  Let $\{\cdot\}$ denote the fractional part and $n \geq 1$ be a fixed integer.
In this short note, we show for any prime $p$ the one-to-one correspondence
$$\sum_{\nu \geq 1} \left\{\frac{n}{p^\nu}\right\} > 1 \quad \iff \quad p \mid
\mathrm{denom}( B_n(x) - B_n ),$$ where $B_n(x) - B_n$ is the $n$th Bernoulli
polynomial without constant term and $\mathrm{denom}(\cdot)$ is its
denominator, which is squarefree.
