How to calculate the number of paths of minimum length possible a knight can take to get from one corner of a chess board to the opposite one? I've written a small Python script to give me the least number of moves it takes a knight to get from one square to any other on a $n{*}n$ chess board.
But then I've wondered how many paths the knight can take from one corner to the opposite one that use the minimal number of moves (i.e. on any $n{*}n$ board in $2*\Big\lceil{\frac{n-1}{3}}\Big\rceil$ moves, e.g. on a $8{*}8$ board in 6 moves).
So, I added another function to the code to "calculate" exactly that number by doing all possible paths from the opposite corner to the start where every step involves a decrease in the minimum number of moves needed to get to that square (which is calculated beforehand with the other function I mentioned at the beginning).
By now I have values for all $n$ from $1$ to $34$ (attached the list below). As you can see, the values get pretty big, which means it takes really long to calculate them using "brute-force" methods. Do you know of any way to calculate that number without the need of a computer trying all the possibilities?

 A: What do you mean by "brute force" methods? Here is one possible method.
Mark the initial cell with a $0$. Mark all the cells you can reach in one step with a $1/1$. Mark any unmarked cell you can reach from a $1/1$ square  with a $2/?$, where $?$ is the number of cells with a $1/1$ mark that see it. At stage $n+1$ mark each of the unmarked cells which sees a cell with an $n/-$ with $n+1/a$, where $a$ is the sum of the second digits of the $n/-$ cells it sees. The first entry for each cell is the number of moves required to get there. The second entry is the number of minimum paths.
There will be ways of making the marking efficient
I think that if you study edge effects you might be able to be more efficient in going from side $N$ to side $N+1$
A: Not an answer, but too long for a comment.
Every third entry starting at the 2 in position 4 is https://oeis.org/A000984 , the sequence of central binomial coefficients
$$
\binom{2 n}{n} = \frac{2n !}{(n!)^2}.
$$
Perhaps you can show that your set of knight paths matches one of the many sets that sequence counts.
The other two sequences corresponding to entries at positions $3k$ and  $3k+2$ are not in OEIS. You may want to submit them and your full sequence for inclusion. You should certainly add the $3k+1$ counts to the entry at A000984 .
A: The sequence $a(n), n\geq 0$
\begin{align*}
1,0,2,2,8,4,6,\color{blue}{108},40,20,858,252,70,5\,596,1\,344,252,\ldots\tag{1}
\end{align*}
giving the wanted number of shortest knight paths in an $(n\times n)$ board from the left bottom corner to the top right corner is stored as A120399 in OEIS.

The following formula for $a(n), n>3$ is stated: Let \begin{align*}
K=K(n)=2\left\lfloor\frac{n+1}{3}\right\rfloor
\end{align*}
be the shortest path length. Then $a(n), n>3$ is given as
\begin{align*}
a(n)&=2(K-2)\binom{K-1}{K/2-2}&n\equiv0\mod(3)\\
a(n)&=\binom{K}{K/2}&n\equiv1\mod(3)\\
a(n)&=(K-2)(K-3)\binom{K-2}{K/2-1}\\
&\qquad+2\left((K-2)\binom{K-1}{K/2-2}-2\binom{K-2}{K/2-3}\right)&n\equiv2\mod(3)\\
&\qquad+2\left(\binom{K-2}{2}\binom{K-2}{K/2-4}-2\binom{K-3}{K/2-5}\right)
\end{align*}

Example: ($n=8$)
We calculate the number of shortest paths for the $(8\times 8)$ board. We obtain
\begin{align*}
K=K(8)=2\left\lfloor\frac{8+1}{3}\right\rfloor=6
\end{align*}
Since $8\equiv2\mod(3)$ we calculate
\begin{align*}
\color{blue}{a(8)}&=4\cdot 3\binom{4}{2}+2\left(4\binom{5}{1}-2\binom{4}{0}\right)\\
&=12\cdot 6 + 2\left(4\cdot 5 -2\right)\\
&=72+36\\
&\,\,\color{blue}{=108}
\end{align*}
in accordance with the blue marked value in the sequence (1). Note the binomial coefficient is set equal to $0$ if the lower index is negative.
