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-   -   probable largest prime. (https://www.mersenneforum.org/showthread.php?t=23014)

sudaprime 2018-02-02 16:22

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 [B]M3= M(M2)[/B])
M7 = 127 is prime, ( or [B]M7= M(M3)[/B] )
M127 = 1.7e38 is prime or( [B]M127= M(M7)[/B]) and it is most likely that,
[B]M(M127) is a prime.[/B]

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.:smile:

paulunderwood 2018-02-02 17:45

Therefore [TEX]M^n(2)[/TEX] [B]is prime[/B] for all n>0. :smile:

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

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=sudaprime;479092]... it is most likely that,
[B]M(M127) is a prime.[/B][/QUOTE]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:
[url]http://www.doublemersennes.org/mm127.php[/url]

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 [i]all[/i] 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

1 Attachment(s)
[attach]17642[/attach]

danaj 2018-02-02 22:22

[QUOTE=CRGreathouse;479115]For comparison, you're 438 times more likely to win [i]all[/i] of the Powerball, Mega Millions, Eurojackpot, and EuroMillions jackpots buying just one ticket each.[/QUOTE]

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=CRGreathouse;479115]For comparison, you're 438 times more likely to win [i]all[/i] of the Powerball, Mega Millions, Eurojackpot, and EuroMillions jackpots buying just one ticket each.

* This can be made precise in the usual way.[/QUOTE]

If I'm that likely to win all those, I might even win without buying tickets!!!



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

[SIZE="1"]though I did not win the huge 570M dkk jackpot. I won 83 dkk (~$14).[/SIZE]

CRGreathouse 2018-02-03 05:59

[QUOTE=VBCurtis;479152]3, 5, and 7 are all prime. By continuing this pattern of adding two every time, all odd numbers are prime. Woot![/QUOTE]

[url]https://xkcd.com/1310/[/url]

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.

LaurV 2018-02-05 08:10

[QUOTE=retina;479159]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.[/QUOTE]

The only problem with that is that you will need 438 weeks, which is about 9 years, and I bet that after you win one of them first time, you will not give a ship anymore, about playing, and you will go to some tropical paradise to live the rest of your life - or better, go to some cold place and open a computing center (cold place is better, it helps with killing the heat that the servers will generate), buy few supercomputers and crunch for some prime numbers... And if you have the guts and play the second time, and win again, then expect CIA, FBI, KGB, Mossad, Yakuza, etc., to start hunting for you to "study" you...


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