The link provided by JavaMan indeed answers this question: How to reverse the $n$ choose $k$ formula?
A summary of the algorithm -- took me while to figure out from that answer, so I'm writing it as an algorithm instead of an explanation of one. If you want to know how this works, I highly recommend visiting that answer and upvoting it, it's very informative -- is as follows. First, the problem is minimizing n
given X
such that:
${n \choose k} = X$
First we determine max(k)
given that we know that: $X > \frac{4^k}{2k+1}$
From there we iterate over 1 to max(k)
. At each step of the iteration, n
, if it exists is contained in the interval:
$\sqrt[k]{k! X} + k \ge n \ge \sqrt[k]{k! X}$
We iterate over that interval and check if ${n \choose k}=X$. We can then easily find the minimum such n
.