33.20 Given a positive integer $c$, let the sequence $x_{1}, x_{2}, \cdots$ satisfy: $x_{1}=c$, and
$$
x_{n}=x_{n-1}+\left[\frac{2 x_{n-1}-(n+2)}{n}\right]+1, n=2,3, \cdots \text {, }
$$

where $[x]$ denotes the greatest integer not exceeding $x$. Find the general term formula of the sequence $\left\{x_{n}\right\}$.
Obviously, 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 $a_{1}=c-1$,
$$
a_{n}=a_{n-1}+\left[\frac{2 a_{n-1}}{n}\right]=\left[\frac{n+2}{n} a_{n 1}\right](n=2,3, \cdots) .
$$

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,
$$
\left[\frac{n+2}{n} u_{n-1}\right]=\left[A \frac{(n+2)}{n} n(n+1)\right]=A \frac{(n+1)(n+2)}{2}=u_{n},
$$

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

thus $\left\{y_{n}\right\}$ also satisfies (1).
Let $\approx_{n}=\left[\frac{(n+2)^{2}}{4}\right], n=1,2, \cdots$. When $n=2 m$ and $m \geqslant 1$,
$$
\left[\frac{n+2}{n} z_{n-1}\right]=\left[\frac{m+1}{m}\left[\frac{(2 m+1)^{2}}{4}\right]\right]=\left[\frac{m+1}{m} m(m+1)\right]=(m+1)^{2}=z_{n} \text {. }
$$

When $n=2 m+1(m \geqslant 1)$,
$$
\left[\frac{n+2}{n} z_{n-1}\right]=\left[\frac{2 m+3}{2 m+1}\left[\frac{(2 m+2)^{2}}{4}\right]\right]=\left[\frac{2 m+3}{2 m+1}(m+1)^{2}\right]-[(m+1)
$$
$\left.(m+2)+\frac{m+1}{2 m+1}\right]=(m+1)(m+2)=\left[\frac{(2 m+3)^{2}}{4}\right]=z_{n}$. Thus $\left\{z_{n}\right\}$ also satisfies (1).
For any non-negative integer $A$, let $v_{n}=u_{n}+y_{n}=A \frac{(n+1)(n+2)}{2}+n, w_{n}=u_{n}+z_{n}=$ $A \frac{(n+1)(n+2)}{2}+\left[\frac{(n+2)^{2}}{4}\right], n=1,2, \cdots$, obviously $\left\{v_{n}\right\}$ 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 $3 \mid a_{1}$, $a_{n}=\frac{a_{1}}{6}(n+$ 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, when $c \equiv 1(\bmod 3)$, $x_{n}=\frac{c-1}{6}(n+1)(n+2)+1$; when $c \equiv 2(\bmod 3)$, $x_{n}=\frac{c-2}{6}(n+1)(n+2)+n+1$; when $c \equiv 0(\bmod 3)$, $x_{n}=\frac{c-3}{6}(n+1)(n+$
2) $+\left[\frac{(n+2)^{2}}{4}\right]+1$.