4. As shown in Figure 1, in a regular hexagon $A B C D E F$ with side length 2, a moving circle $\odot Q$ has a radius of 1, and its center moves on the line segment $C D$ (including the endpoints). $P$ is a moving point on and inside $\odot Q$. Let the vector $\overrightarrow{A P}=m \overrightarrow{A B}+n \overrightarrow{A F}(m, n \in \mathbf{R})$. Then the range of $m+n$ is . $\qquad$
4. $[2,5]$.

From the given information, we have
$$
\begin{array}{l}
\overrightarrow{A P} \cdot \overrightarrow{A B}=4 m-2 n, \\
\overrightarrow{A P} \cdot \overrightarrow{A F}=-2 m+4 n . \\
\text { Then } m+n=\frac{1}{2}(\overrightarrow{A P} \cdot \overrightarrow{A B}+\overrightarrow{A P} \cdot \overrightarrow{A F}) \\
=\frac{1}{2} \overrightarrow{A P} \cdot(\overrightarrow{A B}+\overrightarrow{A F})=\frac{1}{4} \overrightarrow{A P} \cdot \overrightarrow{A D} .
\end{array}
$$

Notice that, $\overrightarrow{A P} \cdot \overrightarrow{A D}$ equals the projection of $\overrightarrow{A P}$ onto $\overrightarrow{A D}$ multiplied by $|\overrightarrow{A D}|(|\overrightarrow{A D}|=4)$.

When point $Q$ is at point $C$ and point $P$ is on $B C$, the projection of $\overrightarrow{A P}$ onto $\overrightarrow{A D}$ is the shortest, which is 2;

When point $Q$ is at point $D$ and point $P$ is on $A D$, the projection of $\overrightarrow{A P}$ onto $\overrightarrow{A D}$ is the longest, which is 5.
In summary, $m+n \in[2,5]$.