smc

For the fraction to be reducible, the greatest common factor of the numerator and the denominator must be greater than $1$. Using the Euclidean algorithm, we compute $$\begin{array}{rl}\gcd \left(5n+6,n-13\right)& =\gcd \left(5n+6-5(n-13),n-13\right)\\ & =\gcd \left(71,n-13\right).\end{array}$$ Since $71$ is prime, it follows that this GCD will be $1$ unless $(n-13)$ is a multiple of $71$, which first occurs when $n=71+13=84$, so the answer is $\boxed{\text{none of these}}$.