Call a sequence of positive integers $(a_n)_{n \ge 1}$ a "CGMO sequence" if $(a_n)_{n \ge 1}$ strictly increases, and for all integers $n \ge 2022$, $a_n$ is the smallest integer such that there exists a non-empty subset of $\{a_{1}, a_{2}, \cdots, a_{n-1} \}$ $A_n$ where $a_n \cdot \prod\limits_{a \in A_n} a$ is a perfect square.

Proof: there exists $c_1, c_2 \in \mathbb{R}^{+}$ s.t. for any "CGMO sequence" $(a_n)_{n \ge 1}$ , there is a positive integer $N$ that satisfies any $n \ge N$, $c_1 \cdot n^2 \le a_n \le c_2 \cdot n^2$.
1. **Upper Bound:**
   - We start by considering the upper bound for the sequence \( (a_n) \).
   - Let \( \lfloor \sqrt{a_n} \rfloor = k \). This implies \( k^2 \le a_n < (k+1)^2 \).
   - For the next term \( a_{n+1} \), we need it to be the smallest integer such that there exists a non-empty subset \( A_{n+1} \subseteq \{a_1, a_2, \ldots, a_n\} \) where \( a_{n+1} \cdot \prod_{a \in A_{n+1}} a \) is a perfect square.
   - If we choose \( a_{n+1} = (k+1)^2 \), then \( \lfloor \sqrt{a_{n+1}} \rfloor = k+1 \).
   - This means that \( \lfloor \sqrt{a_n} \rfloor \) increases by at most 1 with each increase in \( n \).
   - Therefore, for large enough \( n \), \( \lfloor \sqrt{a_n} \rfloor \le 2n \).
   - Hence, \( a_n \le (2n)^2 = 4n^2 \).
   - We can take \( c_2 = 4 \).

2. **Lower Bound:**
   - Consider the prime factorization of the squarefree part of \( a_i \) for \( 1 \leq i \leq 2021 \).
   - Let \( k \) be the number of distinct primes dividing the squarefree part of these \( a_i \).
   - Define \( S \) as the set of at most 2021 vectors in \( \mathbb{F}_2^k \) representing the prime factorization of the squarefree part of \( a_1, \ldots, a_{2021} \).
   - Let \( T \) be the set of at most \( 2^{2021} \) vectors which can be written as the sum of the vectors in a (possibly empty) subset of \( S \).
   - The sum of any two elements \( t_1, t_2 \in T \) is also in \( T \), since if \( t_1 \) is the sum of \( s_1 \subseteq S \) and \( t_2 \) is the sum of \( s_2 \subseteq S \), then \( t_1 + t_2 \) is the sum of \( s_1 \oplus s_2 \subseteq S \).
   - Thus, the squarefree part of any \( a_n \) can be represented by a vector in \( T \), since it should equal the squarefree part of the product of some number of squarefree terms, which in turn can be represented by the sum of some elements of \( T \) by induction.
   - Now let \( c_1 = 2^{-2021 \cdot 100} \).
   - If there exists some large positive integer \( n \) where \( a_n \leq c_1 n^2 \), then by the increasing condition, there are at most \( \sqrt{c_1} n \) elements in the sequence that occur before \( n \) with any given squarefree part.
   - Therefore, there are at most \( 2^{2021} \sqrt{c_1} n < n \) elements before the \( n \)-th term, which is a contradiction.
   - Hence, \( a_n \ge c_1 n^2 \).

The final answer is \( \boxed{ c_1 \cdot n^2 \le a_n \le c_2 \cdot n^2 } \) where \( c_1 = 2^{-2021 \cdot 100} \) and \( c_2 = 4 \).