4. Arrange $n(n \geqslant 2)$ coins in a row. If there is a coin with the head side up, then one can select one of them, and flip the consecutive odd number of coins (which can be one) starting from this coin from left to right simultaneously (flipping means turning a head-up coin to head-down, or a head-down coin to head-up), which is called one operation. When all coins are head-down, the operation stops. If all coins are head-up at the beginning, is there a way to perform $\left[\frac{2^{n+1}}{3}\right]([x]$ denotes the greatest integer not exceeding the real number $x$) operations?
（Gu Bin provided the problem）
4. For each arrangement of the coins, define the corresponding sequence $c_{1} c_{2} \cdots c_{n}$: if the $i$-th coin from left to right is heads up, then $c_{i}=1$; otherwise, $c_{i}=0$.
Thus, the initial sequence is $\underbrace{11 \cdots 1}_{n \uparrow}$.
Each time, consider the first 1 from right to left, and starting from this 1, perform an operation on the longest sequence of odd-numbered digits, and let the total number of operations in this scheme be $a_{n}$.
Next, we use mathematical induction to prove that $a_{n}=\left[\frac{2^{n+1}}{3}\right]$.
When $n=1$, it is easy to see that $a_{1}=1=\left[\frac{2^{2}}{3}\right]$.
Assume that when $n=k$, the proposition holds, i.e.,
$$
a_{k}=\left[\frac{2^{k+1}}{3}\right] \text {. }
$$

Then, when $n=k+1$,
(1) If $k$ is odd, i.e., $\underbrace{11 \cdots 1}_{k+1 \uparrow}$ becomes $1 \underbrace{00 \cdots 0}_{k \uparrow}$ after $a_{k}$ operations, and then after one more operation, it can become $0 \underbrace{11 \cdots 10}_{k-11}$, and then after $a_{k}-1$ operations, it becomes $\underbrace{00 \cdots 0}_{k \uparrow}$ (because the process of $\underbrace{11 \cdots 1}_{k \uparrow}$ becoming $\underbrace{00 \cdots 0}_{k \uparrow}$ involves the first step being $\underbrace{11 \cdots 10}_{k-1 \uparrow}$), therefore,
$$
\begin{array}{l}
a_{k+1}=2 a_{k}=2\left[\frac{2^{k+1}}{3}\right] \\
=2 \times \frac{2^{k+1}-1}{3}=\frac{2^{k+2}-2}{3}=\left[\frac{2^{k+2}}{3}\right] ;
\end{array}
$$
(2) If $k$ is even, i.e., $\underbrace{11 \cdots 1}_{k+1 \uparrow}$ becomes $1 \underbrace{00 \cdots 0}_{k \uparrow}$ after $a_{k}$ operations, and then after one more operation, it becomes $0 \underbrace{11 \cdots 1}_{k \uparrow}$, and finally, after $a_{k}$ more operations, it becomes $\underbrace{00 \cdots 0}_{k+1 \uparrow}$, thus we have
$$
\begin{array}{l}
a_{k+1}=2 a_{k}+1=2\left[\frac{2^{k+1}}{3}\right]+1 \\
=2 \times \frac{2^{k+1}-2}{3}+1=\frac{2^{k+2}-1}{3}=\left[\frac{2^{k+2}}{3}\right] .
\end{array}
$$

By mathematical induction, we know that $a_{n}=\left[\frac{2^{n+1}}{3}\right]$.
Therefore, there exists an operation scheme that meets the requirements.