Although what I propose may not be the "classic" sieve algorithm, in truth you only need to test up to the primes which are less than the square root of the target number. In other words, you need not divide by any primes greater than 4 when testing 16. Based on the associative property of multiplication, if you can show that no numbers <= sqrt(n) evenly divide n, then you have also shown that there are no numbers => sqrt(n) which evenly divide n.
While this may add a few chars to your code, it reduces the compute time enormously when you get to large n's.
Update: Actually, it doesn't look like ANY of us are implenting the true Sieve algorithm. The algorithm can be found here: http://www.math.utah.edu/~alfeld/Eratosthenes.html. It does in fact use the sqrt(n) modification, but doesn't do dividetesting ... instead it does "weeding out" of multiples. Same result, different technique.
Update 2: my mistake ... Ase is indeed doing the correct Sieve algorithm by using his bit vector technique.
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