Solving a complicated vector equation I need to solve the following equation for $x$: 
$$x + a x(x)=l,$$
where $a$ is a constant and $l$ and $x$ are vectors. By $x(x)$, I mean that every element in $x$ is shifted according to itself or in other words $x$ is applied to itself. For example let's say $x=[x_1, x_2, x_3, x_4, x_5]=[-4, 1, 0, 2, 1]$, then $x(x)=[1, 2, 0, 1, -4]$. So, $x_1$ is basically shifted by $x_1$ and thus it is located at last in $x(x)$; $x_2$ is shifted by $x_2$ and thus it is located at first. 
Note that the elements of $x$ are not integers. This is only an example for better understanding the notation. $x$'s elements are "real" numbers. In case of real values, interpolation will be performed for calculating $x(x)$.
Also, note that the direction of shift is (+) $\leftarrow$ (-).
Any comments or ideas is very very much appreciated. Also, I was wondering if there is a standard mathematics notation for what I have shown here as $x(x)$.       
 A: If I well understand (?), your problem is to find the solution of the linear system (*):
$$
A \vec x= \vec l \iff
\begin{bmatrix}
1&a&0&0&\cdots&0\\
0&1&a&0&\cdots&0\\
0&0&1&a&\cdots&0\\
\cdots\\
a&0&0&0&\cdots&1\\
\end{bmatrix}
\begin{bmatrix}
x_1\\
x_2\\
x_3\\
\cdot\\
x_n
\end{bmatrix}=
\begin{bmatrix}
l_1\\
l_2\\
l_3\\
\cdot\\
l_n
\end{bmatrix}
$$
Tis is the only notation in my mind and the solution is given by $\vec x= A^{-1} \vec l$ where the problem is to find the inverse matrix $A^{-1}$. Fortunately this can be done with a general rule. For $n=5$ you can see here  (and you can extend the solution by induction if you want).

(*) the system is:
$$
\begin {cases}
x_1+ax_2=l_1\\
x_2+ax_3=l_2\\
x_3+ax_4=l_3\\
\cdot \\
\cdot \\
x_n+ax_1=l_n\\
\end {cases}
$$
is this a good interpretation of your problem? 
A: Partial answer with a general angle of attack to the problem:
I'm going to assume $l$ is known and we solve for $x$, further assumptions below:
If we denote the vectors $x$ and $l$ as a $n*1$ matrix, one can perform elementary row operations to it, as you describe it with "shifting". The standard notation is $ T_i,_j $, with $i$ and $j$ switching places. If the "shift" is greater than $n$, modular arithmetic solves it for us, namely it is $x_k$ modulo $n$.
we know that $ax(x) = (l-x)$, so for example
$a
\begin{bmatrix}
q-x_1 \\
p-x_2\\
r-x_3
\end{bmatrix}
$ = $(T_1(y_1)T_2(y_2)T_3(y_3))$
$
\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
\end{bmatrix}
$,
where $y_k = n+1-x^r_k$, and $x^r$ is the remainder of $x/n$, and all row operations must be evaluated before applying them, hence the brackets.
I think you could find a more analytical form for righthandside matrix, but I hope this will help.
