13. The sequence $\left\{a_{n} \mid\right.$ is defined as: $a_{0}=0, a_{1}=a_{2}=1$ $a_{n+1}=a_{n}+a_{n-1}(n \in \mathbf{N})$, find the greatest common divisor of $a_{2002}$ and $a_{1998}$.
13. First, from $\left(a_{n+1}, a_{n}\right)=\left(a_{n}, a_{n-1}\right)$, it is easy to prove that
$$
\left(a_{n+1}, a_{n}\right)=1(n \in \mathbf{N}) \text {. }
$$

Furthermore, using mathematical induction, we can prove that $a_{m}=a_{n} a_{m-n+1}+a_{n-1} a_{m-n} \quad(m \geqslant n)$.
From the above equation, it is clear that the greatest common divisor (gcd) of $a_{m-n}$ and $a_{n}$ must be a divisor of the gcd of $a_{m}$ and $a_{n}$. The gcd of $a_{m}$ and $a_{n}$ must also be a divisor of $a_{n-1} a_{m-n}$. By $\left(a_{n+1}, a_{n}\right)=1$, it is known that it is also a divisor of the gcd of $a_{m-n}$ and $a_{n}$.
Therefore, $\left(a_{m}, a_{n}\right)=\left(a_{m-n}, a_{n}\right)$.
Let $(m, n)=q$,
it is easy to deduce that $\left(a_{m}, a_{n}\right)=\cdots=\left(a_{0}, a_{q}\right)=a_{q}$.
Thus, $\left(a_{2002}, a_{1998}\right)=a_{(2002,1998)}=a_{2}=1$.
Therefore, the greatest common divisor of $a_{2002}$ and $a_{1998}$ is 1.