Quote:
Originally Posted by Primeinator
How would 2^n < k in of itself make a number riesel? I thought that any number of the form k*2^n 1 is a riesel number.

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. Let's choose 15648132147819545 for example.
15648132147819545=k*2^n1
15648132147819546=k*2^n
15648132147819546 has only one factor of 2, so we set n to 1
2*7824066073909773=k*2
7824066073909773=k
So 15648132147819545=7824066073909773*2^11
In case you're wondering, there is an identical requirement for Proth (k*2^n
+1) numbers.
http://en.wikipedia.org/wiki/Proth_number