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 2013-06-17, 15:03 #1046 Rodrigo     Jun 2010 Pennsylvania 32·103 Posts no factor for M3321935797 from 2^82 to 2^83 [mfakto 0.12-Win mfakto_cl_barrett92_2] Several objectives having been achieved , I'm done for the time being. Rodrigo
2013-06-18, 07:45   #1047
ET_
Banned

"Luigi"
Aug 2002
Team Italia

22·1,193 Posts

Quote:
 Originally Posted by Rodrigo no factor for M3321935797 from 2^82 to 2^83 [mfakto 0.12-Win mfakto_cl_barrett92_2] Several objectives having been achieved , I'm done for the time being. Rodrigo

Thank you Rodrigo!

Luigi

 2013-08-21, 01:22 #1048 lavalamp     Oct 2007 London, UK 22×7×47 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.)
 2013-08-21, 01:57 #1049 LaurV Romulan Interpreter     Jun 2011 Thailand 23·1,117 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 2p-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.
 2013-08-21, 11:31 #1050 lavalamp     Oct 2007 London, UK 22×7×47 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 no-one 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.
2013-08-22, 04:48   #1051
LaurV
Romulan Interpreter

Jun 2011
Thailand

100010111010002 Posts

Quote:
 Originally Posted by lavalamp 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 no-one 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.
Well, many people (myself included) put a lot of work into this, indeed. I would dare to say more, as we already took all expos below 1 billion (under 4G29=2^32 on James' site) to about 2^65 (2^60 on James' site), and we did not find any new Wf prime (remark that 1 Billion is about 2^30), and considering that all mersene factors are 2kp+1, I would dare to say that not such a number below (approx) 60 bits exists.

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^p-1 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:
2^6-1=3^2*7
2^20-1= 3*5^2*11*31*41
2^21-1=7^2*127*337
2^155-1=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 counter-example. But how to prove this theoretically, I have no freaking idea...

Last fiddled with by LaurV on 2013-08-22 at 04:56

 2013-08-23, 13:51 #1052 wblipp     "William" May 2003 New Haven 1001001001002 Posts I suppose your definition of trivial also covers 3^3 divides 2^18-1 3^4 divides 2^54-1 3^n divides 2^(2*3^(n-1))-1 I suppose the Cunningham book has a name for these.
2013-08-23, 16:39   #1053
LaurV
Romulan Interpreter

Jun 2011
Thailand

100010111010002 Posts

Quote:
 Originally Posted by wblipp I suppose your definition of trivial also covers 3^3 divides 2^18-1 3^4 divides 2^54-1 3^n divides 2^(2*3^(n-1))-1 I suppose the Cunningham book has a name for these.
Sure! Given an odd x, denote with y the smaller number such as x divides 2^y-1, then if the exponent n contains x and y, then m=2^n-1 is divisible by x^2. For a prime x, y will always be a factor of x-1. For a composite x, computing y is a bit more complicate, but it can always be done fast.

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: 2^330-1 is divisible by 33^2, or 2^54-1 is divisible by 9^2, or 2^500-1 is divisible by 25^2).

Last fiddled with by LaurV on 2013-08-23 at 16:54

 2013-11-05, 17:44 #1054 Le Comte   Feb 2013 28 Posts [Tue Nov 05 14:55:00 2013] UID: Andres Aitsen/i7-920, no factor for M3321928171 from 2^83 to 2^84 [mfaktc 0.20 barrett87_mul32_gs]
 2018-08-16, 22:58 #1055 clowns789     Jun 2003 The Computer 1100000002 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.
 2018-09-16, 10:22 #1056 clowns789     Jun 2003 The Computer 27·3 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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