, 2025

First, we will prove by induction that $a_n = (a_1 - 1)n! + n$ for all $n \ge 1$. The base case $a_1$ is trivial. Now suppose that the closed form holds for some $a_n$. Then
$$\begin{align*} a_{n+1} &= (n + 1) (((a_1 - 1)n! + n) - n + 1) \\ a_{n+1} &= (n + 1) ((a_1 - 1)n! + 1) \\ a_{n+1} &= (a_1 - 1)(n + 1)! + (n + 1) \end{align*}$$
Hence, the proof by induction is complete. For the sake of clarity, let $c = a_1 - 1$.
Let $p$ be a prime number dividing $\gcd(a_n, a_{n+1})$ for $n \ge 2$. Assume $p \le n$. Then
$cn!+n \equiv n \pmod p$ however, we have $c(n+1)!+n+1 \equiv n+1 \pmod p$ thus we conclude
$p > n$. We have $a_{n+1} = (n+1)(cn!+1)$ so then $p \mid (n+1)(cn!+1)$. Assume $p$
divides $cn!+1$. Then $cn!+n \equiv cn!+1 \pmod p$, which implies $n \equiv 1 \pmod p$, which is a
contradiction since $p > n$. So either $\gcd(a_n, a_{n+1}) = 1$ or $p = n+1 = \gcd(a_n, a_{n+1})$.
Now suppose $p = n + 1$, notice that in this case $p \ge 3$. Then, according to
Wilson's theorem $p \mid n! + 1$ so we have $p \mid n! + 1 \mid cn! + c$, which implies $p $\mid$
cn! + c + n - n $\implies p \mid$ c - n $\implies p \mid$ c + 1$, which is however, equal to$a_1$. The chain of
thoughts is reversible so we obtain $\gcd(a_{p-1}, a_p) = p$ if and only if $p \mid a_1$. Therefore,
the answer is that $k$ is the biggest odd prime divisor of $a_1$ or 1 if $a_1$ is a power of 2.