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Old 2005-10-02, 06:19   #1
Citrix
 
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Jun 2003

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Default 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?

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Last fiddled with by Citrix on 2005-10-02 at 06:20
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Old 2005-10-02, 18:07   #2
jinydu
 
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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...
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Old 2005-10-04, 08:08   #3
geoff
 
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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?
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