20200911, 23:00  #1 
Mar 2016
5^{2}×11 Posts 
factors of Mersenne numbers
If f  Mp and p is prime then
a) f  2n²1 b) but not f=2n²1 a) is clear for me but b) is a guess (I checked some numbers) The first and fastest counterexample is welcome (Or a proof ?) Greetings, there must be somewhere a new Mersenne Prime Bernhard 
20200912, 01:05  #2 
Jul 2003
wear a mask
2^{2}·19^{2} Posts 

20200912, 01:09  #3  
Feb 2017
Nowhere
2×3^{2}×199 Posts 
Quote:
First example with prime f: p = 3, n = 2; f = 2*2^{2}  1 = 7 divides 2^{3}  1 = 7. Second example with prime f: p = 5, n = 4; f = 2*4^{2}  1 = 31 divides 2^{5}  1 = 31 48 other examples known where f = 2^{p}  1 is prime. Hey, you didn't say f had to be a proper divisor  or prime. I checked the factor table of 2^{n}  1, odd n < 1200, and did not find any examples with f prime other than f = M_{p} I didn't check for composite proper factors of the form 2*n^{2}  1. Last fiddled with by Dr Sardonicus on 20200912 at 01:10 Reason: xingif posty 

20200912, 06:07  #4  
Romulan Interpreter
Jun 2011
Thailand
21254_{8} Posts 
Quote:
Here, n is any integer. What he tries to do for a while by now, is to find a previously unknown, non trivial, factor of a mersenne number in gimps' range, based on the fact that all mersenne numbers with odd exponent are the form 2n^21, and based on the fact that the series s(n)=2n^21 for integer n, is a divisibility series. So, if you write all of them in a row, from 1 to infinity, you can sieve them, and extract the factors. In this series, mersenne numbers with prime exponents appear at indexes powers of two. For example, m=M11=2^111=2*((2^5)^2)1 appears at index 2^5=32. In the same time, m is divisible by 23, therefore s(32) is divisible by 23. But, because this is a divisibility series, if s(x) is divisible by some q for some particular x, then s(x+q), s(xq), s(x+aq), s(xaq), s(x), ......, etc, are all divisible by q, for any integer a. Therefore, s(3223)=s(9) is also divisible by 23. Indeed, s(9)=2*9^21=161, which is 7*23, and it is much smaller than 2047. When you sieve this series with primes, after sieving with just few primes (here, 7 is the third odd prime) you find 23 at index 9, and you just factored M11. This works fine for small numbers, but it becomes laborious for the range where we work now with TF  to have any chance to find a 75 bit factor, you have to sieve and parse this string up to 2^50 terms or so. Current gimps methods are much faster. But he might get lucky, as I told him in the past many times. I think is is still searching this, but he didn't come yet with a previously unknown, non trivial factor, in gimps range (i.e. exponent smaller than 1G). So, the second question in fact, asks for which n the order of 2 in 2*n^21 is prime, which is indeed true only for primes (this thread is more like misc math) Last fiddled with by LaurV on 20200912 at 06:22 

20200913, 02:05  #5 
Aug 2006
2^{2}·1,483 Posts 

20200913, 04:53  #6 
Mar 2016
5^{2}×11 Posts 
f should be a proper divisor from the non prime Mp,
where Mp is a Mersenne number with prime p. 
20200913, 06:12  #7 
Aug 2006
1011100101100_{2} Posts 

20200913, 09:34  #8  
"Jeppe"
Jan 2016
Denmark
2^{5}×5 Posts 
Quote:
It is true that M(p) = 2^p  1, for odd p, can be written as s(n); you take n = 2^{(p1)/2}. For example, M(101) = 2^101  1 = 2*(2^50)^2  1 = s(2^50). Not sure if the relation n = 2^{(p1)/2} is related to masser's question. /JeppeSN 

20200914, 17:36  #9 
Mar 2016
5^{2}×11 Posts 
You did not understand the algorithm. Have a look at : http://devalco.de/quadr_Sieb_2x%5E21.php I improved the page and hope that the content is well described, if you have improvements, tell it to me. 
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