20051229, 16:59  #12  
"Robert Gerbicz"
Oct 2005
Hungary
2·3^{6} Posts 
Quote:
Let n=2^247636 then n*2^5092031=2^7568391 is the 32. Mersenne prime. 

20051229, 18:20  #13  
Jun 2003
2×3^{3}×7×13 Posts 
Quote:


20051230, 13:02  #14  
"Jason Goatcher"
Mar 2005
3×7×167 Posts 
Quote:
Quote:
Either way, I'm having a lot of fun with this, so until I talk to B2(the guy who started me on this and should know the answer, since he's the head honcho of Riesel Sieve) I'm going to keep on trucking. I figure either axn1 is right or B2 had a mental moment and gave me the wrong equation. Last fiddled with by jasong on 20051230 at 13:03 

20051230, 13:08  #15 
Jul 2005
2×193 Posts 
A quick program I knocked up shows that sieving is pretty efficient.
To rule out a specific value of n you need to find a factor, i.e. x=2^509203 n*x  1 = 0 (mod p) For any odd prime p. (Since x is even, n*x will be even so n*x1 will always be odd, so 2 can never be a factor). Rearranging this: n*x = 1 (mod p) n = x^1 (mod p) So, for each p do the following: Calculate the modular inverse of x (mod p). The resulting number is the value of n such that p is a factor of n*x  1. You can then rule out this value of n, plus any additional multiple of p. i.e. n, n+p, n+2p, n+3p, etc. Yves Gallot's proth.exe can do this as well (faster than my program not surprisingly)... After sieving to an appropriate depth you can then PRP the remaining candidates. 
20051230, 16:04  #16  
Jan 2005
Caught in a sieve
5·79 Posts 
Quote:
Last fiddled with by Ken_g6 on 20051230 at 16:05 

20051230, 16:26  #17 
Jan 2005
Caught in a sieve
5·79 Posts 
I found a counterexample:
3*2^7020381 is prime (from the 321 project). 702038*2^31 = 5616303 = 3*7*11*41*593 
20051230, 18:12  #18  
Jun 2003
2×3^{3}×7×13 Posts 
Ooh! Too many posts to reply to
@Greenbank: You have just discovered NewPGen's main sieiving algorithm. The "k*b^n1" sieve of NewPGen will do just that. And it's bloody fast! @Ken_g6: You'll see some posts up where jasong said: Quote:
Now, the implied part of n & k both being odd! That's a tricky one. Lets say that we allow n to be even, and that n*2^k1 is a counter example (in R. Gerbicz's case, n=2^247636.) Basically that means that, our conjucturer, in order to still keep the conjucture going _could_ claim that the actual expression should be (n/2)* 2^(k+1)  1 and then try to find a prime of the form (k+1) *2^x  1. Long story short, in order to avoid any confusion and conclusively disprove the conjucture, I assume that both n & k should be odd! (Phew!) @jasong: No sites. Just what you see in the posts. But finding an odd n such that n*2^5092031 should put this conjucture for rest once and for all. Last fiddled with by axn on 20051230 at 18:15 

20051230, 20:23  #19  
"Robert Gerbicz"
Oct 2005
Hungary
2·3^{6} Posts 
Quote:
But it isn't hard to prove the conjecture: Let k is a positive integer, a=2^(k+1) and b=2^k1 then a and b are relative primes. So by Dirichlet's theorem ( see: http://mathworld.wolfram.com/DirichletsTheorem.html ) the arithmetic progression a*m+b contains an infinite number of primes, here a*m+b=2^(k+1)*m+2^k1=(2*m+1)*2^k1=n*2^k1 where n=2*m+1 is odd! So your conjecture is true. 

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