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Old 2004-08-04, 19:06   #1
Citrix
 
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Jun 2003

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Default New Strategy

I was thinking, that since the new PRP is really fast compared to the old one, it might be better to take k=214519, which is the heaviest k and most likely to produce a prime, and PRP it. If there are some bugs only a few numbers may produce false residues, the rest can be used for double check. But if we find a prime, our work will be really reduced.

What do you all think?

Citrix
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Old 2004-08-05, 16:36   #2
ValerieVonck
 
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Citrix ok for me.
When my range for RieselSieve is completed, I think I will re-join your search.

Could you please make a list with the work to do?

k weight of K #tests
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Old 2004-08-05, 16:55   #3
ltd
 
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I think we should stay with the strategy we use at the moment. The advantage of crunching all k at the same time is that when finding a prime only a some other PC calclulate a residue for the same K and so there are not so many wasted results before we switch from one k to the next.

Lars
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Old 2004-08-05, 19:02   #4
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My computers will crunch all the k's.

k=214519 is for people who want to use PRP3. Soon LLR4 will support +1 numbers also and LLR4 will be faster than PRP3.

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Old 2004-10-31, 11:47   #5
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How large does n have to become before it is worthwhile using P-1 before the prp test?

If one n was tested to a higher level and no prime found then it could be an opportunity to introduce P-1 testing, and maybe get some of the factoring people interested in the project.
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Old 2004-10-31, 12:25   #6
ltd
 
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I can not tell you when it will really makes sence to start p-1 testing again but i make tests with p-1 from time to time. The last tests have been made with n around 1.25 Mil. At that level p-1 is not a choise.

With the sieving level we have reached i would expect that p-1 will be interesting at around n=1.7 Mil. But until the time we reach that with PRP tesing we should have sieved much deeper then that.

I think we will really catch up at a level around n=2.4 Mil.

But that are only my guesses. I will make some p-1 tests from time to see if i am correct.

Lars
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