20140418, 03:16  #12  
"Serge"
Mar 2008
San Diego, Calif.
10100010100111_{2} Posts 
Quote:


20140418, 03:30  #13 
Apr 2014
5×17 Posts 

20140418, 04:47  #14 
Romulan Interpreter
"name field"
Jun 2011
Thailand
10358_{10} Posts 
Well, you didn't get anything unusual. First, all the composite mersenne numbers are 7 (mod 8) so they must have at least one factor which is "78", and they always have an odd number of "78" factors, otherwise their product would not be 78. A composite mersenne can not have an even number of factors which are 78, because in this case their product two by two will be 18, so it can't be 2^n1. In the same time, they may have, or may not have factors which are "18". These factors can came in arbitrary number, including zero. Also, a composite mersenne whose prime exponent p is 34 can have a factor which is 2p+1 [edit: two times bigger than p]. But a composite mersenne whose prime exponent p is 14, can not have a factor which is 2p+1, because in this case the factor will be 38, for example (take the pen!) which is impossible. The smallest possible factor can be 6p+1 in this case [edit: 6 times higher, the minimum factor]. There is no factor which is 4p+1, nor for p=14 neither for p=34, because in this case they can't be 1 or 7 mod 8. [edit: also both "2p+1 factors of mersenne numbers with p=34 exponents" and "6p+1 factors of mersenne numbers with p=14 exponents" are "78 factors" i.e. the smallest possible factors are always 78, statistically]
Therefore, if you take "all factors below a million", you will have in this set:
Last fiddled with by LaurV on 20140418 at 05:02 
20140418, 11:28  #15 
"Forget I exist"
Jul 2009
Dartmouth NS
20511_{8} Posts 
tapion have you looked at the thread about k values in mersenne factors ? , also have you considered that once 2kp+1 is eliminated you can say (2kp+1)(2jp+1) > 4kjp^2+2kp+2jp+1 > a new k value of (2kp+1)*j+k for all values j>=0. edit: is also eliminated;
Last fiddled with by science_man_88 on 20140418 at 12:02 
20140418, 12:09  #16 
Apr 2014
5·17 Posts 
I'm aware of all of those things, @LaurV and scienceman. Note that primes = 2p+1 where p is also prime are Mersenne factors = 7 mod 8, if p are Sophie Germain primes = 3 mod 4. This 1:1 correspondence links them to the distribution of Sophie Germain primes (which occur evenly distributed among p = 1 and 3 mod 4). Then by taking the ratio among primes = 7 mod 8 that are Mersenne factors to those that aren't, we can multiply the expected factors that are = 7 mod 8 by the ratio again to get the expected number of factors = 1 mod 8. Then since Mersenne factors can only be = 1 or 7 mod 8, add the two estimations together and you get the number of expected Mersenne factors less than n.
Since Mersenne primes can only be = 7 mod 8, we should be able to discern a constant and a logarithmic factor that when multiplied by the expected number of Mersenne factors = 7 mod 8 will yield an approximation of Mersenne primes less than n. Note that Wagstaff came up with an estimation for Mersenne primes, his approximation was e^y/ln(2)*ln(ln(x)), where y is the Mascheroni constant. But there may be a way to state this as a ratio of primes = 7 mod p that are Mersenne primes, as opposed to just Mersenne factors. Last fiddled with by tapion64 on 20140418 at 12:11 
20140418, 15:08  #17 
"Bob Silverman"
Nov 2003
North of Boston
1DD3_{16} Posts 

20140418, 16:12  #18 
Aug 2006
2^{2}×3×499 Posts 
How could that be? We don't even know whether there are infinitely many Mersenne primes, so it's consistent with current knowledge that all but finitely many Mersenne exponents are equivalent mod 8.

20140418, 16:49  #19  
Apr 2014
5×17 Posts 
Quote:
@Silverman the absent, the original question was clearly not a conjecture, it was a statement of a pattern, that led to the conjecture which links the distribution of Mersenne factors to the twin prime (or ShahWilson, technically) constant. As far as I can see, this is 'new', in that I can't find a published paper or textbook that asserts the same. There's clearly heuristical evidence for the conjecture, and there's a solid mathematical basis for it coming from the distribution of Sophie Germain primes. EDIT: I did realize a mistake in the language I used. There's not a 1:1 correspondence from Sophie Germain primes = 3 mod 4 to primes = 7 mod 8, it's one way only. Clearly not all primes = 7 mod 8 have corresponding Sophie Germain primes (p1)/2. But it doesn't affect the conjecture, as it only depends on the ratio of those which do have corresponding Sophie Germain primes. Last fiddled with by tapion64 on 20140418 at 17:07 

20140418, 17:02  #20  
"Bob Silverman"
Nov 2003
North of Boston
7635_{10} Posts 
Quote:
The question was about factors of 2^p1, for general p and such p are clearly equidistributed mod 8 by e.g. PNT for AP. There are certainly infinitely many Mersenne numbers and infinitely many prime factors of Mersenne numbers. I am assuming, of course, that the definition of Mersenne number is 2^p1 for prime p. I have also seen it defined as 2^n1, for general n. And of course, I am being pedantic in saying that we do know that there are infinitely many Mersenne primes. We just have no proof.. 

20140418, 18:00  #21  
"Bob Silverman"
Nov 2003
North of Boston
3×5×509 Posts 
Quote:
et.al. from the 80's in which they proved the first case of FLT was true for infinitely many primes based upon a generalization of the Sophie Germain criteria. They got counts for a given prime p of the number of primes of the form 2p+1, 4p+1, 6p+1, ... 2kp+1, k = 1.....oo. Note that this preceded Wiles' result. 

20140418, 21:02  #22 
∂^{2}ω=0
Sep 2002
República de Califo
2^{2}·2,939 Posts 
Not so  we don't take kindly to useless/misleading/bogus stats. Look closely and you'll see that statistics pervades every area of this forum, whether it be roundoff error handling in floatingpointbased modmul math, or the largescale distribution of primes of various kinds, or p1, ecm and NFS factorization algorithms.

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