20110216, 00:47  #45 
"Phil"
Sep 2002
Tracktown, U.S.A.
2141_{8} Posts 
Zuzu made some more careful calculations that I have now repeated, so I now estimate that we would have had about a 14.5% chance of finding the last prime in the range 51462959092394 instead of the 17% I wrote above. He also wrote on the Seventeen or Bust Forum that when this project started, at n = 1.4 million, we would have predicted a 0.9% chance of finding all the prps by 9092394. I haven't repeated that particular calculation, but it sounds like it is in the right ballpark. We have been incredibly lucky! Hopefully, some of this luck will rub off on Seventeen or Bust now.
A few other interesting tidbits:
Fortunately, with a billion computers, this would only take 300 years, as the tests can be trivially distributed. Maybe we should start another project! 
20110216, 01:09  #46 
A Sunny Moo
Aug 2007
USA
2^{2}×11^{2}×13 Posts 
I don't suppose an ECPP proof could be distributed easily? (I know it can be distributed amongst multiple cores of the same computer, but is there any reason why they have to be on the same computer?) Because if it could be done, then that might be the next step for this project: working through the unproven PRPs from the bottom up. (Perhaps, by the time the biggest ones are reached, computers will be sufficiently faster that the proofs will be within reach by ECPP, or by some faster method if it becomes available by then.)

20110216, 01:13  #47 
Feb 2009
27_{16} Posts 
What are we waiting for? I am sure when our successors perfect quantum computing they will appreciate the 0.000001% head start. ;)

20110216, 02:49  #48 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
1000010110111_{2} Posts 
IMHO that strongly suggests something more than luck. Has anyone tested ranges to check if 2^n+k produces more primes than expected over any chosen range? It wouldn't be hard to search low n over a broad k range (even if well outside what was needed to prove the conjecture), e.g. such that you can expect 100 or more primes, and compare expected primes to observed to see if the trend holds up.
Last fiddled with by TimSorbet on 20110216 at 02:49 
20110216, 04:33  #49  
"Phil"
Sep 2002
Tracktown, U.S.A.
19·59 Posts 
Quote:
By the way, I added a few more names in the "thanks to ..." section of post 38. I can't believe I left out Justin (enderak), as I even mentioned him in the post! Also, Alex, Nathan, and Robert. If anyone else spots any oversights, please let me know! 

20110216, 06:22  #50  
"Gary"
May 2007
Overland Park, KS
10111110001100_{2} Posts 
Quote:
What about finding some factors of 2^9092392+40292 ? Would that help with the PRP test for 2^9092392+40291 ? The factor DB has only 2^2*3 for it and I wouldn't know how to begin factoring such a large number. 

20110216, 13:49  #51 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
It would help strengthen a P+1 PRP check. I've done 10 curves at B1=11000 but not found any prime factor >3 yet. I use mprime with
Code:
ECM2=1,2,9092390,10073,11000,0,90,"3" Last fiddled with by akruppa on 20110406 at 13:17 
20110216, 15:32  #52 
"William"
May 2003
Near Grandkid
5^{3}·19 Posts 
Don't you need to account for partial coverings? Some k's are much more likely than others to have primes.

20110216, 16:22  #53  
Jun 2003
5460_{10} Posts 
Quote:
Quote:
My conclusion was that how many primes a series produces is only weakly predictive of where the first prime would be. So one doesn't necessarily help with the other. Of course, it wouldn't surprise me if the analysis was deeply flawed. 

20110216, 22:19  #54 
"Phil"
Sep 2002
Tracktown, U.S.A.
10001100001_{2} Posts 
That's why I suggested fixing n and searching a range of k's, with enough k's, we would expect the weights to average out. On the other hand, maybe someone thinks that these particular k's were for some reason, more likely to yield primes at low n. That would be difficult to test, it basically would require extending this project!

20110216, 22:27  #55 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
10B7_{16} Posts 
I am going to search 3<=k<10K (k odd) for 10K<=n<=20K. From doing some calculations on portions of the range sieved to 1M, I expect approximately 10000 primes to be in this range. I'll have a more exact figure when sieving is complete. We'll see how it turns out. If anyone feels my bounds are a bad representation, they can search elsewhere (and, if they have a good reason why, I might be inclined to stop searching this).
Last fiddled with by TimSorbet on 20110216 at 22:30 
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