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Old 2007-01-21, 13:10   #23
Citrix
 
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Quote:
Originally Posted by michaf View Post
I couldn't notice anything quicker about pfgw's factoring routines then there is in srsieve (quite the opposite, actually).

What's pfgw's command to find "91268055041 | 22^134217728+1" quickly?
Try this
http://www.underbakke.com/AthGFNsv/

For PFGW see the pfgw documentation.
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Old 2007-01-21, 13:10   #24
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Quote:
Originally Posted by michaf View Post
I couldn't notice anything quicker about pfgw's factoring routines then there is in srsieve (quite the opposite, actually).

What's pfgw's command to find "91268055041 | 22^134217728+1" quickly?
If p is a prime divisor of b^(2^n)+1 then p=2 or p=k*2^(n+1)+1 (where k is an integer).
Using this you can get a much faster sieve.

Last fiddled with by R. Gerbicz on 2007-01-21 at 13:12 Reason: typo
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Old 2007-01-21, 13:15   #25
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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]
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Old 2007-01-21, 22:26   #26
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Quote:
Originally Posted by Citrix View Post
wooot... WOW :>
Much heat has been generated before I was pointed at this program :>

Thanks... this is amazingly quick!

Last fiddled with by michaf on 2007-01-21 at 22:27
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Old 2007-01-22, 19:33   #27
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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.
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Old 2007-01-23, 00:47   #28
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YOu will have to try prime95 for p-1/ECM after that limit is reached.

Last fiddled with by Citrix on 2007-01-23 at 00:48
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Old 2007-01-23, 15:08   #29
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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 P-1
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Old 2007-11-22, 06:41   #30
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Default Eliminate k=22 & 484 and n=0

Quote:
Originally Posted by michaf View Post
That was indeed what I was thinking :)
_and_ it took some 24 hours to get to that sieving point :>
We do not include k's that can be reduced when determining the Riesel and Serpinksi conjectures. I am currently compiling a site of all known primes of the form k*10^n-1 at gbarnes017.googlepages.com/primes-kx10n-1.htm. I would never include multiples of 10 for k's because they can always be reduced to (k/10)*10^(n+1)-1. For the same reason, RPS does not include multiples of 2 for k's base 2. If we included multiples of 10 for base 10, we'd still have no primes for k=450 and k=4500 because there is only one prime at n=1 for k=45. (Riesel k base 10 is k=10176. 3 k's still need primes to prove conjecture.)

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 2007-11-22 at 06:57
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Old 2007-11-22, 15:35   #31
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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.
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Old 2007-11-24, 19:59   #32
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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.
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Old 2007-11-24, 23:37   #33
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Quote:
Originally Posted by masser View Post
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.

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:
However, perhaps we need a new definition? We could extend the pattern of definitions above in this way:

2'. k > 0 is a base 5 Sierpinski number if k*5^n+1 is not prime for any integer n >= -1.

2''. k > 0 is a base 5 Sierpinski number if k*5^n+1 is not prime for any integer n >= -2.

and so on. This leads to the obvious definition:

3. k > 0 is a base 5 Sierpinski number if k*5^n+1 is not prime for any integer n.

I think this definition better fits your idea that only candidates that determine unique sequences should be considered.

For the purposes of this project it makes no difference whether we use definition 2 or definition 3.

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 base-22 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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