Number of Ways to Fill a Matrix with symbols subject to Weird contsraint.

Answerified comment as requested:

If I'm counting correctly, the first few values (starting with $n=1$) are $4$, $25$, $120$, $497$, $1924$, $7265$, $27288$, $102745$, $388692$, $1477721$, $5643064$, $21632785$, $83204260$, $320932177$. This sequence isn't in OEIS.

Here's the code I used.

I don't know how amcalde did it, but after thinking about it some more, I have a derivation, which I've translated to match this formulation of the problem, rather than the one I posted.

Note: this answer is wrong, though I feel like I covered every case without double-counting. I only post it because I think it's close, and maybe someone can see what went wrong with it too arrive at the correct derivation.

(update: I thought I had it for a second, after I got the correct result for the first couple values. It's not the same form amcalde got, though and they aren't equivalent)

We attack it case-by-case...

In the following, "opposite symbol" means diagonally opposite, so $a$ and $d$ are opposites, as are $b$ and $c$.

for cases 1-5, we assume that each symbol is configured in a rectangle (i.e. the number of a given symbol in each row is either $0$ or $k$ for some $k\le n$ and likewise (with possibly different $k$) for each column):

Case 1: there is only one unique symbol in the matrix. there are $4$ of these.

Case 2: every column (or every row) contains only one unique symbol, with two unique symbols in the array as a whole. there are $n-1$ ways to choose the first column of the second symbol, times $4$ ways to choose the symbols, times $2$ to count the same thing for rows for a total of $8n-8$

Case 3:some columns (or rows) contain (the same) 2 distinct symbols, and the rest contain only a single (different) distinct symbol. using similar counting to the above, there are $8(n-1)^2$ of these.

Case 4: there are four distinct symbols in the matrix and no symbol shares a column or row with its opposite. there are $(n-1)^2$ of these.

Case 5: there are four distinct symbols and two opposite symbols appear together in at least one column or row (but not both, because of the rectangle requirement). There are $4(n-1){{n-1} \choose 2}=2(n-1)(n-1)(n-2)=(2n-4)(n-1)^2$ of these.

all other configurations belong to case 6.

Case 6: at least one column and at least one row contains two opposite symbols. In other words, at least one symbol fails to form a rectangle shape.

This is the tough one, but suppose for a second that we're talking about $a$ and $d$. Each such case will correspond to a reflection $a \iff c$ and $b \iff d$, so we'll just multiply by two when we're done. Now imagine drawing a line between every $a$ and $d$ that appear in the same row or column. this will form a stairstep-shaped line going from lower right to upper left. If there are $k$ rows and $r$ columns consisting of $a$ and $d$, then this is a classic stars-and-bars problem with ${k+r} \choose r$ possibilities for the shape of the stairstep. The number of different places a stairstep can be placed is a source of potential pitfalls. To simplify it we assume at first that our stairstep starts with a horizontal line (we'll double our result again, and start our sum at $k=0$), and count the number of a stairstep that ends with a vertical line (we'll also start at $ r=0 $) can be placed. There are $n-r-1$ horizontal positions. There are also $n-k-1$ vertical positions, in all but one of which the stairstep can optionally be extended with an additional horizontal line, so the total number of possible stairsteps is

$$4\sum_{k=0}^{n-2}\left((2n-2k-3)\sum_{r=0}^{n-2}\left({{r+k} \choose r}(n-r-1)\right)\right)$$$$=4\sum_{k=0}^{n-2}\left((2n-2k-3){{k+n} \choose {k+2}}\right)$$$$=4\left(3{{2n-1} \choose n}-2n^2+n-2\right)=6{{2n} \choose n}-8n^2+4n-8$$

Adding it all up we get $$6{{2n} \choose n}-8n^2+4n-8+(2n+5)(n-1)^2+8n-8+4= 6{{2n} \choose n}+2n^3-7n^2+4n-7$$