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:

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]

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.

[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.

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.

[attach]17642[/attach]

[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!

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!

[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]

[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]

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.