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2007-05-10, 01:13   #4
Citrix

Jun 2003

5·317 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).