I proposed this question at MO 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.
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 $I = (-\infty,0)$.
