With the usual $ q- $ratings

$[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

$$ b (n, k, r, q) = det left (q ^ {r binom {i-j} 2} frac {[2i+k+1]_q} {[i+j+k]_q} binom {i + j + k} {i-j + 1} _q right) _ {i, j = 0} ^ {n-1}. $$

We can show that $ b (n, k, 1, q) = binom {2n + k-1} {n} _q. $ So $ b (n, k, 1, q) $ has positive coefficients as a polynomial in $ q $ for each positive integer k $

Calculations suggest that $ 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.

An idea how to prove that?