1999 Putnam A6

The sequence $(a_n)_{n\geq 1}$ is defined by $a_1=1, a_2=2, a_3=24,$ and, for $n\geq 4$,

$$a_n = \frac{6a_{n-1}^2a_{n-3} -

8a_{n-1}a_{n-2}^2}{a_{n-2}a_{n-3}}.$$

Show that, for all n, $a_n$ is an integer multiple of $n$.

序列 $(a_n)_{n\geq 1}$ 由 $a_1=1, a_2=2, a_3=24,$ 定义，对于 $n\geq 4$，

$$a_n = \frac{6a_{n-1}^2a_{n-3} -

8a_{n-1}a_{n-2}^2}{a_{n-2}a_{n-3}}.$$

证明，对于所有 n，$a_n$ 是 $n$ 的整数倍。