Inverting 5. A sequence whose terms are all 0 or 1 is called a $0, 1$ sequence. Let $A$ be a finite $0,1$ sequence. Denote $f(A)$ as the $0,1$ sequence obtained by changing every 1 in $A$ to 0,1 and every 0 to 1,0. For example, $f((1,0,0,1))=(0,1,1,0$, $\cdots)$ (n times).

Question: In $f^{4}((1))$, how many terms are 0?
Let $f^{\mathrm{n}}((1))=f^{\mu}$. The number of pairs of consecutive terms in $f^{n}$ that are 0,0 or 0,1 are denoted as $g_{\mathrm{n}}, h_{\mathrm{n}}$.

According to the problem, the 0,0 pairs in $f^{2}$ must be derived from the 0,1 pairs in $f^{n-1}$ after the transformation $f$.
The 0,1 pairs in $f^{\mathrm{n}-1}$ must be derived from the 1 or 0,0 pairs in $f^{\mathrm{n}-2}$ after the transformation $f$.
$f^{\mathrm{n}-2}$ has $2^{\mathrm{n}-2}$ terms, half of which are 1.
Therefore, $g_{\mathrm{a}}=h_{\mathrm{n}-1}=2^{\mathrm{n}-3}+g_{\mathrm{n}-2}$.
This is a recursive formula:
$$
\begin{array}{ll}
g_{\mathrm{n}}=2^{\mathrm{D}-3}+g_{\mathrm{n}-2}, & \\
g_{\mathrm{n}-2}=2^{\mathrm{n}-5}+g_{\mathrm{n}-4}, & \\
\cdots \cdots, \\
g_{4}=2+g_{2}, g_{2}=1, & \text { (if } 2 \mid n \text { ) } \\
g_{8}=2^{0}+g_{1}, g_{1}=0 . & \text { (if } 2+n \text { ) }
\end{array}
$$

Adding all these equations, we get:
$$
\begin{array}{l}
g_{\mathrm{n}}=2^{\mathrm{n}-3}+2^{\mathrm{n}-5}+\cdots+\left\{\begin{array}{l}
2^{0}+g_{1} \\
2^{1}+g_{2}
\end{array}\right. \\
=\left\{\begin{array}{l}
\frac{2^{n-1}-1}{4-1}+0, \text{ if } 2+n, \\
\frac{2\left(2^{\mathrm{n}-2}-1\right)}{4-1}+1, \text{ if } 2 \mid n .
\end{array}\right. \\
\end{array}
$$

Thus, $g_{u}=\frac{1}{3}\left(2^{\mathrm{n}-1}-(-1)^{y-1}\right)$.