20070509, 04:15  #1 
"Jason Goatcher"
Mar 2005
5·701 Posts 
DC project for Brier problem?
I know people think I'm an idiot for suggesting this, but I'm not saying we should work on it the same way Riesel Sieve and Seventeen or Bust are doing it.
Before I explain my idea, I'll explain the Brier problem, so everyone knows. Most people know that the Riesel and Sierpenski conjecture involves numbers where k*2^n+1 and K*2^n1 have ks that will always yield a composite, no matter what you plug in as n. They're basically trying to find n's that yield each k below the number prime. Riesel Sieve is down to 68 k's and Seventeen or Bust is down to 7 k's. The Brier problem involves finding the lowest number that is both a Riesel number and a Sierpenski number. Okay, now that that's out of the way... I suggest taking candidate k's and coming up with an automated way of plugging them into the NewPGen engine. Basically, it would be k*2^n+or1, with about 1000 n and sieving up to a low p like 5000. This way we could get rid of numbers as quickly as possible. Any combination where all the ns disappeared would be tried a second time with 10,000 ns, then 100,000 ns. If 100,000 ns worked, then they would try +1 if the first try had been 1, or viceversa. There's probably a lot of other stuff we could add that would expedite things, but that's the basic idea. What do you guys think? 
20070509, 16:02  #2 
Mar 2004
Belgium
837_{10} Posts 
It is an interesting idea but first try make a list of the k's you want to test.
Several other projects did also check these kind of numbers. I think you have go to for a distributed approach like BOINC Primegrid / RieselSieve. First sieve a whole lot of numbers to 1000T+ and then start crunching away. For the moment, I am doing your 9*2^n1 from 0 to 33M+ & 3^16 till 4M. I do not know that it is still necessary to search / sieve candidates for 9*2^n1. Regards Cedric Last fiddled with by ValerieVonck on 20070509 at 16:02 
20070509, 23:44  #3  
"Erling B."
Dec 2005
2·41 Posts 
Quote:
and yours too if you like later on. 

20070510, 01:13  #4  
Jun 2003
3^{2}×5^{2}×7 Posts 
Quote:


20070510, 23:01  #5 
"Jason Goatcher"
Mar 2005
5×701 Posts 
I'm hoping there's a way to disqualify the vast number of choices. While I don't understand modular arithmetic, except at the most basic level, I would think it could be used to disqualify numbers much more quickly than sieving each number, probably 1,000s of times more quickly.

20070511, 08:53  #6  
"Curtis"
Feb 2005
Riverside, CA
1000110001101_{2} Posts 
Quote:
Curtis 

20070511, 22:28  #7  
Jun 2003
3^{2}·5^{2}·7 Posts 
Quote:
It is possible to do this, but you will have to restrict your self to certain kind of k's. For example you can exclude all k's that are multiple of 3's since there probability to be a brier number is low. Basically instead of testing all the k's, you are only testing low weight sierpinski and riesel numbers together. But if you wanted to test each and every k till 10^27, I don't think that will be feasible with current technology. 

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