Recursive Primes?

[jinydu]

"The next four, MM13, MM17, MM19 and MM31 have known factors. And that's the extent of our knowledge."

So there goes the hope that all double Mersennes are prime. But MM127 is more than a double Mersenne, its a quintuple Mersenne. So maybe there is hope...

[Terrence Law]

Originally Posted by philmoore

As noted on Chris Caldwell's webpage linked by nfortino, and also on Will Edgington's status page for iterated Mersenne numbers:
http://www.garlic.com/~wedgingt/MMPstats.txt,
Landon Curt Noll has already checked all possible factors of this number less than 1.02*10^51. If you are really serious about looking for factors of MM127, I would suggest a distributed effort, similar to the distributed effort Tony Forbes has organized for MM61. These numbers seem to be tough, and I would guess that the best chance for finding a factor of one of these numbers lies with MM89, MM107, or MM127.

But I found a factor for (MM61+2)/3, which equals to (2^(2^61-1)+1)/3 equals to (2^2305843009213693951+1)/3

The actual factor is 576*(2^61-1)+1 equals to 1328165573307087715777.

[Terrence Law]

Originally Posted by ewmayer

MM31 already has 3 known factors. Tony Forbes' code is extremely efficient, and will work on MM89, MM127 etc., not just on MM61.

Viewed statistically, any of these numbers will only have a chance ~50-60% of having a factor small enough to be found via sieve-based factoring. That means that if a small factor is not found for them, we may never know the status of colossi like MM61, MM89, MM127 without some radical breakthrough in factoring algorithms or the mathematical theory of the Mersennes. Even quantum computing will likely not work on such large numbers, because the sheer number of qubits in each (and AFAIK know QC needs to deal with the entire number, which is one big one disadvantage it has over simple sieving) is so large.

How about trying to verify the New Mersenne Conjecture for MM61?

[Terrence Law]

Originally Posted by jinydu

"The next four, MM13, MM17, MM19 and MM31 have known factors. And that's the extent of our knowledge."

So there goes the hope that all double Mersennes are prime. But MM127 is more than a double Mersenne, its a quintuple Mersenne. So maybe there is hope...

All right. Let's make it into 2^(2^(2^...)-1)-1)-1)-1, a centuple Mersenne, which is C104 or C113 for testing. Assuming that all the lower terms are prime, keep testing higher, this tantalizing sequence.

[Terrence Law]

Originally Posted by jinydu

I know this sounds obvious, but:

M(2) = (2^2)-1 = 3 Prime
M(3) = (2^3)-1 = 7 Prime
M(7) = (2^7)-1 = 127 Prime
M(127) is prime according to Mathematica 5.0

Also M(M(127)) is prime, according to the museum and the compiling library and the C++ computer procedures.

M(M(M(127))) is prime, according to the tables and the enabling cookies of the computer disk.

[Axel Fox]

Sorry to reply to a message so early in this thread, but I think you have a slight misconception one of your messages

MM127 = M(2^127 - 1) = 2(2^127 - 1) - 1

which is not the same as

2^(2^127) - 2

because this is definately not prime. Since 2^(2^127) is a power of 2 and is therefor an even number, so is 2^(2^127) - 2 and this means that 2 is a factor of 2^(2^127) - 2

MM127 is actually (2^(2^127))/2 - 1

[nfortino]

Originally Posted by Axel Fox

Sorry to reply to a message so early in this thread, but I think you have a slight misconception one of your messages

MM127 = M(2^127 - 1) = 2(2^127 - 1) - 1

which is not the same as

2^(2^127) - 2

because this is definately not prime. Since 2^(2^127) is a power of 2 and is therefor an even number, so is 2^(2^127) - 2 and this means that 2 is a factor of 2^(2^127) - 2

MM127 is actually (2^(2^127))/2 - 1

Your statements are true (besides a ^ missing in the first expression), but I was actually making my comment based on the fact that 2^(2^127)-2=2*(MM(127)). Since factors of the form 2kp+1 are odd, any factor of that form dividing 2^(2^127)-2 must divide MM(127). I should point out that my statement that exploiting this fact would allow faster factoring is probably false, as it requires 126 squarings as opposed to 125 squarings and 125 multiplications by 2 for the direct method. I don't remember the logic I used to come to my original statement, but it was probably faulty.

[Axel Fox]

Ah, ok. Now I understand.

Sorry for the typo in my first line and the misunderstanding on my part.