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Old 2005-04-21, 03:17   #1
Dougy
 
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Question The New Mersenne Conjecture

Does anyone have any current information regarding the testing status of the New Mersenne Conjecture? Or any other information regarding this, that I could include in my project?

Let p be any odd natural number. If two of the following conditions hold, then so does the third:
p = 2^k+/-1 or p = 4^k+/-3
2^p-1 is a prime
(2^p+1)/3 is a prime.
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Old 2005-04-21, 06:18   #2
akruppa
 
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Chris Caldwell has info about the NMC at http://www.utm.edu/research/primes/m...onjecture.html

Alex
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Old 2005-04-21, 13:01   #3
R.D. Silverman
 
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Quote:
Originally Posted by akruppa
Chris Caldwell has info about the NMC at http://www.utm.edu/research/primes/m...onjecture.html

Alex

I was present when John Selfridge first posed this conjecture.

His remarks included the somewhat obvious statement that the conjecture
was clearly true, that we know all the instances where it is true, and that
it is impossible of proof.

It is clear, even on *very* loose probabilistic grounds that 2^n-1 and
(2^n+1)/3 are simultaneously prime only finitely often.

This conjecture is another instance of what Richard Guy calls his "Strong
Law of Small Numbers".

Even John says that the conjecture is a minor curious coincidence.
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Old 2005-04-22, 12:53   #4
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Quote:
Originally Posted by Dougy
.... Or any other information regarding this, that I could include in my project ?
What is the subject of your project ?
Tony
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Old 2005-04-22, 17:58   #5
ewmayer
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Quote:
Originally Posted by R.D. Silverman
I was present when John Selfridge first posed this conjecture.

His remarks included the somewhat obvious statement that the conjecture
was clearly true, that we know all the instances where it is true, and that
it is impossible of proof.
Impossible of proof or highly improbable, in the sense that it's extremely unlike that there is another (as yet undiscovered) Mersenne or (2^p + 1)/3 prime whose exponent is of the requisite type?

Quote:
It is clear, even on *very* loose probabilistic grounds that 2^n-1 and
(2^n+1)/3 are simultaneously prime only finitely often.

This conjecture is another instance of what Richard Guy calls his "Strong
Law of Small Numbers".

Even John says that the conjecture is a minor curious coincidence.
That latter point bothers me slightly - call me pedantic, but I don't believe in formally conjecturing something unless one has serious mathematically-based reasons to believe it may be true (and in a nontrivial fashion), or at least a sufficiently large number of experimental data points to indicate something interesting is going on. The risk of this kind of thing becoming a time-wasting distraction for others is great if (a) the conjecture comes from someone with authority, and (b) the conjecture is of a kind that, even if is false, is difficult to show as such. (Compare the NMC to the all-Fermat-numbers-are-prime conjecture, for instance - the latter was not proven false for many years, but it was clear from the beginning that if false, it would likely be shown to be false relatively quickly as larger Fermats were studied and subjected to trial factoring - as it happens, the very next case beyond the largest known-prime one proved to be composite, as have all the others whose status has been established one way or another).

Don't get me wrong - I also know John and have great respect for him and his work, that's why something the NMC seems somewhat out of character for him (at least to me). But unlike Bob I wasn't there when he proposed it, so perhaps in the interim it's taken on more gravitas than it deserves or JS ever intended for it.
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Old 2005-04-23, 00:07   #6
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My project is on Mersenne numbers, with a focus on Euler's work.
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Old 2005-05-29, 17:28   #7
akruppa
 
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Mere curious coincidence or not, I have been trying to prove a few probable primes from the NMC table for a while, particularly (2^42737+1)/3. I just found a new prime factor of 2,5342L, an algebraic factor of (2^42737+1)/3-1:

p48 = 352752377673314068757006632344836735032049556013

2,5342L has lot of prime factors, 8 of them (179 digits worth) found so far. The cofactor is down to 626 digits and still composite.

Alex

I should add: this discovery was very lucky, the factor was found with B1=3M (p40 parameters)! The group order is

2^2 3^3 167 193 911 1789 17327 69191 91867 2614223 2650897 81465869

with sigma=6006597770531109.

Last fiddled with by akruppa on 2005-05-29 at 17:32
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Old 2005-10-15, 18:15   #8
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I think the conjecture can be extended to 1G, by just simple sieving. Find a factor for 2^p+1 or 2^p-1. Either way the p will be eliminated. It probably can be incorporated into the LMH program code, if anyone is interested.


Secondly, If you assume that gaussian mersennes and mersennes have the same distribution of primes and you extend the above to them then the conjecture fails for p=29.

Anyway these are the only known p so far for which Gaussian mersenne and their counterpart produce a prime.

p = 3, 5, 7, 11, 29 and 283.

Or can a new conjecture be made about Gaussian mersennes?

I am at present trying to extend the list


Citrix

Last fiddled with by Citrix on 2005-10-15 at 18:17
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Old 2005-10-15, 20:01   #9
ewmayer
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Quote:
Originally Posted by Citrix
I think the conjecture can be extended to 1G, by just simple sieving. Find a factor for 2^p+1 or 2^p-1. Either way the p will be eliminated. It probably can be incorporated into the LMH program code, if anyone is interested.
What do you think the odds are of finding an explicit factor for all candidates with p < 10^9?
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Old 2005-10-15, 22:41   #10
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Quote:
Originally Posted by ewmayer
What do you think the odds are of finding an explicit factor for all candidates with p < 10^9?

very low, will take very very long.


Citrix
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Old 2005-10-16, 03:45   #11
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Quote:
Originally Posted by ewmayer
What do you think the odds are of finding an explicit factor for all candidates with p < 10^9?

But it can be incorporated into the LMH code at no extra computational cost. So if you are interested they can look into it.

Citrix
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