probable largest prime.

sudaprime · 2018-02-02, 16:22

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.

paulunderwood · 2018-02-02, 17:45

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

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

science_man_88 · 2018-02-02, 17:53

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.

retina · 2018-02-02, 17:57

Quote:

Originally Posted by sudaprime

... 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.

CRGreathouse · 2018-02-02, 18:47

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.

Uncwilly · 2018-02-02, 22:15

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danaj · 2018-02-02, 22:22

Quote:

Originally Posted by CRGreathouse

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!

VBCurtis · 2018-02-03, 02:52

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!

ATH · 2018-02-03, 05:00

Quote:

Originally Posted by CRGreathouse

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).

CRGreathouse · 2018-02-03, 05:59

Quote:

Originally Posted by VBCurtis

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/

retina · 2018-02-03, 06:02

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.

2018-02-02, 16:22	#1
sudaprime Feb 2018 1₁₆ 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.

2018-02-02, 17:45	#2
paulunderwood Sep 2002 Database er0rr 2×5×467 Posts	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

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2018-02-02, 17:53	#3
science_man_88 "Forget I exist" Jul 2009 Dartmouth NS 2·5²·13² Posts	The double mersenne numbers follow 2x^2+4x+1 if I knew how to apply this a certain number of times easily, we could find polynomials that these Catalan mersennes are on.

2018-02-02, 18:47	#5
CRGreathouse Aug 2006 2²×3×499 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 2kM127 = 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.

2018-02-02, 22:15	#6
Uncwilly 6809 > 6502 """"""""""""""""""" Aug 2003 101×103 Posts 10101100001111₂ Posts

2018-02-03, 02:52	#8
VBCurtis "Curtis" Feb 2005 Riverside, CA 2·5·7·83 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!

2018-02-03, 06:02	#11
retina Undefined "The unspeakable one" Jun 2006 My evil lair 3·37·61 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 multi-multi-billionaire. Oh well, nothing is perfect.