Re: [math-fun] Inverse factorial?

(For my own peace of mind, I suppose I'll have to figure out how that relates to the off-the-cuff inverse factorial heuristic...)

1: x! = N/e (yours) 2: x log x = log N (from 1, ignoring a lower-order term) 3: log x + log log x = log log N (log of 2) 4: x (log log N - log log x) = log N (substituting 3 into 2) 5: x log log N = log N (discarding a lower-order term) 6: x = log N / log log N (rearranging) -- g