20050913, 10:20  #1 
Feb 2004
Paris
3 Posts 
Some news about Home Prime ?
Hello
On http://www.worldofnumbers.com/topic1.htm, it seems nothing new. Has anyone any info about this ? It seems that the work has ended on January 3, 2004 (by Alex Kruppa and Paul Leyland) Looks like the c204 which is 34636914551761683256158051843633814787706289345767962219592\ 92066545246725876130493435583943733963381945857837752697856\ 75210636696425094776859733305947996048061499249566197147212\ 934512427988113420226762897 could be GMPECMable even if 5500 curves at B1=11M, and 9000 curves at B1=44M has already been done. Any more results about this ? Cheers MoZ 
20050913, 12:20  #2 
Jul 2004
Potsdam, Germany
3·277 Posts 
Chances are that there exists a factor < ~60 digits, which can be found by ECM, given enough computing power (over time).
But for most, the benefit/effort ratio is not high enough to continue this factorization attempt. For example, a lot of Cunningham numbers have obtained way less effort, but are generally considered "more valuable" (AFAIK, the Home Primes are merely a recreational project with no further use). Having said that, I don't want to dictate where people have to put their efforts on. I think that everyone should choose their own favorites, based on their individual preferences. So please, only take this posting as decision support. 
20050913, 12:33  #3 
"Nancy"
Aug 2002
Alexandria
100110100011_{2} Posts 
>Chances are that there exists a factor < ~60 digits
I'm not too sure about that... Alex 
20050913, 13:09  #4 
Jul 2004
Potsdam, Germany
3×277 Posts 
Well, I planned to add "but are not very high", but forgot to do it lateron.

20050913, 14:16  #5  
Nov 2003
1110100100100_{2} Posts 
Quote:
ECM curve counts. My paper "A Practical Analysis of ECM" shows how to do this. Suppose the number we are trying to factor is a randomly chosen integer [i.e. chosen uniformly at random from all odd numbers of the same size] Dickman's function tells us the distribution (in terms of size) of its prime factors. This gives a Bayesian prior. Then, the ECM failures yield further information [a sample]. We then use Bayes' Theorem to derive a posterior distribution, from which we can compute an expected value. 

20060220, 20:20  #6  
"Jason Goatcher"
Mar 2005
3·7·167 Posts 
Quote:
Just kidding. :D 

20060228, 12:02  #7 
May 2005
Lyon
1000_{2} Posts 
Hello,
For information: hp49: i have run 2280 @ b1=11e7 on hp49(100).c204 (with ecm 5.0.3) I had done them in 2004 and at the beginning of 2005. Sean Irvine 100 * B1= 11e7 Details:  http://euclide.euclide.free.fr/hp49/ (it is same the link that I had already posted a few months ago, it didn't evolve  http://www.angelfire.com/falcon2/hom...oendprime.html hp146/273: 7704 @ B1=43e6 3 @ B1 = 11e7 p+1 1 @ B1 = 10^9 p1 3 @ B1 = 10^9 hp242: p+1 1 @ B1 = 10^9 p1 3 @ B1 = 2*10^9 hp300: p+1 1 @ B1 = 10^9 p1 3 @ B1 = 2*10^9 hp312: 4402 @ B1 = 11e6 7779 @ B1= 43e6 370 @ B1 = 11e7 p+1 1 @ B1 = 10^9 p1 3 @ B1 = 10^9 hp495: p+1 1 @ B1 = 10^9 p1 3 @ B1 = 2*10^9 hp858: p+1 1 @ B1 = 10^9 p1 3 @ B1 = 2*10^9 All without results :( Alex p.s : Sorry for my so bad english :( Last fiddled with by [Leo_01] on 20060228 at 12:03 
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