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Let $[n]_q=\frac{1-q^n}{1-q}$ with $[0]_q=0$. Recall the $q$-factorials $[n]_q!=[1]_q[2]_q\cdots[n]_q$ (with $[0]_q!=1$) and the $q$-binomials $$\binom{n}k_q=\frac{[n]_q!}{[k]_q!\,[n-k]_q!}.$$

Now, consider the polynomials $$W_n(q):=\frac{1-q^{3n}}{1-q^{2n}}\binom{2n}n_q.$$ Examples. $W_1(q)=q^2+q+1$ and $W_2(q)=q^6 + q^5 + 2q^4 + q^3 + 2q^2 + q + 1$.

QUESTION. Is this true? The number of monomials in $W_n(q)$ equals $n^2+n+1$.

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    $\begingroup$ Is it obvious the $W_n(q)$ are indeed polynomials in $q$? $\endgroup$
    – Sam Hopkins
    Commented Nov 8, 2021 at 18:53
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    $\begingroup$ Don't you just compute $\deg W_n$? $\endgroup$
    – Ilya Bogdanov
    Commented Nov 8, 2021 at 19:03
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    $\begingroup$ @SamHopkins: it's not obvious. But, similar techniques proving Gaussian polynomials and $q$-Catalan polynomials should do it. $\endgroup$
    – T. Amdeberhan
    Commented Nov 8, 2021 at 19:11
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    $\begingroup$ @IlyaBogdanov: it's related, but one needs to show there are no internal zeroes (missing terms). $\endgroup$
    – T. Amdeberhan
    Commented Nov 8, 2021 at 19:35
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    $\begingroup$ Is it true that all coefficients of $W_n$ are non-negative? $\endgroup$
    – Pietro Majer
    Commented Nov 8, 2021 at 19:52

1 Answer 1

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Using the fact that $1-q^{3n}=(1-q^{2n})+q^{2n}(1-q^n)$, we can write $$W_n(q)=\binom{2n}{n}_q+q^{2n}\binom{2n-1}{n-1}_q$$ and then the result follows from the fact that the degree of $W_n(q)$ is $n^2+n$, together with the fact that $q$-binomial coefficients are polynomials in $q$ with positive coefficients (unimodal even).

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