4. Let $A B C D E F$ be a regular hexagon. A frog starts at vertex $A$ and can jump to one of the two adjacent vertices at each step. If it reaches point $D$ within 5 jumps, it stops jumping; if it does not reach point $D$ within 5 jumps, it stops after 5 jumps. How many different possible sequences of jumps can the frog make from the start to the stop?

The number of different possible sequences of jumps is $\qquad$.
4. 26 As shown in the figure, it is clear that the frog cannot reach point $D$ from point $A$ by jumping 1 time, 2 times, or 4 times. Therefore, the frog's jumping methods are only of the following two scenarios: (1) reaching point $D$ in 3 jumps, with 2 ways to do so;
(2) stopping after 5 jumps (the first 3 jumps do not reach point $D$).
Note that among the $2^{3}$ ways of the first 3 jumps, 2 ways reach point $D$, so there are $2^{3}-2=6$ ways for the first 3 jumps.
The last 2 jumps have $2^{2}$ ways, thus there are $6 \times 2^{2}=24$ ways.
From (1) and (2), we know that there are a total of $2+24=26$ ways.