There are $n$ coins in a row, $n\geq 2$. If one of the coins is head, select an odd number of consecutive coins (or even 1 coin) with the one in head on the leftmost, and then flip all the selected coins upside down simultaneously. This is a $move$. No move is allowed if all $n$ coins are tails.
Suppose $m-1$ coins are heads at the initial stage, determine if there is a way to carry out $ \lfloor\frac {2^m}{3}\rfloor $ moves
1. **Representation of Coin States**:
   - Represent each state of the coins by a binary number where heads are converted to \(1\)s and tails to \(0\)s. For example, \(THTT = (0100)_2 = 4\) and \(HTTH = (1001)_2 = 9\).

2. **Definition of Functions**:
   - Let \(f(x)\) denote the maximum number of moves that the state \(x = (x_{n-1} x_{n-2} \cdots x_0)_2\) can be carried out.
   - Let \(g(x) := x_0 + x_1 + \cdots + x_{n-1}\) be the number of heads in the state \(x\).

3. **Weight Function**:
   - Define the weight function \(w_i\) as follows:
     \[
     w_i := \begin{cases}
     \frac{2^{i+1} + 1}{3}, & \text{if } 2 \mid i \\
     \frac{2^{i+1} - 1}{3}, & \text{otherwise}
     \end{cases}
     \]

4. **Claim**:
   - We claim that \(f(x) \geq \sum_{i=0}^{n-1} x_i w_i\).

5. **Base Case**:
   - For \(x = 0\) and \(x = 1\), we have \(f(0) \geq 0\) and \(f(1) \geq 1\), which are trivial.

6. **Induction Hypothesis**:
   - Suppose for all \(x < k\), the claim holds.

7. **Induction Step**:
   - For \(x = k\), select \(i\) such that \(k_i = 1\) and \(k_0 = \cdots = k_{i-1} = 0\). Let \(j = i \pmod{2}\).
   - We can flip the \(l\)-th coin counted from the right for \(j \leq l \leq i\), and the state becomes \(k - 2^j\).
   - Since \(j = 0\) or \(1\), the base case \(x = 0, 1\) is enough. Therefore, \(f(k) \geq f(k - 2^j) + 1\).

8. **Weight Function Analysis**:
   - For \(j = 0\) (i.e., \(2 \mid i\)):
     \[
     w_i = \frac{2^{i+1} + 1}{3} = \frac{2^i + 2^{i-1} + \cdots + 2^1}{3} + 1 = \frac{2^i - 1}{3} + \frac{2^{i-1} + 1}{3} + \frac{2^{i-2} - 1}{3} + \cdots + \frac{2^1 + 1}{3} + 1 = w_j + \cdots + w_{i-1} + 1
     \]
   - For \(j = 1\) (i.e., \(2 \nmid i\)):
     \[
     w_i = \frac{2^{i+1} - 1}{3} = \frac{2^i + 2^{i-1} + \cdots + 2^2}{3} + 1 = \frac{2^i + 1}{3} + \frac{2^{i-1} - 1}{3} + \frac{2^{i-2} + 1}{3} + \cdots + \frac{2^2 - 1}{3} + 1 = w_j + \cdots + w_{i-1} + 1
     \]

9. **Combining Results**:
   - Therefore, there is a formula:
     \[
     w_i = w_j + \cdots + w_{i-1} + 1 \quad \text{(1)}
     \]
   - Hence,
     \[
     f(k) \geq f(k - 2^j) + 1 \overset{\text{induction hypothesis}}{\geq} \sum_{l=0}^{n-1} (k - 2^j)_l w_l + 1 = \sum_{l=i+1}^{n-1} k_l w_l + \sum_{l=j}^{i-1} w_l + 1 \overset{\text{(1)}}{=} \sum_{l=i}^{n-1} k_l w_l = \sum_{l=0}^{n-1} k_l w_l
     \]
   - Thus, the claim holds.

10. **Final Step**:
    - Since for all \(i\), there is \(w_{i+1} \geq w_i\), we have:
      \[
      f(x) \geq \sum_{i=0}^{n-1} x_i w_i \geq \sum_{i=0}^{g(x)-1} w_i \overset{\text{(1)}}{=} \begin{cases}
      w_{g(x)} - 1, & \text{if } 2 \mid g(x) \\
      w_{g(x)}, & \text{otherwise}
      \end{cases} = \left\lfloor \frac{2^{g(x)+1}}{3} \right\rfloor
      \]
    - This completes the proof.

The final answer is \( \boxed{ \left\lfloor \frac{2^m}{3} \right\rfloor } \).