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As usual, denote $[n]_q=1+q+\cdots+q^{n-1}=\frac{\,\,1-q^n}{1-q}$ and $[n]_q!=[1]_q[2]_q\cdots[n]_q$. Furthermore, we write $$\binom{n}k_q=\frac{[n]_q!}{[k]_q!\cdot[n-k]_q!}.$$ As a follow up on this ...

Let $C_n=\frac1{n+1}\binom{2n}n$ be the all-familiar Catalan numbers. Then, the following identity has received enough attention in the literature (for example, Lagrange Inversion: When and How): \...

Let $n\ge 2r$ be positive integers. Is there a closed form for following finite summation involving in q-binomial coefficients $$\sum_{s=0}^r(-1)^sq^{\frac{s(s+1)}{2}}{n-2r+s\brack n-2r}_q{n\brack r-...