20031214, 08:51  #1 
Dec 2003
Hopefully Near M48
6DE_{16} Posts 
Recursive Primes?
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 
20031214, 10:24  #2  
"Patrik Johansson"
Aug 2002
Uppsala, Sweden
5^{2}×17 Posts 
Re: Recursive Primes?
Quote:


20031214, 12:05  #3 
Nov 2003
3×5×11 Posts 
Actually, you have discovered the Catalan sequence. It is discussed at the bottom of http://www.utm.edu/research/primes/mersenne/ Obviously, it is impossible to check 2^{M127}1 for primality.

20031214, 13:35  #4 
Nov 2002
2×37 Posts 
but we could start a sub project to try to factor MM127.
I'm currently writing a program to fator this huge number i will then post in the forum when i have completed my work!! 
20031214, 14:36  #5 
Nov 2003
A5_{16} Posts 
You've probably already thought of this, but by using the successive squaring method, it is much faster to test if 2^{2[sup]127}[/sup] == 2 mod p instead of testing if 2^{M127} == 1 mod p. Because you are only testing numbers of the form 2k*M127+1, which are odd, any factor you find of 2^{2[sup]127}[/sup]2 will be a factor of MM127.

20031215, 10:12  #6 
Dec 2003
Hopefully Near M48
2×3×293 Posts 
(2^M127)1 has over 5.12*10^37 digits!
Only two months ago, I used to think that 10^100 was large beyond comprehension because it overloaded my calculator. 
20031215, 11:40  #7 
Dec 2003
Hopefully Near M48
6DE_{16} Posts 
The new biggest number I can handle is a little bigger than 1.9202*10^646,456,887.
The MaxNumber appears to be exactly 2^([2^31][352+1.6017072196434280233747626345680681683204815701770939995185472071170806884765625*10^16]) So MM31 is "just" out of reach (actually MM31 is over 10^106 times larger). "but we could start a sub project to try to factor MM127. I'm currently writing a program to fator this huge number" How could we do that? 
20031215, 13:24  #8 
Nov 2002
112_{8} Posts 
by usind large integer libraries like GMP, miracl or giantint!!! they can hand numbers up to 10 million digits
BUT we dont need to calculate MM127 as a number because we only want to fin a factor( A LLtest is with the todays computers out of reach) For factoring we use the modular exponation( so the number we are working with is 2*digits of factor maximum!!! 
20031215, 18:06  #9 
"Phil"
Sep 2002
Tracktown, U.S.A.
2137_{8} Posts 
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. 
20031215, 18:57  #10 
∂^{2}ω=0
Sep 2002
República de California
11695_{10} Posts 
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 ~5060% of having a factor small enough to be found via sievebased 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. 
20031215, 19:19  #11  
"Phil"
Sep 2002
Tracktown, U.S.A.
3·373 Posts 
ewmayer said:
Quote:


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