Finding a relationship for $\prod_{i=0}^{n-1}x-i$ based on the solution of $1/x=x-n$ Let $\xi_n$ be the positive solution of the equation
$$x=\frac{1}{x}+n\tag{1}\label{1}$$
The sequence $\{a_n\}_{n=1}^{\infty}$ is defined as:
$$a_n=\xi_n(\xi_n-1)\cdots(\xi_n-n+1)\tag{2}\label{2}$$
Now, since (1) implies $\xi_n^2=1+n \xi_n$, it is safe to say that $a_n$ can be ultimately expressed as a linear function of $\xi_n$.
In other words, if we expand $a_n$ in terms of $\xi_n$, all of the $\xi^m,m>1$ terms can be eliminated and replaced with lower degrees. And if we repeat this process, we would finally have:
$$a_n=b_n\xi_n+c_n$$
The problem is, I am having trouble in finding a general formula for $b_n$ and $c_n$. I tried to expand the polynomial (2) and simplify it according to (1), but it became way too complex. I also tried to calculate $b_n,c_n$ for some $n$ in order to find a pattern, but had no luck. Here is the result for $n=1,...,10$:
$$
\begin{array}{c|lcr}
n & \xi_n & b_n & c_n \\
\hline
1 & \frac{\sqrt{5}+1}{2} & 1 & 0 \\
2 & \sqrt{2}+1 & 1 & 1 \\
3 & \frac{\sqrt{13}+3}{2} & 3 & 0 \\
4 & \sqrt{5}+2 & 8 & 4 \\
5 & \frac{\sqrt{29}+5}{2} & 35 & 0 \\
6 & \sqrt{10}+3 & 162 & 54 \\
7 & \frac{\sqrt{53}+7}{2} & 1001 & 0 \\
8 & \sqrt{17}+4 & 6656 & 1664 \\
9 & \frac{\sqrt{85}+9}{2} & 53865 & 0 \\
10 & \sqrt{26}+5 & 467500 & 93500 \\
\end{array}$$
Well, apparently $c_n=0$ for odd values of $n$. But I am stuck there for now and can't go further. Any ideas or hints would be appreciated.
Edit: 
I forgot to mention that if $c_n\ne 0$ (i.e. $n$ is even) then it looks like $b_n=\frac{n}{2} c_n$. Although I have no clue why?
 A: Notice that if $i+j = n$ then
$$(x-i)(x-j) = x^2 - xn + ij = 1 + ij$$
We can use this to derive a formula for $a_n$ by grouping the terms $\{1,n-1\}$, $\{2,n-2\}$ and so on. If $n$ is even then we find
$$\prod_{i=0}^{n-1}(x-i) = x\left(x - \frac{n}{2}\right)\prod_{i=1}^{\frac{n}{2}-1}(x-i)(x-n+i) = \left(\frac{n}{2}x + 1\right)\prod_{i=1}^{\frac{n}{2}-1}(1+i(n-i))$$
If $n$ is odd then 
$$\prod_{i=0}^{n-1}(x-i) = x\prod_{i=1}^{\frac{n-1}{2}}(x-i)(x-n+i) = x\prod_{i=1}^{\frac{n-1}{2}}(1+i(n-i))$$
The expressions above explains the relations you have found; $\frac{b_n}{c_n} = \frac{n}{2}$ for even $n$ and $c_n = 0$ for odd $n$.
A: You have
$$
\begin{gathered}
  \xi _n ^1  = \xi _n ^1  \hfill \\
  \xi _n ^2  = 1 + n\,\xi _n  = \xi _n ^0  + n\,\xi _n ^1  \hfill \\
  \xi _n ^3  = \xi _n ^1  + n\,\xi _n ^2  = \xi _n ^1  + n\,\left( {1 + n\,\xi _n } \right) \hfill \\
  \quad  \vdots  \hfill \\
  \xi _n ^q  = \xi _n ^{q - 2}  + n\,\xi _n ^{q - 1} \quad \;\left| {\;\xi _n ^0  = 1,\xi _n ^1  = \text{sol}(x - 1/x - n)} \right. \hfill \\ 
\end{gathered} 
$$
Solve this recurrence, and throw it into
$$
a_{\,n}  = \xi _n ^{\,\underline {\,n\,} }  = \sum\limits_{0\, \leqslant \,k\, \leqslant \,n} {\left( { - 1} \right)^{\;n - k} \left[ \begin{gathered}
  n \hfill \\
  k \hfill \\ 
\end{gathered}  \right]\xi _n ^k } 
$$
