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?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.