Return to 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 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 $1+q^{2^{j_t}}$ is absent from the LHS of (3) ensures that (3) indeed emulates (2), naturally.

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.

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.

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 $1+q^{2^{j_t}}$ is absent from the LHS of (3) ensures that (3) indeed emulates (2), naturally.

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? $$\prod_{k=1}^{t-1} (1+q^{2n_{j_k}}) \qquad \text{divides} \qquad C_n(q), \tag3$$ 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.

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.