Quote:
Originally Posted by MiniGeek
I was mistaken about every number being represented by some k*2^n1. Every odd number can be represented by k*2^n1 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.
http://en.wikipedia.org/wiki/Proth_number

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?