It is well known that if $p$ is prime, Stirling numbers of the first and second kind, $s_1(p,k)$ and $s_2(p,k)$, are divisible by $p$ if $1\le k\le p-1$ (Lagrange ; easiest is working in $\mathbb F_p$ with the factorization of $X^p-X$). But is the converse true, i.e. is $n$ prime if all the $s_1(n,k)$ (or $s_2(n,k)$) are divisible by $n$ for $1\le k\le n-1$? I checked it for $n<1000$ (hard to do much more with simple programs) ; the similar well-known result for binomial coefficients is not hard to prove using the explicit formula, but I have no idea for Stirling numbers.

It is well known that if $p$ is prime, Stirling numbers of the first and second kind, $s_1(p,k)$ and $s_2(p,k)$, are divisible by $p$ if $1<k\le p-1$ (Lagrange ; easiest is working in $\mathbb F_p$ with the factorization of $X^p-X$). But is the converse true, i.e. is $n$ prime if all the $s_1(n,k)$ (or $s_2(n,k)$) are divisible by $n$ for $1<k\le n-1$? I checked it for $n<1000$ (hard to do much more with simple programs) ; the similar well-known result for binomial coefficients is not hard to prove using the explicit formula, but I have no idea for Stirling numbers.