Example 3 The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=1, a_{n+1}=a_{n}+$ $\frac{1}{\left[a_{n}\right]}$, where $\left[a_{n}\right]$ denotes the greatest integer not exceeding $a_{n}$. Find the general term formula of the sequence $\left\{a_{n}\right\}$.
This is a problem I devised in 2002. At first glance, it seems unrelated to grouped sequences, but if you write out the first few terms and arrange them in a right-angled triangular table, you will find it similar to Example 2.
$$
\begin{array}{l}
a_{1}=\frac{0}{1}+1 \\
a_{2}=\frac{0}{2}+2 \quad a_{3}=\frac{1}{2}+2 \\
a_{4}=\frac{0}{3}+3 \quad a_{5}=\frac{1}{3}+3 \quad a_{6}=\frac{2}{3}+3 \\
a_{7}=\frac{0}{4}+4 \quad a_{8}=\frac{1}{4}+4 \quad a_{9}=\frac{2}{4}+4 \quad a_{10}=\frac{3}{4}+4
\end{array}
$$

Using mathematical induction, it can be proven that the $k$-th row is
$$
\begin{array}{l}
a_{\frac{k(k-1)}{2}+1}=\frac{0}{k}+k, a_{\frac{k(k-1)}{2}+2}=\frac{1}{k}+k, \cdots, \\
a_{\frac{k(k-1)}{2}+k}=\frac{k-1}{k}+k .
\end{array}
$$

Suppose $a_{n}$ is in the $k$-th row. Using the method from Example 2, we can find $k=\left[\frac{1+\sqrt{8 n-7}}{2}\right]$. Therefore,
$$
a_{n}=\frac{n-\frac{k(k-1)}{2}-1}{k}+k=\frac{n-1}{k}+\frac{k+1}{2} \text {, }
$$

where $k=\left[\frac{1+\sqrt{8 n-7}}{2}\right]$.