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With the usual $q-$notations $[n]_q=1+q+\cdots+q^{n-1}=\frac{\,\,1-q^n}{1-q},$ $[n]_q!=[1]_q[2]_q\cdots[n]_q$ and $\binom{n}k_q=\frac{[n]_q!}{[k]_q!\cdot[n-k]_q!}$ let $$b(n,k,r,q)=\det\left(q^{r\binom{i-j}2}\frac{[2i+k+1]_q}{[i+j+k]_q}\binom{i+j+k}{i-j+1}_q\right)_{i,j=0}^{n-1}.$$

It can be shown that $b(n,k,1,q)=\binom{2n+k-1}{n}_q.$ Therefore $b(n,k,1,q)$ has positive coefficients as a polynomial in $q$ for each positive integer $k.$

Computations suggest that also $b(n,k,0,q)$ and $b(n,k,2,q)=q^{n(n+k-1)}b(n,k,0,1/q)$ have positive coefficients.

Any idea how to prove this?

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