Consider a binary string $s$ of length $2n$, a reduction step consists in removing a pair of consecutive 0's or of consecutive 1's from $s$. Such steps can be iterated until the final string is alternating single 0's and 1's. Let $c_n$ be the number of binary strings $s$ that reduce to the empty string.
- For $n=1$, this would be the case for $00$ and $11$, so $c_1 = 2$.
- For $n=2$, this would be the case for $0000$, $0011$, $1100$, $0110$, $1001$, $1111$, so $c_2 = 6$.
Looking at the first values, it seems that $c_n = {2n \choose n}$ but I have no clue how to prove this.
Another variant of this is to consider binary strings with $2n_0$ 0's and $2n_1$ 1's, and then it seems that the number of strings that reduce to the empty string is ${n_0 + n_1 \choose n_0 } ^2$ but again I don't have a good way to prove this.
