Let $a_{n}=A u^{3}+B n^{2}+C n+D$, from $a_{1}=1$, $a^{2}=2, a_{3}=4, a_{4}=p$ (any number) we get the system of equations:
$$
\left\{\begin{array}{c}
A+B+C+D=1 \\
8 A+4 B+2 C+D=2 \\
27 A+9 B+3 C+D=4 \\
64 A+16 B+4 C+D=\rho
\end{array}\right.
$$
Given the sequence $1,2,4,32-t, 135 \cdots 4 t, \cdots$. (where $t$ is any parameter)
Let $t=0$ in (5), we get
$$
a_{n}=n\left(n^{3}-\frac{35}{6} n^{2}+\frac{21}{2} n-\frac{14}{3}\right).
$$

The sequence is $1,2,4,32,135, \cdots$.
From this, we can see that for a sequence with a finite number of terms and no specific properties (such as arithmetic or geometric) explicitly stated, its general term formula is not unique. Therefore, the terms following these finite terms can be varied. More interestingly, by combining these general term formulas, many new general term formulas can be formed, for example, combining formulas (2), (4), and (6):
As long as $p+q+r=1$, the sequence
$$
a_{n}=p \times(2)+q \times(4)+r \times(6)
$$

has the first three terms as $1,2,4$, and the first five terms are:
$$
1,2,4,7+q+25 r, 11+4 q+124 r,
$$
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Readers can naturally think of various other transformation methods.
In middle school textbooks, a sequence is often given, and we are asked to “write a general term formula for the sequence.” Now we can easily understand the profound meaning of “a” in this context, and we can also easily explain why different students often write different general term formulas for the same given terms.

The general term formula for a sequence is not limited to integer functions and exponential functions of $n$; it can also be given by rational functions of $n$. For example, let $a_{n}=\frac{P}{n}+Q+R n$, and based on $a_{1}=1, a_{2}=2, a_{3}=4$, we can find
$$
\therefore a_{n}=\frac{3}{n}-4 \frac{1}{2}+\frac{5 n}{2}.
$$

This sequence is $1,2,4,6 \frac{1}{4}, 8 \frac{3}{5}, \cdots$, and its first three terms are also $1,2,4$.
$$
\begin{array}{l}
\text { Solving } A=\frac{1}{6}(p-9), \\
B=\frac{1}{2}(15-2 p). \\
C=\frac{1}{6}(11 p-80), D=8-p. \\
+\frac{1}{6}(11 p-80) n+(8-p) \\
\end{array}
$$

Thus, $a_{n}=\frac{1}{6}(p-7) n^{3}+\frac{1}{2}(15-2 p) n^{2}$.
From (3), we get the sequence $1, 2, 4, p, 4 p-17, \cdots$,
Let $p=8$ in (3), we get
$$
a_{n}=\frac{1}{6} n\left(n^{2}-3 n+8\right).
$$

Thus, the sequence is $1,2,4,8,15, \cdots$.
Here we can see that the first four terms of sequence (4) and (1) are the same, but the fifth term is different. If $p=\sqrt{3}$, then from (3) we get the sequence:
$$
1,2,4, \sqrt{3}, 4 \sqrt{3}-17, \cdots
$$

The terms after the fourth term are irrational.
At this point, the problem is satisfactorily solved, but careful readers may ask, can $a_{n}$ be set as a polynomial of higher degree to ensure that the first three terms are $1, 2, 4$? The answer is yes, for example, let $a_{n}=n^{4}+B n^{3}+C n^{2}+D n+t$, using a similar method we get.
$$
\begin{array}{l}
\therefore \quad a_{n}=n^{4}-\frac{1}{6}(t+35) n^{3}+(t \\
\left.+\frac{21}{2}\right) n^{2}-\frac{1}{6}(11 t+28) n+t,
\end{array}
$$