Thread: Riesel primes
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Old 2009-07-18, 15:23   #5
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Feb 2005
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Originally Posted by Mini-Geek View Post
I was mistaken about every number being represented by some k*2^n-1. Every odd number can be represented by k*2^n-1 if 2^n can be less than k.

In case you're wondering, there is an identical requirement for Proth (k*2^n+1) numbers.
Now that makes sense. And of course, with the exception of two, being odd is a necessity for any number x being a Riesel prime.

However, you can still get an odd number by having n < k. A simple example:

3*2^2 -1 = 11 = ! prime.
15*2^3 - 1 = 119 = ! prime

Any number of k*2^n -1 will be odd, given k is odd. For n > k a prime can turn up. The first is k =3 and n = 6 which yields the following
3*2^6 - 1 = 191 = ! prime.

Finally, n = k is also a possibility. 3*2^3 -1 = 23 = ! prime.

Apparently, all three scenarios are possible?
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