In a purely algebraic way, I've just stumbled onto the fact that


$$2^{2k} = \sum\_{i+j=k} \binom{2i}{i}\binom{2j}{j},$$


i.e. that the self-convolution of the sequence $\binom{2k}{k}$ is the sequence $2^{2k}$. This is the type of identity for which I would expect there to be a beautiful, simple, direct, and probably well-known combinatorial argument. I just spent some time looking for it but couldn't find it.



> 
> Is there a direct combinatorial proof of the above?
> 
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Geometric remark: such an argument would partition the vertices of an even-dimensional hypercube into some sort of combinatorially meaningful classes of size $\binom{2i}{i}\binom{2j}{j}$. For example, for $k=2$, the vertices of the $4$-cube would be partitioned into classes of size $6=1\cdot 6$, $4=2\cdot 2$, and $6=6\cdot 1$.



Direct combinatorial proof that $2^{2k} = \sum \binom{2i}{i}\binom{2j}{j}$?
Yes, this has an elementary combinatorial interpretation, because
$$
{2i \choose i} 2^{-2i} {2j\choose j}2^{-2j}
$$
(for $i+j=k$) is the probability that the time of the last return to the starting point of a random walk of length $2k$ equals $2i$. This takes some work to show, but is completely elementary; see for example Theorem 3.1 of my (undergraduate level) [notes on random walks](http://www2.math.ou.edu/~cremling/teaching/ln.html) (I should also add that this material is taken straight from Feller's book).


Now your identity is an immediate consequence because after division by $2^{-2k}$ the RHS computes the probability that the last return time equals something.


