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Old 2005-12-29, 09:30   #5
jasong
 
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"Jason Goatcher"
Mar 2005

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Quote:
Originally Posted by axn1




Can you clarify this? when you say "the other form(n*2^k-1) also has a prime", do you mean n*2^k-1 _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*2^509203-1 for some x. By the conjucture, you will expect to have a prime 509203*2^x-1. 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^n-1 can be found prime for a certain k, supposedly there's an n(possibly a different one) that makes k*2^n-1 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.

Last fiddled with by jasong on 2005-12-29 at 09:31
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