Actually that helps a lot. I was writing a basic Sieve of Eratosthenes to generate prime numbers, and was using that equation to estimate an upper bound on the

Actually that helps a lot. I was writing a basic Sieve of Eratosthenes to generate prime numbers, and was using that equation to estimate an upper bound on the size of the sieve array if looking for at least k primes. Or more simply, if I wanted at least k primes, x would be a loose upper bound on the numbers I would have to look at. So in short, approximation is good enough. I wrote the following to implement it: def approximate(k, times) return k if times < 1 return k * Math.log(approximate(k, times - 1)) end For values of k I am looking at (1E4 up to 1E9), recursing 10 times approximated to within 0.01 of the actual value, which is definitely good enough for what I need. Thanks everyone for your replies. I'd still be interested in an exact answer, but at least this solves today's problem.