II. A sequence of numbers where all terms are 0 or 1 is called a 0,1 sequence. Let $A$ be a finite 0,1 sequence, and let $f(A)$ denote the 0,1 sequence obtained by changing each 1 in $A$ to 0,1 and each 0 to 1,0. For example, $f((1,0,0,1))=(0,1,1,0,1,0,0,1)$, and let $f^{n}(A)$ denote $f(f(\cdots f(A) \cdots))(n$ times $f)$.
Try to find: In $f^{n}((1))$, how many pairs of consecutive terms are 0?

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Let's denote $f^{n}((1))=f^{n}$. The number of consecutive pairs $(0,0)$ and $(0,1)$ in $f^{n}$ are denoted as $g_{n}, h_{n}$, respectively. According to the problem, the $(0,0)$ pairs in $f^{n}$ must be derived from the $(0,1)$ pairs in $f^{n-1}$; the $(0,1)$ pairs in $f^{n-1}$ must be derived from the $(1,0)$ or $(0,0)$ pairs in $f^{n-2}$. Since $f^{n-2}$ has $2^{n-2}$ terms, half of which are 1, we have
$$
g_{n}=h_{n-1}=2^{n-3}+g_{n-2} .
$$

Thus, we have
$$
\begin{array}{c}
g_{n}=2^{n-3}+g_{n-2}, \\
g_{n-2}=2^{n-5}+g_{n-4}, \\
\cdots, \\
g_{4}=2+g_{2}, g_{2}=1 \text {, (if } 2 \mid n \text { ). } \\
g_{3}=2^{0}+g_{1}, g_{1}=0, \text { (if } 2 \nmid n \text { ). }
\end{array}
$$

When $2 \mid n$,
$$
\begin{aligned}
g_{n} & =2^{n-3}+2^{n-5}+\cdots+2^{1}+g_{2} \\
& =\frac{2\left(2^{n-2}-1\right)}{4-1}+1 \\
& =\frac{1}{3}\left(2^{n-1}+1\right) .
\end{aligned}
$$

When $2 \nmid n$,
$$
\begin{aligned}
g_{n} & =2^{n-3}+2^{n-5}+\cdots+2^{0}+g_{1}=\frac{2^{n-1}-1}{4-1}+0 \\
& =\frac{2^{n-1}-1}{3} .
\end{aligned}
$$

Therefore,
$$
g_{n}=\frac{1}{3}\left[2^{n-1}-(-1)^{n-1}\right] .
$$