Thread: Riesel primes
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Old 2009-07-18, 12:34   #4
Mini-Geek
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"Tim Sorbera"
Aug 2006
San Antonio, TX USA

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Quote:
Originally Posted by Primeinator View Post
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^n-1. Every odd number can be represented by k*2^n-1 if 2^n can be less than k. Let's choose 15648132147819545 for example.
15648132147819545=k*2^n-1
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*2^1-1

In case you're wondering, there is an identical requirement for Proth (k*2^n+1) numbers.
http://en.wikipedia.org/wiki/Proth_number
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