Summation equation Recently encountered the following problem:
$$x_n = x_1 - 2 a \sum\limits_{i = 1}^{n-1} \left( 1 - \frac{a}{x_i} \right)$$,
where $n>1$, and $x_1$ is known. The task is to find $x_n$ for arbitrary n. It resembles an integral equation, but in a discrete form. What are the possible tricks to solve for that?
P.S.: Of course, one can try to solve it iteratively, but then the answer becomes cumbersome quickly. Although no one promised me that this particular problem has a nice and compact answer at all.
 A: Thanks for the suggestions, most likely its impossible to solve it analytically in a discrete form, but in a continuous form - yes, you can make a good estimation.
First of all by considering $y_{n+1}-y_{n}$ we can get the following differential equation:
$$ y'(x)=-2a+\frac{2a^2}{y(x)}, \quad y(x=1)=x_1,$$
the solution of which is given by:
$$ y(x) = a \left( W\left(\exp \left( -2 x + \frac{c_1}{a} -1 - \ln(a) \right) \right) - 1 \right), $$
where $W(x)$ is the Lambert W function, and the constant $с_1$ is found from the initial condition $y(x=1)=x_1$ (If anyone knows how to write it analytically -  I would be grateful).
I compared a continuous solution with a discrete one, and in the limit I am interested in $x_1 \gg a$ it works for small $n$.

It is not that precise when $x_1 \sim a$, but very good at $x_1 \gg a$. Moreover, it seems that after that linearly decaying part, the continuous solution shows the average between $x_n$ for even $n$, and odd $n$. The only thing left is to estimate $n$ at which the linear decay stops.
