Thread: factors of Mersenne numbers View Single Post
2020-09-12, 06:07   #4
LaurV
Romulan Interpreter

Jun 2011
Thailand

896210 Posts

Quote:
 Originally Posted by masser What is n?
Quote:
 Originally Posted by Dr Sardonicus Hey, you didn't say f had to be a proper divisor -- or prime.
His "framework" is known, he is "barking at the same tree" for few years by now.

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^2-1, and based on the fact that the series s(n)=2n^2-1 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=2^11-1=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(x-q), s(x+aq), s(x-aq), s(-x), ......, etc, are all divisible by q, for any integer a. Therefore, s(32-23)=s(9) is also divisible by 23. Indeed, s(9)=2*9^2-1=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^2-1 is prime, which is indeed true only for primes

(this thread is more like misc math)

Last fiddled with by LaurV on 2020-09-12 at 06:22