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#1 |
Dec 2010
2×37 Posts |
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F33, F34, and F35 are the smallest Fermat numbers (of unknown primality) that could be prime.
I'm assuming that various researchers have attempted to factor them and that their attempts have been considerably deep. Given they have effectively ruled out many small potential factors, what are the "probabilities" that these numbers are prime? I'm most interested in the "probability" that F33 is prime given that it has no (relatively) small factors. Last fiddled with by siegert81 on 2018-07-17 at 22:36 |
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#2 |
Sep 2003
2·5·7·37 Posts |
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Lack of small factors doesn't mean much. After all, for Mersenne numbers in the typical ranges we work on, we find small factors for only about 60% of them. For the rest we have to do Lucas-Lehmer tests, but nearly all of them test composite and Mersenne primes are exceedingly rare.
For Fermat numbers the counterpart to the Lucas-Lehmer test is the Pépin test, but F33 has an exponent of about 8.6 billion so it's just too big to do this test on the current generation of computers. I'd imagine that the odds of finding a Fermat prime decrease as the exponent increases, similar to what is observed for Mersenne numbers. For Mersennes we have the Wagstaff conjecture which matches the existing data pretty well. At each order of magnitude the number of Mersenne primes is roughly the same: we expect to find as many primes with exponent between 1,000 and 10,000 as the number with exponent between 1 million and 10 million or between 1 billion and 10 billion. At some point in the future it may become practical to do ECM testing on F33, but not yet. The largest Fermat number for which ECM testing is practicable is F29, with an exponent of about 537 million, which already has a known small factor = 1120049 × 229+2 + 1. I have been doing some ECM testing on it to try to find a second factor, but it takes about 120GB of memory and about a week to do a single ECM curve for F29. Last fiddled with by GP2 on 2018-07-18 at 04:26 |
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#3 | |
Einyen
Dec 2003
Denmark
3·1,151 Posts |
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http://www.prothsearch.com/fermat.html#Search For F33: k*235 searched to k=4.8*1017 and k*236 searched to k=7*1017 and lower for k*237, k*238 etc. Now: 7*1017 * 236 is about 295 so 95 bits. Compare that to F33 = 28,589,934,592 + 1. So it has been factored (almost) up to the 90,000,000th root of F33. For a Mersenne number in the 90M range the 90,000,000th root is 2, so that would correspond to no factoring at all on a 90M exponent. For F34 and F35 it is even worse: ~295 vs 217,179,869,184 and 234,359,738,368 Last fiddled with by ATH on 2018-07-18 at 04:05 |
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#4 | |
Einyen
Dec 2003
Denmark
3×1,151 Posts |
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#5 |
Banned
"Luigi"
Aug 2002
Team Italia
10011000001112 Posts |
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Not as efficient as a LL test, but quite better than a single core in my tests.
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#6 |
Banned
"Luigi"
Aug 2002
Team Italia
130716 Posts |
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BTW I just restarted my trial-work on n=37...
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#7 | |
Sep 2003
2·5·7·37 Posts |
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The machine is an r4.4xlarge on AWS, with eight cores. The "R" instance types have much more memory than the "C" types but slower processors. I'm using the other seven cores to do ECM on small Mersenne numbers because I worried that the Fermat ECM wouldn't use all eight cores cost-effectively. I didn't actually try it though, maybe I will. |
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#8 | |
"Rashid Naimi"
Oct 2015
Remote to Here/There
41·59 Posts |
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I would appreciate if someone would confirm that F33 is the smallest Fermat number with unknown primality status.
I would also appreciate if someone would work out the number of decimal-digits of F33. I keep running out of memory computing it. Thank you advance. ETA Quote:
Are F31 and F32 known to be composite? Last fiddled with by a1call on 2018-07-18 at 21:25 |
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#9 | |||
Sep 2003
2×5×7×37 Posts |
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F31 has the factor 5463561471303 × 233 + 1 F32 has the factor 1479 × 234 + 1 |
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#10 |
"Rashid Naimi"
Oct 2015
Remote to Here/There
97316 Posts |
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Thank you for taking the time to reply with such details.
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#11 | |
"Rashid Naimi"
Oct 2015
Remote to Here/There
45638 Posts |
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How is that not meaningless, given that there are infinite number of Fermat numbers? Thank you in advance. |
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