Example 6 Let the terms of the sequence $\left\{a_{n}\right\}$ be
$$
1,2.3,4,2,3,4,5,3.4,5,6, \cdots \text {. }
$$

Find $a_{n}$ and $S_{n}$, and calculate the values of $a_{80}$ and $S_{1921}$.
Given the sequence $1,2,3,4,5,6,7,8,9,10,11,12$, $\cdots$ has a general term of $n$, and the sequence $0,0,0,0,3,3,3,3,6,6$. $6,6, \cdots$ has a general term of $3\left[\frac{n-1}{4}\right]$. Subtracting the corresponding terms, we get the sequence $1,2,3,4,2,3,4,5,3,4,5,6 \cdots$ with the general term:
$$
\begin{array}{l}
n-3\left[\frac{n-1}{4}\right] . \\
\text { Therefore, } a_{n}=n-3\left[\frac{n-1}{4}\right], \\
a_{80}=80-3\left[\frac{80-1}{4}\right]=23 .
\end{array}
$$

To find the sum of the first $n$ terms, this problem can be viewed as:
Let the sequence $\left\{a_{n}\right\}$ have the first four terms as $1,2,3,4$, and $a_{n+4}$ $=a_{n}+1(n=1,2, \cdots)$, find $S_{n}$.
Thus, the solution can be as follows:
Let the undetermined constants $i, m \in \mathbf{N}, k$ be non-negative integers, $i \in$ $\{1,2,3,4\}, m$ be the segment order number, such that $n=4 m+i$, then
$$
\begin{array}{l}
S_{n}=10+14+18+\cdots+(4 m+6) \\
+(m+1) i+k \\
=2 m(m+4)+(m+1) i+k \\
=2 \cdot \frac{n-i}{4}\left(\frac{n-i}{4}+4\right)+\left(\frac{n-i}{4}+1\right) i+k \\
=\frac{n^{2}+16 n+i^{2}-8 i+8 k}{8} \\
=\left\{\begin{array}{l}
\frac{n^{2}+16 n-8}{8}-\frac{i^{2}+8 i-8 k-8}{8} \\
\frac{n^{2}+16 n}{8}-\frac{i^{2}+8 i-8 k}{8} .
\end{array},\right. \\
\because 0<\frac{i^{2}+8 i-8 k-8}{8}<1(i=1,2,3 ; k= \\
\end{array}
$$
$0,1,3)$,
$$
0=\frac{i^{2}+8 i-8 k}{8}<1(i=4, k=6),
$$
$\therefore$ The sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$ is
$$
\begin{array}{l}
S_{n}=\left\{\begin{array}{l}
{\left[\frac{n^{2}+16 n-8}{8}\right], \text { when } 4+n \text {, }} \\
{\left[\frac{n^{2}+16 n}{8}\right], \text { when } 41 n \text { . }}
\end{array} \quad(n \in \mathbf{N})\right. \\
\therefore S_{1921}=\left[\frac{1921^{2}+16 \cdot 1921-8}{8}\right](\because 411921) \\
=\left[\frac{3720969}{8}\right]=465121 \text {. } \\
\end{array}
$$

Note 1: To solve this problem, we cleverly set the undetermined constants $m$ and $i$, but the value of $m$ is set but not solved. The key to solving the problem is to first express $n$ using $m$ and $i$, and simultaneously express the sum $S_{n}$ using $m$, $i$, and $k$ (non-zero integers), then substitute $m=\frac{n-i}{4}$ back, and after calculation, decompose the right-hand side of $S_{n}$ into the difference of two terms $f(n)$ $-g(i, k)$, making it satisfy the inequality $0 \leqslant g(i$, $k)<1$, thus using $[x]$ to find $S_{n}=[f(n)](n \in$ N).

One of the most commonly used methods to find the general term formula of a sequence is this approach.