20031210, 21:53  #1 
3^{5}·5 Posts 
Is 2^{mersenne prime}1 prime?
why the new prime number as an exponent for 2 ^X  1 does not work as the largest number?
As I understand the exponent must be prime which this new number is, so why does it not work? 
20031210, 21:56  #2 
Jun 2003
The Texas Hill Country
10001000001_{2} Posts 
That the exponent is prime is a necessary, but not sufficient condition.
2^(M40) 1 may be prime. But it is too large for us to test. 
20031210, 22:00  #3 
Aug 2002
2^{2}·5·13 Posts 
Not all prime numbers will generate Mersenne primes. The range we are working on now is somewhere between 20 and 22 million ( 33 million + for the oines that are goiing for the big price). 77 million is about the top end of the range we can work on with todays client.
All these numbers have 8 digits in the exponent. The newly found prime has 6,320,430 digits, and a potential Mersenne generated from that would be unbelievably huge. PM 
20031212, 04:07  #4 
"William"
May 2003
New Haven
3×787 Posts 
Note that we usually indicate exponentiation with "^", although it also works to write [ sup ]exponent[ /sup ] so it shows as^{exponent}.
Then the questions are "why not just calculate 2^{M40}1?" and "Are we calculating 2^{M40}1?" The core issues are that a prime exponent is necessary to get a prime but not sufficient  that is, not all prime exponents generate prime numbers, so we must test each one, and 2^{M40}1 is much too big to be tested by any known methods even using all the computing power in existence. So no, we are not testing that. We are mostly testing prime exponents a little bit larger than the exponent that lead to the recent prime. Some people are doublechecking smaller exponents because about 2% of tests have errors. A few people are checking prime exponents for numbers that have over 10 million digits. Last fiddled with by ewmayer on 20051228 at 21:42 Reason: Edited to reflect merging of 2 similar threads 
20031213, 03:06  #5  
"Richard B. Woods"
Aug 2002
Wisconsin USA
2^{2}×3×641 Posts 
Re: so is it so easy?
Quote:
2^{20996011}1 is the largest known Mersenne prime now. (We're pretty sure there are other Mersenne primes that we don't yet know.) Before we proved that 2^{20996011}1 is prime, the largest known Mersenne prime was 2^{13466917}1, and before that one was found to be prime, a still smaller number was the largest known Mersenne prime. But not all Mersenne primes have been discovered in numerical order. At least once in the past, someone found a Mersenne prime that was smaller than the largest known Mersenne prime as of that date ... because the discoverer of the thenlargest one had skipped testing some smaller candidates. GIMPS has not yet finished testing all the possible Mersenne primes between 2^{13466917}1 and 2^{20996011}1. Because GIMPS participants use a wide variety of computers, some much faster than others, and for other reasons, the tests are not necessarily completed in numerical order. It is possible that the next Mersenne prime to be discovered will be smaller than 2^{20996011}1 (and there's even a small chance that the next Mersenne prime to be discovered will be less than 2^{13466917}1). So, as well as testing numbers larger than 2^{20996011}1, we have to finish testing (or even retest, if there is an error in a test) many of the smaller numbers before we can conclude that the next Mersenne prime is larger than 2^{20996011}1. Does this answer your questions? 

20031213, 04:52  #6 
10001111101101_{2} Posts 
what i mean is that : ( i dont now how u did the power on this thing)
why dont we just calculate this: ....209960111 ..2 2 we know its a prime for sure, because 2p1 is a prime...... so why dont we just use the largest we know now as a power of 2 then 1... and just calculate that? 
20031213, 05:21  #7  
"William"
May 2003
New Haven
3·787 Posts 
Quote:


20031213, 05:37  #8  
Sep 2003
3^{2}×7×41 Posts 
Quote:
1) if P is not prime, then 2^{P}1 can't be prime. 2) if P is prime, then 2^{P}1 might be prime, but not necessarily (in fact, the odds are very low). Second of all: It's hopeless to even think about doing a LucasLehmer test on such an enormous number. There isn't enough time between now and the end of the universe, and not enough atoms in the universe to build a computer that could do it. Even 2^{2[sup]61}1[/sup]1 is far too big (see http://www.ltkz.demon.co.uk/ar2/mm61.htm). PS, To do exponents, just use [ sup ] and [ /sup ] (except without the blanks next to the brackets). 

20031213, 17:55  #9 
2·3^{3}·5·19 Posts 
o, i get it now.... :>
but how about the money? do we share if i find it out or its just mine? 
20031213, 18:03  #10 
Nov 2003
3×5×11 Posts 
We share. The rules are posted at http://www.mersenne.org/prize.htm

20031214, 23:26  #11 
Oct 2002
Lost in the hills of Iowa
700_{8} Posts 
If GIMPS was ranked as a supercomputer on the "Top 500 supercomputers" list, IIRC we'd be somewhere in the top 10 or 15 *already*.
To be fair, if all of the other large projects like SETI and Distributed.Net got ranked on that list, we'd drop a few places  I suspect SETI and D.Net EACH harness more overall CPU power than the Earth Simulator (the highestpower supercomputer on that list).... So just adding one "supercomputer" wouldn't make things a lot faster  in effect, we're *ALREADY* a supercomputer, of a widelydistributed nature.... 
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