Mersenne Sequence

[TheMawn]

http://www.youtube.com/watch?v=09JslnY7W_k

Is this something new to anyone here? I feel like there might be something to this but I lack the time and probably knowledge to dig in more, myself. On the other hand, I would very much like to hear from anyone else who has already looked into it.

EDIT: Specifically the bit on the Mersenne sequence. This idea that each new Mersenne composite contains a new prime number not yet seen in the sequence.

[LaurV]

Whoaaa... she is talking about mersenne numbers but doesn't know if 127 is prime or not... she lost half of my respect , but she can paint very sexy handlebottoms, aaa... sorry, mandelbrottoms, aaa.. that... mandelbrots... with handlers ... she got the respect back...

About the mersenne series producing new primes, I don't think that is unproven. I could personally squeeze a kind of proof out of my pumpkin. It is clear that all the prime exponents will produce new primes, as two mersenne numbers with relatively-prime exponents do not have the same factor. Not only for primes, but for relatively primes too. For example, 2^35-1 and 2^33-1 do not share the same factors, as 35 and 33 do not share the same factors. This is easy to prove. What would be left is that every "step" where there is a mersenne with a composite exponent, introduces a new prime.

For this, I would start by observing that every odd number, not necessarily prime, divides some mersenne number. This is easy to prove. For example, 3 divides 2^2-1, 5 divides 2^4-1, 7 divides 2^3-1, 9 divides 2^6-1, 11 divides 2^10-1.... 21 divides 2^6-1, 55 divides 2^20-1 and so on. Therefore, every prime divides some mersenne number m=2^n-1.

Then, we say that if p divides 2^x-1, then p divides 2^{kx}-1 for any k. Let z be the smaller exponent such as p divides 2^{kz}-1. We can easily prove that p divides only 2^{kz}-1 for all natural k. This means, there are no other mersenne numbers, out of this string, which are divisible by p.

Now, let's take a composite exponent c=kz, the correspondent mersenne number m=2^{kz}-1 can be decomposed as (2^k-1)*(2^z-1)*(something). The "something" is a sum of powers, what remains after decomposition. So the factors of m are either factors of the first two parenthesis, which already appeared in the mersenne sequence, or they are factors of the (something), which means they did not appear in the sequence. Because, remember, there are no other (smaller) 2^q-1 divisible by p, with q<kz.

Unless the (something) is void, or a factor (proper or not) of the product of first parenthesises, which is impossible, from the elementary algebra, we will always have new prime number in that string. Now, 6 is a special exception, where the "something" is 3, for example 2^6-1=2^{2*3}-1=(2^2-1)(2^3-1)(2+1)=3*7*3. This does not happen further in the string. In fact, after a while, the "something" is always bigger than the product of all the other factors, so we only have to "investigate", eventually, only to that limit.

Someone may think that being a perfect number has something to do with it, but not. For example:

gp > factorint(2^28-1)
%1 =
[3 1]
[5 1]
[29 1]
[43 1]
[113 1]
[127 1]
gp > getp(3)
%2 = 2
gp > getp(5)
%3 = 4
gp > getp(29)
%4 = 28
gp > getp(43)
%5 = 14
gp > getp(113)
%6 = 28
gp > getp(127)
%7 = 7

So here, 3, 5, 43, 127 are not new, they appeared at terms respectively 2, 4, 14, 7, but 29 and 113 are new, they didn't appear till term 28. Also, both of 29 and 113, only will divide mersenne numbers of the form 2^{28*k}-1 for any natural k, and no others.