So, I *think* most of our hypothetical problems go away with a simple adjustment to f(n):f(n) = floor(log_M((n-1)/B))
Correct. And nice. :-) Equivalently, f(n) = ceil(log_M (n / B)). If f(n) = c, it means B*(M^(c-1)) < n <= B*(M^(c)). So f(n) = 0 means n <= B. --------------------------------------------------------------------- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
