3. Given a sequence of positive integers $a_{1}, a_{2}, \cdots, a_{n}, \cdots$, where $a_{1}=2^{2006}$, and for each positive integer $i, a_{i+1}$ is equal to the square of the sum of the digits of $a_{i}$. Find the value of $a_{2006}$.
3. Since the remainders when $2, 2^{2}, 2^{3}, 2^{4}, 2^{5}, 2^{6}, \cdots$ are divided by 9 are $2, 4, 8, 7, 5, 1, \cdots$, therefore, the remainders when $2^{m+6}$ and $2^{m}$ are divided by 9 are equal.

But $2006 = 334 \times 6 + 2$, so the remainder when $a_{1}$ is divided by 9 is 4. Thus, the remainder when the sum of the digits of $a_{1}$ is divided by 9 is 4.

Since the remainders when $a_{2}$ and $4^{2}$ are divided by 9 are equal, the remainder when $a_{2}$ is divided by 9 is 7.

Since the remainders when $a_{3}$ and $7^{2}$ are divided by 9 are equal, the remainder when $a_{3}$ is divided by 9 is 4, and the remainder when the sum of the digits of $a_{3}$ is divided by 9 is 4.
On the other hand, $a_{1} = 2^{2006} < 2^{3 \times 666} < 10^{666}$;
the sum of the digits of $a_{1}$ does not exceed $9 \times 669 = 6021$, so $a_{2} \leqslant 6021^{2} < 37 \times 10^{6}$;
the sum of the digits of $a_{2}$ does not exceed $9 \times 7 + 2 = 65$, so $a_{3} \leqslant 65^{2} = 4225$;
the sum of the digits of $a_{3}$ does not exceed $9 \times 3 + 3 = 30$, so $a_{4} \leqslant 30^{2}$.

Since $a_{4}$ is the square of the sum of the digits of $a_{3}$, $a_{4}$ is the square of a number that leaves a remainder of 4 when divided by 9.

Since $a_{4} \leqslant 30^{2}$, $a_{4}$ is one of $4^{2}, 13^{2}, 22^{2}$, i.e., one of $16, 169, 484$.
Since $a_{5}$ is one of $49, 256$, then
$$
a_{6} = 169, a_{7} = 256, a_{8} = 169, \cdots \text{.}
$$

By analogy, we get $a_{20006} = 169$.