Mersenne Digest Tuesday, June 22 1999 Volume 01 : Number 586 ---------------------------------------------------------------------- Date: Sun, 20 Jun 1999 20:41:14 -0400 (EDT) From: lrwiman <[EMAIL PROTECTED]> Subject: Re: Mersenne: Question about speed... >> Sorry to bother you people with this but can anybody tell me why my celeron >> 400 all of a sudden slows to (almost) half speed when it reaches the >> 1,000,000,000 (actually a little more, i guess it's around 2^30) mark in >> factoring numbers??? My p233mmx slows about 15 percent. > I'm just guessing, but it could be that the Celeron's smaller cache gets > filled up. Doubtful, since the LL remainder never goes above Mp. The only thing that inherantly increases is the iteration count, which takes up about log_2(p) bits. Not that much... - -Lucas Wiman P.S. 2^30 is ~100,000,000 not ~1,000,000 ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Sun, 20 Jun 1999 21:35:12 -0400 From: Jud McCranie <[EMAIL PROTECTED]> Subject: Re: Mersenne: Question about speed... At 08:41 PM 6/20/99 -0400, lrwiman wrote: > >Doubtful, since the LL remainder never goes above Mp. The only thing that >inherantly increases is the iteration count, which takes up about log_2(p) >bits. Not that much... >P.S. 2^30 is ~100,000,000 >not ~1,000,000 Actually 2^30 is ~ 1,000,000,000. 2^20 is ~ 1,000,000. He said 1,000,000,000 but he may have meant 1,000,000. Prime95 can't handle 1,000,000,000 yet. Anyhow, the Celeron cache is only 128KB, which is 1.34 million bits. So the Celeron cache size may be the reason for the slowdown at about 1,000,000 bits. +----------------------------------------------+ | Jud "program first and think later" McCranie | +----------------------------------------------+ ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Sun, 20 Jun 1999 21:00:41 PDT From: david campeau <[EMAIL PROTECTED]> Subject: Re: Mersenne: Question about speed... > >Sorry to bother you people with this but can anybody tell me why my celeron >400 all of a sudden slows to (almost) half speed when it reaches the >1,000,000,000 (actually a little more, i guess it's around 2^30) mark in >factoring numbers??? My p233mmx slows about 15 percent. The numbers >currently being factored are in the 9.7 million range (shoudn't matter, it >happened also in the 7.5 million range...). > >Thanks in advance... > >Otto. > Prime95 is trying to find factor in the 2^63 range, soo it take a bit longer then 2^62 1,000,000,000 * 2^32 ~= 2^62 David ______________________________________________________ Get Your Private, Free Email at http://www.hotmail.com ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Mon, 21 Jun 1999 10:53:54 +0200 From: "Steinar H. Gunderson" <[EMAIL PROTECTED]> Subject: Mersenne: Own-compiled versions of GIMPS software Hi, I've hit an interesting problem, and George didn't seem to have the answer. I think I've asked the list before, but now the problem has become a little more serious... I'm currently using a P2/448 (overclocked), and I needed/wanted to add my own hooks (for FTP, GUI output and some other minor things). Now, since George can't possibly include the security code algorithms in the source (and I couldn't possibly think of trying to crack them), the security codes are being set to zero (00000000), and thus PrimeNet won't credit the work to my account. Now, since my version for some reason is faster than George's (2-3% on LL tests, 6-7% on factoring; I have no clue why), I don't really like going back to a slower version. Any suggestions on how to improve the speed? (The machine is idle all the time. Earlier, I tried ReCache, and it didn't help much. The program is started with `-m' and `-w' parameters, and I'm using the ELF version, and the source.zip, both downloaded directly from the link at www.mersenne.org. v18.1 used in both cases, Linux kernel v2.3.6 used, but the same `symptoms' existed when I was using the 2.2.x kernels. gcc 2.7.2.3 and glibc 2.0.111 used. Sigh.) /* Steinar */ ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Mon, 21 Jun 1999 10:17:58 -0400 From: George Woltman <[EMAIL PROTECTED]> Subject: Re: Mersenne: Question about speed... Hi, At 12:42 AM 6/21/99 +0200, Otto Bruggeman wrote: >Sorry to bother you people with this but can anybody tell me why my celeron >400 all of a sudden slows to (almost) half speed when it reaches the >1,000,000,000 (actually a little more, i guess it's around 2^30) mark in >factoring numbers??? My p233mmx slows about 15 percent. The numbers >currently being factored are in the 9.7 million range (shoudn't matter, it >happened also in the 7.5 million range...). This is normal. The factoring code uses different algorithms to find factors below 2^62, from 2^62 to 2^63, and from 2^63 to 2^64. What you have noticed is prime95 switching over to the 2^62 to 2^63 code which is slower than the below 2^62 code. Hope that helps, George ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Mon, 21 Jun 1999 13:31:34 -0500 From: Amy and Shane Sanford <[EMAIL PROTECTED]> Subject: Re: Mersenne: Own-compiled versions of GIMPS software >Now, since my version for some reason is faster than George's (2-3% on LL >tests, 6-7% on factoring; I have no clue why), I don't really like going >back to a slower version. I have played around with the Prime95 V18.1 & found a simular improvement in compiling on a more recent compiler (MS VC++ 5.0 & 6.0). The orignal code indicates it is a MS VC++ 4.x program -- which had a terrible optimizer. Interestingly enough after recompiling the Recache program doesn't seem to help at all so maybe there is a difference in the way it is utilizing ram. I also noticed a couple of things in the GUI that was slowing it down by another % or so. All in all I was able to squeeze about 4%-5% extra speed out of the LL. However I never allowed any of those results to be reported to Primenet because I have not found a way to do enough QA testing on it to completely satisfy myself. It would be very nice if we could put together a QA procedure that we could test these possible performance enhancers on our own then if everything seemed to be ok to send the changes to George & QA team for possible inclusion in a new version. Shane Sanford ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Mon, 21 Jun 1999 20:07:50 +0200 From: "Otto Bruggeman" <[EMAIL PROTECTED]> Subject: Mersenne: Thanks for the answers... Hi, Thanks for answering my question... But this brings me to another one : Wouldn't it be better to first do factoring till 2^62 and then do the the ones above 2 ^62 ??? Just for Celerons and other high clock multiplyer processors, or is that impossible with the the program ??? I think this cause half of the factors i found are below 2^62. I don't know how it is with all the factors that are found but isn't this worth looking into ??? Just my two cents to get more results faster.... Again thanks in advance for replies... Otto ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Mon, 21 Jun 1999 23:42:55 +0200 From: Cyril Flaig <[EMAIL PROTECTED]> Subject: Mersenne: Once again factoring Well, my question is very simple. How can we see that 2*k*p+1 is composite? Or is it composite for any special k like k=4*n ? ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Mon, 21 Jun 1999 22:54:29 +0100 From: Gordon Spence <[EMAIL PROTECTED]> Subject: Mersenne: Re: Distribution of Factors > >Date: Mon, 14 Jun 1999 15:21:52 -0700 >From: Will Edgington <[EMAIL PROTECTED]> >Subject: Mersenne: Re: factoring 10^7 digits (was LL and factoring & quitting) > > >>We will of course have to check factors considerably further than > >>we are doing on our current exponent range (due to the increased > >>LL iteration time.) > >Yup. And don't forget that the larger the exponent, the fewer the >possible factors in a given range (e.g., from 0 to 2^40 or 0 to 2^63). > Ok, I'll ask the stupid question, I stopped maths at the year before university, WHY is this the case? regards G Gordon Spence, Nokia IP Telephony Applications Engineer Grove House, Waltham Way, [EMAIL PROTECTED] White Waltham, Maidenhead, http://www.nokiaiptel.com/ Berkshire, SL6 3TN, UK. Office: +44 1628 827204 GSM: +44 385 576623 ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Mon, 21 Jun 1999 23:03:13 +0100 From: Gordon Spence <[EMAIL PROTECTED]> Subject: Mersenne: Re: Poachers Beware >------------------------------ > >Date: Sat, 19 Jun 1999 17:53:11 -0500 >From: Ken Kriesel <[EMAIL PROTECTED]> >Subject: Mersenne: Poachers beware > >I request that anyone running exponents that are not assigned to them >by either the IPS or George Woltman pause, and ask George for a >nonoverlapping assignment. Absolutely agree. I kept quiet until now, but the more I have thought about it, the more annoyed I got at what was done. I have already found one prime, so my name is in the history books, I would have been extremely p****ed off if a poacher had stolen my exponent and beat me to it. I do not wish anyone to risk having this happen to them. [snip] > (In my darker moments, it occurs to me that this cpu >farm would be powerful for factoring more fully, the LLtests of someone >who's been poaching numbers assigned to others.) I have access for a while to about 3 dozen quad P3-450 boxes. I had thought about taking *ALL* the numbers that Aaron is running for his testing and running them on these machines. So that when he checks them in, he finds that they have all been tested already..... BUT, I like your idea even better. Find a factor of numbers he has already tested so that he *loses* the credit for those exponents. What a _great_ idea.....Now what was that url for the program to factor to 95 bits.... regards G finder of 36th Mersenne prime M2976221 Gordon Spence, Nokia IP Telephony Applications Engineer Grove House, Waltham Way, [EMAIL PROTECTED] White Waltham, Maidenhead, http://www.nokiaiptel.com/ Berkshire, SL6 3TN, UK. Office: +44 1628 827204 GSM: +44 385 576623 ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Mon, 21 Jun 1999 23:03:15 +0200 From: "Steinar H. Gunderson" <[EMAIL PROTECTED]> Subject: Mersenne: Re: Factoring and M38 On Sun, Jun 20, 1999 at 12:10:35AM -0400, [EMAIL PROTECTED] wrote: >A) The 38th Mersenne prime discovered has an exponent in the neighborhood of >6,900,000. How interesting, I've been testing an exponent in the vicinity of 6,700,000 for a few months now... Perhaps somebody just beat me to the computer speed? (Of course, assuming the `islands' theory holds. I won't participate in that discussion.) >B) We *are* missing a Mersenne prime between 3021377 and the 38th discovered >prime (all that's been disclosed about it is that it has an exponent between >6 and 7 million). I believe that this missing prime's exponent is in the >neighborhood of 4,700,000. In other words, it will be reported either very soon (since all new LL tests go over at least 7 million now) or not in a long, long time (since double-checks still are in the vicinity of 3.5 million)? >Time will tell to see if I'm right. :-D Ptoo bad (C) won't be >confirmed/rejected for a few years. Guesses are free. I suggest that everybody makes their own guess (in public), and the one who comes closest gets... errr, well... becomes happy. One guess per person, no poachers :-) (ie. the one who makes his/her wish first, gets to `keep' it, and nobody else should be able to guess on the same number.) (I'll do my guess soon, but I must of course consult the IPS list of factors and completed LL tests.) /* Steinar */ ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Mon, 21 Jun 1999 18:47:45 -0400 From: Jud McCranie <[EMAIL PROTECTED]> Subject: Re: Mersenne: Re: Distribution of Factors At 10:54 PM 6/21/99 +0100, Gordon Spence wrote: >>Yup. And don't forget that the larger the exponent, the fewer the >>possible factors in a given range (e.g., from 0 to 2^40 or 0 to 2^63). >> > >Ok, I'll ask the stupid question, I stopped maths at the year before >university, WHY is this the case? Because a factor of Mp must be of the form 2*k*p+1 (actually only half of those are possible), and as p increases the number of potential factors of that form <= X decreases. +----------------------------------------------+ | Jud "program first and think later" McCranie | +----------------------------------------------+ ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Mon, 21 Jun 1999 17:54:44 -0500 From: "Willmore, David" <[EMAIL PROTECTED]> Subject: RE: Mersenne: Re: Poachers Beware Listen, everyone, please stop. It has been requested on this group that this topic be dropped. Please, please, have the decency to let this topic go. If you *must* discuss it, keep it off the list in private emails. Please remember that a message written in anger is *never* on topic--well, in alt.flame, sure..... Cheers, David > ---------- > From: Gordon Spence[SMTP:[EMAIL PROTECTED]] > Sent: Monday, June 21, 1999 5:03 PM > To: [EMAIL PROTECTED] > Subject: Mersenne: Re: Poachers Beware > > >------------------------------ > > > >Date: Sat, 19 Jun 1999 17:53:11 -0500 > >From: Ken Kriesel <[EMAIL PROTECTED]> > >Subject: Mersenne: Poachers beware > > > >I request that anyone running exponents that are not assigned to them > >by either the IPS or George Woltman pause, and ask George for a > >nonoverlapping assignment. > > Absolutely agree. I kept quiet until now, but the more I have thought > about > it, the more annoyed I got at what was done. I have already found one > prime, so my name is in the history books, I would have been extremely > p****ed off if a poacher had stolen my exponent and beat me to it. > > I do not wish anyone to risk having this happen to them. > > [snip] > > > (In my darker moments, it occurs to me that this cpu > >farm would be powerful for factoring more fully, the LLtests of someone > >who's been poaching numbers assigned to others.) > > I have access for a while to about 3 dozen quad P3-450 boxes. I had > thought > about taking *ALL* the numbers that Aaron is running for his testing and > running them on these machines. So that when he checks them in, he finds > that they have all been tested already..... > > BUT, I like your idea even better. Find a factor of numbers he has already > tested so that he *loses* the credit for those exponents. What a _great_ > idea.....Now what was that url for the program to factor to 95 bits.... > > regards > > G > > finder of 36th Mersenne prime M2976221 > > > Gordon Spence, Nokia IP Telephony > Applications Engineer Grove House, Waltham Way, > [EMAIL PROTECTED] White Waltham, Maidenhead, > http://www.nokiaiptel.com/ Berkshire, SL6 3TN, UK. > Office: +44 1628 827204 GSM: +44 385 576623 > ________________________________________________________________ > Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm > ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Tue, 22 Jun 1999 01:10:10 +0200 From: "Steinar H. Gunderson" <[EMAIL PROTECTED]> Subject: Re: Mersenne: Re: Poachers Beware On Mon, Jun 21, 1999 at 11:03:13PM +0100, Gordon Spence wrote: >I have access for a while to about 3 dozen quad P3-450 boxes. I had thought >about taking *ALL* the numbers that Aaron is running for his testing and >running them on these machines. So that when he checks them in, he finds >that they have all been tested already..... > >BUT, I like your idea even better. Find a factor of numbers he has already >tested so that he *loses* the credit for those exponents. What a _great_ >idea.....Now what was that url for the program to factor to 95 bits.... Come on, if you think what Aaron has done is bad, are you any better yourself? Revenge is not the idea here. We are all trying to be a team. /* Steinar */ ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Mon, 21 Jun 1999 20:07:50 +0200 From: "Otto Bruggeman" <[EMAIL PROTECTED]> Subject: Mersenne: Thanks for the answers... Hi, Thanks for answering my question... But this brings me to another one : Wouldn't it be better to first do factoring till 2^62 and then do the the ones above 2 ^62 ??? Just for Celerons and other high clock multiplyer processors, or is that impossible with the the program ??? I think this cause half of the factors i found are below 2^62. I don't know how it is with all the factors that are found but isn't this worth looking into ??? Just my two cents to get more results faster.... Again thanks in advance for replies... Otto ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Mon, 21 Jun 1999 18:33:05 -0500 From: "Willmore, David" <[EMAIL PROTECTED]> Subject: RE: Mersenne: Re: Distribution of Factors > From: Jud McCranie[SMTP:[EMAIL PROTECTED]] > At 10:54 PM 6/21/99 +0100, Gordon Spence wrote: > >>Yup. And don't forget that the larger the exponent, the fewer the > >>possible factors in a given range (e.g., from 0 to 2^40 or 0 to 2^63). > >> > > > >Ok, I'll ask the stupid question, I stopped maths at the year before > >university, WHY is this the case? > > Because a factor of Mp must be of the form 2*k*p+1 (actually only half of > those > are possible), and as p increases the number of potential factors of that > form > <= X decreases. > Ahhh, because the smallest factor must be <= the sqrt() of the number! Sorry, Gordon, I was wondering the same thing when you asked this. :) Cheers, David ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Mon, 21 Jun 1999 19:39:40 -0400 (EDT) From: lrwiman <[EMAIL PROTECTED]> Subject: Re: Mersenne: Re: Distribution of Factors >>>Yup. And don't forget that the larger the exponent, the fewer the >>>possible factors in a given range (e.g., from 0 to 2^40 or 0 to 2^63). >>> >> >>Ok, I'll ask the stupid question, I stopped maths at the year before >>university, WHY is this the case? > Because a factor of Mp must be of the form 2*k*p+1 (actually only half of > those are possible), and as p increases the number of potential factors of th > at form <= X decreases. Actually, it 2*k*p+1 must be ==+/-1 mod 8 which is 2/8=1/4. This can be further reduced by checking for divisibility by 3, and 5. So thats 1/(15*p) of numbers that we are actually checking. - -Lucas Wiman ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Tue, 22 Jun 1999 02:05:18 +0200 From: "Steinar H. Gunderson" <[EMAIL PROTECTED]> Subject: Mersenne: (My guess at) the M38 exponent is... (No, nothing official here, calm down!) And now, for my official guess at M38! :-) There are (at the time of writing) 59303 exponents between 6,000,000 and 6,999,999 (that's the only thing George has said about the exponent, except for the number of digits) that has not been checked out by IPS. Of these, my guess will be: 6264277 The one who comes closest will of course be the winner. If two are exactly as good (very improbable -- sorry, poaching is not allowed here), they'll share the honour. /* Steinar */ ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Mon, 21 Jun 1999 19:52:59 -0500 From: Ken Kriesel <[EMAIL PROTECTED]> Subject: RE: Mersenne: Re: Distribution of Factors At 06:33 PM 1999/06/21 -0500, "Willmore, David" <[EMAIL PROTECTED]> wrote: >Ahhh, because the smallest factor must be <= the sqrt() of the number! >Sorry, Gordon, I was wondering the same thing when you asked this. :) Um, no. For exponents of current interest, p~=7,000,000, Mp=2^p-1 has p bits, and sqrt(Mp) has p/2 bits, or ~3,500,000, while factoring "only" goes to ~63 bits. Factoring depth is determined by the lesser of: factor<=sqrt(Mp) how far it pays to go (when factoring further yields less than LLtesting) numerical limit of available code Ken ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Mon, 21 Jun 1999 21:16:08 -0400 From: Jud McCranie <[EMAIL PROTECTED]> Subject: RE: Mersenne: Re: Distribution of Factors At 06:33 PM 6/21/99 -0500, Willmore, David wrote: >Ahhh, because the smallest factor must be <= the sqrt() of the number! Yes, but that doesn't matter here. We are checking for divisors less than some relatively small limit (much smaller than the number itself). For a given limit, there are fewer possible divisors as p increases. +----------------------------------------------+ | Jud "program first and think later" McCranie | +----------------------------------------------+ ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Mon, 21 Jun 1999 21:35:03 -0400 From: Jud McCranie <[EMAIL PROTECTED]> Subject: Re: Mersenne: Re: Distribution of Factors At 07:39 PM 6/21/99 -0400, lrwiman wrote: >Actually, it 2*k*p+1 must be ==+/-1 mod 8 which is 2/8=1/4. This can be >further reduced by checking for divisibility by 3, and 5. >So thats 1/(15*p) of numbers that we are actually checking. Yes, but the crucial thing in answering his question is the factor of p. +----------------------------------------------+ | Jud "program first and think later" McCranie | +----------------------------------------------+ ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Mon, 21 Jun 1999 20:14:05 -0400 From: George Woltman <[EMAIL PROTECTED]> Subject: Re: Mersenne: Thanks for the answers... Hi Otto, At 08:07 PM 6/21/99 +0200, Otto Bruggeman wrote: >Wouldn't it be better to first do factoring till 2^62 and then do the the >ones above 2 ^62 ??? This oft asked for feature is already implemented in v19. If an exponent is already factored to 2^55, then it does all 16 passes to 2^56, then all 16 passes to 2^57, etc. Also, the output no longer mentions passes, rather it mentions percent complete. Regards, George ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Tue, 22 Jun 1999 06:56:42 +0100 From: "Brian J. Beesley" <[EMAIL PROTECTED]> Subject: Re: Mersenne: Once again factoring > Well, my question is very simple. How can we see that 2*k*p+1 is composite? > Or is it composite for any special k like k=4*n ? There is another constraint - 2kp+1 must be congruent to 1 or 7 modulo 8. If p is congruent to 1 modulo 8, then for k = 0 (mod 4), 2kp+1 = 1 (mod 8); k = 1 (mod 4) gives 2kp+1 = 3 (mod 8); k = 2 (mod 4) gives 2kp+1 = 5 (mod 8) and k = 3 (mod 4) gives 2kp+1 = 7 (mod 8). So only values of k congruent to 0 or 3 (mod 4) need to be considered. Working similarly, if p = 3 mod 8 then only look at k = 0 (mod 4) or 1 (mod 4); p = 5 mod 8 gives k = 0 or 3 (mod 4) & p = 7 (mod 8) gives k = 0 or 3 (mod 8). That cuts out half the cases straight away. Incidentally, since this gives a nice "try two, then skip two" pattern, independent of the value of p mod 8, we never actually need to calculate 2kp+1 mod 8, this is another timesaver. Now we can filter out multiples of small primes & reduce the number of contenders needing to be trial factored some more. A good way of doing this (eliminates multiples of all the small primes 3, 5, 7, 11, 13, 17) is to contruct a sieve table with 255,255 elements then index into it with 2kp+1 mod 255255. This "modulo" operation can be done by adding 2p mod 255255, or 6p mod 255255, depending on whether we're on a "try next" or "try next but 3" cycle, and subtracting 255255 if possible. When we've eliminated enough candidates, just try the division. It actually doesn't matter if the trial divisor is composite, it's a question of whether it costs more to try the division anyway or to go on to eliminate the candidate factor by proving it prime. Regards Brian Beesley ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Tue, 22 Jun 1999 03:20:50 -0400 (EDT) From: lrwiman <[EMAIL PROTECTED]> Subject: Mersenne: More 10,000,000+ digit factors I have found 1,821 new factors in the range of Brian's 10,000,000+ digits. All of the other primes in this range have been tested through 2^46. They are avalaible at: http://www.tasam.com/~lrwiman/fact46 or http://www.tasam.com/~lrwiman/fact46.gz - -Lucas Wiman ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Fri, 18 Jun 1999 09:17:38 +0200 From: "Steinar H. Gunderson" <[EMAIL PROTECTED]> Subject: Re: Mersenne: Finite or infinite? On Fri, Jun 18, 1999 at 01:00:20PM +1200, Halliday, Ian wrote: >ourworld.compuserve.com, the full URL of which nobody ever seemed to >remember. Now I'm in the same boat as S Gunderson, who has switched to >double checking because he doesn't like having to wait months for a result. You're certainly in the same boat as me, but I've still only got _one_ PC (a P60) on double-checking! My other (a PII/448) is currently doing a mixture of LL/factoring/double-check (as an experiment), and a third (a P233) is doing LL work. In addition, when summer is over, I'll reinstall Prime95 at my school, and get those 40 486s back. I guess you must have only confused me with somebody else, since I once participated in a thread on this topic. No harm done :-) /* Steinar */ ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Tue, 22 Jun 1999 13:14:28 +0200 From: "Steinar H. Gunderson" <[EMAIL PROTECTED]> Subject: Mersenne: Re: Poachers Beware On Tue, Jun 22, 1999 at 12:23:47AM +0100, Gordon Spence wrote: >I think the point is to hammer into Aarons skull that what he did was wrong >period. By doing the same yourself? No, let us please let this discussion drop. >Anyway if it turns up some new factors, then that is advancing scientific >knowledge and the team gains anyway ;-) The team does not gain from this. /* Stenar */ ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Tue, 22 Jun 1999 13:21:33 +0200 From: "Steinar H. Gunderson" <[EMAIL PROTECTED]> Subject: Re: Mersenne: Once again factoring On Tue, Jun 22, 1999 at 06:56:42AM +0100, Brian J. Beesley wrote: >There is another constraint - 2kp+1 must be congruent to 1 or 7 >modulo 8. Here comes something that has been confused me for a long time. Isn't 1 modulo 8 = 1? You are always talking about `1 modulo something'. My definition of `modulo' (a mod b is the remainder of a/b) gives: 0 mod 4 = 0 1 mod 4 = 1 2 mod 4 = 2 3 mod 4 = 3 4 mod 4 = 0 5 mod 4 = 1 etc., and by this follows that 1 mod x = 1 (at least for x >= 1). But it looks like I've missed something important... /* Steinar */ ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Tue, 22 Jun 1999 08:31:43 -0400 From: Jud McCranie <[EMAIL PROTECTED]> Subject: Re: Mersenne: Once again factoring At 01:21 PM 6/22/99 +0200, Steinar H. Gunderson wrote: >Here comes something that has been confused me for a long time. Isn't >1 modulo 8 = 1? You are always talking about `1 modulo something'. Almost always it is something modulo something else is congruent to 1. +----------------------------------------------+ | Jud "program first and think later" McCranie | +----------------------------------------------+ ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Tue, 22 Jun 1999 06:08:22 -0700 From: Paul Leyland <[EMAIL PROTECTED]> Subject: RE: Mersenne: Once again factoring > From: Steinar H. Gunderson [mailto:[EMAIL PROTECTED]] > On Tue, Jun 22, 1999 at 06:56:42AM +0100, Brian J. Beesley wrote: > >There is another constraint - 2kp+1 must be congruent to 1 or 7 > >modulo 8. > > Here comes something that has been confused me for a long time. Isn't > 1 modulo 8 = 1? You are always talking about `1 modulo something'. > My definition of `modulo' (a mod b is the remainder of a/b) gives: > > 0 mod 4 = 0 > 1 mod 4 = 1 > 2 mod 4 = 2 > 3 mod 4 = 3 > 4 mod 4 = 0 > 5 mod 4 = 1 > > etc., and by this follows that 1 mod x = 1 (at least for x >= 1). But > it looks like I've missed something important... You are missing the implied parentheses. You seem to be wrongly interpreting the expression as (2kp) + (1 modulo 8). Interpret it as (2kp+1) modulo 8 and it might make more sense. For example, if p = 5, then: 2*1*5+1 = 11 == 3 mod 8 2*2*5+1 = 21 == 5 mod 8 2*3*5+1 = 31 == 7 mod 8 2*4*5+1 = 41 == 1 mod 8 2*5*5+1 = 51 == 3 mod 8 and so on. Paul ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Tue, 22 Jun 1999 13:44:15 -0400 From: Marc Getty <[EMAIL PROTECTED]> Subject: Mersenne: PC Magazine Press > For those of you who read PC Magazine, there is a short column by Bill > Machrone in the July 1999 issue on page 85 that talks about GIMPS, Aaron > Blosser, and the US West episode. Not a detailed examination of what > happened but some good press on why you might want to participate in such a > search by a "responsible research organization." There are also two other items relating to the US West v. Aaron Blosser episode available at PC Magazine's web site. First is a short uninformed tidbit: http://www.zdnet.com/pcmag/issues/1721/367991.htm And second is the apology and retraction for the above: http://www.zdnet.com/products/stories/reviews/0,4161,388805,00.html - -Marc Marc Getty [EMAIL PROTECTED] Department of Dental Informatics, Temple University http://www.temple.edu/dentistry/di/ 215-707-8192 ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Tue, 22 Jun 1999 20:55:19 -0700 From: Lang Pal <[EMAIL PROTECTED]> Subject: Re: Mersenne: Mersenne 3/2 conjecture I recommend to replace the expression " (3/2)^n " by " (PI*SQRT(2))/3". Then we receive 1,480960969 instead of 3/2 and the correlation coefficient would be almost just the same (0,996117397). Why then to replace? I think, there must be some relation between the Mersenne problem and the partition function (considering the Hardy-Ramanujan-Rademacher formula). Other factors must be found somehow and once in the (I hope "next" ) future. Best regards Paul La'ng Budapest, Hungary Jud McCranie wrote: > If you take the following comma delimited file into a spreadsheet, and > graph it (say with a line chart) it shows the relationship of Mersenne > exponents to their index, for the first 37 Mersenne primes. The first > column is the log of (3/2)^n, the second column is the log of the exponent > of the nth Mersenne prime...... > +----------------------------------------------+ > | Jud "program first and think later" McCranie | > +----------------------------------------------+ > > ________________________________________________________________ > Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Tue, 22 Jun 1999 15:29:33 -0400 From: Jud McCranie <[EMAIL PROTECTED]> Subject: Re: Mersenne: Mersenne 3/2 conjecture At 08:55 PM 6/22/99 -0700, Lang Pal wrote: >I recommend to replace the expression " (3/2)^n " by " >(PI*SQRT(2))/3". Then >we receive >1,480960969 instead of 3/2 and the correlation coefficient would be >almost >just the same >(0,996117397). Why then to replace? I think, there must be some >relation >between the Mersenne problem and the partition function (considering >the >Hardy-Ramanujan-Rademacher formula). Other factors must be found >somehow and >once in the (I hope "next" ) future. That sounds very interesting. 3/2 happens to be close to the value based on the current data, but otherwise I don't know of any justification for 3/2. +----------------------------------------------+ | Jud "program first and think later" McCranie | +----------------------------------------------+ ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Tue, 22 Jun 1999 20:45:29 +0100 From: "Brian J. Beesley" <[EMAIL PROTECTED]> Subject: Re: Mersenne: Once again factoring On 22 Jun 99, at 20:14, Cyril Flaig wrote: > Could someone explain me how the sieve table works? OK. If you can follow C code fragments then the following should be at least as clear as a whole heap of words. int i; char * sieve; char * sptr; sieve = (char *) malloc(255255); for ( sptr=sieve, i=0 ; i<255255 ; sptr++, i++ ) * sptr = 1; for ( sptr=sieve, i=0 ; i<255255 ; sptr+=3, i+=3 ) * sptr = 0; for ( sptr=sieve, i=0 ; i<255255 ; sptr+=5, i+=5 ) * sptr = 0; for ( sptr=sieve, i=0 ; i<255255 ; sptr+=7, i+=7 ) * sptr = 0; for ( sptr=sieve, i=0 ; i<255255 ; sptr+=11, i+=11 ) * sptr = 0; for ( sptr=sieve, i=0 ; i<255255 ; sptr+=13, i+=13 ) * sptr = 0; for ( sptr=sieve, i=0 ; i<255255 ; sptr+=17, i+=17 ) * sptr = 0; The above is initialization code which only needs to be executed once. We now have a table pointed to by sieve in which the j'th element is non-zero only if j is not divisible by 3, 5, 7, 11, 13 or 17 (the six smallest odd prime numbers. 2 does not worry us since a factor of 2^p-1 obviously must be odd) - provided 0 <= j < 255255. Now 3 * 5 * 7 * 11 * 13 * 17 = 255255, so, if we want to see if a candidate factor f is divisible by any of 3, 5, 7, 11, 13, 17, all we have to do is to look in the (f % 255255)'th element of the sieve table. i.e. go on to do further checks on the factor if sieve[f % 255255] != 0. If you like pure pointer notation then instead read * (sieve + (f % 255255)) != 0. Got it? Regards Brian Beesley ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Tue, 22 Jun 1999 20:48:56 +0100 From: "Brian J. Beesley" <[EMAIL PROTECTED]> Subject: Re: Mersenne: Mersenne 3/2 conjecture On 22 Jun 99, at 20:55, Lang Pal wrote: > I recommend to replace the expression " (3/2)^n " by " > (PI*SQRT(2))/3". Then > we receive > 1,480960969 instead of 3/2 and the correlation coefficient would be > almost > just the same > (0,996117397). Why then to replace? Well, we're dealing with experimental data, and both look about equally good. On balance I prefer the argument with pi in it. Pi deserves a place in most fundamental laws... (Much waving of arms ... obviously a beautiful solution isn't always right) Regards Brian Beesley ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Tue, 22 Jun 1999 15:28:40 -0600 From: "Aaron Blosser" <[EMAIL PROTECTED]> Subject: RE: Mersenne: Mersenne 3/2 conjecture > Well, we're dealing with experimental data, and both look about > equally good. On balance I prefer the argument with pi in it. Pi > deserves a place in most fundamental laws... > > (Much waving of arms ... obviously a beautiful solution isn't always > right) Besides, any equation just looks much cooler if you throw in a pi or an e or something. Just the word "transcendental" makes an equation seem otherworldly! :-) ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ Date: Tue, 22 Jun 1999 18:32:38 -0500 From: JON STRAYER <[EMAIL PROTECTED]> Subject: RE: Mersenne: $1000 supercomputer > > "HAL-300GrW1, a "hypercomputer" that is said to be 60,000 times as fast > > as a 350-MHz Pentium, and many times as fast as IBM's supercomputer > > Pacific Blue." > From the same article: > is on its high-end hypercomputer line, HAL. The > HAL-300GrW1 has a price tag of about $26 million, > If we could get one of these, we could use the money better for 60,000 > PII-350's anyway:) Only if we could get the PII-350's for $433 each. ________________________________________________________________ Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm ------------------------------ End of Mersenne Digest V1 #586 ******************************
