Since Genefer is "only" a probable prime test, that number still needs to be checked for primality using PFGW. However, the odds of Genefer falsely reporting it as prime are astronomically small, so this is essentially a necessary formality.

This is really big news. Assuming it checks out, it's the largest prime number ever found at PrimeGrid and the 11th largest prime overall.

Details to follow...
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My lucky number is 75898⁵²⁴²⁸⁸+1

Mighty exciting!
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Twitter: IainBethune
Proud member of team "Aggie The Pew". Go Aggie!
3073428256125*2^1290000-1 is Prime!

Mighty exciting!

Hopefully the challenge will churn up another :)
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Hopefully the challenge will churn up another :)

Hopefully the challenge will churn up another :)

Or two. :)

Two PRPs and the challenge did not start yet!

I don't know which app is faster now, specially with such huge primes, but if LLR is still faster, you could use it instead of PFGW.

Larger.

Looks like there's a new PRP at n=19. It's the 3rd on this range since the project has been ported to BOINC.

Well congratulations to the finder on another Mega (and record) GFN. What is really amazing is that according to Yves Gallot's method for calculating the expected number of GFNs in a range, we would only have expected 1 GFN for N=524288, and instead we have found 4 - which has about a 1.5% probability.

So either we have just been exceedingly lucky with N=524288, or there are in fact more GFNs at high n than predicted, which would be exciting indeed for the GFN-WR project!
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Twitter: IainBethune
Proud member of team "Aggie The Pew". Go Aggie!
3073428256125*2^1290000-1 is Prime!

What is really amazing is that according to Yves Gallot's method for calculating the expected number of GFNs in a range, we would only have expected 1 GFN for N=524288, and instead we have found 4 - which has about a 1.5% probability.

Hi all, first of all congratulations on your search.
GPU computing is a new step for the prime number search, and you are the avant-garde!

A lucky man wrote that “The theory behind the math for those predictions is suspect” but my guess is B = 3,380,000 :-)
I think that any B in [2012 ; 10^7] is a good choice.
Note that C_n ~ n log 2 and Li(B) ~ B / log B. Then B = 2^n is the “generic” estimate.
Here, B is smaller because C_22 = 17.3 and 22 log 2 = 15.9.

“It's time for a new formula.”…

On one hand, the results: for B in [2 ; 500000]
n = 16, primes: 9, expected: 7,
n = 17, primes: 4, expected: 3.5,
n = 18, primes: 4, expected: 2,
n = 19, primes: 4, expected: 1.
Another interesting result is with Mersenne primes:
p in [10^7 ; 5.10^7], primes: 9, expected: 4.
Then the number of huge primes is larger than expected!

On the other hand, the expected number of primes is based on the Bateman-Horn conjecture (see wikipedia:Bateman-Horn_conjecture).
This conjecture is not based on “what we've seen in the past” but is a generalization of some theorems.
If the polynomial is x, the conjecture is the prime number theorem and if it is ax+b, the conjecture is Dirichlet's theorem.
It is practical. C is linked to the sieve process: it is the ratio of the numbers that pass through the sieve.
The conjecture is just the mathematical form of: the probability of a number n = f(x), which has no small factor, being prime depends on its size but not on f.

Then, what? We are lucky, may be. Or an unknown property was found, that would be very interesting!!!

The results of sieving are a first indicator. What are they for n =18, 19 and 22?

The results of sieving are a first indicator. What are they for n =18, 19 and 22?

As of 21 August 2012 at 0030 UTC:

The basic property for the estimate of the number of GF primes is:
let p = k*2^(n + 1) + 1 be prime. Then it divides 2^n/p GF numbers of the form b^2^n + 1.

If we sieve R odd numbers, for all p < p_max, the expected candidates that left are
prod_{p < p_max} (1 - 2^n/p) * R.

By Mertens' 3rd theorem:
prod_{p < p_max} (1 - 2^n/p) = 2 * C_n / (e^gamma * log(p_max)),
where gamma is Euler's constant.

If we sieve [2, B_max],
Candidates / (B_max/2) = 2 * C_n / (e^gamma * log((p_max))

and finally we have

Candidates = e^-gamma * C_n * B_max / log((p_max).

We obtain for B_max = 500000:
n C_n p_max ratio Expected candidates
18 13.0 2.510E+18 0.345 86140
19 13.1 1.855E+19 0.332 82890
20 14.5 1.985E+19 0.366 91609
21 16.1 1.935E+19 0.407 101775
22 17.4 2.120E+19 0.439 109768
and B_max = 1E8:
n C_n p_max ratio B_max Expected candidates
18 13.0 2.510E+18 0.345 17228044
19 13.1 1.855E+19 0.332 16577985
20 14.5 1.985E+19 0.366 18321722
21 16.1 1.935E+19 0.407 20355000
22 17.4 2.120E+19 0.439 21953527

The statistics of sieving is close to estimate :-)