20180202, 16:22  #1 
Feb 2018
1 Posts 
probable largest prime.
complexions in computing ristricts calculating the probable largest prime, in Mersenne's series.
the series goes like this. M3 = 7 is prime. (M2 = 3, starting from 2, is prime and M3= M(M2)) M7 = 127 is prime, ( or M7= M(M3) ) M127 = 1.7e38 is prime or( M127= M(M7)) and it is most likely that, M(M127) is a prime. practically it takes ages to devolop a machine to calculate M(M127) then it takes more to test whether its a prime or not. we can not test this at this time. 
20180202, 17:45  #2 
Sep 2002
Database er0rr
2^{2}·919 Posts 
Therefore is prime for all n>0.
This could be similar to the mistake Fermat made for Last fiddled with by paulunderwood on 20180202 at 17:49 
20180202, 17:53  #3 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
The double mersenne numbers follow 2*x^2+4*x+1 if I knew how to apply this a certain number of times easily, we could find polynomials that these Catalan mersennes are on.

20180202, 17:57  #4 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
6,143 Posts 
I expect the likelihood of it being prime is the opposite of what you claim. Being such a large number I find it more likely to be composite. I base this on my unmathematical observation that there are a great many possible numbers that could be a divisor. I see no reason to put any credence into an apparent progression length of only four.

20180202, 18:47  #5 
Aug 2006
1761_{16} Posts 
I don't know if more recent work has been done, but Double Mersennes Prime Search has searched k < 111e15:
http://www.doublemersennes.org/mm127.php This means that MM127 has no prime factors below 2*k*M127 = 3.777... * 10^55, which in turn means that it's exp(gamma)*log(3.777... * 10^55) ~ 228 times more likely to be likely than an average number of its size.* This raises the 'probability' of it being prime from 1/log(MM127) to 228/log(MM127) ~ 228/(M127 * log 2) which is a little less than 2 in 10^36. For comparison, you're 438 times more likely to win all of the Powerball, Mega Millions, Eurojackpot, and EuroMillions jackpots buying just one ticket each. * This can be made precise in the usual way. 
20180202, 22:15  #6 
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
7·37^{2} Posts 

20180202, 22:22  #7 
"Dana Jacobsen"
Feb 2011
Bangkok, TH
2^{2}·227 Posts 

20180203, 02:52  #8 
"Curtis"
Feb 2005
Riverside, CA
4789_{10} Posts 
Since 3 makes a pattern according to the OP, here is a way to find an even larger prime:
3, 5, and 7 are all prime. By continuing this pattern of adding two every time, all odd numbers are prime. Woot! 
20180203, 05:00  #9  
Einyen
Dec 2003
Denmark
3×5×11×19 Posts 
Quote:
I did actually win the Eurojackpot tonight, seriously!!! though I did not win the huge 570M dkk jackpot. I won 83 dkk (~$14). 

20180203, 05:59  #10  
Aug 2006
3^{2}·5·7·19 Posts 
Quote:


20180203, 06:02  #11 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
6,143 Posts 
Okay, so to summarise this thread it appears that the way to prove MM127 is prime is to simply win all of the Powerball, Mega Millions, Eurojackpot, and EuroMillions jackpots 438 times. That seems doable. The only downside is becoming a multimultibillionaire. Oh well, nothing is perfect.

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