20040623, 11:15  #12 
May 2003
1100000_{2} Posts 
Ok, you guys got me thinking a bit hard, my probability and combinatorics skills seem to be a bit rusty. Sorry for the confusion.

20040623, 11:25  #13 
May 2003
2^{5}·3 Posts 
BTW
BTW :
If you guys are right (and after some thinking, I think you are) then the probability of finding a factor between X and Y bits is (approximately) (YX)/(Y1) Maybe this should be put on the site under the 1/X thingy, to avoid this discussion to repeat itself in the future. 
20040623, 13:00  #14  
"Mark"
Feb 2003
Sydney
23D_{16} Posts 
Quote:


20040624, 00:58  #15  
Dec 2003
Hopefully Near M48
6DE_{16} Posts 
Quote:
If it was THE correct formula, there would be a 100% chance of finding a factor between 2 (2^1) and 4 (2^2), which obviously isn't true. 

20040624, 01:30  #16 
"Juan Tutors"
Mar 2004
230_{16} Posts 
So the trial factoring part is essentially settled with this approximation:
ASSUMING that, through trial factoring, the probability of finding a factor between 2^x and 2^(x+1) is 1/x for any Mersenne number, so that the probability of NOT finding a factor is (x1)/x, the probability of findng a factor between 2^62 and 2^68 is 1(61/62)*(62/63)*...*(66/67) = 161/67 = 6/67 = 8.96%. I have no reason to believe whether it is actually higher or lower than this. I won't even ask the data people to do some sort of check on that. However, I would guess that, for this range, the number of possible factors of M_p declines linearly as growth in the prime p in M_p, since possible factors must be of the form 2kp+1. This is good enough though, and I'm too busy to try to find a more rigorous formula. Plus, I would have trouble arguing that any simpleenough formula would be accurate for such a small range. Now about the second part of the question, I don't know if anyone should bother. In about 4 weeks I will start a new P1 factoring on M33844211. I'll set it my RAM allocation to 200 MB and see what happens. 
20040624, 01:56  #17 
May 2003
2^{5}·3 Posts 
Well, I'm not really interested that much in a correct formula either, nor do the others in this discussion, I think. I just like to discuss about mathematical stuff and maybe learn something out of it here and there.
Good luck on the P1 factoring 
20040926, 02:37  #18 
"Juan Tutors"
Mar 2004
2^{4}×5×7 Posts 
From the statement on www.mersenne.org that the probability of finding a factor between 2^x and 2^(x+1) is 1/x, we got the formula that the probability of finding a factor between 2^x and 2^y is 1  (x1)/(y1). I was just wondering whether anyone has created a script to check whether that estimate has held up so far. My reason is pretty lame...I'm a little to the left on the bell curve, b/c it took me 19 TFs to find a factor. I also haven't found any through 12 P1 trials

20040926, 03:32  #19 
Dec 2003
Hopefully Near M48
2·3·293 Posts 
I've noticed that, unfortunately, the Primenet account report doesn't keep track of the number of TF's attempted, only the number of factors found and the number of CPU years. Why is that?

20040926, 08:06  #20 
Aug 2002
Termonfeckin, IE
101011010000_{2} Posts 
Yes, do a search on the forum and there are some posts from over a year ago where the numbers were verified and came pretty close to acurrate. But remember these are probabilities and you could be lucky or unlucky. If you are factoring numbers from 2^62>2^67 then the probability that you won't find a factor in 19 tries is about 22%
jinydu: That was a design decision taken several years ago. And yes we all think it was unfortunate. But then hindsight is 20/20. 
20040926, 09:47  #21 
Dec 2003
18_{16} Posts 
Of the 18 exponents (all in the 25M26M range) I found in the recent weeks 6 factors, one in 62, three in 63 and two in 65 bit range.
Best, Matthias C. Noch 
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