jmc

The expression $\frac{n+1}{13-n}$ is negative if $n$ is less than $-1$ or greater than $13$, so $n$ must be between 0 and $12$ inclusive. Also, since $\frac{n+1}{13-n}$ is prime, it must be greater than or equal to 2. Solving $$\begin{array}{rl}\frac{n+1}{13-n}& \ge 2\\ n+1& \ge 26-2n\\ 3n& \ge 25\\ n& \ge 8\frac{1}{3},\end{array}$$ we only have to check 9, 10, 11, and 12. (The multiplication by $13-n$ is justified since we have already know that $n$ is less than 13). We find that when $n=\boxed{12},$ the expression equals 13.