4. Given the sequence $\left\{a_{n}\right\}$ composed of positive integers, and it is an increasing sequence, with
$$
a_{n+2}=a_{n+1}+2 a_{n}\left(n \in \mathbf{Z}_{+}\right) \text {. }
$$

If $a_{5}=52$, then $a_{7}=$ ( ).
(A) 102
(B) 152
(C) 212
(D) 272
4. C.

From the problem, we know that $a_{5}=5 a_{2}+6 a_{1}=52$.
Given that $a_{1}$ and $a_{2}$ are positive integers,
$$
6 \mid \left(52-5 a_{2}\right) \Rightarrow 6 \mid \left(a_{2}-2\right) \text {. }
$$

Let $a_{2}=6 k+2(k \in \mathbf{N})$.
Notice that, $52>5 a_{2}=30 k+10$.
Thus, $k=0$ or 1.
If $k=0$, then $a_{2}=2, a_{1}=7$, which contradicts the increasing nature of the sequence $\left\{a_{n}\right\}$.
Therefore, $k=1, a_{2}=8, a_{1}=2$.
Hence, $a_{3}=12, a_{4}=28, a_{6}=108, a_{7}=212$.