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Generalized Fermat Prime Search :
Gaps between GF primes
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Yves Gallot Volunteer developer
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Assuming Bateman-Horn conjecture, I computed the number of pairs of GF primes and the density of the gaps between GF primes.
This is an extension of Hardy-Littlewood k-tuple conjecture for pairs or of the twin prime conjecture (these conjectures are for "GFN-0" primes).
You can download the paper here.
I compared the estimated numbers of pairs and gaps with the lists of GFN-1, GFN-2 and GFN-3 computed by Kellen (the b-range is 2-2G).
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.
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Ravi FernandoProject administrator
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Very interesting as always Yves.
I notice that the predictions look extremely accurate for pairs, but less so for gaps. (E.g. for GFN-1 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"? | |
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Yves GallotVolunteer developer
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I notice that the predictions look extremely accurate for pairs, but less so for gaps. (E.g. for GFN-1 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 Bateman-Horn 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.
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Excellent analysis and fantastic paper Yves! Beautiful work as always!
Still loving how high the W1(2210) value is (~4.2559). This makes me curious if it is possible to calculate the limit for the maximum W1(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 W1(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 :) | |
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Yves GallotVolunteer developer
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Still loving how high the W1(2210) value is (~4.2559). This makes me curious if it is possible to calculate the limit for the maximum W1(a) value, or if there is a theoretical maximum value at all.
Edit: I suppose that W1(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 * p1 * p2 * ... pn, where pi are the primes of the form 4 k + 1, we have w1(pi, a) = 2.
Then W1(a) >= prod pi (1 - 2/pi) / (1 - 4/pi) ~ 1.25357 log pn.
The constant can be computed with Dirichlet series but the most important is that the product is divergent. Then W1(a) is unbounded. | |
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 BurVolunteer tester
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What value is on the x-axis of the pairs?
For gaps I assume it's the size of the gap in regard to the exponent?
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1281979 * 2^485014 + 1 is prime ... no further hits up to: n = 5,700,000 | |
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I think the x-axis is the difference in base b between the two generalized Fermat primes considered.
For example where it says GFN-3 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 GFN-3 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 | |
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Yves GallotVolunteer developer
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At least, that is my interpretation.
You're right!
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BurVolunteer tester
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Ok, thanks. More intricate than I thought...
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1281979 * 2^485014 + 1 is prime ... no further hits up to: n = 5,700,000 | |
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Generalized Fermat Prime Search :
Gaps between GF primes |
