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The present problem comes from further consideration of my earlier questions, from here and here.

Start with the following variants of Catalan triangles $\frac{2k+1}{n+k+1}\binom{2n}{n-k}$. Now, define the functions (proven polynomials) $$F_n(q)=\prod_{j=1}^n\frac{1+q^{2j-1}}{(1-q)(1-q^2)}\sum_{k=0}^n\frac{(-q)^k(1+q^{2n+1})}{1+q^{2k+1}}\frac{2k+1}{n+k+1}\binom{2n}{n-k}.$$

QUESTION. Are the polynomials $F_n(q)$ palindromic?

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    $\begingroup$ Your title says unimodal while the text of your question asks about palindromicity. Of course, both are interesting/reasonable. $\endgroup$
    – Sam Hopkins
    Commented Feb 25, 2021 at 18:09
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    $\begingroup$ @SamHopkins: thank you. Edited accordingly now. $\endgroup$
    – T. Amdeberhan
    Commented Feb 25, 2021 at 18:36
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    $\begingroup$ It seems empirically that $F_n(1)$ is the Euler number $E_{2n}$, which equals the number of alternating permutations of $1,2,\dots,2n$. Do the coefficients have a combinatorial interpretation in terms of Euler numbers or something else counted by $E_{2n}$? The sequence of coefficients for $n=2,3,4$ are (2,1,2), (5,6,14,11,14,6,5), and (14,28,76,107,172,185,221,185,172,107,76,28,14). $\endgroup$
    – Richard Stanley
    Commented Feb 25, 2021 at 20:56
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    $\begingroup$ It is easy to see that $F_{n}\left(\frac{1}{q}\right)=q^{n(n+1)} F_{n}(q)$. So it suffices to show that the polynomial $F_{n}(q)$ is of order $n(n+1)$ to show that it is palindromic. $\endgroup$
    – Johannes Trost
    Commented Feb 26, 2021 at 12:37
  • $\begingroup$ @JohannesTrost: do you mean $q^{n(n+1)}F_n(1/q)=F_n(q)$? A polynomial satisfying this is palindromic whatever its degree. $\endgroup$
    – Richard Stanley
    Commented Mar 4, 2021 at 2:09

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