Factoring Double mersennes

Citrix · 2005-10-02, 06:19

MMn= 2^(Mn)-1

Based on http://www.primepuzzles.net/conjectures/conj_015.htm
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

jinydu · 2005-10-02, 18:07

Quote:

Originally Posted by Citrix

MMn= 2^(Mn)-1

Based on http://www.primepuzzles.net/conjectures/conj_015.htm
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

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

geoff · 2005-10-04, 08:08

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?

2005-10-02, 06:19	#1
Citrix Jun 2003 646₁₆ Posts	Factoring Double mersennes MMn= 2^(Mn)-1 Based on http://www.primepuzzles.net/conjectures/conj_015.htm 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 Last fiddled with by Citrix on 2005-10-02 at 06:20

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2005-10-04, 08:08	#3
geoff Mar 2003 New Zealand 13×89 Posts	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?