三角形顶点随机游走

设 $a_n$ 为从 A 出发走 $n$ 步回到 A 的路径数，$b_n$ 为从 A 出发走 $n$ 步到达 B 的路径数（到 C 同样也是 $b_n$）。

递推：

• 要在 $n+1$ 步回到 A，前一步必须在 B 或 C：$a_{n+1}=2b_n$。
• 要在 $n+1$ 步到 B，前一步在 A（走 A→B）或在 C（走 C→B）：$b_{n+1}=a_n+b_n$。

初值 $a_0=1,b_0=0$。

逐步算到 $n=8$：

$$a_1=0,\ b_1=1; \ a_2=2,\ b_2=1; \ a_3=2,\ b_3=3; \ a_4=6,\ b_4=5; \ a_5=10,\ b_5=11; \ a_6=22,\ b_6=21; \ a_7=42,\ b_7=43; \ a_8=2b_7=86.$$

因此答案为 $\boxed{86}$。

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英文解析

Let $a_n$ be the number of paths starting from A and returning to A in $n$ steps, and let $b_n$ be the number of paths starting from A and reaching B in $n$ steps (the number of paths reaching C is also $b_n$).

Recurrence relations:

• To return to A in $n+1$ steps, the previous step must have been at B or C: $a_{n+1}=2b_n$.
• To reach B in $n+1$ steps, the previous step must have been at A (taking the step A→B) or at C (taking the step C→B): $b_{n+1}=a_n+b_n$.

Initial values: $a_0=1, b_0=0$.

Step-by-step calculation up to $n=8$:

$$a_1=0,\ b_1=1; \ a_2=2,\ b_2=1; \ a_3=2,\ b_3=3; \ a_4=6,\ b_4=5; \ a_5=10,\ b_5=11; \ a_6=22,\ b_6=21; \ a_7=42,\ b_7=43; \ a_8=2b_7=86.$$

Thus, the answer is $\boxed{86}$.