4. As shown in Figure 2, in the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, the dihedral angle $A-B D_{1}-A_{1}$ is $\alpha, A B: B C: C C_{1}=1: 1: 2$. Then $\tan \alpha=$ $\qquad$
4. $2 \sqrt{6}$.

As shown in Figure 5, connect $A C$, $A D_{1}$, $B D_{1}$, $A_{1} B$, and $C D_{1}$.

Since $A_{1} D_{1} / / B C$, it is known that points $A_{1}$, $B$, $C$, and $D_{1}$ are coplanar.

Draw $C E \perp B D_{1}$ at point $E$, intersecting $A_{1} B$ at point $F$, and connect $A E$.
By symmetry, we have
$$
A E \perp B D_{1}, A E=E C.
$$

Therefore, $\angle A E F$ is the plane angle of the dihedral angle $A-B D_{1}-A_{1}$, i.e., $\alpha=\angle A E F$.
Let $A B=a$. Then $B C=a$, $C C_{1}=2 a$. Thus,
$$
A D_{1}=\sqrt{5} a, A C=\sqrt{2} a, B D_{1}=\sqrt{6} a.
$$

In the right triangle $\triangle A B D_{1}$, $A E=\frac{A B \cdot A D_{1}}{B D_{1}}=\frac{\sqrt{5}}{\sqrt{6}} a$.
In $\triangle A C E$, by the cosine rule we get
$$
\begin{array}{l}
\cos \angle A E C=\frac{A E^{2}+C E^{2}-A C^{2}}{2 A E \cdot C E} \\
=\frac{2 A E^{2}-A C^{2}}{2 A E^{2}}=-\frac{1}{5}.
\end{array}
$$

Thus, $\cos \alpha=\cos \angle A E F$
$$
\begin{array}{l}
=\cos \left(180^{\circ}-\angle A E C\right)=-\cos \angle A E C=\frac{1}{5} \\
\Rightarrow \sin \alpha=\sqrt{1-\cos ^{2} \alpha}=\frac{2 \sqrt{6}}{5} \\
\Rightarrow \tan \alpha=\frac{\sin \alpha}{\cos \alpha}=2 \sqrt{6}.
\end{array}
$$