Filling an $n\times n$ board This problem has been bothering me for quite a while now.
Consider an $n\times n$ chessboard, with $n$ being an odd positive integer. In the middle square of the board, a $0$ is placed. Starting with that board, how many ways are there to fill the board up with non-negative integers (greater than or equal to $0$) such that no two squares that share an edge are filled with numbers whose difference is greater than $1$.
The problem is easy to do for $n=3$ by simple casework, and I've found the answer for that case to be $433$, but I am clueless on what to do for larger $n$. Even getting the number of ways for $n=5$ seems very hard.
And, before anyone asks, this problem came from my head, not a book or anything of the sorts, so I don't have any clues to how this can be found.
Thanks!
 A: This is not an answer but a note how to compute the number of ways $\mathcal{N}_n$ for $n = 5$.
We can split the $5\times 5$ board into $8$ right angled triangles. Let us label the states in one of the triangle as follows:
$$\begin{array}{|ccccc|}
\hline
* & * & * & * & e\\
* & * & * & c & d\\
* & * & 0 & a & b\\
* & * & * & * & *\\
* & * & * & * & *\\
\hline
\end{array}$$
It is easy to check given the constraint, there are


*

*5  possibilities for $\verb/ab/$ - $00, 01, 10, 11, 12$.

*12 possibilities for $\verb/ce/$ - $00, 01, 02, 10, 11, 12, 13, 20, 21, 22, 23, 24$.


Given any legal combination of $a, b, c, e$, the number of possibilities for $d$ can be
counted by brute force. The result is summarized by following table.
$$\begin{array}{r|lllll}
& & & \verb/ab/ & & \\
\verb/ce/ & 00 & 01 & 10 & 11 & 12\\
\hline
00 & 2 & 2 & 2 & 2 & 1\\
01 & 2 & 2 & 2 & 2 & 1\\
02 & 1 & 1 & 1 & 1 & 1\\
10 & 2 & 2 & 2 & 2 & 1\\
11 & 2 & 3 & 2 & 3 & 2\\
12 & 1 & 2 & 1 & 2 & 2\\
13 & 0 & 1 & 0 & 1 & 1\\
20 & 0 & 0 & 1 & 1 & 1\\
21 & 0 & 0 & 1 & 2 & 2\\
22 & 0 & 0 & 1 & 2 & 3\\
23 & 0 & 0 & 0 & 1 & 2\\
24 & 0 & 0 & 0 & 0 & 1\\
\end{array}$$
For the other 7 triangles, it is clear each of them have a similar set of possibilities.
If we pick 8 admissible configurations, one for each of the 8 triangles and glue them together. We can construct a legal configuration for the whole board provided the configurations of the triangles match at their boundaries. 
A consequence of this is if we define $M_5$ as the $12\times 5$ matrices with entries in above table, the total number of ways for $n = 5$ can be evaluated as a trace!
$$\mathcal{N}_5 = \text{Tr} \left( (M_5^T M_5 )^4 \right) = 178383613
$$
As a double check, for the easier case $n = 3$, the corresponding $M_3$ has entries given by
the table:
$$\begin{array}{r|rl}
& \quad\rlap{\verb/ a/} & \\
\verb/c/ & 0 & 1\\
\hline
0 & 1 & 1\\
1 & 1 & 1\\
2 & 0 & 1
\end{array}
$$
This leads to
$$\mathcal{N}_3 = \text{Tr}\left(
\begin{bmatrix}1 & 1 & 0\\1 & 1 & 1\end{bmatrix}
\begin{bmatrix}1 & 1\\1 & 1\\0 & 1\end{bmatrix}
\right)^4
= \text{Tr}\begin{bmatrix}2 & 2\\ 2 & 3\end{bmatrix}^4
= 433
$$
as expected.
Notes


*

*This way of counting $\mathcal{N}_n$ is inspired by the general technique Transfer-matrix method for solving problems in statistical mechanics. The key is break the configuration into units similar to each other.
Compute the possibilities for individual units and represent them as matrices.
Finally, convert the original sum into the trace of product of these matrices.

