(2) 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 the sequence $\left\{x_{n}\right\}$.
Solution: Clearly, when $n \geqslant 2$,
$$
x_{n}=x_{n-1}+\left[\frac{2\left(x_{n-1}-1\right)}{n}\right]
$$

Let $a_{n}=x_{n}-1$, then
$$
\begin{array}{l} 
a_{1}=c-1 \\
a_{n}=a_{n-1}+\left[\frac{2 a_{n-1}}{n}\right]=\left[\frac{n+2}{n} a_{n-1}\right], n=2,3, \cdots
\end{array}
$$

Let $u_{n}=A \frac{(n+1)(n+2)}{2}, n=1,2, \cdots$, where $A$ is a non-negative integer. Since when $n \geqslant 2$,
$$
\begin{aligned}
{\left[\frac{n+2}{n} u_{n-1}\right]=} & {\left[A \frac{(n+2)}{2 n} n(n+1)\right]=} \\
& A \frac{(n+1)(n+2)}{2}=u_{n}
\end{aligned}
$$

Thus, the sequence $\left\{u_{n}\right\}$ satisfies (1).
Let $y_{n}=n, n=1,2, \cdots$, then when $n \geqslant 2$,
$$
\begin{aligned}
{\left[\frac{n+2}{n} y_{n-1}\right]=} & {\left[\frac{(n+2)(n-1)}{n}\right]=} \\
& {\left[n+1-\frac{2}{n}\right]=n=y_{n} }
\end{aligned}
$$

Thus, $\left\{y_{n}\right\}$ also satisfies (1).
Let $z_{n}=\left[\frac{(n+2)^{2}}{4}\right], n=1,2, \cdots$. When $n=2 m$ and $m \geqslant 1$,
$$
\begin{aligned}
{\left[\frac{n+2}{\dot{n}} z_{n-1}\right]=} & {\left[\frac{m+1}{m}\left[\frac{(2 m+1)^{2}}{4}\right]\right]=\left[\frac{m+1}{m} m(m+1)\right]=. } \\
& (m+1)^{2}=z_{n}
\end{aligned}
$$

When $n=2 m+1(m \geqslant 1)$,
$$
\begin{aligned}
{\left[\frac{n+2}{n} z_{n-1}\right]=} & {\left[\frac{2 m+3}{2 m+1}\left[\frac{(2 m+2)^{2}}{4}\right]\right]=} \\
& {\left[\frac{2 m+3}{2 m+1}(m+1)^{2}\right]=} \\
& {\left[(m+1)(m+2)+\frac{m+1}{2 m+1}\right]=} \\
& (m+1)(m+2)=\left[\frac{(2 m+3)^{2}}{4}\right]=z_{n}
\end{aligned}
$$

Thus, $\left\{z_{n}\right\}$ also satisfies (1).
For any non-negative integer $A$, let
$$
\begin{array}{c}
v_{n}=u_{n}+y_{n}=A \cdot \frac{(n+1)(n+2)}{2}+n \\
w_{n}=u_{n}+z_{n}=A \cdot \frac{(n+1)(n+2)}{2}+\left[\frac{(n+2)^{2}}{4}\right], n=1,2, \cdots
\end{array}
$$

Clearly, $\left\{v_{n}\right\}$ and $\left\{w_{n}\right\}$ both satisfy (1).
Since $u_{1}=3 A, y_{1}=1, z_{1}=\left[\frac{9}{4}\right]=2$, when 3 divides $a_{1}$,
$$
a_{n}=\frac{a_{1}}{6}(n+1)(n+2)
$$

When $a_{1} \equiv 1(\bmod 3)$,
$$
a_{n}=\frac{a_{1}-1}{6}(n+1)(n+2)+n
$$

When $a_{1} \equiv 2(\bmod 3)$,
$$
a_{n}=\frac{a_{1}-2}{6}(n+1)(n+2)+\left[\frac{(n+2)^{2}}{4}\right]
$$

In summary,
$$
\begin{array}{c}
x_{n}=\frac{c-1}{6}(n+1)(n+2)+1, c \equiv 1(\bmod 3) \\
x_{n}=\frac{c-2}{6}(n+1)(n+2)+n+1, c \equiv 2(\bmod 3) \\
x_{n}=\frac{c-3}{6}(n+1)(n+2)+\left[\frac{(n+2)^{2}}{4}\right]+1, c \equiv 0(\bmod 3)
\end{array}
$$