Reverse factorial

Question

Well, we all know that if N is given it\'s easy to calculate N!. But what about the inverse?

N! is given and you are about to find N - Is that possible ? I\'m curio

Answer 1

Yes. Let's call your input x. For small values of x, you can just try all values of n and see if n! = x. For larger x, you can binary-search over n to find the right n (if one exists). Note hat we have n! ≈ e^(n ln n - n) (this is Stirling's approximation), so you know approximately where to look.

The problem of course, is that very few numbers are factorials; so your question makes sense for only a small set of inputs. If your input is small (e.g. fits in a 32-bit or 64-bit integer) a lookup table would be the best solution.

(You could of course consider the more general problem of inverting the Gamma function. Again, binary search would probably be the best way, rather than something analytic. I'd be glad to be shown wrong here.)

Edit: Actually, in the case where you don't know for sure that x is a factorial number, you may not gain all that much (or anything) with binary search using Stirling's approximation or the Gamma function, over simple solutions. The inverse factorial grows slower than logarithmic (this is because the factorial is superexponential), and you have to do arbitrary-precision arithmetic to find factorials and multiply those numbers anyway.

For instance, see Draco Ater's answer for an idea that (when extended to arbitrary-precision arithmetic) will work for all x. Even simpler, and probably even faster because multiplication is faster than division, is Dav's answer which is the most natural algorithm... this problem is another triumph of simplicity, it appears. :-)