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 2007-05-09, 04:15 #1 jasong     "Jason Goatcher" Mar 2005 3×7×167 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^n-1 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+or-1, 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 vice-versa. 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?
 2007-05-09, 16:02 #2 ValerieVonck     Mar 2004 Belgium 7×112 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^n-1 from 0 to 33M+ & 3^16 till 4M. I do not know that it is still necessary to search / sieve candidates for 9*2^n-1. Regards Cedric Last fiddled with by ValerieVonck on 2007-05-09 at 16:02
2007-05-09, 23:44   #3
japelprime

"Erling B."
Dec 2005

25×3 Posts

Quote:
 Originally Posted by CedricVonck For the moment, I am doing your 9*2^n-1 from 0 to 33M+ & 3^16 till 4M. I do not know that it is still necessary to search / sieve candidates for 9*2^n-1. Regards Cedric
Keep on sieveing as long you feel like doing it. I made a small homepage with my range 7*2^n+-1. Maybe we can host all this bunch of number somewhere in one place when we know what to do with them . I can keep it all here at the moment: http://rafteikning.is/~prime/
and yours too if you like later on.

2007-05-10, 01:13   #4
Citrix

Jun 2003

22×397 Posts

Quote:
 Originally Posted by jasong 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^n-1 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+or-1, 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 vice-versa. 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?
Good idea, but it will take too long to do this. Smallest brier number has 27 digits. Even if you could eliminate 10^10 candidates per second then it would take ~10^17 sec. (Which is huge, more than the age of the planet earth).

2007-05-10, 23:01   #5
jasong

"Jason Goatcher"
Mar 2005

3×7×167 Posts

Quote:
 Originally Posted by Citrix Good idea, but it will take too long to do this. Smallest brier number has 27 digits. Even if you could eliminate 10^10 candidates per second then it would take ~10^17 sec. (Which is huge, more than the age of the planet earth).
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.

2007-05-11, 08:53   #6
VBCurtis

"Curtis"
Feb 2005
Riverside, CA

14FB16 Posts

Quote:
 Originally Posted by jasong 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.
Citrix is merely pointing out that you're missing the scale of this smallest-known Brier number. Say you rule out all but one in 10 trillion candidates... that leaves you with many trillions of k's to then test somehow. Your hope for disqualifying enough to make your idea worthwhile is what was done to establish the 27-digit Brier number as the suspected lowest (I think)... no form of exhaustive search can be completed in your lifetime, I believe.

-Curtis

2007-05-11, 22:28   #7
Citrix

Jun 2003

22×397 Posts

Quote:
 Originally Posted by jasong 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^n-1 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+or-1, 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 vice-versa. 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?

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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