For each integer $a_{0}>1$, define the sequence $a_{0}, a_{1}, \cdots$ as follows: for any $n \geqslant 0$,
$a_{n+1}=\left\{\begin{array}{ll}\sqrt{a_{n}}, & \sqrt{a_{n}} \text { is an integer; } \\ a_{n}+3, & \text { otherwise. }\end{array}\right.$
Find all $a_{0}$ such that there exists a number $A$ for which $a_{n}=A$ for infinitely many $n$.
1. $a_{0}$ is all multiples of 3.

Since $a_{n+1}$ is determined solely by $a_{n}$, the sequence $\left\{a_{n}\right\}$ has infinitely many equal terms if and only if the sequence is eventually periodic, which is equivalent to the sequence being bounded.

Notice that perfect squares modulo 3 are congruent to 0 or 1. If there is some term $a_{k} \equiv 2(\bmod 3)$, then $a_{k}$ is not a perfect square,
$$
a_{k+1}=a_{k}+3 \text {, and } a_{k+1} \equiv 2(\bmod 3) \text {, }
$$

By induction, for $m \geqslant k$, we have $a_{m+1}=a_{m}+3$, and the sequence is strictly increasing from the $k$-th term onward, with no upper bound.

In particular, if $a_{0} \equiv 2(\bmod 3)$, such an $a_{0}$ does not satisfy the condition.
If $3 \mid a_{k}$, then regardless of
$$
a_{k+1}=\sqrt{a_{k}} \text { or } a_{k+1}=a_{k}+3 \text {, }
$$

we have $3 \mid a_{k+1}$.
By induction, for $m \geqslant k, a_{m}$ is always a multiple of 3.
In particular, if $3 \mid a_{0}$, then every term in the sequence is a multiple of 3.

Assume $3 \mid a_{0}$, and choose a perfect square $N^{2}>$ $a_{0}$, and $3 \mid N$.
Next, we show that for each $n$, $a_{n} \leqslant N^{2}$.
Proof by contradiction. Suppose there exists a term greater than $N^{2}$, take the smallest $k$ such that $a_{k}>N^{2}$.
Since $3 \mid a_{k}$, then $a_{k} \geqslant N^{2}+3$.
And $a_{k-1} \leqslant N^{2}, a_{k}-a_{k-1} \leqslant 3$, then
$a_{k-1}=N^{2}$.
Thus, by definition, $a_{k}=N$, which is a contradiction.
Therefore, there does not exist $a_{k}>N^{2}$.
Hence, the sequence $\left\{a_{n}\right\}$ is bounded, and such an $a_{0}$ satisfies the requirement.
Now consider the case $a_{0} \equiv 1(\bmod 3)$.

It is easy to see that if $3 \nmid a_{k}$ and $a_{k}>1$, then
$3 \nmid a_{k+1}$, and $a_{k+1}>1$.
Therefore, every term in the sequence $\left\{a_{n}\right\}$ is not divisible by 3 and is greater than 1.

Assume the sequence $\left\{a_{n}\right\}$ is bounded, i.e., it is eventually periodic, then each term is congruent to 1 modulo 3. Take the maximum term $a_{k}$ in the final period, then $a_{k}$ is a perfect square. Otherwise,
$$
a_{k+1}=a_{k}+3>a_{k} \text {. }
$$

Let $a_{k}=N^{2}$.
Then $a_{k+1}=N$
$$
\begin{array}{l}
\Rightarrow N \equiv 1(\bmod 3), \text { and } N>1 \\
\Rightarrow N \geqslant 4 .
\end{array}
$$

Since there exists $j>k$, such that
$$
a_{j}=a_{k}=N^{2}>(N-2)^{2} \text {, }
$$

and $a_{k+1}=N \leqslant(N-2)^{2}$ (note $N \geqslant 4$), take the smallest $l>k+1$, such that $a_{l}>(N-2)^{2}$, hence
$$
\begin{array}{l}
a_{l-1} \leqslant(N-2)^{2}<a_{l} \\
\Rightarrow a_{l}=a_{l-1}+3 .
\end{array}
$$

By $a_{l} \equiv a_{l-1} \equiv 1(\bmod 3)$, we have
$$
a_{l-1}=(N-2)^{2} \equiv 1(\bmod 3),
$$

Thus, by definition $a_{l}=N-2$, which is a contradiction.
Therefore, $a_{0} \equiv 1(\bmod 3)$ does not satisfy the requirement.
The conclusion is proved.