- If p:

1) is not the 1st element of the set of primes;

2) does not equal 2;

3) equals either minimum 3 or other primes;

(2^p+1)/3=p*

(p*-->prime can be equal p or not).

The problem: if p=17 then p*=17*a (a--> any integer).

The question: is there any prime as 17?

At the end, I can say that "1" is not prime and when p=1 then p* also equals 1 and if we noticed that the conditions (1, 2, 3) contain the numbers which are used in the formula.

Sincerely,

By Samir Musali - From: samir.musali
> If p:

I'm not sure I completelf follow you, but what you've written looks remarkably similar to the conjecture here:

> 1) is not the 1st element of the set of primes;

> 2) does not equal 2;

> 3) equals either minimum 3 or other primes;

> (2^p+1)/3=p*

> (p*-->prime can be equal p or not).

>

> The problem: if p=17 then p*=17*a (a--> any integer).

>

> The question: is there any prime as 17?

>

> At the end, I can say that "1" is not prime and when p=1

> then p* also equals 1 and if we noticed that the conditions

> (1, 2, 3) contain the numbers which are used in the

> formula.

http://primes.utm.edu/glossary/xpage/NewMersenneConjecture.html

Phil