This question extends my earlier MO post for which I’m grateful for answers and useful comments.

The Catalan numbers $ C_n=\frac1{n+1}\binom{2n}n$ satisfy the following well-known arithmetic property: $ $ \text{$ C_{1,n}$ is odd iff $ n=2^j-1$ for some $ j$ }.\tag1$ $ The *$ 2$ -adic valuation* of $ x\in\mathbb{N}$ is the highest power $ 2$ dividing $ x$ , denoted by $ \nu(x)$ . Let $ s(x)$ stand for the *sum of the binary digits* of $ x$ . Then, we have the fact that $ $ \nu(C_{1,n})=s(n+1)-1. \tag2$ $ Let $ n+1=n_r2^r+n_{r-1}2^{r-1}+\cdots+n_12+n_0$ be the binary expansion of $ n+1\in\mathbb{N}$ , for some $ n_j\in\{0,1\}$ . Further, denote by $ (n+1)^*=\{n_{j_1},n_{j_2},\dots,n_{j_t}\}$ the non-zero digits ordered as $ j_1>j_2>\cdots>j_t$ . **Note:** $ \#(n+1)^*=s(n)$ .

One version of the $ q$ -Catalan polynomials $ C_n(q)$ is given in the manner $ $ C_n(q)=\frac1{[n+1]_q}\binom{2n}n_q;$ $ where $ [0]_q:=1, [n]_q=\frac{1-q^n}{1-q}=1+q+\cdots+q^{n-1}$ and $ \binom{n}k_q=\frac{[n]_q!}{[k]_q![n-k]_q!}$ . Here $ [n]_q!=[1]_q[2]_q\cdots[n]_q$ .

Working in the spirit of (1) and (2), I was curious to find a possible $ q$ -analogue.

QUESTION 1.Is this true? If so, how does the proof go? $ $ \prod_{k=1}^{t-1} (1+q^{2^{j_k}}) \qquad \text{divides} \qquad C_n(q), \tag3$ $ and no other such factors divide it!

**REMARK.** In view of the fact that the term $ 1+q^{2n_t}$ is absent from the LHS of (3) ensures that (3) indeed emulates (2), naturally.