## “Oddity” of $q$-Catalan polynomials: Part II

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 $$_q:=1, [n]_q=\frac{1-q^n}{1-q}=1+q+\cdots+q^{n-1}$$ and $$\binom{n}k_q=\frac{[n]_q!}{[k]_q![n-k]_q!}$$. Here $$[n]_q!=_q_q\cdots[n]_q$$.

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.