20070121, 13:10  #23  
Jun 2003
1,579 Posts 
Quote:
http://www.underbakke.com/AthGFNsv/ For PFGW see the pfgw documentation. 

20070121, 13:10  #24  
"Robert Gerbicz"
Oct 2005
Hungary
1,429 Posts 
Quote:
Using this you can get a much faster sieve. Last fiddled with by R. Gerbicz on 20070121 at 13:12 Reason: typo 

20070121, 13:15  #25 
Jun 2003
2×3^{2}×269 Posts 
And srsieve (as well as PFGW) has a modular sieve option that will restrict itself to factors of the correct form. [Not sure if sr2sieve has this option implemented]

20070121, 22:26  #26  
Jan 2005
737_{8} Posts 
Quote:
Much heat has been generated before I was pointed at this program :> Thanks... this is amazingly quick! Last fiddled with by michaf on 20070121 at 22:27 

20070122, 19:33  #27 
Jan 2005
479 Posts 
Citrix,
do you happen to know if athGFNSieve can sieve beyond about 9159662798383349761? Two instances I've had upto that point, and then give up on me. Process is still using CPU power, but no more updates either on screen or in file. 
20070123, 00:47  #28 
Jun 2003
62B_{16} Posts 
YOu will have to try prime95 for p1/ECM after that limit is reached.
Last fiddled with by Citrix on 20070123 at 00:48 
20070123, 15:08  #29 
Jan 2005
479_{10} Posts 
Any idea on what good B1, B2 values might be for P1?
I cannot do anything on stage 2 (insufficient memory). What is usually needed to get say 30 digit factors? I know how ecm works with that regard, but no idea on what to do with P1 
20071122, 06:41  #30  
May 2007
Kansas; USA
10100000100111_{2} Posts 
Eliminate k=22 & 484 and n=0
Quote:
Also, don't include n=0 when proving these types of conjectures. n=0 is prime for the same k's in all bases and hence is redundant and unnecessary. As Citrix stated, eliminate k=22 and k=484 from your search because they are multiples of 22. 484*22^n+1 would simply be 22^(n+2)+1. As stated by Phil, k=22 can only be prime if n is a power of 2. The same of course applies to any k that is a power of 22. And any k that is a multiple of 22 can be reduced...i.e. 44*22^n+1 = 2*22^(n+1)+1. The conjecture is proven if a prime is found for all k's < then the first known Sierpinski # that are both: (1) Not multiples of the base. (2) Do not have the same factor for every occurrence of n. (i.e. every n has a factor of 3 for all k==2 mod 3 for base 22 so clearly you wouldn't count those as a Sierpinski #'s). Gary Last fiddled with by gd_barnes on 20071122 at 06:57 

20071122, 15:35  #31 
Jul 2003
wear a mask
2^{2}·3·127 Posts 
Please see
http://www.mersenneforum.org/showthread.php?t=4832  the issue of whether n=0 should be counted or not is layed out very well there. ... and don't dictate what others should do. 
20071124, 19:59  #32 
Jan 2006
Hungary
100001100_{2} Posts 
Michaf, are you still working on this numbers? If not, do you mind if I take over?
In the thread there is a post stating that you tested all until n = 18000. How far did you get? Willem. 
20071124, 23:37  #33  
May 2007
Kansas; USA
19·541 Posts 
Quote:
I'm sorry, Masser, Michaf, and everyone else. Masser was right. I came across as a bit of a 'dictator' there. (Well, actually a LOT of a dictator.) Masser, I checked out the thread that you posted here. It is a good one but I guess I am a little bit confused now. In the post, doesn't Geoff state the following?: Quote:
I can only speculate and maybe you can confirm that we want to use the above base 5 defintion for all bases. Is that your thinking? If so, here is where I'm confused: Based on this definition, wouldn't we eliminate k=22 from testing for the base22 Sierpinski conjecture because 22*22^0+1=23, which of course is prime? And wouldn't it also eliminate k=484? Because 484*22^(1)+1=23, which is prime again? I'm asking because I saw a suggestion in the 'base 6 to 18' thread that we need one place where all of the information is brought together for the conjectures for all bases. What I would like to do is put together a web page solely dedicated to showing all of the bases (perhaps up to 50 or so), their respective known conjectured Riesel and Sierpinski numbers, and what k's are left to find a prime to prove the conjectures for each base (for the reasonable ones, obv not base 7). If I can understand this k=22 and k=484 issue for base 22, that would be very helpful. Thank you, Gary 

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