Desperately seeking primes II.

So this new algorithm can figure out if a certain number is a prime in O(log(n)^6) as opposed to O(log(n)^12). So that would mean O(log(10^(10^7))^6) = O(10^(6*10^7))= a constant multiplied by a number with 6 million digits of operations instead of a constant multiplied by a number with 12 million digits of operations in order to find a prime that is worth it. Yes, that is quite an improvement.