Just 2 Rows

It turns out that $f(n) = \frac1{n+1} \binom{2n}{n},$ the $n$ th Catalan number . To see this, we produce a bijection between filled grids and valid parenthetical expressions with $n$ left parentheses and $n$ right parentheses. It is well-known that the latter is counted by the Catalan numbers (see the wiki for a proof).

The bijection is very straightforward: given a filled grid, place the left parentheses in the positions given by the numbers in the top row (which must be in ascending order reading from left to right), and the right parentheses in the positions given by the numbers in the bottom row (which must be in ascending order reading from left to right). It should be clear how to go back and forth between filled grids and parenthetical expressions. For example, the expression $(()()())$ corresponds to the grid $\begin{matrix} 1&2&4&6 \\ 3&5&7&8 \end{matrix}.$

The point is that a valid parenthetical expression must have the property that the $k$ th left parenthesis comes before the $k$ th right parenthesis, for every $k.$ This shows that every number in the bottom row of the corresponding grid is larger than its upstairs neighbor, and the numbers in each row are already in ascending order by construction, so a valid parenthetical expression corresponds to a grid with the given property. The converse is just as clear.

Finally, we have $\lim_{n\to\infty} \frac{f(n)}{f(n-1)} = \lim_{n\to\infty} \frac{4n-2}{n+1} = \fbox{4}.$

I love this problem, and @Patrick Corn 's answer says is all! @Patrick Corn

I wonder if this can be generalized to the $m \times n$ case? (Not just $m \times 2$ )