Is there a simple formula for $\binom{2n}{n} \pmod{n^3}$?

Question

Is there a simple formula for the following? $$f(n) = \binom{2n}{n} \pmod{n^3}$$

I know $f(n) = 2$ iff $n$ is prime and greater than $3$, but I don't know anything about composite numbers.

Answer 1

Just a comment.

Numbers appear to be in this form:

\begin{aligned} &{2 n \choose n} \mod n^3 \equiv {2 k \choose k} + \frac{t n^3}{k^3} \\\\ & \text{for }kp \quad p \in \mathbb{P} \text{ prime}, 1\geq t > k^3, k \in \mathbb{N}, k > 1 \end{aligned}

For $k=1, {2p \choose p} \equiv 2 \bmod p^3$

which, as stated in the comments, is a reformulation of Wolstenholme Theorem.