Define the functions
$$F\_n(q)=\frac1{(1-q)^{2n}}\sum\_{k=0}^n(-q)^k\frac{2k+1}{n+k+1}\binom{2n}{n-k}
\prod\_{j=0,\,j\neq k}^n\frac{1+q^{2j+1}}{1+q}.$$
The numbers $\frac{2k+1}{n+k+1}\binom{2n}{n-k}$ belong to a family of Catalan triangle of which the special case
$k=0$ yields the Catalan numbers $C\_n=\frac1{n+1}\binom{2n}n$. 


I have raised [this question](https://mathoverflow.net/questions/302028/positivity-of-certain-polynomial-coefficients) earlier which has not yet received an answer. Hence, I would like to propose a (milder) problem. My hope here is to see a better proof than my lengthy argument for



> 
> $F\_n(q)$ are all polynomials.
> 
> 
> 



Prove that these are polynomials
First notice that $\frac{2k+1}{n+k+1}\binom{2n}{n-k} = \frac{2k+1}{2n+1}\binom{2n+1}{n-k}$.


To prove that $F\_n(q)$ is a polynomial it is enough to show that
$$\sum\_{k=0}^n(-1)^k(2k+1)\binom{2n+1}{n-k}\frac{q^k}{1+q^{2k+1}}$$
has the zero $q=1$ of multiplicity $2n$.
Plugging $q=e^t$, one needs to show that
\begin{split}
&\frac{e^{-t/2}}{2}\sum\_{k=0}^n(-1)^k(2k+1)\binom{2n+1}{n-k}\mathrm{sech}(\frac{2k+1}2t)\\
=& \frac{e^{-t/2}}{2}\sum\_{k=0}^n(-1)^k(2k+1)\binom{2n+1}{n-k} \sum\_{l\geq 0} \frac{E\_{2l}(\frac{2k+1}2t)^{2l}}{(2l)!}\\
=& \frac{e^{-t/2}}{2} \sum\_{l\geq 0} \frac{E\_{2l}t^{2l}}{2^{2l}(2l)!}\sum\_{k=0}^n (-1)^k\binom{2n+1}{n-k} (2k+1)^{2l+1}
\end{split}
has the zero $t=0$ of multiplicity $2n$, where $E\_{2l}$ are [Euler numbers](https://en.wikipedia.org/wiki/Euler_number). So, the problem boils down to proving that
$$(\star)\qquad\sum\_{k=0}^n (-1)^k\binom{2n+1}{n-k} (2k+1)^{2l+1} = 0\quad\text{for all }l<n.$$


**Proof of $(\star)$.**


Let us notice that the l.h.s. of $(\star)$ is nothing else but $(-1)^n$ times the coefficient of $x^{2n}$ in the polynomial
$$f(x):=(1-x^2)^{2n+1} \frac{A\_{2l+1}(x)}{(1-x)^{2l+2}}=(1+x)^{2n+1}(1-x)^{2n-2l-1} A\_{2l+1}(x),$$
where $A\_{2l+1}(x)$ is [Eulerian polynomial](https://en.wikipedia.org/wiki/Eulerian_number). It is crucial to notice that both $(1+x)^{2n+1}$ and $A\_{2l+1}(x)$ are [palindromic polynomials](https://en.wikipedia.org/wiki/Reciprocal_polynomial) and so is their product, while $(1-x)^{2n-2l-1}$ is an antipalindromic polynomial. It follows that $f(x)$ is also antipalindromic, and since $\deg f(x)=4n$, we have $[x^{2n}]\ f(x)=0$.
QED


