20060325, 08:13  #1 
Cranksta Rap Ayatollah
Jul 2003
641 Posts 
search for MMM127 small factors?
Would the allure of MM127's possible primality be diminished if a CatalanMersenne number was found to be composite? Has anyone searched for small factors of the CM numbers greater than MM127?
mild tangent: I'm not aware if attempting to find small factors of these monstrous numbers is even feasible, so, if I'm talking out of my arse, leave your inflammatory comments at the door. Frankly, I'm too busy with my other math classes to go learning computational mathematics from scratch and buying $70 books to do so, nor is the question compelling enough for me to go through that much toil. If you want to make toast but you're out of bread, you don't go out and plant wheat. 
20060325, 12:44  #2 
Aug 2002
Buenos Aires, Argentina
2^{2}×337 Posts 
Travis,
It's not a bad idea. Since MM127 is almost surely composite, MMM127 factors should not have the form 2*k*MM127+1. My program to compute factors of googolplexplex could be adapted to perform trial division by small factors (say less than 10^12). If MMM127 has such a small factor, we are sure that MM127 is not prime. 
20060325, 13:09  #3 
Aug 2002
Buenos Aires, Argentina
2504_{8} Posts 
What I wrote in my previous post is not correct. If MM127 is equal to the product of primes a*b*c*..*z, then after dividing MMM127 by 2^a1, 2^b1, etc. (which are very large numbers whose factors have the form 2*k*a+1, 2*k*b+1, which are out of reach) we get the primitive factor.
The factors of this primitive factor have the form 2*k*MM127+1. So if we cannot find a factor of MM127 we also cannot expect to find a factor of MMM127. Last fiddled with by alpertron on 20060325 at 13:10 
20060325, 15:50  #4  
Nov 2003
2^{2}×5×373 Posts 
Quote:


20060325, 17:05  #5 
Cranksta Rap Ayatollah
Jul 2003
1201_{8} Posts 
Ahh, okay.
Thanks to both :) 
20060502, 02:23  #6 
Jun 2003
Pa.,U.S.A.
2^{2}·7^{2} Posts 
one assurence numerically speaking
If one could show x(x1)! evenly and with multiples of 5,
you will be quaranteed of finding factors . So also for m^x 127. 
20060502, 16:28  #7  
∂^{2}ω=0
Sep 2002
República de California
3·5^{3}·31 Posts 
Quote:


20060611, 15:38  #8 
Bemusing Prompter
"Danny"
Dec 2002
California
2376_{10} Posts 
I think that testing one trial factor of MMM127 (2k * MM127 +1) takes roughly the same amount of computation as LLtesting MM127.

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