Question 199
Let $a_1, a_2, a_3, a_4, \dots$ be a sequence of non-negative integers such that
$$
a_{n+2} = \binom{a_{n+1}}{a_n},
$$
where
$$
\binom{m}{k} = \begin{cases} \frac{m!}{k!(m-k)!} & \text{if } m \ge k, \\ 0 & \text{otherwise.} \end{cases}
$$
Prove that such a sequence tends to infinity if and only if $2 \le a_1 \le a_2 - 2$.
$$
a_{n+2} = \binom{a_{n+1}}{a_n},
$$
where
$$
\binom{m}{k} = \begin{cases} \frac{m!}{k!(m-k)!} & \text{if } m \ge k, \\ 0 & \text{otherwise.} \end{cases}
$$
Prove that such a sequence tends to infinity if and only if $2 \le a_1 \le a_2 - 2$.