"Oddity" of $q$-Catalan polynomials: Part II

Question

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: $\text{$C_{1,n}$ is odd iff $n=2^j-1$ for some $j$}$.

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 $$\nu(C_{1,n})=s(n+1)-1. \tag1$$ 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.

Answer 1

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.