10. (20 points) Given the sequence $A: a_{1}, a_{2}, \cdots, a_{n}$, satisfying $a_{i} \in\{0,1\}(i=1,2, \cdots, n)$. Define the transformation $T$: $T$ changes each 1 in the sequence $A$ to 0,1, and each 0 to 1,0.
If $A_{0}$ is 0,1,
$$
A_{k}=T\left(A_{k-1}\right)(k=1,2, \cdots),
$$

and let $a_{k}$ be the number of pairs of consecutive 0s in $A_{k}$. Find the expression for $a_{k}$.
10. Let $A_{k}$ contain $b_{k}$ pairs of $(0,1)$.

Since the $(0,0)$ pairs in $A_{k+1}$ can only be derived from the $(0,1)$ pairs in $A_{k}$, we have
$$
a_{k+1}=b_{k} \text {. }
$$

Thus, the $(0,1)$ pairs in $A_{k+1}$ can be generated through two ways:
(1) from the 1s in $A_{k}$;
(2) from the $(0,0)$ pairs in $A_{k}$.
Since $A_{0}: 0,1$, by the definition of transformation $T$, the number of 0s and 1s in $A_{k}$ is always equal, and there are a total of $2^{k+1}$ of them.
Therefore, $b_{k+1}=a_{k}+2^{k}$.
Then $a_{k+2}=a_{k}+2^{k}$.
From $A_{0}: 0,1 \Rightarrow A_{1}: 1,0,0,1$
$$
\Rightarrow A_{2}: 0,1,1,0,1,0,0,1 \text {. }
$$

Thus, $a_{0}=0, a_{1}=1$.
When $k$ is even,
$$
\begin{array}{l}
a_{k}=a_{k-2}+2^{k-2} \\
=\cdots \\
=a_{2}+2^{2}+2^{4}+\cdots+2^{k-2} \\
=1+2^{2}+2^{4}+\cdots+2^{k-2} \\
=\frac{1}{3}\left(2^{k}-1\right) ;
\end{array}
$$

When $k$ is odd,
$$
\begin{array}{l}
a_{k}=a_{1}+2^{t}+2^{n}+\cdots+2^{k-2} \\
=1+2^{1}+2^{3}+\cdots+2^{k-2} \\
=\frac{1}{3}\left(2^{k}+1\right) .
\end{array}
$$

In summary, $a_{k}=\left\{\begin{array}{ll}\frac{1}{3}\left(2^{k}+1\right), & k \text { is odd; } \\ \frac{1}{3}\left(2^{k}-1\right), & k \text { is even. }\end{array}\right.$