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#78 | |
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Serpentine Vermin Jar
Jul 2014
3,313 Posts |
Quote:
![]() EDIT: By "done lately and confirmed" I mean there are people who do sometimes run new tests of previously discovered primes - the server doesn't accept them, but people do the tests, I guess just to see what happens, or whatever. Last fiddled with by Madpoo on 2018-12-19 at 03:19 |
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#79 | |
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"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
152B16 Posts |
Quote:
In PRP3, for the series of all known Mersenne exponents, I'm half way through M49 now, having started from M2, running the exponents big enough for gpuOwL V5.0-9c13870, and the lower exponents were run on prime95 v29.4b8. After the gpuowl run completes (and perhaps with some delay if I've guessed M51 right), run times will go up at https://www.mersenneforum.org/showpo...6&postcount=10 Haven't seen anything out of the ordinary yet. That's encouraging but not definitive. If they all pass, it does not necessarily mean the existing codes are without flaw. If one fails, it does not necessarily mean the existing codes are flawed, it could be a transient hardware issue.There's a sliver of a chance of an error occurring outside the loop of iterations that are guarded for accuracy by the Gerbicz check. No disrespect to any of the authors (or users, or hardware designers for that matter; some tasks are just very hard). I've fixed code that failed nonreproducibly at low rates. Ppb or lower frequency bugs can be hard to determine exist, much less identify and resolve. Current assignments can be of order 1018 core clocks to complete. |
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#80 | |
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Sep 2003
5·11·47 Posts |
Quote:
So I would expect that the counts above will never be exactly 50/50 even theoretically. If you slightly alter the λ of a Poisson distribution then you slightly alter your counts at k=0. We could look at the factors themselves, rather than distinct exponents with or without factors. Based on the same dataset I used before (frozen on 2018-12-01), I get: Code:
all f f <= 65 bits p=1 mod 4, f=1 mod 8 9846883 9304449 p=3 mod 4, f=1 mod 8 9833742 9310151 p=1 mod 4, f=7 mod 8 10655959 10119410 p=3 mod 4, f=7 mod 8 11565454 11045807
If p=3 mod 4 and 2p+1 is prime, then it's a factor of the Mersenne number with exponent p, and f=2p+1=7 mod 8. And if p=1 mod 4 then the smallest possible factor is 6p+1, and again in this case there will be f=6p+1=7 mod 8. Last fiddled with by GP2 on 2018-12-19 at 08:29 |
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#81 | |
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"Mihai Preda"
Apr 2015
3·457 Posts |
Quote:
So the "systematic error" could be only in the form of undiscovered primes (below the current wavefront). If that would be the case, then the density of MPs would be even farther away from what's theoretically predicted. |
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#82 | |
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Dec 2008
you know...around...
2×5×67 Posts |
Quote:
Not sure how good of a comparison they are, but generalized repunits may provide enough data for some empirical analysis. Some time ago I searched for primes p such that 2kp+1 give less primes than average for small k, thus making it more probable that (b^p-1)/(b-1) is prime. The output was not that significant, as far as I remember, but it was just a thought. |
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#83 | |
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Sep 2003
50318 Posts |
Quote:
And apart from Wagstaff (b=−2), I've been doing ECM lately to find more factors for generalized repunits for bases 3, 5, 7, 11, 12 and −3, −5, −7, −10, −11, −12. I've contributed the factors to FactorDB and found some new PRP cofactors, But it's mostly very unexplored territory. For instance, for base b=3 I'm finding lots of small factors unknown by FactorDB at a wavefront of a mere p=63k or so, and at even lower p for the other bases. So right now there just aren't that many known primes or semiprimes yet for these repunits, and the statistics are too low to be meaningful. |
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#84 | ||
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Sep 2003
5·11·47 Posts |
As mentioned earlier, filtering by non-tiny p is in practice equivalent to filtering by the asymmetricity of the factors that make up the semiprime.
We can only find relatively small factors, so beyond the Cunningham project range where nearly everything is fully factored, every Mersenne semprime or Wagstaff semiprime that we are capable of discovering is necessarily composed of two highly asymmetric factors. These were the counts when we filtered Mersenne semiprimes by size of the exponent p: Quote:
Code:
Mersenne semiprimes
filter by asymmetric bit length:
len(small factor) < 0.05 *
len(Mersenne number)
mod 4
p=1 15
p=3 29
mod 8
p=1 8
p=3 11
p=5 7
p=7 18
Quote:
Code:
Wagstaff semiprimes
filter by asymmetric bit length:
len(small factor) < 0.05 *
len(Mersenne number)
mod 4
p=1 21
p=3 11
mod 8
p=1 12
p=3 5
p=5 9
p=7 6
Last fiddled with by GP2 on 2018-12-21 at 20:35 |
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#85 |
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Sep 2003
5×11×47 Posts |
All factors of Mersenne numbers are 1 or 7 (mod 8). But all factors in general are 1 or 5 (mod 6), except for 2 and 3.
Combining the two gives: All factors of Mersenne numbers are 1 or 7 or 17 or 23 (mod 24). The various congruence classes are not equally likely (factor data frozen as of December 1): Code:
[(1, 170925), (7, 190532), (17, 185312), (23, 207834)] 0M < p < 10M [(1, 135127), (7, 150156), (17, 146802), (23, 165115)] 10M < p < 20M [(1, 83674), (7, 95232), (17, 92200), (23, 104402)] 990M < p < 1000M Code:
p=1 (mod 4)
[(1, 85562), (7, 102644), (17, 92628), (23, 88934)] 0M < p < 10M
[(1, 67370), (7, 81196), (17, 73487), (23, 69787)] 10M < p < 20M
[(1, 41802), (7, 51688), (17, 46104), (23, 44159)] 990M < p < 1000M
p=3 (mod 4)
[(1, 85363), (7, 87888), (17, 92684), (23, 118900)] 0M < p < 10M
[(1, 67757), (7, 68960), (17, 73315), (23, 95328)] 10M < p < 20M
[(1, 41872), (7, 43544), (17, 46096), (23, 60243)] 990M < p < 1000M
Empirically, all Mersenne numbers with odd exponent are 7 mod 24 (that is, the values of the Mersenne numbers themselves, not the values of the exponents). Therefore a Mersenne semiprime must consist of either a (1, 7) mod 24 pair of factors or a (17, 23) mod 24 pair of factors. So the frequency asymmetries of 1,7,17,23 mod 24 factors maybe can start to explain why p=1 (mod 4) and p=3 (mod 4) seem to produce semiprimes with different likelihood? Last fiddled with by GP2 on 2018-12-29 at 16:56 |
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#86 |
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Sep 2003
5·11·47 Posts |
Likewise, all factors of Wagstaff numbers are 1 or 3 (mod 8) and consequently 1, 11, 17, 19 (mod 24).
For all known factors (as of a couple of days ago), we also have asymmetries: Code:
all
[(1, 520596), (11, 807965), (17, 638798), (19, 672074)]
p=1 (mod 4)
[(1, 257566), (11, 514932), (17, 316467), (19, 279297)]
p=3 (mod 4)
[(1, 263030), (11, 293033), (17, 322331), (19, 392777)]
Last fiddled with by GP2 on 2018-12-29 at 16:46 |
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#87 | |
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"Forget I exist"
Jul 2009
Dumbassville
20C016 Posts |
Quote:
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#88 |
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"Jeppe"
Jan 2016
Denmark
23×3×7 Posts |
I do not know what "empirically" means in the preceding posts, but for any odd \(p\) we have \[2^p\equiv(-1)^p = -1 \pmod 3\] and also for any \(p\ge 3\) we have \[2^p=2^3\cdot 2^{p-3}\equiv 0\cdot 2^{p-3}=0 \pmod 8\] so if we combine, whenever \(p\) is both odd and at least three, then \(2^p\equiv 8 \pmod {24}\). Therefore \(2^p-1\equiv 7 \pmod {24}\) and \(2^p+1\equiv 9 \pmod {24}\).
/JeppeSN Last fiddled with by JeppeSN on 2018-12-29 at 18:41 |
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