On a congruence involving $q$-Catalan numbers

TL;DR

This paper develops a $q$-analogue of a known congruence involving sums of Catalan numbers, extending previous results and based on a specific $q$-congruence established by the author and Petrov.

Contribution

The paper introduces a new $q$-analogue of Sun--Tauraso's congruence for Catalan number sums, expanding the scope of $q$-congruence theory.

Findings

Established a $q$-analogue of Sun--Tauraso's congruence

Extended Tauraso's $q$-congruence results

Connected $q$-congruences with classical Catalan number properties

Abstract

Based on a $q$-congruence of the author and Petrov, we set up a $q$-analogue of Sun--Tauraso's congruence for sums of Catalan numbers, which extends a $q$-congruence due to Tauraso.

Equations

$$C_n=\left(\genfrac{}{}{0ex}{}{2n}{n}\right)\frac{1}{n+1}=\left(\genfrac{}{}{0ex}{}{2n}{n}\right)-\left(\genfrac{}{}{0ex}{}{2n}{2n+1}\right).$$

$$\sum _{k=0}^{p-1}\left(\genfrac{}{}{0ex}{}{2k}{k}\right)\equiv \left(\frac{p}{3}\right)(\mathrm{mod}{p}^2),$$

$$\sum _{k=0}^{p-1}{C}_k\equiv \frac{3}{2}\left(\frac{p}{3}\right)-\frac{1}{2}(\mathrm{mod}{p}^2),$$

$$\left[\genfrac{}{}{0ex}{}{n}{k}\right]={\left[\genfrac{}{}{0ex}{}{n}{k}\right]}_q=\{\begin{array}{ll}\frac{(q;q{)}_n}{(q;q{)}_k(q;q{)}_{n-k}}& \text{if 0⩽k⩽n},\\ 0& \text{otherwise,}\end{array}$$

$${\Phi }_n(q)=\prod _{\begin{array}{c}1\le k\le n\\ (n,k)=1\end{array}}(q-e^{2k\pi i/n}).$$

$$\sum _{k=0}^{n-1}{q}^k\left[\genfrac{}{}{0ex}{}{2k}{k}\right]\equiv \left(\frac{n}{3}\right)q^{\frac{n^2-1}{3}}(\mathrm{mod}{\Phi }_n(q{)}^2),$$

$$C_n(q)=\frac{1}{[n+1{]}_q}\left[\genfrac{}{}{0ex}{}{2n}{n}\right]=\left[\genfrac{}{}{0ex}{}{2n}{n}\right]-q\left[\genfrac{}{}{0ex}{}{2n}{n+1}\right].$$

$$\sum _{k=0}^{n-1}{q}^k{C}_k(q)\equiv \{\begin{array}{ll}{q}^{\lfloor n/3\rfloor }& \text{if n≡0,1(mod3)}\\ -1-q^{(2n-1)/3}& \text{if n≡2(mod3)}\end{array}(\mathrm{mod}{\Phi }_n(q)),$$

$$\sum _{k=0}^{n-1}{q}^k{C}_k(q)\equiv \{\begin{array}{ll}-q^{\frac{n^2-1}{3}}-q^{\frac{n(2n-1)}{3}}& \text{if n≡2(mod3),}\\ q^{\frac{n^2-1}{3}}-\frac{n-1}{3}(q^n-1)& \text{if n≡1(mod3).}\end{array}$$

$$\sum _{k=0}^{n-1}{q}^{k+1}\left[\genfrac{}{}{0ex}{}{2k}{k+1}\right]\equiv \{\begin{array}{ll}{q}^{\frac{n(2n-1)}{3}}& \text{if n≡2(mod3),}\\ \frac{n-1}{3}(q^n-1)& \text{if n≡1(mod3).}\end{array}$$

$$\sum _{k=1}^{n-1}\left(\frac{k-1}{3}\right)\frac{(-1{)}^k{q}^{\frac{1}{3}\left(2k^2-k\left(\frac{k-1}{3}\right)\right)-\frac{k(k-1)}{2}}}{1-q^k}\equiv \{\begin{array}{ll}0& \text{if n≡2(mod3)},\\ \frac{n-1}{6}& \text{if n≡1(mod3)}.\end{array}$$

$$\sum _{k=1}^{n-1}(-1{)}^k\left(\frac{k-1}{3}\right)\frac{q^{\frac{1}{3}\left(2k^2-k\left(\frac{k-1}{3}\right)\right)-\frac{k(k-1)}{2}}}{1-q^k}=\sum _{k=0}^{\lfloor \frac{n-3}{3}\rfloor }\frac{(-1{)}^k{q}^{\frac{(k+1)(3k+2)}{2}}}{1-q^{3k+2}}-\sum _{k=1}^{\lfloor \frac{n-1}{3}\rfloor }\frac{(-1{)}^k{q}^{\frac{k(3k+5)}{2}}}{1-q^{3k}}.$$

$$\sum _{k=0}^{n-1}\frac{(-1{)}^k{q}^{\frac{(k+1)(3k+2)}{2}}}{1-q^{3k+2}}-\sum _{k=1}^n\frac{(-1{)}^k{q}^{\frac{k(3k+5)}{2}}}{1-q^{3k}}\equiv 0(\mathrm{mod}{\Phi }_{3n+2}(q)).$$

$$\sum _{k=0}^{n-1}\frac{(-1{)}^k{\omega }^{\frac{(k+1)(3k+2)}{2}}}{1-{\omega }^{3k+2}}$$

$$=\sum _{k=1}^n\frac{(-1{)}^{n-k}{\omega }^{\frac{k(3k-1)}{2}+\frac{(3n+2)(n+1)}{2}-(3n+2)k}}{1-{\omega }^{3n-3k+2}}$$

$$=\sum _{k=1}^n\frac{(-1{)}^k{\omega }^{\frac{k(3k+5)}{2}}}{1-{\omega }^{3k}},$$

$$\sum _{k=0}^{n-1}\frac{(-1{)}^k{\omega }^{\frac{(k+1)(3k+2)}{2}}}{1-{\omega }^{3k+2}}-\sum _{k=1}^n\frac{(-1{)}^k{\omega }^{\frac{k(3k+5)}{2}}}{1-{\omega }^{3k}}=0,$$

$$\sum _{k=0}^{n-1}\frac{(-1{)}^k{\zeta }^{\frac{(k+1)(3k+2)}{2}}}{1-{\zeta }^{3k+2}}-\sum _{k=1}^n\frac{(-1{)}^k{\zeta }^{\frac{k(3k+5)}{2}}}{1-{\zeta }^{3k}}=\frac{n}{2}.$$

$$\sum _{k=0}^{n-1}\frac{(-1{)}^k{\zeta }^{\frac{(k+1)(3k+2)}{2}}}{1-{\zeta }^{3k+2}}$$

$$=\sum _{k=n+1}^{2n}\frac{(-1{)}^k{\zeta }^{\frac{k(3k-1)}{2}+(3n+1)(2n-2k+1)}}{1-{\zeta }^{-3k}}$$

$$=-\sum _{k=n+1}^{2n}\frac{(-1{)}^k{\zeta }^{\frac{k(3k+5)}{2}}}{1-{\zeta }^{3k}},$$

$$\sum _{k=0}^{n-1}\frac{(-1{)}^k{\zeta }^{\frac{(k+1)(3k+2)}{2}}}{1-{\zeta }^{3k+2}}-\sum _{k=1}^n\frac{(-1{)}^k{\zeta }^{\frac{k(3k+5)}{2}}}{1-{\zeta }^{3k}}=-\sum _{k=1}^{2n}\frac{(-1{)}^k{\zeta }^{\frac{k(3k+5)}{2}}}{1-{\zeta }^{3k}}.$$

$$\sum _{k=0}^{n-1}\frac{(-1{)}^k{\zeta }^{\frac{(k+1)(3k+2)}{2}}}{1-{\zeta }^{3k+2}}-\sum _{k=1}^n\frac{(-1{)}^k{\zeta }^{\frac{k(3k+5)}{2}}}{1-{\zeta }^{3k}}$$

$$=-\sum _{k=1}^{2n}\frac{(-1{)}^{2n+1-k}{\zeta }^{\frac{(2n+1-k)(6n-3k+8)}{2}}}{1-{\zeta }^{3(2n+1-k)}}$$

$$=-\sum _{k=1}^{2n}\frac{(-1{)}^{1-k}{\zeta }^{\frac{(3k-1)(k-2)}{2}+(3n+1)(2n+3-2k)}}{1-{\zeta }^{1-3k}}$$

$$=-\sum _{k=1}^{2n}\frac{(-1{)}^k{\zeta }^{\frac{k(3k-1)}{2}}}{1-{\zeta }^{3k-1}}.$$

$$\sum _{k=1}^{2n}\frac{(-1{)}^k{\zeta }^{\frac{k(3k-1)}{2}}}{1-{\zeta }^{3k-1}}=-\frac{n}{2}.$$

$$\sum _{k=0}^{n-1}{q}^k\left[\genfrac{}{}{0ex}{}{2k}{k+1}\right]=\sum _{k=0}^{n-1}\left(\frac{n-k-1}{3}\right)q^{\frac{1}{3}\left(2(n-k{)}^2-(n-k)\left(\frac{n-k-1}{3}\right)-3\right)}\left[\genfrac{}{}{0ex}{}{2n}{k}\right],$$

$$1-q^{2n}=(1+q^n)(1-q^n)\equiv 2(1-q^n)(\mathrm{mod}{\Phi }_n(q{)}^2).$$

$$\left[\genfrac{}{}{0ex}{}{2n}{k}\right]$$

$$\equiv 2(1-q^n)\frac{(1-q^{-1})\cdots (1-q^{-k+1})}{(1-q)(1-q^2)\cdots (1-q^k)}(\mathrm{mod}{\Phi }_n(q{)}^2)$$

$$=2(q^n-1)\frac{(-1{)}^k{q}^{-\frac{k(k-1)}{2}}}{1-q^k}.$$

$$\sum _{k=0}^{n-1}{q}^{k+1}\left[\genfrac{}{}{0ex}{}{2k}{k+1}\right]$$

$$=\left(\frac{n-1}{3}\right)q^{\frac{1}{3}\left(2n^2-n\left(\frac{n-1}{3}\right)\right)}+\sum _{k=1}^{n-1}\left(\frac{n-k-1}{3}\right)q^{\frac{1}{3}\left(2(n-k{)}^2-(n-k)\left(\frac{n-k-1}{3}\right)\right)}\left[\genfrac{}{}{0ex}{}{2n}{k}\right]$$

$$\equiv \left(\frac{n-1}{3}\right)q^{\frac{1}{3}\left(2n^2-n\left(\frac{n-1}{3}\right)\right)}$$

$$+2(q^n-1)\sum _{k=1}^{n-1}\left(\frac{n-k-1}{3}\right)\frac{(-1{)}^k{q}^{\frac{1}{3}\left(2(n-k{)}^2-(n-k)\left(\frac{n-k-1}{3}\right)\right)-\frac{k(k-1)}{2}}}{1-q^k}(\mathrm{mod}{\Phi }_n(q{)}^2).$$

$$\sum _{k=1}^{n-1}\left(\frac{n-k-1}{3}\right)\frac{(-1{)}^k{q}^{\frac{1}{3}\left(2(n-k{)}^2-(n-k)\left(\frac{n-k-1}{3}\right)\right)-\frac{k(k-1)}{2}}}{1-q^k}$$

Full text

On a congruence involving $q$-Catalan numbers

Ji-Cai Liu

Department of Mathematics, Wenzhou University, Wenzhou 325035, PR China

jcliu2016@gmail.com

Abstract. Based on a $q$-congruence of the author and Petrov, we set up a $q$-analogue of Sun–Tauraso’s congruence for sums of Catalan numbers, which extends a $q$-congruence due to Tauraso.

Keywords: $q$-congruences; $q$-Catalan numbers; cyclotomic polynomials

MR Subject Classifications: 11B65, 11A07, 05A10

1 Introduction

In combinatorics, the Catalan numbers are a sequence of natural numbers, which play an important role in various counting problems. The $n$th Catalan number is given by the following binomial coefficient:

[TABLE]

The closely related numbers are the central binomial coefficients ${2n\choose n}$ for $n\geq 0$.

Both Catalan numbers and central binomial coefficients satisfy many interesting congruences (see, for instance, [7, 8]). In 2011, Sun and Tauraso [8] proved that for primes $p\geq 5$,

[TABLE]

where $\left(\frac{\cdot}{p}\right)$ denotes the Legendre symbol.

In the past few years, $q$-analogues of congruences ($q$-congruence) for indefinite sums of binomial coefficients as well as hypergeometric series attracted many experts’ attention (see, for example, [3, 2, 4, 5, 6, 9, 10]). It is worth mentioning that Guo and Zudilin[5] developed an interesting microscoping method to prove many $q$-congruences.

In order to discuss $q$-congruences, we first recall some $q$-series notation. The $q$-binomial coefficients are defined as

[TABLE]

where the $q$-shifted factorial is given by $(a;q)_{n}=(1-a)(1-aq)\cdots(1-aq^{n-1})$ for $n\geq 1$ and $(a;q)_{0}=1$. Moreover, the $q$-integers are defined by $[n]_{q}=(1-q^{n})/(1-q)$, and the $n$th cyclotomic polynomial is given by

[TABLE]

Recently, the author and Petrov [6] established a $q$-analogue for (1.1) as follows:

[TABLE]

which was originally conjectured by Guo [2] and generalises a $q$-congruence of Tauraso [9]. There are several natural $q$-analogues of Catalan numbers (see [1]). Here and throughout the paper, we consider the following $q$-analogue of Catalan numbers:

[TABLE]

In 2012, Tauraso[9] obtained a weak $q$-version of (1.2) as follows:

[TABLE]

where $\lfloor x\rfloor$ denotes the integral part of real $x$. In this note, we aim to set up a $q$-analogue of (1.2) as well as another related $q$-congruence for sums of binomial coefficients.

Theorem 1.1

For any positive integer $n$, the following holds modulo $\Phi_{n}(q)^{2}$:

[TABLE]

In order to prove (1.5), we shall establish the following $q$-congruence.

Theorem 1.2

For any positive integer $n$, the following holds modulo $\Phi_{n}(q)^{2}$:

[TABLE]

It is clear that (1.5) can be directly deduced from (1.3), (1.4) and (1.6). The remainder of the paper is organized as follows. We first set up a preliminary result in the next section, and prove Theorem 1.2 in Section 3.

2 An auxiliary result

Lemma 2.1

For any positive integer $n$, the following holds modulo $\Phi_{n}(q)$:

[TABLE]

Proof. Note that

[TABLE]

We shall distinguish two cases to prove (2.1).

Case 1 $n\equiv 2\pmod{3}$. This case is equivalent to

[TABLE]

Let $\omega$ be a primitive $(3n+2)$th root of unity. Letting $k\to n-k$ in the following sum gives

[TABLE]

where we have used the fact that $\omega^{\frac{(3n+2)(n+1)}{2}}=(-1)^{n+1}$. Thus,

[TABLE]

which is equivalent to (2.2).

Case 2 $n\equiv 1\pmod{3}$. Let $\zeta$ be a primitive $(3n+1)$th root of unity. It suffices to show that

[TABLE]

Note that

[TABLE]

where we replace $k$ by $2n-k$ in the first step. Thus,

[TABLE]

Furthermore, letting $k\to 2n+1-k$ on the right-hand side of (2.4) gives

[TABLE]

An identity due to the author and Petrov [6, (2.4)] says

[TABLE]

Then the proof of (2.3) follows from (2.5) and (2.6). $\square$

3 Proof of Theorem 1.2

Now we are in a position to prove Theorem 1.2. We recall the following identity:

[TABLE]

which was proved by Tauraso in a more general form (see [9, Theorem 4.2]). Since $1-q^{n}\equiv 0\pmod{\Phi_{n}(q)}$, we have

[TABLE]

It follows that for $1\leq k\leq n-1$,

[TABLE]

Multiplying both sides of (3.1) by $q$ and substituting (3.2) into the right-hand side of (3.1), we arrive at

[TABLE]

Furthermore,

[TABLE]

where we set $k\to n-k$ in the first step. Thus,

[TABLE]

We complete the proof of (1.6) by combining (2.1) and (3.4).

Acknowledgments. This work was supported by the National Natural Science Foundation of China (grant 11801417).