the set of numbers that we are checking out at GIMPS. Has anyone proven that an infinite number these are NOT prime (this is VERY likely to be true)? If so, how can one prove this easily (it is probably not possible to indentify a subset that is NOT prime as above because then we would not consider all of them for GIMPS)? Thanks, Benny --- Benny Van Houdt, University of Antwerp Dept. Math. and Computer Science PATS - Performance Analysis of Telecommunication Systems Research Group Universiteitsplein, 1 B-2610 Antwerp Belgium email: [EMAIL PROTECTED] Unsubscribe list info -- http://www.scruz.net/~luke/signup.htm
Re: Mersenne: question
Now lets only focus on the set 2^p - 1 with p prime, i.e., the set of numbers that we are checking out at GIMPS. Has anyone proven that an infinite number these are NOT prime (this is VERY likely to be true)? It's nice to be able to say this, but it is in the FAQ. Check it out at http://www.tasam.com/~lrwiman/FAQ-mers The last three questions (those in section 4) are pertinate to these questions. I would ask that before anyone sends a question to the list, please check the FAQ if it deals with basic questions involving the following: (1) Basics of a mersenne number (what is it, how many digits, etc...) (2) The Lucas-Lehmer test (repeating LL remainders, modular arithmetic, etc...) (3) Factoring Mersenne numbers (how is it done, how do we sieve, etc... (4) Distribution of Mersenne primes and numbers (how many mersenne primes, and how factors and mersenne primes are distributed) If you still have questions after reading the relevant sections of the FAQ, by all means send them to the list!! Thank you, Lucas Wiman Unsubscribe list info -- http://www.scruz.net/~luke/signup.htm
Re: Mersenne: M38 = M6972593
On Mon, Jul 05, 1999 at 09:50:42PM -0700, Eric Hahn wrote: (Note to Scott - create a dummy non-zero residue a stick it in the cleared exponents report). Too late!! The Cleared Exponents Report reads: I think he meant `next time' :-) /* Steinar */ Unsubscribe list info -- http://www.scruz.net/~luke/signup.htm
Re: Mersenne: M38 = M6972593
NOW it does, after the official announcement Remember when Roland found M37? Someone found a 0x000 residue in the report and beat George to the punch, so Scott modified the reports so that they would NOT post a zero residue automatically. So THIS time, when word came that we'd found a potential prime, some enterprising person immediately grabbed the "assigned exponents" file, and the "cleared exponents" file, and by the process of elimination, deduced the prime number because it was the ONLY candidate listed as "assigned" but was not EITHER cleared as non-prime or still in progress. George was telling Scott to correct for this 'leak' so that a really determined person could not do a comparison-elimination to deduce a prime number find before George announces it. Of course, Curt Noll's web page made that a pointless exercise... ;-) At 09:50 PM 7/5/99 -0700, you wrote: (Note to Scott - create a dummy non-zero residue a stick it in the cleared exponents report). Too late!! The Cleared Exponents Report reads: 6972593 62 P 0x 01-Jun-99 13:57 nayan precision-mm Unsubscribe list info -- http://www.scruz.net/~luke/signup.htm Unsubscribe list info -- http://www.scruz.net/~luke/signup.htm
Re: Mersenne: question
At 10:47 AM 7/6/99 +0200, Benny.VanHoudt wrote: Now lets only focus on the set 2^p - 1 with p prime, i.e., the set of numbers that we are checking out at GIMPS. Has anyone proven that an infinite number these are NOT prime (this is VERY likely to be true)? If so, how can one prove this easily (it is probably not possible to indentify a subset that is NOT prime as above because then we would not consider all of them for GIMPS)? If 2p+1 is prime then it divides 2^p-1. If it has been proven that there are in infinite number of prime pairs p and 2p+1 then this proves that there are an infinite number of 2^p-1 that is not prime when p is prime. These are called Sophie Germain primes, and it has been proven that there are an infinite number of them, therefore there are an infinite number of composites of the form 2^p-1 when p is prime. +--+ | Jud "program first and think later" McCranie | +--+ Unsubscribe list info -- http://www.scruz.net/~luke/signup.htm
Re: Mersenne: question
If 2p+1 is prime then it divides 2^p-1. If it has been proven that there are in infinite number of prime pairs p and 2p+1 then this proves that there are an infinite number of 2^p-1 that is not prime when p is prime. These are called Sophie Germain primes, and it has been proven that there are an infinite number of them, therefore there are an infinite number of composites of the form 2^p-1 when p is prime. This is not quite right. The primes must be ==3 mod 4. For example, 29 is prime and ==1 mod 4, but 59 does not divide 2^29-1. I'm not sure whether or not it has been proven whether or not there are an infinity of Sophie Germain primes of the form 4*n+3. I imagine there would be, as there are an infinity of primes in the form 4*n+1 and 4*n+3. -Lucas Wiman Unsubscribe list info -- http://www.scruz.net/~luke/signup.htm
Re: Mersenne: question
On 6 Jul 99, at 11:22, Jud McCranie wrote: If 2p+1 is prime then it divides 2^p-1. Only if p (and therefore 2p+1 also) are congruent to 3 (modulo 4). If it has been proven that there are in infinite number of prime pairs p and 2p+1 then this proves that there are an infinite number of 2^p-1 that is not prime when p is prime. True... These are called Sophie Germain primes, and it has been proven that there are an infinite number of them, Can you please supply a reference to this proof? Chris Caldwell's Prime Pages show this as a conjecture (with a strong heuristic argument). See http://www.utm.edu/research/primes/lists/top20/SophieGermain.html In any case, proving that there an infinite number of S-G primes congruent to 3 (modulo 4) is, presumably, a bit harder - though it would seem very likely to be true - possibly a bit _less_ likely than there being only a finite number of composite Mersenne numbers with prime exponents, though! Regards Brian Beesley Unsubscribe list info -- http://www.scruz.net/~luke/signup.htm
Re: Mersenne: question
Jud McCranie [EMAIL PROTECTED] writes: At 10:47 AM 7/6/99 +0200, Benny.VanHoudt wrote: Now lets only focus on the set 2^p - 1 with p prime, i.e., the set of numbers that we are checking out at GIMPS. Has anyone proven that an infinite number these are NOT prime (this is VERY likely to be true)? If so, how can one prove this easily (it is probably not possible to indentify a subset that is NOT prime as above ^ because then we would not consider all of them for GIMPS)? If 2p+1 is prime then it divides 2^p-1. If it has been proven that there are in infinite number of prime pairs p and 2p+1 then this proves that there are an infinite number of 2^p-1 that is not prime when p is prime. These are called Sophie Germain primes, and it has been proven that there are an infinite number of them, therefore there are an infinite number of composites of the form 2^p-1 when p is prime. Please leave adequate white space in your right margin. Benny's twice-indented (indentified?) text twice still reads well, but three of Jud's indented lines wrap around on my 80-character screen. It is _conjectured_ that p and 2p+1 are simultaneously prime infinitely often, but I have seen no proof. This is related to the twin prime conjecture, in which p and p+2 are simultaneously prime. More generally, if f1(x) and f2(x) are irreducible polynomials with integer coefficients such that i) f1(x) and f2(x) approach +infinity as x - +infinity [This excludes constant polynomials, and 3 - 7*x.]; ii) For each prime q there exists an integer n such that q does not divide the product f1(n)*f2(n) [This excludes f1(x) = x and f2(x) = x + 1.]; then the conjecture predicts infinitely many integers n for which f1(n) and f2(n) are simultaneously prime. A variation of this conjecture extends the result to more than two polynomials. Even the one-polynomial result is unproven when its degree exceeds 1: are there infinitely many primes of the form x^2 + 1? Peter Montgomery Unsubscribe list info -- http://www.scruz.net/~luke/signup.htm
Re: Mersenne: question
At 06:55 PM 7/6/99 +0100, Brian J. Beesley wrote: Can you please supply a reference to this proof? Chris Caldwell's Prime Pages show this as a conjecture (with a strong heuristic argument). No, I was wrong about it having been proven. +--+ | Jud "program first and think later" McCranie | +--+ Unsubscribe list info -- http://www.scruz.net/~luke/signup.htm
Re: Mersenne: LL Factoring DE Crediting
On Tue, Jun 29, 1999 at 12:12:28PM +0200 (OK, late reply, it suddenly struck me that I hadn't replied...), Sturle Sunde wrote: number which is tested already, you climb by pushing someone else down. That isn't very likely to happen, is it? Am I the only one who doesn't trial-factor random LL tested exponents? :-) If I don't factor far enough, that will eventualy happen to me. H, perhaps... (Of course, factoring is a good idea in general; it saves you from wasting times on LL tests.) Therefore I think that Georges formula, just counting LL-results for every Mersenne without a factor in the database and give credit to the people who tested those numbers, is a beautiful solution. But what if a person had a load (say any number for the discussion, 1000 might be a bit extreme, but still _possible_) of 486s only, and didn't want them to LL test becuase that takes _ages_? (As we've discussed earlier, some Dells have problems with flickering during LL tests, which I've experienced myself the last two weeks.) (Then again, it saves some problems -- people with P6-class CPUs (that have almost twice as much `factoring speed' as `LL speed' per cycle (in P90 CPU year)) won't be tempted to do factoring only, and run up the ranks :-)) As long as you do get assigned normal, first-time LL tests (double-checks are already trial-factored with only a marginal chance of a factor missed, so you get an unfair advantage -- no factoring has to be done), George's setup is perfect. When things get a bit more complex, IPS is better, IMHO. /* Steinar */ Unsubscribe list info -- http://www.scruz.net/~luke/signup.htm
Mersenne: Spreading the Word
Well, I've fully updated my GIMPS/PrimeNet banners page. I don't know if anyone but Mr. Kurowski uses them (which is really great of him!), but it's worth a shot. As usual, they're up-to-date for the time being. Those statistics do change fast. They are all 40x400 pixels, 256 color .GIF format. Despite the sluggishness of the gallery page, each individual banner loads quickly. I have 36 versions in total. GIMPS8.GIF in particular looks nifty. If you have a web page, *please* use them somewhere so GIMPS gets more members! There's even a little one (at the bottom of the gallery page) that just says 2^P - 1. For those interested: The gallery page can be found at http://mersenne.cjb.net/ The FTP directory of the banners is at ftp://members.aol.com/stl137/bannerz/ There is now a .ZIP file of all the banners on the gallery page. S.T.L. Unsubscribe list info -- http://www.scruz.net/~luke/signup.htm
Mersenne: SJ Mercury News
For those of you who are interested, the San Jose Mercury News has published the story. http://www.mercurycenter.com/premium/scitech/docs/prime06.htm Unsubscribe list info -- http://www.scruz.net/~luke/signup.htm
Re: Mersenne: SJ Mercury News
Eric Hahn wrote: For those of you who are interested, the San Jose
Mercury News has published the story.
http://www.mercurycenter.com/premium/scitech/docs/prime06.htm
Yes, they did but I was disappointed in the article. No mention of
the GIMPS site! {8-[ All those SETI plugs! {8-| Lets wait and see if
our numbers go up in six weeks or so. spike
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