In a message dated 1/23/2009 9:12:19 P.M. Eastern Standard Time,  
wst...@gmail.com writes:

That is  not the point of lucas-lehmer. The point is to determine
whether or not 2^p  - 1 is a prime number more quickly than ... say
checking for divisibility  by primes up to sqrt(2^p - 1).



OK, I stand corrected.  But what if 2**p-1 is huge, will this function  
really save significant processing time?
 
Also, I used this predicate method as follows:
 
for p in range(3,10000,2):
    if lucas_lehmer(p):
        print p, 2**p-1.
 
On this range of odds from 3 to 10000, p was always prime, and all prime  
numbers p gave 2**p-1 prime.  I thought that if 2**p-1 is prime, then p is  
prime, but not all prime values of p will give 2**p-1 prime.
 
HTH,
A.  Jorge Garcia
calcp...@aol.com
http://calcpage.tripod.com

Teacher  & Professor
Applied Mathematics, Physics & Computer  Science
Baldwin Senior High School & Nassau Community College



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