Tricky two-dimensional recurrence relation I would like to obtain a closed form for the recurrence relation
$$a_{0,0} = 1,~~~~a_{0,m+1} = 0\\a_{n+1,0} = 2 + \frac 1 2 \cdot(a_{n,0} + a_{n,1})\\a_{n+1,m+1} = \frac 1 2 \cdot (a_{n,m} + a_{n,m+2}).$$
Even obtaining a generating function for that seems challenging. Is there a closed form for the recurrence relation or at least for the generating function? Alternatively, is there a closed form for $a_{n,0}$?

Here is the table for $a_{n,m}2^m$ (since these are integers):
$$\begin{pmatrix}
1 & 5 & 14 & 35 & 82 & 186 & 412 & 899 & 1938 \\
 0 & 1 & 5 & 15 & 40 & 98 & 231 & 527 & 1180 \\
 0 & 0 & 1 & 5 & 16 & 45 & 115 & 281 & 660 \\
 0 & 0 & 0 & 1 & 5 & 17 & 50 & 133 & 336 \\
 0 & 0 & 0 & 0 & 1 & 5 & 18 & 55 & 152 \\
 0 & 0 & 0 & 0 & 0 & 1 & 5 & 19 & 60 \\
 0 & 0 & 0 & 0 & 0 & 0 & 1 & 5 & 20 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 5 \\
 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 
\end{pmatrix}$$
The top row does not give a hit in OEIS, but I conjecture that it is given by the following:
$$
a_{n,0} =  \frac{(4 n+3) \Gamma \left(\frac{n+1}{2}\right)}{\sqrt{\pi } \Gamma \left(\frac{n}{2}+1\right)}-2 
$$
if $n$ is even, and
$$a_{n,0} =
\frac{(4n+5) \Gamma \left(\frac{n}{2}+1\right)}{\sqrt{\pi } \Gamma
   \left(\frac{n+3}{2}\right)}-2
$$
if $n$ is odd.
