How to solve this multivariate recursion?:
$$a(m, n, k) = 2a(m-1, n-1, k-1) + a(m-1, n-1, k) + a(m-1, n, k-1) + a(m, n-1, k-1)$$
where $m, n, k$ are positive integers.
Edit:
$a(0,0,0) = a(1,0,0) = a(0,1,0) = a(0,0,1) = 1$
$a(m,0,0) = a(0,m,0) = a(0,0,m) = 0, m > 1$
$a(m,n,k) = a(m,k,n) = a(n,m,k) = a(n,k,m) = a(k,m,n) = a(k,n,m)$
Since there is only one equation that relates the four variables $m$, $n$, $k$ and $a(m,n,k)$ the recursion is under-dertermined. This means that there are probably very many solutions - even if we assume that $a(0,0,0)$ is fixed. In order to come up with concrete solutions we can add up to three additional equations that the function $a$ shall satisfy.
For example let's pick \begin{equation} a(m-1,n,k) = a(m,n-1,k) = a(m,n,k-1) = t\cdot a(m,n,k), \end{equation} where we leave $t\in\mathbb R$ as an unkown constant. With these the first equation (the original recursion formula) can now be simplified to \begin{equation} a(m,n,k) = 2t^3a(m,n,k) + 3t^2a(m,n,k). \end{equation} This equation holds in particular if $t$ is chosen as a root of the polynomial $2t^3+3t^2-1$, i.$\,$e. $t=\frac12$ or $t=-1$. We pick the first choice $t=\frac12$ and use the three chosen equations to derive an explicit solution for $a(m,n,k)$: \begin{equation} a(m,n,k) = t^{m+n+k}a(0,0,0) = \left(\frac12\right)^{m+n+k}a(0,0,0). \end{equation}