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#1 |
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Aug 2010
Kansas
547 Posts |
Hi Math guys!
Well, I've been participating in this project in various forms for, oh, a little over a year, picking up number theory as I go. I was wondering, though, if there was a formula for how many numbers would have to be LL tested to give a 95% chance of a world record prime being included (top 5000). For the sake of simplicity, we'll assume it uses random k values at n~=1M, pre-sieved to 3 trillion using psieve (just for standardization purposes). About how many candidates would be needed to assure with 95% certainty the inclusion of at least 1 prime? I've noted the results turned in for various drives, but wanted to prevent the factoring in of "high weight" k values. Assume odd k values, preventing the re-testing i.e. 6*2^1001000 and 3*2^1001001 would only be tested as the latter. Thanks (Whether or not there are any solutions)! Johannes Schuck |
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#2 |
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Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
10000101010112 Posts |
The LL test is only for Mersenne numbers, but I still get what you're asking. See http://www.mersenneforum.org/showthr...311#post233311 (some explanation) and http://www.mersenneforum.org/showthr...457#post233457 (latest version, download this one). The script (JAR file, you'll need Java installed) works off of sieved files. If you don't have any of those, you can manually make a sieve file with 1 value, and use the numbers it outputs to figure out the number you'd need to test.
To have a 95% chance of a prime being found, you want to expect about 3 primes, which is around 3x the "on average, 1 in x candidates should be prime" figure. Last fiddled with by Mini-Geek on 2012-01-18 at 00:17 |
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#3 |
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Aug 2010
Kansas
547 Posts |
Thanks! the first thread's spreadsheet says about 40,000 for the scenario I was thinking of!
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