Doubts, scientific breakthroughs and conspiracy theories about Wagstaff Conjecture (from Lucky13) - Page 7

Chara34122 · 2018-12-17, 17:28

About Wagstaff 3 mod 4 prevalence. If p= 1 mod 4 prime is Sophie Germain prime, than 2p+1 divides (2^p+1)/3 or not ? 59 divides (2^29+1)/3 for example. Is this correct as the case with 3 mod 4 SG prime for Mersenne number?

And of course congrats with the find!

kriesel · 2018-12-17, 17:31

Quote:

Originally Posted by GP2

See the tables in my previous two posts.

Thank you for the very interesting series of posts. Sincerely.

GP2 · 2018-12-17, 21:13

Sorry if I'm hijacking this thread and chasing mirages, but... rather than looking at all fully-factored Mersenne numbers, let's look specifically at Mersenne semiprimes.

They seem to behave the opposite way from Mersenne primes.

Below is the list of the 325 currently-known fully-factored or probably-fully-factored Mersenne numbers. Alongside each exponent is a count of factors.

By convention, we always omit the cofactor from our factor lists even when it is a prime or probable prime. So the exponents listed below with a "1" next to them represent semi-primes, which strictly speaking have two factors.

Code:

If we look at only the exponents p of the 83 semiprimes we get the following:

Code:

             all   p>1200  p>10k  p>100k
     p=1      34     15     10      7
     p=3      49     29     20     14

So for exponents of Mersenne semiprimes there seems to be a marked prevalence of p=3 vs. p=1, whereas for exponents of Mersenne primes it is the opposite.

At higher values of p, the prevalence seems to be 2-to-1.

However, perhaps what is being selected for is not the size of p, but the asymmetricity of the two factors that make up the semiprime. In the Cunningham range below 1200, everything is fully-factored, but for higher exponents, we are only capable of finding very asymmetric semiprimes with one very small factor and an enormous cofactor.

GP2 · 2018-12-17, 21:31

Quote:

Originally Posted by GP2

Sorry if I'm hijacking this thread and chasing mirages, but... rather than looking at all fully-factored Mersenne numbers, let's look specifically at Mersenne semiprimes.

Let's do the same thing for Wagstaff semiprimes.

Below is the list of the 252 currently known fully-factored or probably-fully-factored Wagstaff numbers.The exponents with a "1" next to them represent semiprimes (one factor, plus the cofactor).

Code:

If we look at only the exponents of the 72 semiprimes we get the following:

Code:

             all   p>1100  p>10k  p>100k
     p=1      40     23     13      5
     p=3      32     12      6      2

So for exponents of Wagstaff semiprimes there seems to be a marked preference of p=1 vs. p=3, whereas for exponents of Wagstaff primes it is the opposite.

At higher values of p, the prevalence seems to be 2-to-1.

Again, what is being selected at higher p might not be the size of p itself, but rather the asymmetricity of the two factors that make up the semiprime.

For Wagstaff numbers, the Cunningham range of everything being fully-factored ends a little sooner (we use 1100 as the cutoff) since they have been a little less thoroughly factored than the Mersenne numbers.

GP2 · 2018-12-18, 00:27

Here is the summary.

If we exclude the tiny exponents (p less than 1000), I would conjecture that:

Mersenne primes are twice as likely to have p=1 rather than p=3 (mod 4)
Mersenne semiprimes are twice as likely to have p=3 rather than p=1 (mod 4)
Wagstaff primes are twice as likely to have p=3 rather than p=1 (mod 4)
Wagstaff semiprimes are twice as likely to have p=1 rather than p=3 (mod 4)

Within p=1 (mod 4), Mersenne primes are twice as likely to have p=1 (mod 8) rather than p=5 (mod 8)
Within p=3 (mod 4), Mersenne semiprimes are twice as likely to have p=7 (mod 8) rather than p=3 (mod 8)
Within p=3 (mod 4), Wagstaff primes are twice as likely to have p=7 (mod 8) rather than p=3 (mod 8)
Within p=1 (mod 4), Wagstaff semiprimes might be twice as likely to have p=1 (mod 8) rather than p=5 (mod 8)

The search for Wagstaff primes and semiprimes has not been as extensive as the search for Mersenne primes and semiprimes.

Code:

Mersenne primes

             filter by
           p greater than:
           10^3 10^4 10^5
mod 4
     p=1    24   19   14
     p=3    12    9    8
mod 8
     p=1    16   13   10
     p=3     6    5    4
     p=5     8    6    4
     p=7     6    4    4

Code:

Mersenne semiprimes

             filter by
           p greater than:
           10^3 10^4 10^5
mod 4
     p=1    16   10    7
     p=3    30   20   14
mod 8
     p=1     8    5    3
     p=3    12    7    5
     p=5     8    5    4
     p=7    18   13    9

Code:

Wagstaff primes

             filter by
           p greater than:
           10^3 10^4 10^5
mod 4
     p=1     8    6    3
     p=3    15   13    8
mod 8
     p=1     6    5    3
     p=3     5    4    2
     p=5     2    1    0
     p=7    10    9    6

Code:

Wagstaff semiprimes

             filter by
           p greater than:
           10^3 10^4 10^5
mod 4
     p=1    24   13    5
     p=3    13    6    2
mod 8
     p=1    14    9    3
     p=3     6    3    2
     p=5    10    4    2
     p=7     7    3    0

GP2 · 2018-12-18, 14:36

Quote:

Originally Posted by GP2

Here is the summary.

The table above omitted M51, which has already been revealed to be 1 (mod 4) and 5 (mod 8).

So it should be updated to:

Code:

Mersenne primes

             filter by
           p greater than:
           10^3 10^4 10^5
mod 4
     p=1    25   20   15
     p=3    12    9    8
mod 8
     p=1    16   13   10
     p=3     6    5    4
     p=5     9    7    5
     p=7     6    4    4

Prime95 · 2018-12-18, 20:08

Quote:

Originally Posted by GP2

I would conjecture that:

Mersenne primes are twice as likely to have p=1 rather than p=3 (mod 4)

A bold conjecture that contradicts Wagstaff's predictions. You'll need great data to back up that claim. Such as data showing that known TF factors are eliminating 3 mod 4 candidates more often than predicted.

GP2 · 2018-12-18, 21:15

Quote:

Originally Posted by Prime95

A bold conjecture that contradicts Wagstaff's predictions. You'll need great data to back up that claim. Such as data showing that known TF factors are eliminating 3 mod 4 candidates more often than predicted.

I claim only that the existing data is consistent with the conjecture. The statistics are admittedly a bit low.

The data is in the four tables in my last post: observed frequencies of p (mod 4) and p (mod 8) for known Mersenne primes and semiprimes and known Wagstaff primes and semiprimes.

I offer no hypotheses or explanations.

PS,
I already posted the data for average number of size-65-bits-or-less factors for Mersenne numbers with p=1 (mod 4) vs. p=3 (mod 4).

(The threshold of 65 bits or less was chosen because thanks to TJAOI we can be pretty sure that we know all factors of this size for all Mersenne numbers with prime exponents up to 1 billion.)

Mersenne numbers in class p=1 are slightly less likely to have factors than those in class p=3, but only slightly. This isn't enough to explain the observed highly asymmetric properties of the primes and semiprimes.

ewmayer · 2018-12-18, 22:11

Quote:

Originally Posted by Prime95

A bold conjecture that contradicts Wagstaff's predictions. You'll need great data to back up that claim. Such as data showing that known TF factors are eliminating 3 mod 4 candidates more often than predicted.

I recall make a similar comment to Richard Crandall and Carl Pomerance fairly early in the days of GIMPS, after we'd found 5 straight M-primes (p = 1398269, 2976221, 3021377, 6972593, 13466917) with p == 1 (mod 4), and the excess of 1s was really going against Wagstaff's statistical prediction. That streak was followed by three straight 3s, but since then, 8 of 9 newly discovered M-primes have had p == 1 (mod 4). So of GIMPS' 17 primes, only 4 have had p == 3 (mod 4), a really striking disparity.

kriesel · 2018-12-19, 01:11

Quote:

Originally Posted by ewmayer

I recall make a similar comment to Richard Crandall and Carl Pomerance fairly early in the days of GIMPS, after we'd found 5 straight M-primes (p = 1398269, 2976221, 3021377, 6972593, 13466917) with p == 1 (mod 4), and the excess of 1s was really going against Wagstaff's statistical prediction. That streak was followed by three straight 3s, but since then, 8 of 9 newly discovered M-primes have had p == 1 (mod 4). So of GIMPS' 17 primes, only 4 have had p == 3 (mod 4), a really striking disparity.

I hope it's either statistical fluctuation, or an actual pattern with mathematical justification TBD, not a consequence of similar effect undiscovered bugs lurking in the primality test codes. Some of the codes have ancestry in common. If there was an issue in common, present in a bit of "DNA in common", despite the best efforts of the authors and extreme care in using differing programs on differing hardware in multiple confirmations by multiple individuals, how could we know?

Madpoo · 2018-12-19, 02:55

Quote:

Originally Posted by Prime95

A bold conjecture that contradicts Wagstaff's predictions. You'll need great data to back up that claim. Such as data showing that known TF factors are eliminating 3 mod 4 candidates more often than predicted.

I ran a quick count of the distinct exponents with any known factors and whether that exponent is 1 or 3 mod 4:

Mod Count
1 14,559,812 (49.301%)
3 14,972,752 (50.699%)

I wouldn't say it's a runaway for the 3 mod 4 exponents being factored. In fact it seems like a pretty basic variation around 50/50 based on an incomplete set.

If I look at the same breakdown of all exponents (below 1 billion) then it looks like (I excluded known primes, including the latest one) - spoiler alert, it's what you would expect, and probably would have come out even closer if I hadn't excluded the known primes which we know skew more (currently) to the "1" side:
Mod Count
1 25,423,460 (49.999%)
3 25,424,023 (50.001%)

EDIT: I also ran it by looking only at factors of 65-bits or less, in case that matters - it does favor the "3" a little more, but not terribly. It is funny that beyond 65-bits, the 1 mod 4's are ahead (by 35,474 exponents out of > 2.14 million)
Mod Count
1 13,591,041 (49.184%)
3 14,042,237 (50.816%)

2018-12-18, 00:27	#71
GP2 Sep 2003 5031₈ Posts	Here is the summary. If we exclude the tiny exponents (p less than 1000), I would conjecture that: Mersenne primes are twice as likely to have p=1 rather than p=3 (mod 4) Mersenne semiprimes are twice as likely to have p=3 rather than p=1 (mod 4) Wagstaff primes are twice as likely to have p=3 rather than p=1 (mod 4) Wagstaff semiprimes are twice as likely to have p=1 rather than p=3 (mod 4) Within p=1 (mod 4), Mersenne primes are twice as likely to have p=1 (mod 8) rather than p=5 (mod 8) Within p=3 (mod 4), Mersenne semiprimes are twice as likely to have p=7 (mod 8) rather than p=3 (mod 8) Within p=3 (mod 4), Wagstaff primes are twice as likely to have p=7 (mod 8) rather than p=3 (mod 8) Within p=1 (mod 4), Wagstaff semiprimes might be twice as likely to have p=1 (mod 8) rather than p=5 (mod 8) The search for Wagstaff primes and semiprimes has not been as extensive as the search for Mersenne primes and semiprimes. Code: Mersenne primes filter by p greater than: 10^3 10^4 10^5 mod 4 p=1 24 19 14 p=3 12 9 8 mod 8 p=1 16 13 10 p=3 6 5 4 p=5 8 6 4 p=7 6 4 4 Code: Mersenne semiprimes filter by p greater than: 10^3 10^4 10^5 mod 4 p=1 16 10 7 p=3 30 20 14 mod 8 p=1 8 5 3 p=3 12 7 5 p=5 8 5 4 p=7 18 13 9 Code: Wagstaff primes filter by p greater than: 10^3 10^4 10^5 mod 4 p=1 8 6 3 p=3 15 13 8 mod 8 p=1 6 5 3 p=3 5 4 2 p=5 2 1 0 p=7 10 9 6 Code: Wagstaff semiprimes filter by p greater than: 10^3 10^4 10^5 mod 4 p=1 24 13 5 p=3 13 6 2 mod 8 p=1 14 9 3 p=3 6 3 2 p=5 10 4 2 p=7 7 3 0 Last fiddled with by GP2 on 2018-12-18 at 00:30

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2018-12-17, 17:28	#67
Chara34122 Nov 2018 Russia 7₁₀ Posts	About Wagstaff 3 mod 4 prevalence. If p= 1 mod 4 prime is Sophie Germain prime, than 2p+1 divides (2^p+1)/3 or not ? 59 divides (2^29+1)/3 for example. Is this correct as the case with 3 mod 4 SG prime for Mersenne number? And of course congrats with the find!