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graNite
Germany4434 Posts
i saw a few math topics recently and noticed there are some good mathematics here. i have a prime number problem and can't solve it. lets look at this progression (is progression the right word?) ![[image loading]](http://s3.imgimg.de/uploads/latex2png1deb35a6php1deb35a6png.png) with ![[image loading]](http://s3.imgimg.de/uploads/latex2pngad1d8bebphpad1d8bebpng.png) it produces: 2 3 7 127 ... how can i prove that every number coming of this progression is prime? if that's not possible, how can i disprove it? ty  | ||
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Bibdy
United States3481 Posts
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graNite
Germany4434 Posts
after 127 comes 2^127-1 ... (which is prime) | ||
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Starfox
Austria699 Posts
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graNite
Germany4434 Posts
Anyone with a supercomp here? Need a5 - a100 :D | ||
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Antisocialmunky
United States5912 Posts
The proof is on wiki. | ||
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gogogadgetflow
United States2583 Posts
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JingleHell
United States11308 Posts
On April 22 2011 08:03 Starfox wrote: You are looking at http://en.wikipedia.org/wiki/Mersenne_prime basically. If you'd be abelt to proof that every step produces a prime you would be the candidate for the next fields medal for sure. :D Isn't there a cash bounty for calculating new Mersenne primes? | ||
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qrs
United States3637 Posts
On April 22 2011 08:03 Starfox wrote: Well...a tiny subset of Mersenne primes, anyway.You are looking at http://en.wikipedia.org/wiki/Mersenne_prime basically. | ||
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stepover12
United States175 Posts
http://en.wikipedia.org/wiki/Catalan's_Mersenne_conjecture#Catalan-Mersenne_number "Although the first five terms (up to M(127)) are prime, no known methods can decide if any more of these numbers are prime (in any reasonable time) simply because the numbers in question are too huge, unless a factor of M(M(127)) is discovered." | ||
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Primadog
United States4411 Posts
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mcc
Czech Republic4646 Posts
On April 22 2011 08:05 graNite wrote: Will do. Anyone with a supercomp here? Need a5 - a100 :D I don't think any computer will help you to do that. a5 = M127 and is a prime, but above that you are too far out. If you would be able to prove your statement, then you would have solved one of important unsolved mathematical problems and I do not think anyone here can really help you with that Basically if I see it correctly you would have proven that there is infinite number of Mersenne primes.EDIT:typo | ||
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ptrpb
Canada753 Posts
FUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUU Just finished a course on this stuff, but I've forgotten most of it sorry man. Best of luck though | ||
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MisterD
Germany1338 Posts
(01:24) gp > a = 2 %1 = 2 (01:24) gp > a = 2 ^ a - 1 %2 = 3 (01:24) gp > a = 2 ^ a - 1 %3 = 7 (01:24) gp > a = 2 ^ a - 1 %4 = 127 (01:24) gp > a = 2 ^ a - 1 %5 = 170141183460469231731687303715884105727 (01:24) gp > a = 2 ^ a - 1 *** length (lg) overflow so .. a5 works, a6 does not work anymore^^ it's a little big. so for a5: (01:24) gp > a %6 = 170141183460469231731687303715884105727 (01:25) gp > factor(a) %7 = [170141183460469231731687303715884105727 1] meaning a5 is still prime. | ||
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Day[9]
United States7366 Posts
So 2^67 isn't prime : ] Cheers! | ||
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Day[9]
United States7366 Posts
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Primadog
United States4411 Posts
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Oracle
Canada411 Posts
But what you might say is that youre looking for an upper bound on n for which this statement holds true? In that case, run a fast fourier transform primality test on a supercomputer with this progession, and hope there exists an n which breaks this progression  | ||
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graNite
Germany4434 Posts
pls solve this one !! :D | ||
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Rtran10
Canada78 Posts
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