In a message dated 1/23/2009 9:12:19 P.M. Eastern Standard Time,
[email protected] 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
[email protected]
http://calcpage.tripod.com
Teacher & Professor
Applied Mathematics, Physics & Computer Science
Baldwin Senior High School & Nassau Community College
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