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Pritam Majumder
Pritam Majumder
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Consider the $n$-element subsets $\{a_1<a_2<\cdots <a_n\}$ of $\{1,\ldots ,2n\}$ satisfying $a_i\geq 2i$ for all $i=1,\ldots ,n$. The number of such subsets is given by $${2n\choose n}-{2n\choose n-1}=\frac{1}{n+1}{2n\choose n},$$ which is the $n$th Catalan Number.

I want to know if the $q$-Catalan number $$\frac{q^{n+1}}{[n+1]_q}{2n\choose n}_q={2n\choose n}_q-{2n\choose n-1}_q$$$$\frac{q^{n}}{[n+1]_q}{2n\choose n}_q={2n\choose n}_q-{2n\choose n-1}_q$$ counts some kind of special $n$-dimensional subspaces inside $\mathbb{F}_q^{2n}$? Note that ${2n \choose n}_q$ is the total number of $n$-dimensional subspaces of $\mathbb{F}_q^{2n}$ ($\mathbb{F}_q$ denotes finite field of order $q$).

Consider the $n$-element subsets $\{a_1<a_2<\cdots <a_n\}$ of $\{1,\ldots ,2n\}$ satisfying $a_i\geq 2i$ for all $i=1,\ldots ,n$. The number of such subsets is given by $${2n\choose n}-{2n\choose n-1}=\frac{1}{n+1}{2n\choose n},$$ which is the $n$th Catalan Number.

I want to know if the $q$-Catalan number $$\frac{q^{n+1}}{[n+1]_q}{2n\choose n}_q={2n\choose n}_q-{2n\choose n-1}_q$$ counts some kind of special $n$-dimensional subspaces inside $\mathbb{F}_q^{2n}$? Note that ${2n \choose n}_q$ is the total number of $n$-dimensional subspaces of $\mathbb{F}_q^{2n}$ ($\mathbb{F}_q$ denotes finite field of order $q$).

Consider the $n$-element subsets $\{a_1<a_2<\cdots <a_n\}$ of $\{1,\ldots ,2n\}$ satisfying $a_i\geq 2i$ for all $i=1,\ldots ,n$. The number of such subsets is given by $${2n\choose n}-{2n\choose n-1}=\frac{1}{n+1}{2n\choose n},$$ which is the $n$th Catalan Number.

I want to know if the $q$-Catalan number $$\frac{q^{n}}{[n+1]_q}{2n\choose n}_q={2n\choose n}_q-{2n\choose n-1}_q$$ counts some kind of special $n$-dimensional subspaces inside $\mathbb{F}_q^{2n}$? Note that ${2n \choose n}_q$ is the total number of $n$-dimensional subspaces of $\mathbb{F}_q^{2n}$ ($\mathbb{F}_q$ denotes finite field of order $q$).

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Pritam Majumder
Pritam Majumder
  • 1.3k
  • 11
  • 22

Consider the $n$-element subsets $\{a_1<a_2<\cdots <a_n\}$ of $\{1,\ldots ,2n\}$ satisfying $a_i\geq 2i$ for all $i=1,\ldots ,n$. The number of such subsets is given by $${2n\choose n}-{2n\choose n-1}=\frac{1}{n+1}{2n\choose n},$$ which is the $n$th Catalan Number.

I want to know if the $q$-Catalan number $$\frac{1}{[n+1]_q}{2n\choose n}_q={2n\choose n}_q-{2n\choose n-1}_q$$$$\frac{q^{n+1}}{[n+1]_q}{2n\choose n}_q={2n\choose n}_q-{2n\choose n-1}_q$$ counts some kind of special $n$-dimensional subspaces inside $\mathbb{F}_q^{2n}$? Note that ${2n \choose n}_q$ is the total number of $n$-dimensional subspaces of $\mathbb{F}_q^{2n}$ ($\mathbb{F}_q$ denotes finite field of order $q$).

Consider the $n$-element subsets $\{a_1<a_2<\cdots <a_n\}$ of $\{1,\ldots ,2n\}$ satisfying $a_i\geq 2i$ for all $i=1,\ldots ,n$. The number of such subsets is given by $${2n\choose n}-{2n\choose n-1}=\frac{1}{n+1}{2n\choose n},$$ which is the $n$th Catalan Number.

I want to know if the $q$-Catalan number $$\frac{1}{[n+1]_q}{2n\choose n}_q={2n\choose n}_q-{2n\choose n-1}_q$$ counts some kind of special $n$-dimensional subspaces inside $\mathbb{F}_q^{2n}$? Note that ${2n \choose n}_q$ is the total number of $n$-dimensional subspaces of $\mathbb{F}_q^{2n}$ ($\mathbb{F}_q$ denotes finite field of order $q$).

Consider the $n$-element subsets $\{a_1<a_2<\cdots <a_n\}$ of $\{1,\ldots ,2n\}$ satisfying $a_i\geq 2i$ for all $i=1,\ldots ,n$. The number of such subsets is given by $${2n\choose n}-{2n\choose n-1}=\frac{1}{n+1}{2n\choose n},$$ which is the $n$th Catalan Number.

I want to know if the $q$-Catalan number $$\frac{q^{n+1}}{[n+1]_q}{2n\choose n}_q={2n\choose n}_q-{2n\choose n-1}_q$$ counts some kind of special $n$-dimensional subspaces inside $\mathbb{F}_q^{2n}$? Note that ${2n \choose n}_q$ is the total number of $n$-dimensional subspaces of $\mathbb{F}_q^{2n}$ ($\mathbb{F}_q$ denotes finite field of order $q$).

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Pritam Majumder
Pritam Majumder
  • 1.3k
  • 11
  • 22

Does $q$-Catalan number count subspaces?

Consider the $n$-element subsets $\{a_1<a_2<\cdots <a_n\}$ of $\{1,\ldots ,2n\}$ satisfying $a_i\geq 2i$ for all $i=1,\ldots ,n$. The number of such subsets is given by $${2n\choose n}-{2n\choose n-1}=\frac{1}{n+1}{2n\choose n},$$ which is the $n$th Catalan Number.

I want to know if the $q$-Catalan number $$\frac{1}{[n+1]_q}{2n\choose n}_q={2n\choose n}_q-{2n\choose n-1}_q$$ counts some kind of special $n$-dimensional subspaces inside $\mathbb{F}_q^{2n}$? Note that ${2n \choose n}_q$ is the total number of $n$-dimensional subspaces of $\mathbb{F}_q^{2n}$ ($\mathbb{F}_q$ denotes finite field of order $q$).