20110517, 14:14  #1 
5117_{8} Posts 
Largest nonmersenne prime
How many digits are there in the largest nonmersenne prime?

20110517, 14:47  #2 
"Tapio Rajala"
Feb 2010
Finland
13B_{16} Posts 
The correct answer would be that there is no upper bound, but largest known nonmersenne prime has 3918990 digits.

20130530, 10:45  #3 
1010000101010_{2} Posts 
Actually, is there a proof that there are infinite nonMersenne primes?

20130530, 10:58  #4 
Tribal Bullet
Oct 2004
3^{2}×5×79 Posts 

20130530, 16:51  #5  
Aug 2012
Mass., USA
13E_{16} Posts 
Quote:
In any case, From Bertrand's Postulate (which is actually a theorem), we know there is always a prime between some integer n and 2n  2 (with n > 3). Choose any integer k (> 1) and pick n = 2^k. Then, there must be a prime in the range 2^k and 2^(k+1)2. Clearly this prime can not be a Mersenne prime since there is no Mersenne number in that interval. Since we have infinite choices for k, and the interval defined for each k is disjoint from the others, there must be an infinite number of nonMersenne primes. 

20130530, 18:10  #6 
May 2013
East. Always East.
11·157 Posts 
Interesting
Cuber Bruce does actually bring up an interesting point. If Bertrand's Postulate is in fact true (I've never really looked into it) then he is correct.
Without that, though, one could have argued that while there is no largest prime, every prime past a certain bound may be a Mersenne number. 
20130530, 20:35  #7  
Mar 2010
2^{6}·3 Posts 
Quote:
Another proof: Assume that every prime is Mersenne from some point. Let P(x) be the number of prime numbers less than n. This number would be less than all powers of 2 less than x plus some fixed number. In other words, P(x)<C*log(x) for some constant C. This is impossible since log(x)=o(x/log(x))=O(P(x)). Last fiddled with by literka on 20130530 at 20:58 Reason: Error in text 

20130530, 20:52  #8  
Aug 2006
5,987 Posts 
Quote:


20130530, 22:36  #9 
Bamboozled!
"๐บ๐๐ท๐ท๐ญ"
May 2003
Down not across
11647_{10} Posts 
Another proof uses Fermat Numbers. F_{n} = 2^{2[sup]n}[/sup]+1 so F_{n}2 is the difference of two squares with factorization (F_{n1}) * (F_{n1}  2), from which we see that no two Fermat numbers share a factor in common, hence the infinitude of primes.

20130530, 23:13  #10 
Aug 2006
13543_{8} Posts 
Sure, but that doesn't give nearly as many nonMersennes as the PNT.

20130531, 04:17  #11 
Romulan Interpreter
"name field"
Jun 2011
Thailand
10,273 Posts 
That would work for mersenne numbers too, as Mx and My do not share a common factor if x and y do not share one.
So, another simpler proof would be to take any odd composite number z=xy, its correspondent Mz will be composite, and at least one of its prime factors is not mersenne number (easy to prove by binary patterning, you need no math for it), so you get a new prime for every combination of distinct x and y odd primes. Ex: start with 3 and 5, you get M15=32767=7*31*151, from which you got a new nonmersenne prime, 151. You will now get a new nonmersenn prime as a factor of M(3*151) and/or M(5*151). You could use M9, or M11 alone, but assuming we don't know which one is prime and which not, we can take the next odd numbers, 7 and 9, they can't share common factors (as they are consecutive odds, difference is 2), make M63, this can't be prime, as its expo is composed, so its split would give a new prime (in fact, 73, 337, etc, many factors) which is not mersenne. Last fiddled with by LaurV on 20130531 at 04:23 
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