Do you really mean "perfect number" or maybe "perfect square"? This would seem to make more sense, since then $a_n=n^2$ is a solution, while on the other hand the sequence of known perfect numbers and certainly grows at least exponentially.

Sorry typo... My bad. Perfect square indeed.

By looking at the exponents $\pmod 2$ we can prove by induction that $$S=\{a_k\}\subset \{k^2\}\cup\{a_1k^2\}\cup\cdots\cup\{a_1a_2k^2\}\cup\cdots\cup\{a_1a_2a_3k^2\}\cup\cdots \cup \{a_1\cdots a_{2021}k^2\}$$Where we take all possible product. We can reduce this union to some union $U=\bigcup_{j=1}^N \{b_jk^2\}$ where $N\leq 2^{2021}$ and all $b_j$ are distinct squarefree numbers (We take the minimal set of $b_j$s with the property that for every product of a subset of $\{a_1,..,a_{2021}\}$ there is a $b_j$ that is equal to this product over some square number). We can also note that $U-S$ must have a finite amount of elements (all $\leq a_{2021}$). By this inclusion, it follows that by taking $S\cap \{1,...,a_n\}$ we have $$\sum_{j=1}^n \left\lfloor\sqrt{\frac{a_n}{b_j}}\right\rfloor-C\leq n\leq \sum_{j=1}^n \left\lfloor\sqrt{\frac{a_n}{b_j}}\right\rfloor$$for some positive constant $C$. Let $B=\sum_{j=1}^N\frac{1}{\sqrt{b_j}}$ (which is bounded above by $S=\prod_{k=1}^{2021}1+\frac{1}{\sqrt{p_i}}$ ). From the second inequality we get $$n\leq \sum_{j=1}^n \left\lfloor\sqrt{\frac{a_n}{b_j}}\right\rfloor\leq\sqrt{a_n}B\implies a_n\geq \frac{n^2}{B^2}$$From the first one we get $$n\geq \sum_{j=1}^N \left\lfloor\sqrt{\frac{a_n}{b_j}}\right\rfloor-C\geq \sqrt{a_n}B-C-N$$$$\implies a_n\leq \frac{(n-C-N)^2}{B^2}$$This imlies for all $\varepsilon>0$ there exists an $M$ such that $a_n\geq \left(\frac{1}{B^2}-\varepsilon\right)n^2\forall n\geq M$. Therefore, the expression is bounded by $\left(\frac{1}{B^2}-\varepsilon\right)n^2$ and $\frac{1}{B^2}n^2$ for all $n$ big enough. Finally, it suffices to note that $B$ is bounded (for all possible choices of $b_j$ which depend on $a_1,...,a_{2021}$) by $S$. Since $1\geq \frac{1}{B^2}\geq \frac{1}{B^2}-\varepsilon\geq \frac{1}{S^2}$ it suffices to choose $c_1=\frac{1}{S^2}$ and $c_2=1$.

An elegant proof that there are infinitely many prime numbers: If the largest prime is P, then we define this sequence: a(n)=n from 1 to P, and for n>P, a(n) is taken the same way as in this problem. Then a(n)=n for all positive integers n. On the other hand, this is a CGMO sequence (just change the 2021 to P in the definition) Then there exists a constant c(1) such that for some N, a(n)≥c(1)n^2 for all n>N, which obviously contradicts the fact that a(n)=n.

cadaeibf wrote: By looking at the exponents $\pmod 2$ we can prove by induction that $$S=\{a_k\}\subset \{k^2\}\cup\{a_1k^2\}\cup\cdots\cup\{a_1a_2k^2\}\cup\cdots\cup\{a_1a_2a_3k^2\}\cup\cdots \cup \{a_1\cdots a_{2021}k^2\}$$Where we take all possible product. We can reduce this union to some union $U=\bigcup_{j=1}^N \{b_jk^2\}$ where $N\leq 2^{2021}$ and all $b_j$ are distinct squarefree numbers (We take the minimal set of $b_j$s with the property that for every product of a subset of $\{a_1,..,a_{2021}\}$ there is a $b_j$ that is equal to this product over some square number). We can also note that $U-S$ must have a finite amount of elements (all $\leq a_{2021}$). By this inclusion, it follows that by taking $S\cap \{1,...,a_n\}$ we have $$\sum_{j=1}^n \left\lfloor\sqrt{\frac{a_n}{b_j}}\right\rfloor-C\leq n\leq \sum_{j=1}^n \left\lfloor\sqrt{\frac{a_n}{b_j}}\right\rfloor$$for some positive constant $C$. Let $B=\sum_{j=1}^N\frac{1}{\sqrt{b_j}}$ (which is bounded above by $S=\prod_{k=1}^{2021}1+\frac{1}{\sqrt{p_i}}$ ). From the second inequality we get $$n\leq \sum_{j=1}^n \left\lfloor\sqrt{\frac{a_n}{b_j}}\right\rfloor\leq\sqrt{a_n}B\implies a_n\geq \frac{n^2}{B^2}$$From the first one we get $$n\geq \sum_{j=1}^N \left\lfloor\sqrt{\frac{a_n}{b_j}}\right\rfloor-C\geq \sqrt{a_n}B-C-N$$$$\implies a_n\leq \frac{(n-C-N)^2}{B^2}$$This imlies for all $\varepsilon>0$ there exists an $M$ such that $a_n\geq \left(\frac{1}{B^2}-\varepsilon\right)n^2\forall n\geq M$. Therefore, the expression is bounded by $\left(\frac{1}{B^2}-\varepsilon\right)n^2$ and $\frac{1}{B^2}n^2$ for all $n$ big enough. Finally, it suffices to note that $B$ is bounded (for all possible choices of $b_j$ which depend on $a_1,...,a_{2021}$) by $S$. Since $1\geq \frac{1}{B^2}\geq \frac{1}{B^2}-\varepsilon\geq \frac{1}{S^2}$ it suffices to choose $c_1=\frac{1}{S^2}$ and $c_2=1$. prove c2＝2 is easier

For the upper bound, take $c_2=4$ and note that if $\lfloor \sqrt{a_n}\rfloor=k$ then $a_{n+1}=(k+1)^2$ works, so the sequence $\lfloor \sqrt{a_n}\rfloor$ increases by at most $1$ with each increase in $n$ and is hence at most $2n$ for large enough $n$. For the lower bound, suppose we have $k$ primes $p_1,\ldots,p_k$ dividing the squarefree part of some $a_i$ for $1 \leq i \leq 2021$. Let $S$ be the set of at most $2021$ vectors in $\mathbb{F}_2^k$ representing the prime factorization of the squarefree part of $a_1,\ldots,a_{2021}$ (so the $j$-th element of the $i$-th vector is $1$ iff $p_j$ divides the squarefree part of $a_j$). 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 $a_n \leq c_1n^2$, 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, so there are at most $2^{2021}\sqrt{c_1}n<n$ elements before the $n$-th: absurd. This finishes the problem. $\blacksquare$