"Richard O'Keefe" <ok at cs.otago.ac.nz> writes: > factors n = [m | m <- [1..n], mod n m == 0] -- saves about 10% time, seems to give the same result: factors n = [m | m <- [1..n `div` 2], mod n m == 0]++[n] (But checking against primes is even faster, it seems) -k -- If I haven't seen further, it is by standing in the footprints of giants