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That's how it works (basically):
You take a small prime number (e.g. 2) and look at all the k,n-pairs left in your list if they are divided by this number. These get kicked out. Then you take the next prime number, 3. Then 5. And so on. Right now, we are at around 1000000000000000. Don't blame me if there are a few 0's too much or too less, I didn't count. Should be around 2^50.
In fact it's no that easy, but your question was not very precise either.
Sieving is very fast, but you never know which k,n-pair will be removed next.
So it can happen that you find a factor for a number that has already been tested, or that will be tested in ten years only. Nevertheless, it is very effective and without it the project would be pointless.
H.
