Search for a general formula from known iterative relation $F$ is a mapping among $\{\theta_{n_1n_2}\}$, with $\eta_{1/2}$ being arbitrary constants involved.
$F: \theta_{n_1n_2} \rightarrow \theta_{n_1+1n_2}+\theta_{n_1n_2+1}+\eta_{1}n_1\theta_{n_{1}-1n_{2}}                                                                    
  +\eta_{2}n_2\theta_{n_{1}n_{2}-1}$
so what is $F^k:\theta_{n_1n_2} \rightarrow 　?$ ($F$ acts $k$ times)
Is there a general formula?
 A: Consider the generating function:
$$H(x,y) := \sum_{i,j} \theta_{i,j} \frac{x^i}{i!} \frac{y^j}{j!}.$$
Extending $F$ by linearity, define
$$F(H)(x,y):= \sum_{i,j} \left(\theta_{i+1,j}+\theta_{i,j+1}+\eta_{1}i\theta_{i-1,j}+\eta_{2}j\theta_{i,j-1}\right) \frac{x^i}{i!} \frac{y^j}{j!}$$
so that the question amounts to finding $n_1!\cdot n_2!\cdot [x^{n_1}y^{n_2}]\ F^{(k)}(H)(x,y)$.
The definition of $F$ implies that
\begin{split}
F(H)(x,y) &= \frac{\partial}{\partial x} H(x,y) + \frac{\partial}{\partial y} H(x,y) + (\eta_1 x + \eta_2 y)H(x,y) \\
&= e^{-\frac{\eta_1}2x^2-\frac{\eta_2}2y^2}\left(\frac{\partial}{\partial x}+\frac{\partial}{\partial y}\right) e^{\frac{\eta_1}2x^2+\frac{\eta_2}2y^2} H(x,y).
\end{split}
Then
$$F^{(k)}(H)(x,y) = e^{-\frac{\eta_1}2x^2-\frac{\eta_2}2y^2}\left(\frac{\partial}{\partial x}+\frac{\partial}{\partial y}\right)^k e^{\frac{\eta_1}2x^2+\frac{\eta_2}2y^2} H(x,y),$$
from where the coefficient of $x^{n_1}y^{n_2}$ can be extracted by standard means as follows.

$$
[x^{n_1}y^{n_2}]\ F^{(k)}(H)(x,y) = 
\sum_{a=0}^{\lfloor n_1/2\rfloor} \frac{(-\eta_1/2)^a}{a!}
\sum_{b=0}^{\lfloor n_2/2\rfloor} \frac{(-\eta_2/2)^b}{b!}
\sum_{c=0}^k \binom{k}c (n_1-2a+c)_c (n_2-2b+k-c)_{k-c}
\sum_{d=0}^{\lfloor (n_1-2a+c)/2\rfloor} \frac{(\eta_1/2)^d}{d!}
\sum_{f=0}^{\lfloor (n_2-2b+k-c)/2\rfloor} \frac{(\eta_2/2)^f}{f!}
\frac{\theta_{n_1-2a+c-2f,n_2-2b+k-c-2f}}{(n_1-2a+c-2f)!(n_2-2b+k-c-2f)!}.
$$
Introducing $s:=a+d$ and $t:=b+f$, we rewrite the above as
$$k!\sum_{c=0}^k 
\sum_{s=0}^{\lfloor (n_1+c)/2\rfloor} \frac{(\eta_1/2)^s}{s!}
\sum_{t=0}^{\lfloor (n_2+k-c)/2\rfloor} \frac{(\eta_2/2)^t}{t!}
C_{n_1,s,c}C_{n_2,t,k-c}
\frac{\theta_{n_1+c-2s,n_2+k-c-2t}}{(n_1+c-2s)!(n_2+k-c-2t)!},
$$
where
$$C_{n,s,c}:=\sum_{a=0}^s \binom{s}{a} \binom{n-2a+c}c (-1)^a = (-1)^n\sum_{a=0}^s \binom{s}{a} \binom{-c-1}{n-2a} (-1)^a$$
is the coefficient of $x^n$ in $(-1)^n (1-x)^s (1+x)^{s-c-1}$. Since $s\leq \lfloor (n_1+c)/2\rfloor$, it follows that $C_{n_1,s,c}=0$ if $s\geq c+1$; and similarly for $C_{n_2,t,k-c}$. Hence, we obtain a closed formula:
$$n_1!n_2!\ [x^{n_1}y^{n_2}]\ F^{(k)}(H)(x,y) = 
n_1!n_2!k!\sum_{c=0}^k 
\sum_{s=0}^{c} \frac{(\eta_1/2)^s}{s!}
\sum_{t=0}^{k-c} \frac{(\eta_2/2)^t}{t!}
C_{n_1,s,c}C_{n_2,t,k-c}
\frac{\theta_{n_1+c-2s,n_2+k-c-2t}}{(n_1+c-2s)!(n_2+k-c-2t)!}.
$$
