Number of ways from top of the black square on chess board to any bottom black square when moving downwards diagonally to black squares only.

Question

I have attempted this problem by drawing out grid like this: \begin{array}{|c|c|c|c|c|c|c|c|} \hline1&&1&&1&&1&\\\hline&2&&2&&2&&1\\\hline2&&4&&4&&3&\\\hline&6&&8&&7&&3\\\hline6&&14&&15&&10&\\\hline&20&&29&&25&&10\\\hline20&&49&&54&&35&\\\hline&69&&103&&89&&35\\\hline \end{array}
This gives me $69+103+89+35=296$ $ways$.
However this method is obviously long and will not be applicable for bigger grids like 20×20 so is there a more systematic way to solve this using combinatorics.
So far i have tried something like : $4$ ways to choose black square on top row. Then $(4×2)-1=7$ ways for choosing a black square on second row (I subtracted $1$ because one of the black square don't lead to 2 black squares but one). Similarly, $(7×2)-1=13$ ways to choose on third row. But for forth row it should be (according to the way I am going) $(13×2)-1=25$ ways but there are 24 ways. So my method stops working.
What would be the way to go about in this question?

Answer 1

The number of diagonal paths from any black square on the top row of a $2n\times 2n$ chessboard to any black square on the bottom row is in OEIS: https://oeis.org/A153336. They give the formula $$ (2n+1)2^{2n-2}-(4n-2)\binom{2n-2}{n-1}. $$ When $n=4$, this indeed evaluates to $296$.