Actually, I will be asking two, but related, questions.



> 
> **Question 1.** Is there some theory about recurrent relations with *several* indices. For example, If we have a relation on $A\_{i;j}$: $$A\_{i+\alpha; j+\beta} = F(A\_{i+\alpha-1;j+\beta -1}, \ldots, A\_{i;j}).$$
> 
> 
> 


And related



> 
> **Question 2.** Let $m, n$, $Q\_{m}^ {n}\in \mathbb{N}$. Then solve given 2-recurrent relation:
>  $$Q\_{m}^{n+1} = 2Q\_{m}^{n} + Q\_{m-1}^{n}$$
>  with the following conditions on $Q$:
>  \begin{align}
> Q\_{j}^{j} &= 1,\\
> Q\_{k}^{l} &= 0,\quad \text{for } k>l.
> \end{align}
> 
> 
> 


I have solved the last problem by pattern recognition followed by induction , two ways that I do not like to prove something. I would like to get more rough method. 




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This relation naturally appeared when I tried to count the number of $m$-dim edges in $n$-dim cube (vertex is $0$-dim, edge - 1-dim. square - $2$-dim and etc). In this context the answer is obvious I think. I will post my solution below. 



Recurrent relation with several indices ( How many $m$-dim cubes in $n$-dim cube )
This was an answer before the typo was edited by the OP on $Q\_m^{n+1}=2Q\_m^n+Q\_{m-1}^{n-1}$.


Let $F(x,y)=\sum\_{n,m\geq0}Q\_n^mx^my^n$ be a generating function. Based on the recurrence relation alone, you should be getting
$$F(x,y)=\frac{P(x,y)}{1-2y-xy^2};$$
for some polynomial $P(x,y)$ which depends on the initial conditions (this, I leave up to you).


*Caveat.* You last initial condition does not sync with the recurrence. For example, $Q(3,3)=2Q(3,2)+Q(2,1)=0$ but supposedly $Q(3,3)=1$ and $Q(3,2)=Q(2,1)=0$. A mismatch!


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After the edit by OP: $Q\_m^{n+1}=2Q\_m^n+Q\_{m-1}^n$. Again, let $F(x,y)=\sum\_{n,m\geq0}Q\_n^mx^my^n$ be a generating function. Based on the recurrence relation alone, you should be getting
$$F(x,y)=\frac{P(x,y)}{1-2y-xy}=P(x,y)\sum\_{n=0}^{\infty}(2+x)^ny^n
=P(x,y)\sum\_{m,n\geq0}\binom{n}m2^{n-m}x^my^n;$$
for some polynomial $P(x,y)$ which depends on the initial conditions ($P=1$, I think). Therefore, $Q\_m^n=\binom{n}m2^{n-m}$.


