Working with cellular automata I came across a system of equations for unknown integers $R\_{k}$ and $C\_{k}$ that looks like this.


$\binom{m}{k}=R\_{k}+C\_{k}+\sum\limits\_{j=1}^{k-1}R\_{j}C\_{k-j}.$


Where 0< k$\leq$ 2m 


(for k>m we take $\binom{m}{k}=R\_{k}=C\_{k}=0$)


Given $R\_{1}$, the system has a unique solution. 


Has anyone seen something similar?
Do you know if it still possible to solve it, if instead of $\binom{m}{k}$ in the left you put something else? 


I just want to know if it's related to something else, and if it's possible to solve more general systems. 



A system of equations for integers
So piggybacking off the two two good answers so far, a system of equations $$a\_k=R\_k+C\_k+\sum\limits\_{j=1}^{k-1}R\_{j}C\_{k-j} \text{ for } 1 \le k \le M$$ is a system of $M$ equations in $2M$ variables which starts out 


$$\begin{align} 
 R\_1 + C\_1 &= a\_1\\\
R\_2+C\_2&= a\_2-R\_1C\_1 \\\
R\_3+C\_3 &= a\_3- R\_2C\_1 - R\_1C\_2 \\\
R\_4+C\_4 &= a\_4- R\_3C\_1 - R\_2C\_2-R\_1C\_3\\\
\end{align}$$


This can be solved top down leading to an a branching cascade of solutions where at stage $j$ we have $$R\_j+C\_j=b\_j$$ where $b\_j$ depends of the previous choices. If the $a\_i$ are non-negative integers and we want the solutions to be of the same form then at that stage we have $\max(0,b\_j+1)$ choices.


The system is equivalent to $$\left(1+\sum\_1^MC\_jx^j\right)\left(1+\sum\_1^MR\_jx^j\right)=1+\sum\_1^Ma\_jx^j+O(x^{M+1})$$ 




---


If we add enough conditions to make it equivalent to $$\left(1+\sum\_1^MC\_jx^j\right)\left(1+\sum\_1^MR\_jx^j\right)=1+\sum\_1^Ma\_jx^j$$ then there is some number of solutions (at most $2^M$) depending on how the right-hand side factors (and the conditions). 


One set of conditions which does this is that $M=2m$ and $R\_j=C\_j=0$ for $j \gt m.$ 


If we say that the $a\_j$ are in $\mathbb{Z}$ and the $R\_j,C\_j$ should be rational then there will be some solutions and they will all be integers (one is $R\_j=0$ and $C\_j=a\_j$). 


As noted, in the given problem the $m+1$ solutions come from $(1+x)^t(1+x)^{m-t}=(1+x)^m$ so they are $C\_j=\binom{t}{j},R\_j=\binom{m-t}{j}$ for some non-negative integer $t \le m.$


It is a disguised form of a [classic problem](http://alexandria.tue.nl/repository/freearticles/597512.pdf) to solve the case that the $a\_k$ are all $1$ for $k \le m$ and the $C\_i,R\_i$ should be nonegative.


