Does this generalisation of Latin squares have a name? I am interested in looking at $n\times n$ tableaux (or matrices) in which (WLOG) each integer in $\{ 1, 2, \ldots, n \}$ occurs exactly $n$ times.  This is a generalisation of a Latin (or even semi-Latin) square, which obviously has this property.  It is a proper generalisation, as there are tableaux with this property that are not semi-Latin, such as $$\left[\begin{matrix} 1&2&2\\3&1&3\\3&1&2\end{matrix}\right].$$  I have counted the number of such tableaux, up to a suitable (for my applications) notion of isomorphism for $n = 2,3,4$, and there are lots of them.  There are $5$ with $n = 2$; $305$ with $n = 3$; and $2630904$, with $n = 4$.  As these generalise Latin squares, it seems like the sort of thing that people in the combinatorics community might have investigated.  Have such tableaux been studied before?  Do they have a name?  (If these have a name, I might do better with Google.)  Many thanks.
EDIT:  For the $2\times 2$ case, there are $6 = {4\choose 2}$ squares, since we can choose any two of the four matrix positions in which to place a $1$, and then the other two spots must contain a $2$.  These are:
$$\left\{
\left[\begin{matrix}1&1\\2&2\end{matrix}\right],
\left[\begin{matrix}1&2\\1&2\end{matrix}\right],
\left[\begin{matrix}1&2\\2&1\end{matrix}\right],
\left[\begin{matrix}2&1\\1&2\end{matrix}\right],
\left[\begin{matrix}2&1\\2&1\end{matrix}\right],
\left[\begin{matrix}2&2\\1&1\end{matrix}\right]\right\}.$$
Now, of these, only the pair $\left[\begin{matrix}1&2\\2&1\end{matrix}\right]$ and $\left[\begin{matrix}2&1\\1&2\end{matrix}\right]$ are equivalent.
The notion of equivalence or isomorphism used is this.  Two such $n\times n$ matrices $(a_{i,j})$ and $(b_{i,j})$ as above, are regarded as essentially the same if there is a permutation $\sigma\in S_{n}$ for which $\sigma( a_{i,j} ) = b_{\sigma i, \sigma j}$, for all $i$ and $j$.
 A: These are called "equi-n-squares."
A: Before your isomorphism, the tableau representation is not important and you can just view it as a list.  As such, you can pick $n$ of the $n^2$ positions for $1$, $n$ of the remaining $n^2-n$ for $2$ and so on.  It becomes $$\binom {n^2}n \binom{n^2-n}n \binom {n^2-2n}n \ldots \binom nn$$ I didn't find 2630904 in OEIS.  Without the isomorphism it is $\frac{(n^2)!}{(n!)^n}$ which is 
A034841.
A: Courtiel and Vaughan, Gerechte designs with rectangular regions, JCD (2012) calls them a gerechte framework.

A gerechte framework is a partition of an $n \times n$ array into $n$ regions of $n$ cells each.

This definition comes from a gerechte design, which is a Latin square together with the entries partitioned into sets of size $n$ such that each part contains each symbol exactly once (like sudoku).

In 1956, W. U. Behrens [4] introduced a specialisation of Latin squares
  which he called “gerechte”. --- Bailey, Cameron, Connelly.  AMM (2008).  (pdf)

W. U. Behrens, Feldversuchsanordnungen mit verbessertem Ausgleich
der Bodenunterschiede, Zeitschrift für Landwirtschaftliches Versuchs-
und Untersuchungswesen 2 (1956), 176-193.
(Unfortunately, I don't have access to this paper, so my trail ends here.)
