Here are two observations. Both struck me as surprising at first and then not so surprising at all.
$a(m,n)$ is the expected number of flips of a fair coin until one gets either $m$ heads or $n$ tails.
Consider the recurrence relation $$b_{m,n}=b(m-1,n)+b(m,n-1)+2^{m+n-2}$$ with $b_{0,n}=b_{m,0}=0.$ Then $$a_{m,n}=\frac{b_{m,n}}{2^{m+n-2}}.$$
I suppose the first observation translates to the behavior of right/up walks in a rectangle. The second observation easily results from the recurrence for the $a_{m,n}.$
Here is a table of the first few values $b_{m,n}:$
$$\left[ \begin {array}{ccccccccc} 1&3&7&15&31&63&127&255&511 \\ 3&10&25&56&119&246&501&1012&2035 \\ 7&25&66&154&337&711&1468&2992&6051 \\ 15&56&154&372&837&1804&3784&7800&15899 \\ 31&119&337&837&1930&4246&9054&18902&38897 \\ 63&246&711&1804&4246&9516&20618&43616&90705 \\ 127&501&1468&3784&9054&20618&45332&97140&204229 \\ 255&1012&2992&7800&18902&43616&97140&210664& 447661\\ 511&2035&6051&15899&38897&90705&204229& 447661&960858\end {array} \right]$$
One observes (and then easily proves) that
- The table reduced $\bmod 2$ is a Sierpinski triangle.
- $b_{1,n}=1\cdot2^n-1.$
- $b_{2,n}=2\cdot 2^{n+1}-(n+4).$
- $b_{3,n}= 3\cdot 2^{n+2}-\frac{n^2+9n+24}2.$.
- $b_{4,n}= 4\cdot 2^{n+3}-\frac{n^3+15n^2+86n+192}6.$
- $b_{5,n}=5\cdot 2^{n+4}-\frac{n^4+22n^3+203n^2+950n+1920}{24}.$