China Western Mathematical Olympiad

The answer is possible.

For any configuration of the coins, we define a corresponding 01-sequence $c_1{c}_2...c_n$ of length $n$ as follows: $c_i=1$, if the $i$-th coin from the left is head, otherwise, $c_i=0$. It is easy to see that the status of the $n$ coins has a one-to-one correspondence to such 01-sequences, so in the following, we will consider this sequence model instead.

Initially, the sequence is $\mathrm{11...11}$, denoted by $1^n$ (with $n$ consecutive digits of 1). Similarly, $\mathrm{00...00}$, denoted by $0^n$ (with $n$ digits of 0). For any 01-sequence with at least a digit "1", consider the following move: locate the first digit "1" from right to left in the sequence, then take the 01-subsequence from left to right starting this "1" of maximal odd length, and change the 01-parity in this subsequence just like flipping the coins in a move. Denote by $a_n$ the total number of moves in the way stated above. We claim: $a_n=\lfloor \frac{2^{n+1}}{3}\rfloor$.

When $n=1$, it is easy to see that $a_1=1=\lfloor \frac{2^2}{3}\rfloor$, proceed by induction. Assume $a_k=\lfloor \frac{2^{k+1}}{3}\rfloor$ holds for $n=k$, i.e., $a_k=\lfloor \frac{2^{k+1}}{3}\rfloor$.