> performs a straightforward trial division: It factors out the highest
> powers of 2 and 3 and then of all numbers <= sqrt(N) which are
> congruent to 1 or 5 mod 6. The point is that this takes
I notice that this description might be somewhat misleading (if N is
composite). Here's another try:
It splits off (_replacing_ N) the highest powers of 2 and 3 and then
successively of all numbers congruent to 1 or 5 mod 6 as long as their
square does not exceed the _current_ N (clearly this trial divisor is
necessarily a prime if it actually divides the current N).
Wolfgang
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wjenkner@inode.at