Example 8 Define the sequence $\left\{a_{n}\right\}$:
$$
\begin{array}{l}
a_{1}=0, \\
a_{n}=a_{\left[\frac{n}{2}\right]}+(-1)^{\frac{n(n+1)}{2}}(n=2,3, \cdots) .
\end{array}
$$

For each non-negative integer $k$, find the number of indices $n$ that satisfy $2^{k} \leqslant n<2^{k+1}$ and $a_{n}=0$.
【Analysis】For any $n \in \mathbf{Z}_{+}$, let $n$ in binary representation, the sum of occurrences of adjacent pairs 00 and 11 be $x_{n}$, and the sum of occurrences of adjacent pairs 01 and 10 be $y_{n}$.
First, prove: $a_{n}=x_{n}-y_{n}$.
In fact, when $n=1$, $x_{1}=y_{1}=0$, so equation (1) holds for $n=1$.

Now assume equation (1) holds for indices $1,2, \cdots, n-1(n \geqslant 2)$.
Next, consider the case for $n$.
If the last two digits of $n$ in binary representation are 00 or 11, then
$n \equiv 0$ or $3(\bmod 4)$.
At this time, $a_{n}=a_{\left[\frac{n}{2}\right]}+1$,
$x_{n}=x_{\left[\frac{n}{2}\right]}+1, y_{n}=y_{\left[\frac{n}{2}\right]}$.
Therefore, equation (1) holds for $n$.
If the last two digits of $n$ in binary representation are 01 or 10, then
$n \equiv 1$ or $2(\bmod 4)$.
At this time, $a_{n}=a_{\left[\frac{n}{2}\right]}-1$,
$x_{n}=x_{\left[\frac{n}{2}\right]}, y_{n}=y_{\left[\frac{n}{2}\right]}+1$.
Therefore, equation (1) holds for $n$.
In summary, equation (1) holds for any $n \in \mathbf{Z}_{+}$.
Next, calculate the number of $n$ in $2^{k} \leqslant n<2^{k+1}$ where $x_{n}$ and $y_{n}$ are equal in binary representation, at this time, $n$ in binary representation is a $k+1$-digit number, denoted as $B_{n}$.

When $k \geqslant 1$, subtract the next digit from each digit of $B_{n}$ from left to right, then take the absolute value, to obtain a $k$-element 0, 1 array $C_{n}$ (for example, if $B_{n}=(1101)_{2}$, then $C_{n}=(011)_{2}$ ).

Notice that, each adjacent pair 00 and 11 in $B_{n}$ becomes a 0 in $C_{n}$, while 01 and 10 become a 1 in $C_{n}$.

Thus, if $x_{n}=y_{n}$, then the number of 1s in $C_{n}$ is equal to the number of 0s.

Conversely, for a $k$-element 0, 1 array $C_{n}=\left(c_{1} c_{2} \cdots c_{k}\right)_{2}$, compute the sum in modulo 2:
$$
b_{1}=1+c_{1}, b_{2}=b_{1}+c_{2}, \cdots, b_{k}=b_{k-1}+c_{k}
$$

where $b_{0}=1$.
Then $B_{n}=\left(b_{0} b_{1} \cdots b_{k}\right)_{2}$ is the binary representation of $n$ satisfying $2^{k} \leqslant n<2^{k+1}$.
This indicates that $B_{n}$ and $C_{n}$ form a one-to-one correspondence.
In summary, the answer to the original problem is equal to the number of $k$-element 0, 1 arrays where the number of 0s and 1s are equal.

Therefore, when $k$ is odd, the answer is 0; when $k$ is even, the answer is $\mathrm{C}_{k}^{\frac{1}{2}}$ (with the convention $\mathrm{C}_{0}^{0}=1$ ).