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 bhelmes 2020-09-11 23:00

factors of Mersenne numbers

If f | Mp and p is prime then

a) f | 2n²-1

b) but not f=2n²-1

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 :cmd: :loco::petrw1:
Bernhard

 masser 2020-09-12 01:05

[QUOTE=bhelmes;556766]If f | Mp and p is prime then

a) f | 2n²-1

b) but not f=2n²-1

[/QUOTE]

What is n?

 Dr Sardonicus 2020-09-12 01:09

[QUOTE=bhelmes;556766]If f | Mp and p is prime then

a) f | 2n²-1

b) but not f=2n²-1

a) is clear for me but b) is a guess (I checked some numbers)

The first and fastest counterexample is welcome (Or a proof ?)
[/QUOTE]
First example: Taking n = 1, f = 1 divides 2[sup]p[/sup] - 1 for any p.

First example with prime f: p = 3, n = 2; f = 2*2[sup]2[/sup] - 1 = 7 divides 2[sup]3[/sup] - 1 = 7.

Second example with prime f: p = 5, n = 4; f = 2*4[sup]2[/sup] - 1 = 31 divides 2[sup]5[/sup] - 1 = 31

48 other examples known where f = 2[sup]p[/sup] - 1 is prime.

Hey, you didn't say f had to be a [i]proper[/i] divisor -- or prime.

I checked the factor table of 2[sup]n[/sup] - 1, odd n < 1200, and did not find any examples with f prime other than f = M[sub]p[/sub]

I didn't check for composite proper factors of the form 2*n[sup]2[/sup] - 1.

 LaurV 2020-09-12 06:07

[QUOTE=masser;556772]What is n?[/QUOTE]

[QUOTE=Dr Sardonicus;556773]
Hey, you didn't say f had to be a [I]proper[/I] divisor -- or prime.
[/QUOTE]

His "framework" is known, he is "barking at the same tree" for few years by now. :razz:

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 [URL="https://en.wikipedia.org/wiki/Divisibility_sequence"]divisibility series[/URL]. 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 :lol:

(this thread is more like misc math)

 CRGreathouse 2020-09-13 02:05

[QUOTE=bhelmes;556766]If f | Mp and p is prime then

a) f | 2n²-1

b) but not f=2n²-1

a) is clear for me but b) is a guess (I checked some numbers)

The first and fastest counterexample is welcome (Or a proof ?)[/QUOTE]

2*2^2-1 | 2^3-1

 bhelmes 2020-09-13 04:53

f should be a proper divisor from the non prime Mp,
where Mp is a Mersenne number with prime p.

 CRGreathouse 2020-09-13 06:12

[QUOTE=bhelmes;556841]f should be a proper divisor from the non prime Mp,
where Mp is a Mersenne number with prime p.[/QUOTE]

In that case, I don't find any examples below the unfactored composite Mp = 2^1213 - 1.

 JeppeSN 2020-09-13 09:34

[QUOTE=LaurV;556786] and based on the fact that the series s(n)=2n^2-1 for integer n, is a [URL="https://en.wikipedia.org/wiki/Divisibility_sequence"]divisibility series[/URL].[/QUOTE]
How? For example, 7 divides 70, but s(7) = 2*7^2 - 1 = 97 does not divide s(70) = 2*70^2 - 1 = 9799. Clearly enough, 97 divides 9797, hence 97 leaves a remainder of 2 when dividing into 9799?

It is true that M(p) = 2^p - 1, for odd p, can be written as s(n); you take n = 2^{(p-1)/2}.

For example, M(101) = 2^101 - 1 = 2*(2^50)^2 - 1 = s(2^50).

Not sure if the relation n = 2^{(p-1)/2} is related to masser's question.

/JeppeSN

 bhelmes 2020-09-14 17:36

[QUOTE=JeppeSN;556848]How? [/QUOTE]

You did not understand the algorithm.
Have a look at :