Quote:
Originally Posted by axn1
Can you clarify this? when you say "the other form(n*2^k1) also has a prime", do you mean n*2^k1 _is_ prime or n*2^<some k> 1 is prime?
At any rate, I highly doubt it if the conjucture is true! All you have to do is find a prime of the form x*2^5092031 for some x. By the conjucture, you will expect to have a prime 509203*2^x1. Unfortunately, since 509203 is a Riesel number, there won't be any such primes.
Am I getting close?

I may have been wrong when I said it could go both ways. If I understood B2 correctly, if n*k^n1 can be found prime for a certain k, supposedly there's an n(possibly a different one) that makes k*2^n1 prime for that same k.
I'm not saying it's true, I'm just going by what B2(the runner of Riesel Sieve) told me, and he indicated it was just a conjecture.