20080613, 11:52  #1 
"Lucan"
Dec 2006
England
2·3·13·83 Posts 
Are Mersenne numbers more likely tobe prime?
I have seen it said often that M=2^k1 (arbitrary k) is a stronger
candidate for primality than other comparably sized numbers. I've never been convinced, and anyway how much more? Gauss says the probaility of an arbitrary positive integer n being prime is 1/ln(n). Wagstaff and results to date say the probability of an arbitrary exponent k yielding a prime M=2^k1 is 2.57/k. (2.57=e^gamma/ln(2)). Now k = log2(M) = ln(M)/ln(2) So 2.57/k = e^gamma/ln(M) gamma=0.5772 so e^gamma=1.78 So if you choose a large integer "at random" the chance of it being prime is only increased by 1.78 if it happens to be a Mersenne number. Comments welcome. (Note that the observation that the number is odd would double the probability of its being prime). David Last fiddled with by davieddy on 20080613 at 12:15 
20080613, 12:40  #2  
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
3×1,423 Posts 
Just thought I'd point out that this isn't exactly right. Since it's 2^k1, k only approximates log2(M).
Quote:
Is this poor English, or British English, or just weird but technically proper English? I don't understand the part after the "and" there. What does this mean? Last fiddled with by MiniGeek on 20080613 at 12:46 

20080613, 12:49  #3  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
11023_{10} Posts 
Quote:
"I've never been convinced and, anyway, how much more?" Alternatively, try splitting it into two sentences: "I've never been convinced. Anyway, how much more?" Paul 

20080613, 12:53  #4 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
3×1,423 Posts 
Is he asking how many more people were never convinced and/or never understood (or rather, as interpreted, stating that many people haven't been convinced/haven't understood)?

20080613, 13:00  #5  
"Lucan"
Dec 2006
England
2·3·13·83 Posts 
Quote:
It was a mild approxiamation. Your logic was fine. (Hence my post) Lazy British Engish. Perhaps "anyway" after the "and " might clarify it. David 

20080613, 13:03  #6 
"Lucan"
Dec 2006
England
2×3×13×83 Posts 

20080613, 13:11  #7  
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
3×1,423 Posts 
Quote:
It's probably the combination of Lazy and British that's confusing me. I can usually tell what people mean with Lazy or British English, but those together is too much. I'm still not really sure what you mean...is this guess correct (from my earlier post): "Is he asking how many more people were never convinced and/or never understood (or rather, stating that many people haven't been convinced/haven't understood)?" 

20080613, 13:22  #8  
(loop (#_fork))
Feb 2006
Cambridge, England
2^{2}·3^{2}·179 Posts 
Quote:
2^p1 for prime p is a fairly strong candidate for primality because of the various Lagrange'sTheorem results about the shape of its factors; for example, for p>50 it's guaranteed not to be divisible by any prime less than 100, whilst 88% of random numbers are divisible by a prime less than 100. 

20080613, 13:23  #9  
"Lucan"
Dec 2006
England
2×3×13×83 Posts 
Quote:
or "I've never been convinced that Mersenne numbers were more likeky to be be prime. If they are, then by what factor?" Which I then proceeded to discuss. I hope to find out how many people agree with me (us?) from response to this thread. Last fiddled with by davieddy on 20080613 at 13:32 

20080613, 13:25  #10 
"William"
May 2003
New Haven
100101000001_{2} Posts 
No. He is asking how much more likely it is that a Mersenne number is prime. The question is being asked in the context of the assumption that "Mersenne numbers are stronger candidates" means they are more likely to be prime.
However, I think Mersenne numbers are stronger candidates because there is cheap powerful primality test. 
20080613, 13:39  #11  
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
3·1,423 Posts 
Quote:
Quote:
After reading the things in this thread, I think that random k is less likely than a random odd number of the same size, but that random p is more likely than a random odd number of the same size, not to mention that it's far easier to test. 

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