On the least common multiple of $q$-binomial coefficients

TL;DR

This paper proves a new identity relating the least common multiple of all $q$-binomial coefficients for a given n to the least common multiple of $q$-integers, extending a known classical identity to the $q$-analogue setting.

Contribution

It establishes a $q$-analogue of Farhi's classical identity, connecting the least common multiples of $q$-binomial coefficients and $q$-integers.

Findings

Proves the identity for the least common multiple of $q$-binomial coefficients.

Extends classical number theory identities to the $q$-analogue context.

Provides a new perspective on the structure of $q$-binomial coefficients.

Abstract

In this paper, we prove the following identity $$ \lcm({n\brack 0}_q,{n\brack 1}_q,...,{n\brack n}_q) =\frac{\lcm([1]_q,[2]_q,...,[n+1]_q)}{[n+1]_q}, $$ where ${n\brack k}_q$ denotes the $q$-binomial coefficient and $[n]_q=\frac{1-q^n}{1-q}$. This result is a $q$-analogue of an identity of Farhi [Amer. Math. Monthly, November (2009)].

Equations

$$\mathrm{lcm}\left({\left[\genfrac{}{}{0ex}{}{n}{0}\right]}_q,{\left[\genfrac{}{}{0ex}{}{n}{1}\right]}_q,\dots ,{\left[\genfrac{}{}{0ex}{}{n}{n}\right]}_q\right)=\frac{\mathrm{lcm}([1{]}_q,[2{]}_q,\dots ,[n+1{]}_q)}{[n+1{]}_q},$$

$$\mathrm{lcm}\left(\left(\genfrac{}{}{0ex}{}{n}{0}\right),\left(\genfrac{}{}{0ex}{}{n}{1}\right),\dots ,\left(\genfrac{}{}{0ex}{}{n}{n}\right)\right)=\frac{\mathrm{lcm}(1,2,\dots ,n+1)}{n+1}.$$

$${\left[\genfrac{}{}{0ex}{}{M}{N}\right]}_q=\{\begin{array}{ll}\frac{[M{]}_q!}{[N{]}_q![M-N{]}_q!},& \text{if 0≤N≤M},\\ 0,& \text{otherwise.}\end{array}$$

$$\mathrm{lcm}\left({\left[\genfrac{}{}{0ex}{}{n}{0}\right]}_q,{\left[\genfrac{}{}{0ex}{}{n}{1}\right]}_q,\dots ,{\left[\genfrac{}{}{0ex}{}{n}{n}\right]}_q\right)=\frac{\mathrm{lcm}([1{]}_q,[2{]}_q,\dots ,[n+1{]}_q)}{[n+1{]}_q}.$$

$${\left[\genfrac{}{}{0ex}{}{n}{k}\right]}_q=\prod _d{\Phi }_d(q),$$

$$\lfloor k/d\rfloor +\lfloor (n-k)/d\rfloor <\lfloor n/d\rfloor$$

$$k\equiv a(\mathrm{mod}d),(n-k)\equiv b(\mathrm{mod}d)$$

$$\mathrm{lcm}\left({\left[\genfrac{}{}{0ex}{}{n}{0}\right]}_q,{\left[\genfrac{}{}{0ex}{}{n}{1}\right]}_q,\dots ,{\left[\genfrac{}{}{0ex}{}{n}{n}\right]}_q\right)=\prod _d{\Phi }_d(q),$$

$$[k{]}_q=\frac{q^k-1}{q-1}=\prod _{d\mid k,d>1}{\Phi }_d(q),$$

$$\frac{\mathrm{lcm}([1{]}_q,[2{]}_q,\dots ,[n+1{]}_q)}{[n+1{]}_q}=\prod _{d\le n,d\nmid (n+1)}{\Phi }_d(q).$$

$$\underset{q\to 1}{\lim }\mathrm{lcm}\left({\left[\genfrac{}{}{0ex}{}{n}{0}\right]}_q,{\left[\genfrac{}{}{0ex}{}{n}{1}\right]}_q,\dots ,{\left[\genfrac{}{}{0ex}{}{n}{n}\right]}_q\right)=\mathrm{lcm}\left(\left(\genfrac{}{}{0ex}{}{n}{0}\right),\left(\genfrac{}{}{0ex}{}{n}{1}\right),\dots ,\left(\genfrac{}{}{0ex}{}{n}{n}\right)\right),$$

$$\underset{q\to 1}{\lim }\frac{\mathrm{lcm}([1{]}_q,[2{]}_q,\dots ,[n+1{]}_q)}{[n+1{]}_q}=\frac{\mathrm{lcm}(1,2,\dots ,n+1)}{n+1}.$$

$${\Phi }_n(1)=\{\begin{array}{ll}p,& \text{if n=pr is a prime power},\\ 1,& \text{otherwise.}\end{array}$$

$$\underset{q\to 1}{\lim }\mathrm{lcm}\left({\left[\genfrac{}{}{0ex}{}{n}{0}\right]}_q,{\left[\genfrac{}{}{0ex}{}{n}{1}\right]}_q,\dots ,{\left[\genfrac{}{}{0ex}{}{n}{n}\right]}_q\right)=\prod _d{\Phi }_d(1),$$

$$\prod _{\text{primes p≤n}}{p}^{{\sum }_{r=1}^{\infty }{\max }_{0\le k\le n}\left\{\lfloor n/p^r\rfloor -\lfloor k/p^r\rfloor -\lfloor (n-k)/p^r\rfloor \right\}}.$$

$$\sum _{r=1}^{\infty }\underset{0\le k\le n}{\max }\left\{\lfloor n/p^r\rfloor -\lfloor k/p^r\rfloor -\lfloor (n-k)/p^r\rfloor \right\}$$

$$=\underset{0\le k\le n}{\max }\sum _{r=1}^{\infty }\left(\lfloor n/p^r\rfloor -\lfloor k/p^r\rfloor -\lfloor (n-k)/p^r\rfloor \right).$$

$$LHS\text{eqref}eq:max=\{\begin{array}{ll}0,& \text{if n=pM+1−1},\\ M-\min \{i:a_i\ne p-1\},& \text{otherwise.}\end{array}$$

$$\lfloor n/p^r\rfloor -\lfloor k/p^r\rfloor -\lfloor (n-k)/p^r\rfloor =\{\begin{array}{ll}0,& \text{if r=1,…,i0},\\ 1,& \text{if r=i0+1,…,M,}\end{array}$$

$$\sum _{r=1}^{\infty }\lfloor n/p^r\rfloor -\lfloor k/p^r\rfloor -\lfloor (n-k)/p^r\rfloor =M-i_0.$$

$$\prod _{\text{primes p≤n}}{p}^{{\max }_{0\le k\le n}{\sum }_{r=1}^{\infty }\left(\lfloor n/p^r\rfloor -\lfloor k/p^r\rfloor -\lfloor (n-k)/p^r\rfloor \right)}=\mathrm{lcm}\left(\left(\genfrac{}{}{0ex}{}{n}{0}\right),\left(\genfrac{}{}{0ex}{}{n}{1}\right),\dots ,\left(\genfrac{}{}{0ex}{}{n}{n}\right)\right).$$

$$\underset{q\to 1}{\lim }\frac{\mathrm{lcm}([1{]}_q,[2{]}_q,\dots ,[n+1{]}_q)}{[n+1{]}_q}=\prod _{d\le n,d\nmid (n+1)}{\Phi }_d(1),$$

$$\frac{\mathrm{lcm}(1,2,\dots ,n+1)}{n+1}.$$

$$\lfloor n/p^{r_i}\rfloor -\lfloor k_i/p^{r_i}\rfloor -\lfloor (n-k_i)/p^{r_i}\rfloor =1fori=1,2,\dots ,m.$$

$$\lfloor n/p^{r_i}\rfloor -\lfloor k/p^{r_i}\rfloor -\lfloor (n-k)/p^{r_i}\rfloor =1fori=1,2,\dots ,m.$$

Full text

On the least common multiple of $q$-binomial coefficients

Victor J. W. Guo

Department of Mathematics, East China Normal University, Shanghai 200062, People’s Republic of China

jwguo@math.ecnu.edu.cn

Abstract. In this paper, we prove the following identity

[TABLE]

where ${n\brack k}_{q}$ denotes the $q$-binomial coefficient and $[n]_{q}=\frac{1-q^{n}}{1-q}$. This result is a $q$-analogue of an identity of Farhi [Amer. Math. Monthly, November (2009)].

Keywords: least common multiple; $q$-binomial coefficient; cyclotomic polynomial

AMS Subject Classifications (2000): 11A07; 05A30

1 Introduction

An equivalent form of the prime number theorem states that $\log{\rm lcm}(1,2,\ldots,n)\sim n$ as $n\to\infty$ (see, for example, [4]). Nair [7] gave a nice proof for the well-known estimate ${\rm lcm}\{1,2,\ldots,n\}\geq 2^{n-1}$, while Hanson [3] already obtained ${\rm lcm}\{1,2,\ldots,n\}\leq 3^{n}$. Recently, Farhi [1] established the following interesting result.

Theorem 1** (Farhi)**

For any positive integer $n$, there holds

[TABLE]

As an application, Farhi shows that ${\rm lcm}\{1,2,\ldots,n\}\geq 2^{n-1}$ follows immediately from (1).

The purpose of this note is to give a $q$-analogue of (1) by using cyclotomic polynomials. Recall that a natural $q$-analogue of the nonnegative integer $n$ is given by $[n]_{q}=\frac{1-q^{n}}{1-q}$. The corresponding $q$-factorial is $[n]_{q}!=\prod_{k=1}^{n}[k]_{q}$ and the $q$-binomial coefficient ${M\brack N}_{q}$ is defined as

[TABLE]

Let ${\rm lcm}$ also denote the least common multiple of a sequence of polynomials in $\mathbb{Z}[q]$. Our main result can be stated as follows:

Theorem 2

For any positive integer $n$, there holds

[TABLE]

2 Proof of Theorem 2

Let $\Phi_{n}(x)$ be the $n$-th cyclotomic polynomial. The following easily proved result can be found in [5, (10)] and [2].

Lemma 3

The q-binomial coefficient ${n\brack k}_{q}$ can be factorized into

[TABLE]

where the product is over all positive integers $d\leq n$ such that $\lfloor k/d\rfloor+\lfloor(n-k)/d\rfloor<\lfloor n/d\rfloor$.

Lemma 4

Let $n$ and $d$ be two positive integers with $n\geq d$. Then there exists at least one positive integer $k$ such that

[TABLE]

if and only if $d$ does not divide $n+1$.

Proof. Suppose that (3) holds for some positive integer $k$. Let

[TABLE]

for some $1\leq a,b\leq d-1$. Then $n\equiv a+b\pmod{d}$ and $d\leq a+b\leq 2d-2$. Namely, $n+1\equiv a+b+1\not\equiv 0\pmod{d}$. Conversely, suppose that $n+1\equiv c\pmod{d}$ for some $1\leq c\leq d-1$. Then $k=c$ satisfies (3). This completes the proof.

*Proof of Theorem 2. * By Lemma 3, we have

[TABLE]

where the product is over all positive integers $d\leq n$ such that for some $k$ ($1\leq k\leq n$) there holds $\lfloor k/d\rfloor+\lfloor(n-k)/d\rfloor<\lfloor n/d\rfloor$. On the other hand, since

[TABLE]

we have

[TABLE]

By Lemma 4, one sees that the right-hand sides of (4) and (5) are equal. This proves the theorem.

3 Theorem 2 is a $q$-analogue of Theorem 1

In this section we will show that

[TABLE]

and

[TABLE]

We need the following property.

Lemma 5

For any positive integer $n$, there holds

[TABLE]

Proof. See for example [6, p. 160].

In view of (4), we have

[TABLE]

where the product is over all positive integers $d\leq n$ such that for some $k$ ($1\leq k\leq n$) there holds $\lfloor k/d\rfloor+\lfloor(n-k)/d\rfloor<\lfloor n/d\rfloor$. By Lemma 5, the right-hand side of (8) can be written as

[TABLE]

We now claim that

[TABLE]

Let $n=\sum_{i=0}^{M}a_{i}p^{i}$, where $0\leq a_{0},a_{1},\ldots,a_{M}\leq p-1$ and $a_{M}\neq 0$. By Lemma 4, the left-hand side of (10) (denoted LHS(10)) is equal to the number of $r$’s such that $p^{r}\leq n$ and $p^{r}\nmid n+1$. It follows that

[TABLE]

It is clear that the right-hand side of (10) (denoted RHS(10)) is less than or equal to LHS(10). If $n=p^{M+1}-1$, then both sides of (10) are equal to [math]. Assume that $n\neq p^{M+1}-1$ and $i_{0}=\min\{i\colon a_{i}\neq p-1\}$. Taking $k=p^{M}-1$, we have

[TABLE]

and so

[TABLE]

Thus (10) holds. Namely, the expression (9) is equal to

[TABLE]

This proves (6). To prove (7), we apply (5) to get

[TABLE]

which, by Lemma 5, is clearly equal to

[TABLE]

Finally, we mention that (10) has the following interesting conclusion.

Corollary 6

Let $p$ be a prime number and let $k_{1},k_{2},\ldots,k_{m}\leq n$, $r_{1}<r_{2}<\cdots<r_{m}$ be positive integers such that

[TABLE]

Then there exists a positive integer $k\leq n$ such that

[TABLE]