Suppose that $a_l = 2l$, $l \ge k$. Let $p$ be the least prime divisor of $k-1$. Then $(l-1, i) = \begin{cases} 1, & 1 \le i < p, \\ p, & i = p, \end{cases}$ and thus
$$(2l + i - 2, l + i - 1) = \begin{cases} 1, & 1 \le i < p, \\ p, & i = p. \end{cases}$$
From (1) we know that
$$a_{l+i-1} = \begin{cases} 2l + i - 1, & 1 \le i < p, \\ 2l + 2p - 2, & i = p. \end{cases}$$
Then $a_{l+p-1} - a_{l+p-2} = (2l+2p-2) - (2l+p-2) = p$ is a prime number, $a_{l+p-1} = 2(l+p-1)$. From the discussion above, we know that there are infinitely many $l \ge k$, such that $a_l = 2l$ and $a_{l+p-1} - a_{l+p-2} = p$ is the least prime divisor of $l-1$.