I am looking at a particular integer sequence, the number of $n\times n$ Young Tableaus (see [OEIS](https://oeis.org/A039622)). In the comments at OEIS, Mitch Harris stated that the same sequence also defines the 



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> Number of linear extensions of the $n\times n$ lattice.
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When looking for some more explanation, I found quite some literature on Boolean Lattices and their linear extensions, but nothing on the $n\times n$ lattice. Not even a definition. Therefore:


1. What is the $n\times n$ lattice?
2. What is a linear extension of such a lattice?
3. Where can I find proof for the statement above?


Alternatively, I take any hints on literature on any of the three questions.


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Linear Extension of the $n\times n$ lattice
1. The $n \times n$ lattice just means the product poset of two chains: $[0,n] \times [0,n]=\{(i,j) | 0 \leq i,j \leq n\}$ where $(i,j) \leq (i',j')$ if and only if $i \leq i'$ and $j \leq j'$.
2. A linear extension of a poset $P$ with $m$ elements is just a bijection $f: P \to \{1,...,m\}$ such that $x \leq y$ in $P$ implies $f(x) \leq f(y)$.
3. If we identify the elements of $[0,n] \times [0,n]$ with the boxes of a square Young diagram, then the condition that $f$ is a linear extension is exactly the same as the condition that the tableau formed by putting the number $f(x)$ in the box corresponding to $x$ is a Young tableau.


This is a special case of a much more general phenomenon. The essential fact is that Young's lattice $Y$ (the set of Young diagrams ordered by inclusion) is the lattice of order ideals of $\mathbb{N} \times \mathbb{N}$. Therefore linear extensions of some ideal $I$ in $\mathbb{N} \times \mathbb{N}$ (in this case $I=[0,n] \times [0,n]$) correspond to maximal chains from $\emptyset$ to $I$ in $Y$. Such chains can naturally be identified with Young tableau by giving boxes the label of the first element of the chain which contains that box.


