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 $W\_2(q)=q^6 + q^5 + 2q^4 + q^3 + 2q^2 + q + 1$.



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> **QUESTION.** Is this true? The number of monomials in $W\_n(q)$ equals $n^2+n+1$.
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A (mild?) question on the number of monomials
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).


