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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 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.

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    $\begingroup$ What is your lengthy argument? In my opinion, you just have to show that the (polynomial) term (in your definition of $F_n(q)$) after $\frac{1}{(1-q)^{2n}}$ has $1$ as root with multiplicity $2n$. This is the shortest argument I can think of. $\endgroup$ – Jose Capco Jun 10 '18 at 7:19
  • $\begingroup$ Don't worry, just try your best. $\endgroup$ – T. Amdeberhan Jun 10 '18 at 23:26
  • $\begingroup$ How did you come across this expression? $\endgroup$ – MTyson Jun 11 '18 at 18:00
  • $\begingroup$ I stumbled on this while working on some $q$-series. $\endgroup$ – T. Amdeberhan Jun 12 '18 at 2:04
  • $\begingroup$ Is your lengthy argument combinatorial in nature? That would seem a nice bonus over the more analytic argument provided... $\endgroup$ – Steven Stadnicki Jun 12 '18 at 23:55
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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. 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. It is crucial to notice that both $(1+x)^{2n+1}$ and $A_{2l+1}(x)$ are palindromic polynomials 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

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  • $\begingroup$ Thanks for this. I'm still hopeful to see alternative proofs. $\endgroup$ – T. Amdeberhan Jun 14 '18 at 15:19

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