Problem 1879

Given distinct positive integers $a_1, a_2, \ldots, a_{2020}$. For $n \ge 2021$, $a_n$ is defined as the smallest number different from $a_1, a_2, \ldots, a_{n-1}$ which does not divide $a_{n-2020} \cdots a_{n-2} a_{n-1}$. We aim to prove that every sufficiently large number appears in the sequence.

### Proof:

Claim: For sufficiently large $n$, the least common multiple (LCM) of a set $S$ of $n$ natural numbers satisfies $\text{lcm}(S) > n^{4040}$.

This claim is intuitive and can be shown through detailed analysis, which is omitted here for brevity.

Claim: The ratio $\frac{a_n}{n}$ is bounded by a constant.

Proof:
Assume, for contradiction, that $\frac{a_n}{n}$ is not bounded. Define a number $t$ as "good" if $\frac{a_t}{t} > \frac{a_s}{s}$ for all $s < t$. Pick a sufficiently large good $t$ such that $\frac{a_t}{t} = r \ge 2$.

Let $T = \{1, 2, \ldots, a_t\} \setminus \{a_1, a_2, \ldots, a_t\}$. Since $|T| \ge rt - t$, by our earlier claim, $\text{lcm}(T) > ((r-1)t)^{4040} > (rt)^{2020} > a_{t-2020} a_{t-2019} \cdots a_{t-1}$. Since $\text{lcm}(T) \nmid a_{t-2020} a_{t-2019} \cdots a_{t-1}$, there exists an element in $T$ smaller than $a_t$ that does not divide $a_{t-2020} a_{t-2019} \cdots a_{t-1}$. This contradicts the definition of $a_t$, as that element would have been chosen instead of $a_t$. Therefore, $\frac{a_n}{n}$ must be bounded by some constant $c$.

Now, assume for contradiction that there exists a sufficiently large number $k \gg c$ that does not appear in the sequence. Let $k$ have a sufficiently large prime power factor, say $p^\alpha \gg c$. For all $t > k$, since $k \neq a_t$, $p^\alpha \mid k \mid a_{t-2020} a_{t-2019} \cdots a_{t-1}$. This implies $q = p^{\lceil \frac{\alpha}{2020} \rceil} \gg c$ divides one of the terms $a_{t-2020}, a_{t-2019}, \ldots, a_{t-1}$. If $a_t$ is divisible by $q$, call $t$ "friendly".

For some $n \gg k$, since at least $n$ numbers in $\{1, \ldots, k + 2020n\}$ are friendly, one of the numbers $a_1, \ldots, a_{k + 2020n}$ is at least $qn > kc + 2020cn$. This contradicts the claim that $\frac{a_n}{n}$ is bounded by $c$.

Therefore, every sufficiently large number must appear in the sequence.

The answer is: \boxed{\text{Every sufficiently large number appears in the sequence}}.