Consider a positive integer $k$, such that $\max(p_1,p_2) \le k \cdot 2021!+1$. One can prove that $p_i \le k \cdot 2021!+1$ for all $i$ by induction, hence the conclusion. If $p_{n-1}=2$ or $p_{n-2}=2$, then $p_n \le (k \cdot 2021!+1)+2+2018$. If $p_{n-1}>2$ and $p_{n-2}>2$, since $p_{n-1}+p_{n-2}+2018$ is even, $p_n \le \frac{p_{n-1}+p_{n-2}+2018}{2} \le (k \cdot 2021!+1)+1009$. In both case, $p_n \le (k \cdot 2021!+1)+2020$. However, $(k \cdot 2021!+1)+1,\cdots, (k \cdot 2021!+1)+2020$ are all composite, hence $p_n \le k \cdot 2021!+1$.

Nordic 2018

Poland 2001

$$b_n=max(p_n,p_{n+1})$$Claim:$b_{n+1}\le b_n+2020.$(easy) exist $k$ $b_1\le 2021!k+1$ no prime number ${2021!k+2,2021!k+3,...,2021!k+2020.}$