I found a recursive formula that looks very similar to Pascal's triangle. Where in Pascal's triangle you have the recurring formula
$$u_{n,k} = u_{n-1,k} + u_{n-1,k-1},$$
which gives the Binomial function $\binom{n}{k}$ with the initial condition $u_{1,0}=u_{1,1} = 1$. In my case the second term is one value lower in the index
$$u_{n,k} = u_{n-1,k} + u_{n-2,k-1} , $$
with the initial condition $u_{1,0} = u_{2,0} = 1 $ and $ u_{2,1} = 2$. 
It looks like the image above (where we miss the top 1). Where n is the height and k is the width. How can one calculate a closed form for this? I've done this bedore with single variable recursion formulas by using the characteristic polynomial, but I'm not sure how to start now.
Hint The entries in the $k$th antidiagonal are binomial coefficients $n \choose k$.