The Delannoy number $D(a,b)$ can be defined as the numbers of paths on $\mathbb Z^2$ from $(0,0)$ to $(a,b)$ using only steps $(0,1)$, $(1,0)$ and $(1,1)$.

It is straightforward to see that they follow the recursion (using either first-step or last step analysis, for example): $$D(a,b)=D(a-1,b) + D(a,b-1)+D(a-1,b-1),$$ where $D(0,b) = D(a,0)=1$.

I came to wonder if closed-form formulas existed for those numbers (actually, it was a riddle presented to me). The formula I came up with first fixes the number of diagonal steps in a given path, then counts the number of arrangements with a multinomial coefficient. This gives: $$D(a,b) = \sum_{i=0}^{a\wedge b}\binom{a+b-i}{a-i,\,b-i,\,i}.$$

However, a second formula is mentioned on MathWorld: $$D(a,b) = \sum_{i=0}^{a\wedge b}2^i\binom{a}{i}\binom{b}{i},$$

and I was wondering if it had a similarly simple combinatorial interpretation.