Chinese Mathematical Olympiad

Suppose that $a_l=2l$, $l\ge k$. Let $p$ be the least prime divisor of $k-1$. Then $(l-1,i)=$ and thus $$(2l+i-2,l+i-1)=\{\begin{array}{ll}1,& 1\le i<p,\\ p,& i=p.\end{array}$$ From (1) we know that $$a_{l+i-1}=\{\begin{array}{ll}2l+i-1,& 1\le i<p,\\ 2l+2p-2,& i=p.\end{array}$$ 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 $l-1$.