13. Use $n$ and $a=[\sqrt{n}]$ to represent $\sum_{k=1}^{n}[\sqrt{k}]$.
13. Obviously, $[\sqrt{k}]=\sum_{j=1}^{a}\left[j^{2} \leqslant k\right]$, (if $a \geqslant[\sqrt{k}]$), thus $\sum_{k=1}^{n}[\sqrt{k}]=\sum_{k=1}^{n} \sum_{j=1}^{a}\left[j^{2} \leqslant k\right]=\sum_{j=1}^{a} \sum_{k=1}^{n}\left[j^{2} \leqslant k\right]=\sum_{j=1}^{a}\left(n+1-j^{2}\right)=(n+1) a-\frac{a(a+1)(2 a+1)}{6}$.