$F$ 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?



Search for a general formula from known iterative relation
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:
$$n\_1!n\_2!\ [x^{n\_1}y^{n\_2}]\ F^{(k)}(H)(x,y) = 
n\_1!n\_2!k!\sum\_{c=0}^k 
\sum\_{s=0}^{c} \frac{(\eta\_1/2)^s}{s!}
\sum\_{t=0}^{k-c} \frac{(\eta\_2/2)^t}{t!}
C\_{n\_1,s,c}C\_{n\_2,t,k-c}
\frac{\theta\_{n\_1+c-2s,n\_2+k-c-2t}}{(n\_1+c-2s)!(n\_2+k-c-2t)!}.
$$


