i am stuck with the following problem: consider the quarter plane $\mathbb{N}_0^2$ with vertices $(i,j)\in\mathbb{N}_0^d$ and edges from each vertex $(i,j)$ to $(i+1,j)$ and $(i,j+1)$, i.e. one "right" edge and one "up" edge.
Now you can turn each vertex off or on. How many different possibilites are there such that $i\in\{1,2,\ldots,\frac{(N+1)(N+2)}{2}-1\}$ vertices are turned off and there is no path from the origin to the diagonal $(n,k)$ satisfying $n+k=N\in\mathbb{N}$ left?
I figured that with $N=2$ you have $5$ vertices and the number of possibilities turning $i$ vertices off and having no path left is $(0,0,1,6,5,1)$, i.e. turning off none or one there is always a path left, turning two off there is one possibility that no path is left, turning three off there are 6 possibilites and so on.
For $N=3$ there are 9 vertices and I calculated the number as $(0,0,1,10,45,x,76,36,9,1)$ where x is unknown. So it seems that the corresponding coefficients of the pascal triangle are added up, i.e. 1,9,36.
However, I am not experienced in combinatorics and have no clue how to get a general formula. It would help me if you point out some literature dealing with this kind of setting! Thank you