Erdős problem 291

More generally, if the leading digit of $n$ in base $p$ is $p-1$ then $p\mid (a_n,L_n)$. There is in fact a necessary and sufficient condition: a prime $p\leq n$ divides $(a_n,L_n)$ if and only if $p$ divides the numerator of $1+\cdots+\frac{1}{k}$, where $k$ is the leading digit of $n$ in base $p$. This can be seen by writing $a_n = \frac{L_n}{1}+\cdots+\frac{L_n}{n}$ and observing that the right-hand side is congruent to $1+\cdots+1/k$ modulo $p$. (The previous claim about $p-1$ follows immediately from Wolstenholme's theorem.)