The motivation to consider the sequences below comes from an efficient way to represent trees on $n$ nodes using $2n$ bits.
Let $n\in\mathbb{N}$ be a positive integer. Let us call $s\in\{0,1\}^{2n}$ a balanced-parentheses (BP) sequence if $\text{card}\big(s^{-1}(\{0\})\big)=\text{card}\big(s^{-1}(\{1\})\big) = n$ and for all $k\in 2n$ we have $$\text{card}\big(s|_k^{-1}(\{1\})\big) \geq \text{card}\big(s|_k^{-1}(\{0\})\big),$$ where $s|_k$ is the restriction of $s$ to $k = \{0,\ldots,k-1\}$.
If $B_n$ is the number of BP sequences in $\{0,1\}^{2n}$, how does $B_n$ grow as $n\to\infty$?
As pointed out in the comments, these are the Catalan numbers. A good starting point is $B_n=\frac{1}{n+1}\binom{2n}{2}$, which is established here. For the asymptotics we can use Stirling's approximation, which gives $B_n=(1+\mathcal O(\frac{1}{n}))\frac{1}{n}\frac{\sqrt{2\pi(2n)}\left(\frac{2n}{e}\right)^{2n}}{2\pi n\left(\frac{n}{e}\right)^{2n}}=(1+\mathcal O(\frac{1}{n}))\frac{1}{\sqrt{\pi}n^{3/2}}4^{n}$.