Thread: Riesel primes
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Old 2009-07-18, 02:54   #2
Mini-Geek
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"Tim Sorbera"
Aug 2006
San Antonio, TX USA

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I'm pretty sure it's only that n is such that 2^n>k (otherwise every number would be a Riesel number, and every prime a Riesel prime).

Last fiddled with by Mini-Geek on 2009-07-18 at 02:55
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