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321 Mega Prime! (2015 Edition, part 2)
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Scott Brown Volunteer moderator
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On 23 June 2015 20:49:21 UTC, PrimeGrid’s 321 Prime Search project found the mega prime:
3*2^11895718-1
The prime is 3,580,969 digits long and enters Chris Caldwell's The Largest Known Primes Database ranked 12th overall. This is the largest known 321 mega prime and the largest prime found to date at PrimeGrid!
The discovery was made by Michael Schulz (Michael Schulz) of Germany using an Intel(R) Core(TM) i5-4570 CPU @ 3.20GHz with 16 GB RAM running Darwin 14.3.0. This computer took about 9 hours and 57 minutes to complete the primality test using LLR.
The prime was verified on 27 June 2015 3:56:03 UTC, by user Matthew of New Zealand using an Intel(R) Core(TM) i7-4510U CPU @ 2.00GHz with 8 GB RAM running Windows 7 Professional. This computer took about 33 hours and 40 minutes to complete the primality test using LLR.
For more details, please see the official announcement.
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For more details, please see the official announcement.
This is a "minus one" type prime.
We can write:
3*2^11895718 - 1 = 3*(2^11895717-1) * 2 + 1
to force it onto "plus" form.
In particular, this prime is -1 (mod 4), not +1 (mod 4).
With knowledge on what factors of (generalized) Fermat numbers look like, it is clear that this prime could not be one. So the section on Fermat divisibility in the above announcement must be a copy-paste mistake.
/JeppeSN
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3*2^11895718 - 1 = 3*(2^11895717-1) * 2 + 1
I'm confused.
Doesn't 3*(2^11895717-1) * 2 + 1 = 3*2^11895718 - 5
It seems you're off by 4.
I think you mean: (3*2^11895717 - 1) * 2 + 1
However, this is not a Proth number as k > 2^n | |
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 Michael GoetzVolunteer moderator
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Yes, the PFGW and Fermat divisor part of the announcement was a copy and paste error.
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My lucky number is 75898524288+1 | |
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3*2^11895718 - 1 = 3*(2^11895717-1) * 2 + 1
I'm confused.
Doesn't 3*(2^11895717-1) * 2 + 1 = 3*2^11895718 - 5
It seems you're off by 4.
I think you mean: (3*2^11895717 - 1) * 2 + 1
However, this is not a Proth number as k > 2^n
Thank you very much, you are entirely correct. That was a typo.
The important thing was that the prime here, when written as k*2^n + 1, must have exponent n at most 1 (exactly n=1 if we require k odd), so it cannot divide a (generalized) Fermat GF(m, b) where m>0.
/JeppeSN | |
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321 Mega Prime! (2015 Edition, part 2) |
