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Old 2008-01-05, 18:47   #100
kar_bon
 
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Default riesel base 6

17459*6^25627-1 is prime!
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Old 2008-01-05, 21:32   #101
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Default another riesel base 6

35965*6^27098-1 is prime.

about 23100 candidates fewer to test!

the last sieve-file contains 258000 candidates for the remaining 17 k's (k=1597 and 9577 extra sieve). so after this 2nd prime there are about 223500 left!
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Old 2008-01-06, 00:40   #102
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Quote:
Originally Posted by kar_bon View Post
here the first prime:

17459*6^25627-1 is prime!

it was the 10th candidate for this k i tested!!!
now about 11400 candidates less to test upto n=1M.
Great! Karsten, are you searching all of the other 17 k's at once? (I know you're doing the lowest 2 k's separately.) If so, when you report any prime, I'll just update all of the ranges searched to the n-value of the prime.


Thanks,
Gary
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Old 2008-01-06, 00:45   #103
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yeah, that's right. all remaining 17 k in one sieve file and llr testing.
i have to sieve much more but this was a blind shot and it works.
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Old 2008-01-06, 10:53   #104
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another one bites the dust:

110784*31^19748-1 is prime

That leaves 17 or bust :>
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Old 2008-01-06, 18:55   #105
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I decided to flood this thread:

51540*31^21120-1 is prime

Now there are 16 candidates left for Riesel base 31

Last fiddled with by michaf on 2008-01-06 at 18:57
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Old 2008-01-06, 23:02   #106
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Default Starting Sierpinski base 6

I will start Sierpinski base 6 and test up to n=30K, if no-one else is working on it?
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Old 2008-01-07, 03:32   #107
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Quote:
Originally Posted by geoff View Post
I will start Sierpinski base 6 and test up to n=30K, if no-one else is working on it?

I was just thinking of that one myself but my resources are quite busy so fire away! The bases divisible by 3 generally drop the k's pretty fast so it shouldn't be too bad even with a high conjecture.

Recommendation: Run PFGW 4 times, 1 each for k == 0 mod 5, 1 mod 5, 2 mod 5, and 3 mod 5 up to n=5K-7K before eliminating multiples of 6 and sieving/LLRing. It'll make for much cleaner pfgw.out files that get pretty huge with a large conjecture such as this one has.


Gary
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Old 2008-01-07, 03:35   #108
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I am going to be a glutton for punishment and start Sierp base 19 from scratch and take all k's up to n=10K. The conjecture is k=765174.


Wish me luck!


Gary

Last fiddled with by gd_barnes on 2008-01-07 at 04:08
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Old 2008-01-07, 06:18   #109
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I've gained access, at least for the short term, to a Pentium 3 1Ghz machine, and though it would probably be mediocre at best for LLR, it should do fine for sieving. I've decided to reserve Riesel Base 30 k=25 for sieving up to n=100K. (I'll be releasing it after the sieving is over.) I've started the sieve with srsieve on my main crunching machine (P4 3.2Ghz), and when it's complete to roughly 0.5-1G (it should take only a few minutes) I'll move it to the Pentium3 and sieve it with sr1sieve.

Since k=25 is a power of 5, no n divisible by 5 can be prime; thus, I'll remove such n after sieving using the method Gary suggested to me in a PM earlier.

Last fiddled with by gd_barnes on 2008-01-07 at 06:33 Reason: No change; accidentally edited and changed back
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Old 2008-01-07, 06:32   #110
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Quote:
Originally Posted by Anonymous View Post
I've gained access, at least for the short term, to a Pentium 3 1Ghz machine, and though it would probably be mediocre at best for LLR, it should do fine for sieving. I've decided to reserve Riesel Base 30 k=25 for sieving up to n=100K. (I'll be releasing it after the sieving is over.) I've started the sieve with srsieve on my main crunching machine (P4 3.2Ghz), and when it's complete to roughly 0.5-1G (it should take only a few minutes) I'll move it to the Pentium3 and sieve it with sr1sieve.

Since k=25 is a power of 5, no n divisible by 5 can be prime; thus, I'll remove such n after sieving using the method Gary suggested to me in a PM earlier.

Yes it is a power of 5, but the analysis is incorrect. k=25 is a perfect square. You should remove all n divisible by 2 because 5^2=25. If it was a perfect 5th power, i.e. k=3^5=243, then you would remove all n divisible by 5.

Analysis: let k=m^2 and n=2q...so m^2*30^(2q)-1 = (m*30^q-1) * (m*30^q+1) hence all even n are composite.


Gary

Last fiddled with by gd_barnes on 2008-01-07 at 06:34
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