mersenneforum.org  

Go Back   mersenneforum.org > Extra Stuff > Miscellaneous Math

Reply
 
Thread Tools
Old 2018-02-02, 16:22   #1
sudaprime
 
Feb 2018

1 Posts
Default 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.
sudaprime is offline   Reply With Quote
Old 2018-02-02, 17:45   #2
paulunderwood
 
paulunderwood's Avatar
 
Sep 2002
Database er0rr

22×23×37 Posts
Default

Therefore M^n(2) is prime for all n>0.

This could be similar to the mistake Fermat made for 2^{2^n}+1

Last fiddled with by paulunderwood on 2018-02-02 at 17:49
paulunderwood is offline   Reply With Quote
Old 2018-02-02, 17:53   #3
science_man_88
 
science_man_88's Avatar
 
"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts
Default

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.
science_man_88 is offline   Reply With Quote
Old 2018-02-02, 17:57   #4
retina
Undefined
 
retina's Avatar
 
"The unspeakable one"
Jun 2006
My evil lair

11·521 Posts
Default

Quote:
Originally Posted by sudaprime View Post
... it is most likely that,
M(M127) is a prime.
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.
retina is online now   Reply With Quote
Old 2018-02-02, 18:47   #5
CRGreathouse
 
CRGreathouse's Avatar
 
Aug 2006

134458 Posts
Default

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.
CRGreathouse is offline   Reply With Quote
Old 2018-02-02, 22:15   #6
Uncwilly
6809 > 6502
 
Uncwilly's Avatar
 
"""""""""""""""""""
Aug 2003
101×103 Posts

100001101000112 Posts
Default

Click image for larger version

Name:	IN B4.jpg
Views:	65
Size:	29.9 KB
ID:	17642
Uncwilly is offline   Reply With Quote
Old 2018-02-02, 22:22   #7
danaj
 
"Dana Jacobsen"
Feb 2011
Bangkok, TH

5×181 Posts
Default

Quote:
Originally Posted by CRGreathouse View Post
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.
So you're saying there's a chance. Keep the dream alive!
danaj is offline   Reply With Quote
Old 2018-02-03, 02:52   #8
VBCurtis
 
VBCurtis's Avatar
 
"Curtis"
Feb 2005
Riverside, CA

29·149 Posts
Default

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!
VBCurtis is offline   Reply With Quote
Old 2018-02-03, 05:00   #9
ATH
Einyen
 
ATH's Avatar
 
Dec 2003
Denmark

2×13×113 Posts
Default

Quote:
Originally Posted by CRGreathouse View Post
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.
If I'm that likely to win all those, I might even win without buying tickets!!!



I did actually win the Eurojackpot tonight, seriously!!!

though I did not win the huge 570M dkk jackpot. I won 83 dkk (~$14).
ATH is offline   Reply With Quote
Old 2018-02-03, 05:59   #10
CRGreathouse
 
CRGreathouse's Avatar
 
Aug 2006

172516 Posts
Default

Quote:
Originally Posted by VBCurtis View Post
3, 5, and 7 are all prime. By continuing this pattern of adding two every time, all odd numbers are prime. Woot!
https://xkcd.com/1310/
CRGreathouse is offline   Reply With Quote
Old 2018-02-03, 06:02   #11
retina
Undefined
 
retina's Avatar
 
"The unspeakable one"
Jun 2006
My evil lair

166316 Posts
Default

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 multi-multi-billionaire. Oh well, nothing is perfect.
retina is online now   Reply With Quote
Reply

Thread Tools


Similar Threads
Thread Thread Starter Forum Replies Last Post
2^9092392+40291 is a probable prime! engracio Five or Bust - The Dual Sierpinski Problem 86 2011-03-31 01:11
Fourth probable prime found, one to go! philmoore Five or Bust - The Dual Sierpinski Problem 22 2010-01-01 00:23
Megadigit probable prime found, our third! philmoore Five or Bust - The Dual Sierpinski Problem 25 2009-09-09 06:48
Record probable prime found! philmoore Five or Bust - The Dual Sierpinski Problem 18 2009-01-28 19:47
need Pentium 4s for 5th largest prime search (largest proth) wfgarnett3 Lounge 7 2002-11-25 06:34

All times are UTC. The time now is 07:47.

Fri Sep 25 07:47:58 UTC 2020 up 15 days, 4:58, 0 users, load averages: 1.06, 1.04, 1.23

Powered by vBulletin® Version 3.8.11
Copyright ©2000 - 2020, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.