Factors of sums involving $q$-binomial coefficients and powers of   $q$-integers

Victor J. W. Guo; Su-Dan Wang

arXiv:1701.07016·math.CO·August 1, 2017

Factors of sums involving $q$-binomial coefficients and powers of $q$-integers

Victor J. W. Guo, Su-Dan Wang

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TL;DR

This paper proves that certain sums involving $q$-binomial coefficients and powers of $q$-integers are Laurent polynomials with integer coefficients, providing a $q$-analogue of a divisibility result on the Catalan triangle and confirming a related conjecture.

Contribution

The authors establish that specific sums involving $q$-binomial coefficients are Laurent polynomials with integer coefficients, extending divisibility results and confirming conjectures in the area.

Findings

01

Sum expressions are Laurent polynomials with integer coefficients.

02

Provides a $q$-analogue of a Catalan triangle divisibility result.

03

Confirms a conjecture by the first author and Zeng.

Abstract

We show that, for all positive integers $n_{1}, \dots, n_{m}$ , $n_{m + 1} = n_{1}$ , and any non-negative integers $j$ and $r$ with $j ⩽ m$ , the expression $\frac{1}{[ n _{1} ]} [n _{1} n _{1} + n _{m}]^{- 1} k = 1 \sum n_{1} [2 k] [k]^{2 r} q^{j k^{2} - (r + 1) k} i = 1 \prod m [n _{i} + k n _{i} + n _{i + 1}]$ is a Laurent polynomial in $q$ with integer cofficients, where $[n] = 1 + q + \dots + q^{n - 1}$ and $[k n] = \prod_{i = 1}^{k} (1 - q^{n - i + 1}) / (1 - q^{i})$ . This gives a $q$ -analogue of a divisibility result on the Catalan triangle obtained by the first author and Zeng, and also confirms a conjecture of the first author and Zeng. We further propose several related conjectures.

Equations87

\frac{1}{[ n _{1} ]} [n _{1} n _{1} + n _{m}]^{- 1} k = 1 \sum n_{1} [2 k] [k]^{2 r} q^{j k^{2} - (r + 1) k} i = 1 \prod m [n _{i} + k n _{i} + n _{i + 1}]

\frac{1}{[ n _{1} ]} [n _{1} n _{1} + n _{m}]^{- 1} k = 1 \sum n_{1} [2 k] [k]^{2 r} q^{j k^{2} - (r + 1) k} i = 1 \prod m [n _{i} + k n _{i} + n _{i + 1}]

\frac{2}{n ^{2}} (n 2 n)^{- 1} k = 1 \sum n (n - k 2 n) k^{2 r + 1} \in Z for r ⩾ 1,

\frac{2}{n ^{2}} (n 2 n)^{- 1} k = 1 \sum n (n - k 2 n) k^{2 r + 1} \in Z for r ⩾ 1,

\frac{2}{n _{1}} (n _{1} n _{1} + n _{m})^{- 1} k = 1 \sum n_{1} k^{2 r + 1} i = 1 \prod m (n _{i} + k n _{i} + n _{i + 1}) \in Z for r ⩾ 0,

\frac{2}{n _{1}} (n _{1} n _{1} + n _{m})^{- 1} k = 1 \sum n_{1} k^{2 r + 1} i = 1 \prod m (n _{i} + k n _{i} + n _{i + 1}) \in Z for r ⩾ 0,

[k n] = ⎩ ⎨ ⎧ \frac{( q ; q ) _{n}}{( q ; q ) _{k} ( q ; q ) _{n - k}}, 0, if 0 ⩽ k ⩽ n, otherwise.

[k n] = ⎩ ⎨ ⎧ \frac{( q ; q ) _{n}}{( q ; q ) _{k} ( q ; q ) _{n - k}}, 0, if 0 ⩽ k ⩽ n, otherwise.

\frac{1}{[ n ] ^{2}} [n 2 n]^{- 1} k = 1 \sum n [2 k] [k]^{2 r} q^{(r + 1) (n - k) + j k^{2}} [n + k 2 n]

\frac{1}{[ n ] ^{2}} [n 2 n]^{- 1} k = 1 \sum n [2 k] [k]^{2 r} q^{(r + 1) (n - k) + j k^{2}} [n + k 2 n]

\frac{1}{[ n _{1} ]} [n _{1} n _{1} + n _{m}]^{- 1} k = 1 \sum n_{1} [2 k] [k]^{2 r} q^{j k^{2} - (r + 1) k} i = 1 \prod m [n _{i} + k n _{i} + n _{i + 1}]

\frac{1}{[ n _{1} ]} [n _{1} n _{1} + n _{m}]^{- 1} k = 1 \sum n_{1} [2 k] [k]^{2 r} q^{j k^{2} - (r + 1) k} i = 1 \prod m [n _{i} + k n _{i} + n _{i + 1}]

B_{n, k} (q) := \frac{[ k ]}{[ n ]} [n - k 2 n], 1 ⩽ k ⩽ n .

B_{n, k} (q) := \frac{[ k ]}{[ n ]} [n - k 2 n], 1 ⩽ k ⩽ n .

[n 2 n]^{- 1} k = 1 \sum n (1 + q^{k}) [k]^{r} B_{n, k}^{s} (q) q^{j k^{2} - (r + s + 1) k /2}

[n 2 n]^{- 1} k = 1 \sum n (1 + q^{k}) [k]^{r} B_{n, k}^{s} (q) q^{j k^{2} - (r + s + 1) k /2}

S_{r} (n; q) = k = 1 \sum n [2 k] [k]^{2 r} q^{(r + 1) (n - k)} [n + k 2 n] .

S_{r} (n; q) = k = 1 \sum n [2 k] [k]^{2 r} q^{(r + 1) (n - k)} [n + k 2 n] .

S_{0} (n; q) = k = 1 \sum n [2 k] q^{n - k} [n - k 2 n]

S_{0} (n; q) = k = 1 \sum n [2 k] q^{n - k} [n - k 2 n]

= [n] [n 2 n] .

[k]^{2} [n + k 2 n] q^{n - k} = [n]^{2} [n + k 2 n] - [2 n] [2 n - 1] [n + k - 1 2 n - 2],

[k]^{2} [n + k 2 n] q^{n - k} = [n]^{2} [n + k 2 n] - [2 n] [2 n - 1] [n + k - 1 2 n - 2],

S_{r} (n; q) = [n]^{2} S_{r - 1} (n; q) - q^{r} [2 n] [2 n - 1] S_{r - 1} (n - 1; q), n = 1, 2, \dots .

S_{r} (n; q) = [n]^{2} S_{r - 1} (n; q) - q^{r} [2 n] [2 n - 1] S_{r - 1} (n - 1; q), n = 1, 2, \dots .

S_{1} (n; q) = [n]^{3} [n 2 n] - q [2 n] [2 n - 1] [n - 1] [n - 1 2 n - 2] = [n]^{2} [n 2 n] .

S_{1} (n; q) = [n]^{3} [n 2 n] - q [2 n] [2 n - 1] [n - 1] [n - 1 2 n - 2] = [n]^{2} [n 2 n] .

S_{r} (n; q) \equiv 0 (mod [n]^{2} [n 2 n]) .

S_{r} (n; q) \equiv 0 (mod [n]^{2} [n 2 n]) .

T_{r} (n; q) = k = 1 \sum n [2 k] [k]^{2 r} q^{(r + 1) (n - k) + k^{2}} [n + k 2 n] .

T_{r} (n; q) = k = 1 \sum n [2 k] [k]^{2 r} q^{(r + 1) (n - k) + k^{2}} [n + k 2 n] .

T_{r} (n; q) \equiv 0 (mod [n]^{2} [n 2 n]) .

T_{r} (n; q) \equiv 0 (mod [n]^{2} [n 2 n]) .

[n _{1} + k n _{1} + n _{2}] [n _{2} + k n _{2} + n _{3}] [n _{3} + k n _{3} + n _{1}] = s = 0 \sum n_{1} - k \frac{q ^{s^{2} + 2 k s} ( q ; q ) _{n_{1} + n_{2} + n_{3} - k - s}}{( q ; q ) _{s} ( q ; q ) _{s + 2 k} ( q ; q ) _{n_{1} - k - s} ( q ; q ) _{n_{2} - k - s} ( q ; q ) _{n_{3} - k - s}},

[n _{1} + k n _{1} + n _{2}] [n _{2} + k n _{2} + n _{3}] [n _{3} + k n _{3} + n _{1}] = s = 0 \sum n_{1} - k \frac{q ^{s^{2} + 2 k s} ( q ; q ) _{n_{1} + n_{2} + n_{3} - k - s}}{( q ; q ) _{s} ( q ; q ) _{s + 2 k} ( q ; q ) _{n_{1} - k - s} ( q ; q ) _{n_{2} - k - s} ( q ; q ) _{n_{3} - k - s}},

S (n_{1}, \dots, n_{m}; r, j, q) = \frac{( q ; q ) _{n_{1} - 1} ( q ; q ) _{n_{m}}}{( q ; q ) _{n_{1} + n_{m}}} k = 1 \sum n_{1} [2 k] [k]^{2 r} q^{j k^{2} - (r + 1) k} C (n_{1}, \dots, n_{m}; k),

S (n_{1}, \dots, n_{m}; r, j, q) = \frac{( q ; q ) _{n_{1} - 1} ( q ; q ) _{n_{m}}}{( q ; q ) _{n_{1} + n_{m}}} k = 1 \sum n_{1} [2 k] [k]^{2 r} q^{j k^{2} - (r + 1) k} C (n_{1}, \dots, n_{m}; k),

C (a_{1}, \dots, a_{l}; k) = i = 1 \prod l [a _{i} + k a _{i} + a _{i + 1}] (a_{l + 1} = a_{1}) .

C (a_{1}, \dots, a_{l}; k) = i = 1 \prod l [a _{i} + k a _{i} + a _{i + 1}] (a_{l + 1} = a_{1}) .

C (n_{1}, \dots, n_{m}; k) = \frac{( q ; q ) _{n_{2} + n_{3}} ( q ; q ) _{n_{m} + n_{1}}}{( q ; q ) _{n_{1} + n_{2}} ( q ; q ) _{n_{m} + n_{3}}} [n _{1} + k n _{1} + n _{2}] [n _{2} + k n _{1} + n _{2}] C (n_{3}, \dots, n_{m}; k) .

C (n_{1}, \dots, n_{m}; k) = \frac{( q ; q ) _{n_{2} + n_{3}} ( q ; q ) _{n_{m} + n_{1}}}{( q ; q ) _{n_{1} + n_{2}} ( q ; q ) _{n_{m} + n_{3}}} [n _{1} + k n _{1} + n _{2}] [n _{2} + k n _{1} + n _{2}] C (n_{3}, \dots, n_{m}; k) .

[n _{1} + k n _{1} + n _{2}] [n _{2} + k n _{1} + n _{2}] = s = 0 \sum n_{1} - k \frac{q ^{s^{2} + 2 k s} ( q ; q ) _{n_{1} + n_{2}}}{( q ; q ) _{s} ( q ; q ) _{s + 2 k} ( q ; q ) _{n_{1} - k - s} ( q ; q ) _{n_{2} - k - s}} .

[n _{1} + k n _{1} + n _{2}] [n _{2} + k n _{1} + n _{2}] = s = 0 \sum n_{1} - k \frac{q ^{s^{2} + 2 k s} ( q ; q ) _{n_{1} + n_{2}}}{( q ; q ) _{s} ( q ; q ) _{s + 2 k} ( q ; q ) _{n_{1} - k - s} ( q ; q ) _{n_{2} - k - s}} .

S (n_{1}, \dots, n_{m}; r, j, q)

S (n_{1}, \dots, n_{m}; r, j, q)

= \frac{( q ; q ) _{n_{2} + n_{3}} ( q ; q ) _{n_{1} - 1} ( q ; q ) _{n_{m}}}{( q ; q ) _{n_{m} + n_{3}}} k = 1 \sum n_{1} s = 0 \sum n_{1} - k \frac{q ^{s^{2} + 2 k s + j k^{2} - (r + 1) k} [ 2 k ] [ k ] ^{2 r} C ( n _{3} , \dots , n _{m} ; k )}{( q ; q ) _{s} ( q ; q ) _{s + 2 k} ( q ; q ) _{n_{1} - k - s} ( q ; q ) _{n_{2} - k - s}}

= \frac{( q ; q ) _{n_{2} + n_{3}} ( q ; q ) _{n_{1} - 1} ( q ; q ) _{n_{m}}}{( q ; q ) _{n_{m} + n_{3}}} l = 1 \sum n_{1} q^{l^{2}} k = 1 \sum l \frac{q ^{(j - 1) k^{2} - (r + 1) k} [ 2 k ] [ k ] ^{2 r} C ( n _{3} , \dots , n _{m} ; k )}{( q ; q ) _{l - k} ( q ; q ) _{l + k} ( q ; q ) _{n_{1} - l} ( q ; q ) _{n_{2} - l}},

\frac{C ( n _{3} , \dots , n _{m} ; k )}{( q ; q ) _{l - k} ( q ; q ) _{l + k}} = \frac{( q ; q ) _{n_{m} + n_{3}}}{( q ; q ) _{n_{3} + l} ( q ; q ) _{n_{m} + l}} C (l, n_{3}, \dots, n_{m}; k),

\frac{C ( n _{3} , \dots , n _{m} ; k )}{( q ; q ) _{l - k} ( q ; q ) _{l + k}} = \frac{( q ; q ) _{n_{m} + n_{3}}}{( q ; q ) _{n_{3} + l} ( q ; q ) _{n_{m} + l}} C (l, n_{3}, \dots, n_{m}; k),

S (n_{1}, \dots, n_{m}; r, j, q) = l = 1 \sum n_{1} q^{l^{2}} [l - 1 n _{1} - 1] [n _{2} - l n _{2} + n _{3}] S (l, n_{3}, \dots, n_{m}; r, j - 1, q) .

S (n_{1}, \dots, n_{m}; r, j, q) = l = 1 \sum n_{1} q^{l^{2}} [l - 1 n _{1} - 1] [n _{2} - l n _{2} + n _{3}] S (l, n_{3}, \dots, n_{m}; r, j - 1, q) .

S (n_{1}, n_{2}; r, j, q) = l = 1 \sum n_{1} q^{l^{2}} [l - 1 n _{1} - 1] [l n _{2}] S (l; r, j - 1, q) .

S (n_{1}, n_{2}; r, j, q) = l = 1 \sum n_{1} q^{l^{2}} [l - 1 n _{1} - 1] [l n _{2}] S (l; r, j - 1, q) .

S (n_{1}, \dots, n_{m}; r, 0, q) = S (n_{1}, \dots, n_{m}; r, m, q^{- 1}) q^{n_{1} n_{2} + n_{2} n_{3} + \dots + n_{m - 1} n_{m} - n_{1} - 2 r} for m ⩾ 2.

S (n_{1}, \dots, n_{m}; r, 0, q) = S (n_{1}, \dots, n_{m}; r, m, q^{- 1}) q^{n_{1} n_{2} + n_{2} n_{3} + \dots + n_{m - 1} n_{m} - n_{1} - 2 r} for m ⩾ 2.

[k m] = d \prod Φ_{d} (q),

[k m] = d \prod Φ_{d} (q),

g cd ([n 2 n], [n]) = 1.

g cd ([n 2 n], [n]) = 1.

\frac{1}{[ n ]} [n 2 n]^{- 1} k = 1 \sum n [2 k] [k]^{r + s - 1} q^{j k^{2} - (r + s + 1) k /2} [n + k 2 n]^{s}

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Full text

**Factors of sums involving $q$ -binomial coefficients

and powers of $q$ -integers**

Victor J. W. Guo1 and Su-Dan Wang2

1School of Mathematical Sciences, Huaiyin Normal University, Huai’an, Jiangsu 223300,

People’s Republic of China

[email protected]

2Department of Mathematics, East China Normal University, Shanghai 200062,

People’s Republic of China

[email protected]

Abstract. We show that, for all positive integers $n_{1},\ldots,n_{m}$ , $n_{m+1}=n_{1}$ , and any non-negative integers $j$ and $r$ with $j\leqslant m$ , the expression

[TABLE]

is a Laurent polynomial in $q$ with integer coefficients, where $[n]=1+q+\cdots+q^{n-1}$ and ${n\brack k}=\prod_{i=1}^{k}(1-q^{n-i+1})/(1-q^{i})$ . This gives a $q$ -analogue of a divisibility result on the Catalan triangle obtained by the first author and Zeng, and also confirms a conjecture of the first author and Zeng. We further propose several related conjectures.

Keywords: $q$ -binomial coefficients; $q$ -Catalan triangle; $q$ -Pfaff-Saalschütz identity; $q$ -Narayana numbers; $q$ -super Catalan numbers

AMS Subject Classifications (2000): 05A30, 05A10, 11B65.

1 Introduction

A few years ago, motivated by the divisibility results in [2, 3, 4, 9, 15, 17, 19, 28], the first author and Zeng [13] proved that

[TABLE]

and more general

[TABLE]

where $n_{m+1}=n_{1}$ .

The first objective of this paper is to give a $q$ -analogue of (1.1) and (1.2). Recall that the $q$ -integers are defined by $[n]=\frac{1-q^{n}}{1-q}$ and the $q$ -shifted factorials (see [7]) are defined by $(a;q)_{0}=1$ and $(a;q)_{n}=(1-a)(1-aq)\cdots(1-aq^{n-1})$ for $n=1,2,\ldots.$ The $q$ -binomial coefficients are defined as

[TABLE]

Our main results can be stated as follows.

Theorem 1.1.

For $j=0,1$ and all positive integers $n$ and $r$ , the expression

[TABLE]

is a polynomial in $q$ with integer coefficients.

Theorem 1.2.

Let $n_{1},\ldots,n_{m},n_{m+1}=n_{1}$ be positive integers. Then for any non-negative integers $j$ and $r$ with $j\leqslant m$ , the expression

[TABLE]

is a Laurent polynomial in $q$ with integer coefficients.

One may wonder why to consider $[2k][k]^{2r}/[2]$ as a $q$ -analogue of $k^{2r+1}$ rather than $[k]^{2r+1}$ . The idea comes from Schlosser’s work [18] on $q$ -analogues of sums of powers of the first $n$ consecutive natural numbers and the first author and Zeng’s work [11] which positively answers Schlosser’s question.

The first author and Zeng [13] also introduced the $q$ -Catalan triangle with entries given by

[TABLE]

The second objective is to give the following congruence related to the $q$ -Catalan triangle.

Theorem 1.3.

Let $n$ be a positive integer. Let $r\geqslant 0$ and $s\geqslant 1$ such that $r\not\equiv s\pmod{2}$ . Then, for $0\leqslant j\leqslant s$ , the expression

[TABLE]

is a Laurent polynomial in $q$ with integer coefficients.

It is easy to see that Theorem 1.3 is a generalization of [13, Theorem 1.4]. Letting $q=1$ in Theorem 1.3, we confirm a conjecture in [13].

The paper is organized as follows. We shall prove Theorems 1.1–1.3 in Sections 2–4, respectively. The $q$ -Pfaff-Saalschütz identity will play an important role in our proof. We give some consequences of Theorem 1.1 in Section 5. Finally, some open problems will be proposed in Section 6.

2 Proof of Theorem 1.1

We first consider the $j=0$ case. Let

[TABLE]

It is easy to see that

[TABLE]

For $r\geqslant 1$ , noticing the relation

[TABLE]

we have

[TABLE]

It follows that

[TABLE]

This proves that $S_{1}(n;q)\equiv 0\pmod{[n]^{2}{2n\brack n}}$ . Applying the recurrence relation (2.1) and by induction on $r$ , we can prove that, for all positive integers $r$ , there holds

[TABLE]

For $j=1$ , let

[TABLE]

It is well known that ${n\brack k}_{q^{-1}}={n\brack k}q^{k(k-n)}$ . Hence, $T_{r}(n;q)=q^{n^{2}+2rn+2n-2r-1}S_{r}(n;q^{-1})$ . By (2.2), we have

[TABLE]

Since both $S_{r}(n;q)$ and $T_{r}(n;q)$ are polynomials in $q$ , we obtain the desired conclusion.

3 Proof of Theorem 1.2

We will need the $q$ -Pfaff-Saalschütz identity (see [7, (II.12)] or [12, 27]):

[TABLE]

where $\frac{1}{(q;q)_{n}}=0$ if $n<0$ . Denote (1.4) by $S(n_{1},\ldots,n_{m};r,j,q)$ . Namely,

[TABLE]

where

[TABLE]

It is easy to see that, for $m\geqslant 3$ ,

[TABLE]

Letting $n_{3}\to\infty$ in (3.1), we get

[TABLE]

Substituting (3.3) and (3.4) into (3.2), we obtain

[TABLE]

where $l=s+k$ . Observing that, for $m\geqslant 3$ ,

[TABLE]

we get the following recurrence relation

[TABLE]

On the other hand, for $m=2$ , applying (3.4) we may deduce that

[TABLE]

We now proceed by induction on $m$ . For $m=1$ , the conclusion follows readily from the proof of Theorem 1.1. Suppose that the expression $S(n_{1},\ldots,n_{m-1};r,j,q)$ is a Laurent polynomial in $q$ with integer coefficients for some $m\geqslant 2$ and $0\leqslant j\leqslant m-1$ . Then by the recurrence (3.5) or (3.6), so is $S(n_{1},\ldots,n_{m};r,j,q)$ for $1\leqslant j\leqslant m$ . It is easy to see that

[TABLE]

Therefore, the expression $S(n_{1},\ldots,n_{m};r,0,q)$ is also a Laurent polynomial in $q$ with integer coefficients. This completes the inductive step of the proof.

4 Proof of Theorem 1.3

Let $\Phi_{n}(x)$ be the $n$ -th cyclotomic polynomial. We need the following result (see [16, Equation (10)] or [5, 14]).

Proposition 4.1.

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

[TABLE]

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

By Proposition 4.1, for any positive integer $n$ , we have

[TABLE]

Let $r+s\equiv 1\pmod{2}$ , $m=s\geqslant 1$ , and $0\leqslant j\leqslant s$ . Setting $n_{1}=\cdots=n_{s}=n$ in Theorem 1.2, we see that

[TABLE]

is a Laurent polynomial in $q$ with integer coefficients. Note that $B_{n,k}(q)$ is a polynomial in $q$ with integer coefficients (see [13, 10]). Hence, $[k]{2n\brack n+k}=[n]B_{n,k}(q)$ is clearly divisible by $[n]$ . It follows that

[TABLE]

is a Laurent polynomial in $q$ with integer coefficients. The proof then follows from (4.1).

5 Some consequences of Theorem 1.2

In this section, we will give some consequences of Theorem 1.2 and confirm some conjectures in [13, Section 7]. Letting $n_{2i-1}=m$ and $n_{2i}=n$ for $i=1,\ldots,r$ in Theorem 1.2 and observing the symmetry of $m$ and $n$ , we obtain

Corollary 5.1.

Let $m$ , $n$ , and $r$ be positive integers, and let $a$ and $j$ be non-negative integers with $j\leqslant 2r$ . Then the expression

[TABLE]

is a Laurent polynomial in $q$ with integer coefficients.

Letting $n_{3i-2}=l$ , $n_{3i-1}=m$ and $n_{3i}=n$ for $i=1,\ldots,r$ in Theorem 1.2, we get

Corollary 5.2.

Let $l$ , $m$ , $n$ and $r$ be positive integers, and let $a$ and $j$ be non-negative integers with $j\leqslant 3r$ . Then the expression

[TABLE]

is a Laurent polynomial in $q$ with integer coefficients.

Taking $m=2r+s$ and letting $n_{i}=n+1$ if $i=1,3,\ldots,2r-1$ and $n_{i}=n$ otherwise in Theorem 1.2, we get

Corollary 5.3.

Let $n$ , $r$ and $s$ be positive integers, and let $a$ and $j$ be non-negative integers with $j\leqslant 2r+s$ . Then the expression

[TABLE]

is a Laurent polynomial in $q$ with integer coefficients.

Let $[n]!=[n][n-1]\cdots[1]$ . It is clear that Theorem 1.2 can be restated as follows.

Theorem 5.4.

Let $n_{1},\ldots,n_{m}$ be positive integers. Let $j$ and $r$ be non-negative integers with $j\leqslant m$ . Then the expression

[TABLE]

where $n_{m+1}=0$ , is a Laurent polynomial in $q$ with integer coefficients.

Letting $n_{1}=\cdots=n_{r}=m$ and $n_{r+1}=\cdots=n_{r+s}=n$ in Theorem 5.4 and noticing the symmetry of $m$ and $n$ , we have

Corollary 5.5.

Let $m$ , $n$ , $r$ and $s$ be positive integers, and let $a$ and $j$ be non-negative integers with $j\leqslant r+s$ . Then the expression

[TABLE]

is a Laurent polynomial in $q$ with integer coefficients.

In particular, the expression

[TABLE]

is a Laurent polynomial in $q$ with integer coefficients.

The $q$ -Narayana numbers $N_{q}(n,k)$ and the $q$ -Catalan numbers $C_{n}(q)$ may be defined as follows:

[TABLE]

It is well known that both $q$ -Narayana numbers and $q$ -Catalan numbers are polynomials in $q$ with non-negative integer coefficients (see [1, 6]). It should be mentioned that the definition of $N_{q}(n,k)$ here differs by a factor $q^{k(k-1)}$ from that in [1]. Motivated mainly by Z.-W. Sun’s work on congruences for combinatorial numbers [20, 21, 22, 23, 24, 25], the first author and Jiang [8] proved that, for $0\leqslant j\leqslant 2r-1$ ,

[TABLE]

Exactly similarly to the proof of (5.1) in [8], we can deduce the following result from Theorem 5.4 by considering four special cases $(n_{1},\ldots,n_{2r})=(n,\ldots,n,n+1,\ldots,n+1),(n+1,\ldots,n+1,n,\ldots,n),(n,n+1,n,n+1,\ldots,n,n+1),(n+1,n,n+1,n,\ldots,n+1,n)$ and noticing that $[n]$ and $[n+1]$ are relatively prime.

Corollary 5.6.

Let $n$ and $r$ be positive integers, and let $a$ and $j$ be non-negative integers with $j\leqslant 2r$ . Then the expression

[TABLE]

is a Laurent polynomial in $q$ with integer coefficients.

6 Some open problems

In this section we propose several related conjectures for further study. Note that some similar conjectures were raised in [8, Section 3].

It is easy to see that the $q$ -super Catalan numbers $\frac{[2m]![2n]!}{[m+n]![m]![n]!}$ are polynomials in $q$ with integer coefficients. Warnaar and Zudilin [26, Proposition 2] proved that the $q$ -super Catalan numbers are in fact polynomials in $q$ with non-negative integer coefficients. The following conjecture related to the $q$ -super Catalan numbers is a refinement of Corollary 5.1 and is also a $q$ -analogue of [13, Conjecture 7.5].

Conjecture 6.1.

Let $m$ , $n$ , $s$ and $t$ be positive integers. Let $r$ be a non-negative integer with $r+s+t\equiv 1\pmod{2}$ and let $j$ be a integer. Then the expression

[TABLE]

is a Laurent polynomial in $q$ , and is a Laurent polynomial in $q$ with non-negative integer coefficients for $0\leqslant j\leqslant s+t$ .

It seems that Corollary 5.6 can be further generalized as follows.

Conjecture 6.2.

Corollary 5.6 is still true for any integer $j$ , and (5.2) is a Laurent polynomial in $q$ with non-negative integer coefficients for $0\leqslant j\leqslant 2r$ .

The following is a generalization of Theorem 1.2.

Conjecture 6.3.

Let $n_{1},\ldots,n_{m},n_{m+1}=n_{1}$ be positive integers. Then for any integer $j$ and non-negative integer $r$ , the expression

[TABLE]

is a Laurent polynomial in $q$ , and is a Laurent polynomial in $q$ with non-negative integer coefficients if $0\leqslant j\leqslant m$ .

We end the paper with the following conjecture. Note that when all the $n_{i}$ ’s are equal to $n$ , it reduces to Corollary 5.6.

Conjecture 6.4.

Let $n_{1},\ldots,n_{m},n_{m+1}=n_{1}$ be positive integers. Then for any integer $j$ and non-negative integer $r$ , the expression

[TABLE]

is a Laurent polynomial in $q$ , and is a Laurent polynomial in $q$ with non-negative integer coefficients if $0\leqslant j\leqslant 2m$ .

Acknowledgments. The first author was partially supported by the National Natural Science Foundation of China (grant 11371144), the Natural Science Foundation of Jiangsu Province (grant BK20161304), and the Qing Lan Project of Education Committee of Jiangsu Province.

Bibliography28

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TL;DR

Contribution

Findings

Abstract

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1 Introduction

Theorem 1.1**.**

Theorem 1.2**.**

Theorem 1.3**.**

2 Proof of Theorem 1.1

3 Proof of Theorem 1.2

4 Proof of Theorem 1.3

Proposition 4.1**.**

5 Some consequences of Theorem 1.2

Corollary 5.1**.**

Corollary 5.2**.**

Corollary 5.3**.**

Theorem 5.4**.**

Corollary 5.5**.**

Corollary 5.6**.**

6 Some open problems

Conjecture 6.1**.**

Conjecture 6.2**.**

Conjecture 6.3**.**

Conjecture 6.4**.**

Theorem 1.1.

Theorem 1.2.

Theorem 1.3.

Proposition 4.1.

Corollary 5.1.

Corollary 5.2.

Corollary 5.3.

Theorem 5.4.

Corollary 5.5.

Corollary 5.6.

Conjecture 6.1.

Conjecture 6.2.

Conjecture 6.3.

Conjecture 6.4.