41st Known Mersenne Prime Reported!!

Originally Posted by davar55

While we're waiting for the EXCITING confirmation:

Has anyone seen THIS conjecture, which is numerically
suggested by and consistent with a simple calculation
with M1 thru "M40": (Check it out)

LIM(n-->infinity) Mn^(1/n) == 3/2

i.e. Mn approx. == K * (3/2)^n for some constant K
i.e. Mn == O((3/2)^N) == BIG-O-OF (3/2)^n

(Possibly) useful for predicting density, if not location,
of MPs, eh?
And the numerical evidence, while brief, is convincing.

Let me try to find analytically the right value for your conjecture, replacing your guess 3/2. The probability that a large number N is prime is 1 / LN(N). Therefore, the probability that a number comparable to 2^N-1 is prime is therefore naively 1/LN(2^N-1). For large N, neglect 1 and it is 1/LN(2^N) = 1/LN(exp(LN(2).N) = 1/(LN(2).N). Write N as exp(P), and the density of primes in P (the logarithmic scale) becomes 1/(LN(2).exp(P)) times exp(P) = 1/LN(2) which would mean that a new prime appears every time you increase P by LN(2), i.e. if you double M. However, 2^N-1 is guaranteed to be odd, which doubles the probability that it is a prime, and assuming no other correlation or conspiracy, it means that you don't need to double M: it is enough to multiply it by SQRT(2).

So therefore I claim that your limits are true, but the correct number is not 3/2, but rather SQRT(2). Let's call it WhateverIsYourName - Motl's theorem, if no one else knows it. ;-)

[Uncwilly]

You folks may want to look at http://www.utm.edu/research/primes/n...tMersenne.html

Quoting from that site: (emphasis mine)

What about the next Mersenne?
So what can we gather from this? One thing is that given one Mersenne prime exponent's p, the next one will fall, on the average, near 1.47576p. But not too near the average much of the time--sometimes the gaps will be small, sometimes large. So the next (possibly the 41st) Mersenne exponent might be about 31,000,000 yielding a Mersenne with about 9.3 million digits. Or it may not.

[patrik]

Originally Posted by tom11784

IIRC the Perl errors on the manual submissions had started before this report of a potential M41, so I would think that means either it was sent directly to George, or it was submitted via PrimeNet...

I think it must have been sent to Primenet some way, since we have it in the summary:

  Factoring only        :   8312          Factored composite    :  13186
  Lucas-Lehmer testing  :  62446          Lucas-Lehmer composite:  74660
  Double-checking LL    :  14131          Double-checked LL     :  39640
                                          Prime, VERIFIED       :      1
                                          Prime, UNVERIFIED     :      1
  ---------------------- -------          ---------------------- -------
                  TOTAL :  84889                          TOTAL : 127488

Also, George said he didn't get the usual mail from the server.

Originally Posted by Prime95

I've downloaded the logs - the server did not email me automatically for some reason. So far the result looks genuine. It was run using version 23 by a top 1000 contributor. If proven correct it will be another world record for GIMPS!!

[tom11784]

maybe the reason he didn't get an email has to do with the Perl issues on the server lately

but he found out soon enough anyways and we'll all know in about a month - sooooo far away

[jinydu]

Does anyone know where I can find an applet that calculates Li(x) (logarithmic integral, as in the Prime Number Theorem) for VERY large x (as large as the numbers we're testing for primality)?

[jinydu]

Originally Posted by Lumídek

Let me try to find analytically the right value for your conjecture, replacing your guess 3/2. The probability that a large number N is prime is 1 / LN(N). Therefore, the probability that a number comparable to 2^N-1 is prime is therefore naively 1/LN(2^N-1). For large N, neglect 1 and it is 1/LN(2^N) = 1/LN(exp(LN(2).N) = 1/(LN(2).N). Write N as exp(P), and the density of primes in P (the logarithmic scale) becomes 1/(LN(2).exp(P)) times exp(P) = 1/LN(2) which would mean that a new prime appears every time you increase P by LN(2), i.e. if you double M. However, 2^N-1 is guaranteed to be odd, which doubles the probability that it is a prime, and assuming no other correlation or conspiracy, it means that you don't need to double M: it is enough to multiply it by SQRT(2).

So therefore I claim that your limits are true, but the correct number is not 3/2, but rather SQRT(2). Let's call it WhateverIsYourName - Motl's theorem, if no one else knows it. ;-)

We can test that assumption. M38 = (2^6972593)-1 is the largest Mersenne prime that we know for sure; i.e. it is known known for sure whether M13466917 is really M39 by size. Also, we have double-checked all the way up to M(8,000,000)

Now, pi(8,000,000) = 539,777
This means that there are 539,777 Mersenne numbers that could be prime, of which only 38 are actually prime. The density is thus about 7.04 * 10^-5.

By comparison Li(M(8,000,000)) / M(8,000,000) ~= 1.80 * 10^-7.

This means that Mersenne numbers are about 391 (not 2) times more likely to be prime than "ordinary" numbers, at least when n < 2^(8 million). However, it should be noted that there is significant error in that estimate because so few Mersenne primes are known (i.e. 38 is a small number, so there is still plenty of room for statistical uncertainty). For instance, the density of Mersenne primes up to M6972592 is significantly different from the density up to M6972593 (i.e. a single Mersenne prime changes the density heavily while a single "ordinary" prime has little effect on the density). Still, I would be quite confident to say that Mersenne numbers less than M(8,000,000) are hundreds of times more likely to prime than ordinary natural numbers less than M(8,000,000)

I think the reason for this is probably because Mersenne numbers have less possible factors than "ordinary" numbers of similar size.

[wpolly]

Could anyone post the 4 candidate exponents along with their residue?

[jinydu]

Just for comparison, I'll make a table.

Let r = (density of Mersenne primes) / (density of regular primes), where (density of Mersenne primes) = (# of Mersenne primes) / (# of prime exponents).

Let n = Mersenne exponent

n-------Mersenne prime density-------Regular prime density----------r
10--------------1.00------------------------0.168----------------5.95
100------------0.400----------------------1.46*10^-2------------27.3
1000---------8.33*10^-2------------------1.44*10^-3------------57.7
10000--------1.79*10^-2------------------1.44*10^-4------------124
10^5---------2.92*10^-3------------------1.44*10^-5------------202
10^6---------4.20*10^-4------------------1.44*10^-6------------291
8*10^6-------7.04*10^-5------------------1.80*10^-7------------390

This table clearly seems to show that Mersenne numbers have an "advantage" over regular numbers, and that this advantage gets larger as the numbers get larger. This means that Mersenne primes do appear to thin out (as everyone probably knows), but not as quickly as regular primes.

In case anyone wants to know, I assumed that Regular prime density = Li (2^n) = (1/ (n*ln 2)) + (1/ (n*ln 2)^2) + (2/ (n*ln 2)^3). I got that formula from http://mathworld.wolfram.com/PrimeNumberTheorem.html.

Double-Checkers: Work hard, and I can add another row to that table!

NOTE: Sorry about the dashes. Its the only way I could make the table display correctly (at least on my browser).

A statistician, would be looking at the least sum of squares.
You know that line between predicted data, and observed data. :surprised
This feature is built into RMA, as an accessory!

[jinydu]

Originally Posted by TTn

A statistician, would be looking at the least sum of squares.
You know that line between predicted data, and observed data. :surprised
This feature is built into RMA, as an accessory!

I didn't really have any predicted data. Was just looking at the observed data to look at some general trends. However, a quick check shows that r does NOT grow linearly with exponentially growing n.

EDIT: A quadratic fit works very well.

x = Log (base 10) n
y = r

y = 9.0921x^2 - 6.3465x + 2.0573, with an R^2 value of 0.9994 (according to Microsoft Excel).

If you prefer natural logarithms:

x = ln (n)
y = r

y = 1.7147x^2 - 2.7543x + 2.0516, with an R^2 value of 0.9994

That's a good fit considering the small sample size of only 38 Mersenne primes.

Ah... hem I was refering to:

http://www.utm.edu/research/primes/n...tMersenne.html

The exact line between these lines, providing it is still a candidate ofcourse.
Or more exactly, the line between general Mersenne primes.
(Download RMA for details at:)

http://www.15k.org/rma/

Shane F.
TTn