Example 9 Given a positive integer $c$, let the sequence $x_{1}, x_{2}, \cdots$ satisfy $x_{1}=c$ and $x_{n}=x_{n-1}+$ $\left[\frac{2 x_{n-1}-(n+2)}{n}\right]+1(n=2,3, \cdots)$, where $[x]$ denotes the greatest integer not exceeding $x$. Find the general term formula of $\left\{x_{n}\right\}$.
Solving the recurrence relation for $x_{n}$ can be transformed into $x_{n}-1=x_{n-1}-1+\left[\frac{2\left(x_{n-1}-1\right)}{n}\right]$. Let $y_{n}=x_{n}-1$, then
$$
y_{1}=c-1, y_{n}=y_{n-1}+\left[\frac{2 y_{n-1}}{n}\right](n=2,3, \cdots) .
$$

Direct calculation yields the following table:
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline \multicolumn{2}{|c|}{$n$} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & $\cdots$ \\
\hline \multirow{4}{*}{$y_{n}(c=3 k+1$ when $)$} & $3 k$ & $6 k$ & $10 k$ & $15 k$ & $21 k$ & $28 k$ & $36 k$ & $\cdots$ \\
\hline$(c=3 k+2$ when $)$ & $3 k+1$ & $6 k+2$ & $10 k+3$ & $15 k+4$ & $21 k+5$ & $28 k+6$ & $36 k+7$ & $\cdots$ \\
\hline & $(c=3 k+3$ when $)$ & $3 k+2$ & $6 k+4$ & $10 k+6$ & $15 k+9$ & $21 k+12$ & $28 k+16$ & $36 k+20$ & $\cdots$ \\
\hline
\end{tabular}

From this table, we conjecture the following formula holds:
$$
y_{n}=\frac{1}{2}(n+2)(n+1) k+\left\{\begin{array}{l}
0 \quad(c=3 k+1 \text { when }) \\
n \quad(c=3 k+2 \text { when }) \\
{\left[\frac{(n+2)^{2}}{4}\right](c=3 k+3 \text { when })}
\end{array} \quad\binom{k=0,1,2, \cdots}{n=1,2,3, \cdots} .\right.
$$

When $n=1$, it is known from the above that (2) holds. Assuming (2) holds for $n=m$, then for $n=m+1$,
if $c=3 k+1$, then by (1) we have
$$
\begin{aligned}
y_{m+1} & =y_{m}+\left[\frac{2}{m+1} y_{m}\right]=\frac{1}{2}(m+2)(m+1) k+\left[\frac{2}{m+1} \cdot \frac{1}{2}(m+2)(m+1) k\right] \\
& =\frac{1}{2}(m+2)(m+1) k+(m+2) k=\frac{1}{2}(m+3)(m+2) k,
\end{aligned}
$$

if $c=3 k+2$, then by (1) we have
$$
\begin{aligned}
y_{m+1} & =y_{m}+\left[\frac{2}{m+1} y_{m}\right] \\
& =\frac{1}{2}(m+2)(m+1) k+m+\left[\frac{2}{m+1}\left(\frac{1}{2}(m+2)(m+1) k+m\right)\right] \\
& =\frac{1}{2}(m+2)(m+1) k+m+(m+2) k+1 \\
& =\frac{1}{2}(m+3)(m+2) k+(m+1),
\end{aligned}
$$

if $c=3 k+3$, then by (1) we have
$$
\begin{aligned}
y_{m+1} & =y_{m}+\left[\frac{2}{m+1} y_{m}\right] \\
& =\frac{1}{2}(m+2)(m+1) k+\left[\frac{(m+2)^{2}}{4}\right]+\left[\frac{2}{m+1}\left(\frac{1}{2}(m+2)(m+1) k+\left[\frac{(m+2)^{2}}{4}\right]\right)\right] \\
& =\frac{1}{2}(m+2)(m+1) k+\left[\frac{(m+2)^{2}}{4}\right]+(m+2) k+\left[\frac{2}{m+1}\left[\frac{(m+2)^{2}}{4}\right]\right] \\
& =\frac{1}{2}(m+3)(m+2) k+\left[\frac{(m+2)^{2}}{4}\right]+\left[\frac{2}{m+1}\left[\frac{(m+2)^{2}}{4}\right]\right] .
\end{aligned}
$$

And when $m=2 t$ is even, we have
$$
\begin{aligned}
& {\left[\left(\frac{(m+2)^{2}}{4}\right)\right]+\left[\frac{2}{m+1}\left[\frac{(m+2)^{2}}{4}\right]\right]=(t+1)^{2}+\left[\frac{2(t+1)^{2}}{2 t+1}\right] } \\
= & (t+1)^{2}+\left[t+1+\frac{t+1}{2 t+1}\right]=(t+1)^{2}+(t+1) \\
= & (t+1)(t+2)=\frac{(m+2)(m+4)}{4}=\frac{(m+3)^{2}-1}{4}=\left[\frac{(m+3)^{2}}{4}\right],
\end{aligned}
$$

When $m=2 t+1$ is odd, we have
$$
\begin{array}{l}
{\left[\frac{(m+2)^{2}}{4}\right]+\left[\frac{2}{m+1}\left[\frac{(m+2)^{2}}{4}\right]\right]=(t+2)(t+1)+\left[\frac{1}{t+1}(t+2)(t+1)\right]=} \\
(t+2)(t+1)+(t+2)=(t+2)^{2}=\frac{(m+3)^{2}}{4}=\left[\frac{(m+3)^{2}}{4}\right] .
\end{array}
$$

Therefore, we always have $y_{m+1}=\frac{1}{2}(m+3)(m+2) k+\left[\frac{(m+3)^{2}}{4}\right]$. This proves that (2) holds for $n=m+1$, thus (2) is proven. From this, we can derive the general term formula for the sequence $\left\{x_{n}\right\}$ as
$$
\begin{aligned}
x_{n} & =y_{n}+1 \\
& =\frac{1}{2}(n+2)(n+1) k+1+\left\{\begin{array}{ll}
0 \quad(c=3 k+1 \text { when }) \\
n \quad(c=3 k+2 \text { when }) \\
{\left[\frac{(n+2)^{2}}{4}\right] \quad(c=3 k+3 \text { when })}
\end{array} \quad\binom{k=0,1,2, \cdots}{n=1,2,3, \cdots} .\right.
\end{aligned}
$$