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• 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
Message 1 of 2 , Jan 15, 2012
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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 ... I m not sure I completelf follow you, but what you ve written looks remarkably similar to the conjecture here:
Message 2 of 2 , Jan 15, 2012
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From: samir.musali
> 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.

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

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

Phil
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