Just curious: Is there a closed formula L(n) for the total number of Latin squares on n symbols? (If not, how 'bout an asymptotic expression?) (Even better: Assume we equate two Latin squares if they differ only by 1) a dihedral transform 2) a permutation of the n symbols 3) two cyclic permutations, applied resp. to the rows and columns. These should lower the count by a factor of roughly 8 n^2 n!. Roughly, since a 3) might yield the same result as a 2). Making this exact seems hard.) --Dan