I proposed [this question at MO](https://mathoverflow.net/questions/378507/laurent-phenomenon) which was resolved neatly by Gerald Edgar in the form
$$
u\_n(x) = 
{2}^{n-1}\prod \_{k=0}^{n-1}(2x+2k+1)
-{2\,n-1\choose n-1}\prod \_{k=0}^{n-1}(x+k).$$
Now that we confirmed that $u\_n(x)$ are all polynomials. I would like to add a follow up:



> 
> **QUESTION.** For $n\geq1$, are the roots of $u\_n(x)$ real negative numbers? It seems to be true.
> 
> 
> 



Real-rooted polynomials
Yes, because $u\_n(x)$ switches sign between each consecutive pair in
$x=0,-1,-2,-3,\ldots,1-n$, and $u\_n(-n) = 0$.


In general, if $P,Q$ are continuous functions each with $n$ simple roots
in an interval $I$, and those roots interlace, then the same argument gives
at least $n-1$ roots of $P-Q$ in $I$, and then an extra root (or even two)
might be forced by the values of $P-Q$ at the endpoints of the interval.
Here $P$ and $Q$ are $2^{n-1} \prod\_{k=0}^{n-1} \, (2x+2k+1)$ and
${2n-1 \choose n-1} \prod\_{k=0}^{n-1} \, (x+k)$, and $I = (-\infty,0)$.


