Solution of a 2D Recurrence sequence Can we solve the following recurrence relation:
$$a_{m,n} = 1 + \frac{a_{m,n-1}+a_{m-1,n}}{2}$$
with $a_{0,n}=a_{m,0}=0$? If not, can we get an estimate of the growth of $a_{m,n}?$
I encountered this question when analyzing a 2D random walk on $R_+^2$ with boundaries at $x = m$ and $y = n$.
 A: Inspired by Robert Israel's answer: Consider the generating function of the double-indexed array $(a_{m,n})$: 
\begin{multline*}
 P(s,t):=\sum_{m,n\ge0}a_{m,n}s^m t^n=\sum_{m,n\ge1}a_{m,n}s^m t^n \\ 
 =\sum_{m,n\ge1}\Big(1 + \frac{a_{m,n-1}+a_{m-1,n}}{2}\Big)s^m t^n
 =\frac{st}{(1-s)(1-t)}+\frac{s+t}2\,P(s,t), 
\end{multline*}
whence
\begin{align*}
 P(s,t)&=\frac{st}{(1-s)(1-t)}\frac1{1-\frac{s+t}2} \\ 
 &=\sum_{m,n\ge1}s^mt^n\sum_{k\ge0}\Big(\frac{s+t}2\Big)^k \\ 
 &=\sum_{m,n\ge1}s^mt^n\sum_{k\ge0}\Big(\frac12\Big)^k \sum_{j=0}^k\binom kj s^j t^{k-j}\\ 
 &=\sum_{p,q\ge1}s^pt^q\sum_{j=0}^{p-1}\sum_{k=j}^{q-1+j}\Big(\frac12\Big)^k \binom kj 
\end{align*}
for $s,t$ in $(-1,1)$, so that 
\begin{equation*}
 a_{p,q}=\sum_{j=0}^{p-1}\sum_{k=j}^{q-1+j}\Big(\frac12\Big)^k \binom kj \tag{0}
\end{equation*}
for natural $p,q$. 

Added: As an illustration of the use of formula (0), let us obtain the asymptotics of $a_{p,q}$ as $p,q\to\infty$. To do that, it is convenient to use the central limit theorem (CLT) of probability theory, applied here to the binomial distribution. Indeed, note that 
\begin{equation*}
 \Big(\frac12\Big)^k \binom kj=P(X_k=j),
\end{equation*}
where $X_k$ has the binomial distribution with parameters $k$ and $1/2$, so that, by the CLT,
\begin{equation*}
 P(X_k\le x)\underset{k\to\infty}\longrightarrow\Phi(g_x(k))
\end{equation*}
uniformly in all real $x$, where $\Phi$ is the standard normal cumulative distribution function
and 
\begin{equation*}
g_x(k):=\frac{x-k/2}{\sqrt{k/4}}=\frac{2x-k}{\sqrt k}=\frac{2x}{\sqrt k}-\sqrt k,  
\end{equation*}
which is decreasing in $k$ for each $x\ge0$. 
Without loss of generality, 
\begin{equation*}
 p\ge q. 
\end{equation*}
Interchanging the order of summation in (0), we have 
\begin{align*}
a_{p,q}&:=\sum _{k=0}^{p+q-2} \sum _{j=\max (0,1+k-q)}^{\min (p-1,k)} P(X_k=j) \\
&=\sum _{k=0}^{p+q-2}P(\max (0,1+k-q)\le X_k\le\min (p-1,k)) \\ 
&=S_1+S_2+S_3=p+T_1-T_2, \tag{1}
\end{align*}
where 
\begin{align*}
S_1&:=\sum _{k=0}^{q-1}P(0\le X_k\le k)=q,  \\  
S_2&:=\sum _{k=q}^{p-1}P(1+k-q\le X_k\le k)
=p-q-\sum _{k=q}^{p-1}P(X_k\le k-q),  \\  
S_3&:=\sum _{k=p}^{p+q-2}P(1+k-q\le X_k\le p-1) \\ 
&=\sum _{k=p}^{p+q-2}[P(X_k\le p-1)-P(X_k\le k-q)], \\  
T_1&:=\sum _{k=p}^{p+q-2}P(X_k\le p-1)
=\sum _{k=p}^{p+q-2}\Phi(g_{p-1}(k))+o(q), \\
T_2&:=\sum _{k=q}^{p+q-2}P(X_k\le k-q)=\sum _{k=q}^{p+q-2}\Phi(g_{k-q}(k))+o(p) \\
&=p-1-U+o(p)=p-U+o(p), \quad U:=\sum _{k=q}^{p+q-2}\Phi(g_{q}(k));  \tag{2}
\end{align*}
here we used the identities $g_{k-q}(k)=-g_{q}(k)$ and $\Phi(-u)=1-\Phi(u)$. 
Let now $A_p$ vary with $p$ so that $A_p\to\infty$, $A_p=o(\sqrt p)$, and 
\begin{equation*}
 k_p:=2p-A_p\sqrt{2p}\in\mathbb Z. 
\end{equation*}
Then for integers $k$ in $[p,k_p]$ we have 
\begin{equation*}
 g_{p-1}(k)=g_p(k)+o(1),\quad 
 g_p(k)= \frac{2p-k}{\sqrt k}\ge\frac{2p-k_p}{\sqrt k_p}\ge A_p\to\infty,  
\end{equation*}
so that $\Phi(g_{p-1}(k))\to1$ uniformly in integers $k$ in $[p,k_p]$ and hence 
\begin{multline*}
 T_{11}:=\sum _{k=p}^{\min(p+q-2,k_p)}P(X_k\le p-1)
 =\sum _{k=p}^{\min(p+q-2,k_p)}\Phi(g_{p-1}(k))+o(q) \\ 
 =\min(p+q-2,k_p)-(p-1)+o(q)+o(q)=q+o(p),  
\end{multline*}
because $p\ge q$ and $A_p\sqrt{2p}=o(p)$. 
Also, 
\begin{equation*}
 0\le T_1-T_{11}=\sum _{k=1+\min(p+q-2,k_p)}^{p+q-2}P(X_k\le p-1)
 \le(p+q-2)-\min(p+q-2,k_p)=o(p), 
\end{equation*}
again because $p\ge q$ and $A_p\sqrt{2p}=o(p)$. So,
\begin{equation*}
 T_1=q+o(p). \tag{3}
\end{equation*}
The term $U$ is estimated similarly to $T_1$. Write 
\begin{equation*}
 U=U_1+U_2+U_3,
\end{equation*}
where 
\begin{align*}
 U_1&:=\sum _{k=q}^{k_q}\Phi(g_{q}(k)), \\ 
 U_2&:=\sum_{k=1+k_q}^{\min(l_q,p+q-2)}\Phi(g_{q}(k)), \\ 
 U_3&:=\sum_{k=1+\min(l_q,p+q-2)}^{p+q-2}\Phi(g_{q}(k)),  
\end{align*}
where 
\begin{equation*}
 l_q:=2q+A_q\sqrt{p+q}. 
\end{equation*}
The term $U_1$ is estimated similarly to, and a bit more simply than, $T_{11}$, and we get 
\begin{equation*}
 U_1=q+o(q). 
\end{equation*}
The term $U_2$ is estimated similarly to $T_1-T_{11}$, and we get 
\begin{equation*}
 U_2=o(q). 
\end{equation*}
For integers $k$ in $[l_q,p+q]$ we have 
\begin{equation*}
 g_q(k)=\frac{2q-k}{\sqrt k}\le\frac{2q-l_q}{\sqrt k}=\frac{-A_q\sqrt{p+q}}{\sqrt k}
 \le-A_q\to-\infty, 
\end{equation*}
so that $\Phi(g_q(k))\to0$ uniformly in integers $k$ in $[l_q,p+q]$ and hence 
\begin{equation*}
 U_3=o((p+q-2)-\min(l_q,p+q-2))=o(p). 
\end{equation*}
So, 
\begin{equation*}
 U=q+o(p). \tag{4}
\end{equation*} 
Collecting the pieces (1)--(4) together, we get 
\begin{equation*}
 a_{p,q}=p+T_1-T_2=p+T_1-p+U+o(p)=p+q-p+q+o(p)=2q+o(p),
\end{equation*}
under the assumption $p\ge q$. Without this assumption, 
\begin{equation*}
 a_{p,q}=2\min(p,q)+o(\max(p,q))
\end{equation*}
as $p,q\to\infty$. 
A: If $P_k(t) = \sum_{m=0}^k a_{m,k-n} t^m$ is the generating function of an ascending antidiagonal, we have
$$P_k(t) = \frac{t^k-t}{t-1} + \frac{1+t}{2} P_{k-1}(t), \ P_0(t) = 0 $$
and this can be solved:
$$ P_k(t) =  \frac{2 t  + 2t^{k+1} - 4 t ((t+1)/2)^k}{(t-1)^2}$$
A: By induction, it is immediate that $a_{m,n}\le 2\min(m,n)$. In addition, when $m$ is fixed and $n>m$ we have $a_{m,n}=2m-P_m(n)/2^n$, where $P_m(x)$ is a polynomial of degree $m$, so $2m$ is an exponentially good bound. For $m=n$, it seems that $a_{n,n}=2n(1-1/\sqrt{\pi n}+o(1/\sqrt{n}))$, but I have not taken the time to prove it.
Added: numerically
$$a_{n,n}=2n(1-1/\sqrt{\pi n}+1/(8\sqrt{\pi n^3})-1/(128\sqrt{\pi n^5})+...)$$
Looks like a well-known expansion, probably in Knuth vol 1.
