20130617, 15:03  #1046 
Jun 2010
Pennsylvania
3A6_{16} Posts 
no factor for M3321935797 from 2^82 to 2^83 [mfakto 0.12Win mfakto_cl_barrett92_2]
Several objectives having been achieved , I'm done for the time being. Rodrigo 
20130618, 07:45  #1047 
Banned
"Luigi"
Aug 2002
Team Italia
3·1,609 Posts 

20130821, 01:22  #1048 
Oct 2007
Manchester, UK
2·3·227 Posts 
Here's a result, not of the usual kind though.
First I verified all of the factors listed for the OBD numbers, no surprise they all checked out. Then I decided to see if any of the squares of the factors were also factors. And... *drumroll* None of them were. Gotta say I'm a little surprised by that, I kind of expected at least one of them to be. Should I have expected such a thing? Is it a particularly unlikely event? From what I understand, smaller factors are much more likely than larger factors. Also, I know that several of the mersennes have lots of known factors, just never the same one twice. Do we just have too small of a sample size here? If I were to run the same analysis on the factors found by GIMPS, what do people expect I would find? (Other than a huge database filled with factors that would take forever to chew through.) 
20130821, 01:57  #1049 
Romulan Interpreter
Jun 2011
Thailand
2634_{16} Posts 
I don't exactly understand, are you just announcing that all mersenne numbers are square free, or what? This is a long time open problem, no one knows the answer, and no one knows how to prove one way or another. The conjecture says that all numbers of the form 2^{p}1 with a prime p are square free. I would dare to say that all numbers of the given form, for ANY p (prime or not) are square free, except trivial factors (the one from algebraic factorization) but some big guns here might kill me for it.

20130821, 11:31  #1050 
Oct 2007
Manchester, UK
2522_{8} Posts 
I see, I was not aware of this conjecture (I am not aware of many things). Reading around I see that any such divisor must also be a Wieferich prime, and at least 52 bits in size.
Still, no saying one won't turn up I suppose. I assume then that various people have already checked the mersenne number factors of GIMPS? I couldn't find any information on that, so if noone knows otherwise, I may have a go at it myself. I did find a post from ~18 months ago by ATH saying there were 33.5 million factors, which isn't so much to chew through as I thought. 
20130822, 04:48  #1051  
Romulan Interpreter
Jun 2011
Thailand
23064_{8} Posts 
Quote:
Most probably (no heuristic or proof, just "feeling") the conjecture is true, and I personally assume that a much larger conjecture is true, saying that "regardless of p being prime or not, m=2^p1 is square free except the trivial case". I call the "trivial" case the "algebraic" tricks like: a) when p contains 2 and 3, then m contains 3^2 b) when p contains 3 and 7, then m contains 7^2 c) when p contains 5 and 31, then m contains 31^2 d) when p contains 7 and 127, then m contains 127^2 e) when p contains 11 and 23, then m contains 23^2 f) when p contains 11 and 89, then m contains 89^2 g) when p contains 4 and 5, then m contains 5^2 g) when p contains 8 and 17, then m contains 17^2 etc. (remark mersenne and fermat numbers, with their factors) ex: 2^61=3^2*7 2^201= 3*5^2*11*31*41 2^211=7^2*127*337 2^1551=31^2*...(square free).... etc. What remains from m after taking out the offender prime completely, is also square free. I would even dare to offer 50 dollars for a counterexample. But how to prove this theoretically, I have no freaking idea... Last fiddled with by LaurV on 20130822 at 04:56 

20130823, 13:51  #1052 
"William"
May 2003
New Haven
2^{6}×37 Posts 
I suppose your definition of trivial also covers
3^3 divides 2^181 3^4 divides 2^541 3^n divides 2^(2*3^(n1))1 I suppose the Cunningham book has a name for these. 
20130823, 16:39  #1053  
Romulan Interpreter
Jun 2011
Thailand
10011000110100_{2} Posts 
Quote:
This was exemplified before by 3 and 2, or 5 and 4, or 23 and 11, or 17 and 8, but it also goes for ANY numbers (not necesarily prime): 9 and 6, 11 and 10, 13 and 12, 15 and 4, 19 and 18, 21 and 6, 23 and 11, 25 and 20, 27 and 18, 29 and 28, 31 and 5, 33 and 10, etc, they all generate high powers on the same idea (ex: 2^3301 is divisible by 33^2, or 2^541 is divisible by 9^2, or 2^5001 is divisible by 25^2). Last fiddled with by LaurV on 20130823 at 16:54 

20131105, 17:44  #1054 
Feb 2013
2 Posts 
[Tue Nov 05 14:55:00 2013]
UID: Andres Aitsen/i7920, no factor for M3321928171 from 2^83 to 2^84 [mfaktc 0.20 barrett87_mul32_gs] 
20180816, 22:58  #1055 
Jun 2003
The Computer
3×131 Posts 
no factor for M3321928171 from 2^84 to 2^85 [mfaktc 0.21 barrett87_mul32_gs]
I'll continue on to 86 bits and will ping William and James once that's complete. 
20180916, 10:22  #1056 
Jun 2003
The Computer
611_{8} Posts 
no factor for M3321928171 from 2^85 to 2^86 [mfaktc 0.21 barrett87_mul32_gs]
As mentioned in the previous post, I'll notify Will and James. This will move us to level 19.07, the first progress (albeit small) in a while. 
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