Factoring Double mersennes
 

MMn= 2^(Mn)-1

Based on [url]http://www.primepuzzles.net/conjectures/conj_015.htm[/url]
if MMn is prime then 2^(Mn)+1/3 will also be. So can we try factoring 2^(Mn)+1/3? Has any work been done on this?

Citrix

[QUOTE=Citrix]MMn= 2^(Mn)-1

Based on [url]http://www.primepuzzles.net/conjectures/conj_015.htm[/url]
if MMn is prime then 2^(Mn)+1/3 will also be. So can we try factoring 2^(Mn)+1/3? Has any work been done on this?

Citrix[/QUOTE]

Well, it's a conjecture. It hasn't been proven yet...

My logic might be letting me down here, but since p=M(n) automatically satisfies the first condition of the NMC when M(n) is a Mersenne prime, finding a factor of (2^M(n)+1)/3, as Citrix suggests, would prove that either the double Mersenne MM(n) is composite or the NMC is false. If we didn't already know that MM(n) was composite wouldn't that be progress?