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Generalized Fermat Prime Search :
Gaps between GF primes
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Yves GallotVolunteer developer Project scientist Send message
Joined: 19 Aug 12 Posts: 820 ID: 164101 Credit: 305,989,513 RAC: 2,326

Assuming BatemanHorn conjecture, I computed the number of pairs of GF primes and the density of the gaps between GF primes.
This is an extension of HardyLittlewood ktuple conjecture for pairs or of the twin prime conjecture (these conjectures are for "GFN0" primes).
You can download the paper here.
I compared the estimated numbers of pairs and gaps with the lists of GFN1, GFN2 and GFN3 computed by Kellen (the brange is 22G).
It will be extended to larger n but the formula becomes more complex when n gets larger and I must first generate the solutions with a computer algebra system.
For n = 15, the actual number of primes is too small to check the conjecture but the fluctuation could be visible for n ~ 10.
 

Ravi FernandoProject administrator Volunteer tester Project scientist Send message
Joined: 21 Mar 19 Posts: 211 ID: 1108183 Credit: 13,846,995 RAC: 7,910

Very interesting as always Yves.
I notice that the predictions look extremely accurate for pairs, but less so for gaps. (E.g. for GFN1 gaps of length 4, the error appears to be about 150000, which would be about 70 standard deviations.) Could there be some error in the formula? Maybe some slight dependence between the events "n^2 + 1 and (n+4)^2 + 1 are both prime" and "(n+2)^2 + 1 is composite"?  

Yves GallotVolunteer developer Project scientist Send message
Joined: 19 Aug 12 Posts: 820 ID: 164101 Credit: 305,989,513 RAC: 2,326

I notice that the predictions look extremely accurate for pairs, but less so for gaps. (E.g. for GFN1 gaps of length 4, the error appears to be about 150000, which would be about 70 standard deviations.) Could there be some error in the formula? Maybe some slight dependence between the events "n^2 + 1 and (n+4)^2 + 1 are both prime" and "(n+2)^2 + 1 is composite"?
I don't like the method for converting pairs into gaps but I didn't find better.
I would have thought that the dependency between pairs would be stronger.
It can be corrected if we compute the number of triplets with BatemanHorn conjecture, etc., but it is increasingly difficult.
What was more surprising to me were the fact that the number of solutions to (x^{2^n} + 1)((x + a)^{2^n} + 1) (mod p) [a fixed] is 2^{n+1} except for a finite set of p's.
There may be a simple reason for it but I don't see it.
 


Excellent analysis and fantastic paper Yves! Beautiful work as always!
Still loving how high the W_{1}(2210) value is (~4.2559). This makes me curious if it is possible to calculate the limit for the maximum W_{1}(a) value, or if there is a theoretical maximum value at all. I suppose that 567970 could possibly have a higher value, but my ability to calculate this is zero :)
Looking forward to seeing the rest!
Edit: I suppose that W_{1}(64090) is also a candidate for having a higher value. With 5, 13 and 17 in 2210 I just had Fermat primes on the brain, but 29 works too :)  

Yves GallotVolunteer developer Project scientist Send message
Joined: 19 Aug 12 Posts: 820 ID: 164101 Credit: 305,989,513 RAC: 2,326

Still loving how high the W_{1}(2210) value is (~4.2559). This makes me curious if it is possible to calculate the limit for the maximum W_{1}(a) value, or if there is a theoretical maximum value at all.
Edit: I suppose that W_{1}(64090) is also a candidate for having a higher value. With 5, 13 and 17 in 2210 I just had Fermat primes on the brain, but 29 works too :)
Yes 64090 = 2 * 5 * 13 * 17 * 29 and let's continue:
If a = 2 * p_{1} * p_{2} * ... p_{n}, where p_{i} are the primes of the form 4 k + 1, we have w_{1}(p_{i}, a) = 2.
Then W_{1}(a) >= prod _{pi} (1  2/p_{i}) / (1  4/p_{i}) ~ 1.25357 log p_{n}.
The constant can be computed with Dirichlet series but the most important is that the product is divergent. Then W_{1}(a) is unbounded.  

BurVolunteer tester
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Joined: 25 Feb 20 Posts: 515 ID: 1241833 Credit: 414,481,880 RAC: 163

What value is on the xaxis of the pairs?
For gaps I assume it's the size of the gap in regard to the exponent?
____________
1281979 * 2^485014 + 1 is prime ... no further hits up to: n = 5,700,000  


I think the xaxis is the difference in base b between the two generalized Fermat primes considered.
For example where it says GFN3 pairs, and where it says 10 on the horizontal axis, we count pairs of numbers
b^(2^3)+1 and (b+10)^(2^3)+1
both of which are prime.
Where it says GFN3 gaps, and 10 on the abscissa axis, we count same kind of pairs but with the additional restriction that the primes must be consecutive, i.e. none of the intermediate candidates (b+2)^(2^3)+1, (b+4)^(2^3)+1, (b+6)^(2^3)+1, (b+8)^(2^3)+1 are allowed to be prime.
At least, that is my interpretation.
/JeppeSN  

Yves GallotVolunteer developer Project scientist Send message
Joined: 19 Aug 12 Posts: 820 ID: 164101 Credit: 305,989,513 RAC: 2,326

At least, that is my interpretation.
You're right!
 

BurVolunteer tester
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Joined: 25 Feb 20 Posts: 515 ID: 1241833 Credit: 414,481,880 RAC: 163

Ok, thanks. More intricate than I thought...
____________
1281979 * 2^485014 + 1 is prime ... no further hits up to: n = 5,700,000  

Message boards :
Generalized Fermat Prime Search :
Gaps between GF primes 