This question extends my [earlier MO post](https://mathoverflow.net/questions/324816/oddity-of-generalized-catalan-numbers) for which I'm grateful for answers and useful comments.


The [Catalan numbers](https://en.wikipedia.org/wiki/Catalan_number) $C\_n=\frac1{n+1}\binom{2n}n$ satisfy: 
$\text{$C\_{1,n}$ is odd iff $n=2^j-1$ for some $j$}$.


The *[$2$-adic valuation](https://en.wikipedia.org/wiki/P-adic_order)* 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](https://oeis.org/A000120)* of $x$. Then, we have the fact that 
$$\nu(C\_{1,n})=s(n+1)-1. \tag1$$
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 $[0]\_q:=1, [n]\_q=\frac{1-q^n}{1-q}$ 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), I was curious to find a possible $q$-analogue. 



> 
> **QUESTION 1.** Is this true? If so, how does the proof go?
>  $$\prod\_{k=1}^{t-1} (1+q^{2^{j\_k}}) \qquad \text{divides} \qquad C\_n(q), \tag2$$
>  and no other such factors divide it!
> 
> 
> 


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



"Oddity" of $q$-Catalan polynomials: Part II
For a positive integer $m$, an $m$-th primitive root $\alpha\_m$
of $-1$ [which is the root of the polynomial $1+q^m$, and when $m$ is a power of two the converse also holds: any root of $q^m+1$ is an $m$-th primitive root
of $-1$] is a root of $[k]\_q$ exactly when $2m$ divides $k$. So, $\alpha\_m$ is a root of $[N!]\_q$ of multiplicity $[\frac{N}{2m}]$. Therefore, $\alpha\_m$ is a root of $C\_n(q)$ of multiplicity $$\left[\frac{2n}{2m}\right]-\left[\frac{n}{2m}\right]-\left[\frac{n+1}{2m}\right].$$ When is this positive? For $m=1$ never, so assume that $m>1$. Denote $n+1=2mk+r$, $0\leqslant r<2m$. Then we have 
$$
\left[\frac{2n}{2m}\right]-\left[\frac{n}{2m}\right]-\left[\frac{n+1}{2m}\right]= 2k-\chi\_{r=0}+\chi\_{r>m}-\left(2k-\chi\_{r=0}\right)-2k=\chi\_{r>m}.
$$
Now if $m=2^a$, and $n+1=2^{j\_1}+2^{j\_2}+\dots+2^{j\_t}$,
$j\_1>j\_2>\dots>j\_t$, the remainder of $n+1$
modulo $2^{a+1}$ is greater than $2^a$ exactly when
$a\in \{j\_1,j\_2,\dots,j\_{t-1}\}$. This implies your claim.


Note also that $C\_n(q)$ is a product of cyclotomic polynomials in certain powers (since so is each $[k]\_q$). All cyclotomic polynomials other than $\Phi\_{2^{a+1}}=1+q^{2^a}$, $a=0,1,\dots$, take value 1 at point 1. Therefore substituting $q=1$ to the $q$-version we get the usual $q=1$ oddity statement.


