# Some new results about $q$-trinomial coefficients

Yifan Chen  DEPARTMENT OF MATHEMATICS, SHANGHAI UNIVERSITY, SHANGHAI 200444,
P. R. CHINA  ∗ Corresponding author<EMAIL_ADDRESS>(X. Wang),
<EMAIL_ADDRESS>(C<EMAIL_ADDRESS>(Y. Chen). ,  Chang Xu  and
Xiaoxia Wang∗

###### Abstract.

In this paper, we present several new congruences on the $q$-trinomial
coefficients introduced by Andrews and Baxter. A new congruence on sums of
central $q$-binomial coefficients is also established.

###### Key words and phrases:

$q$-trinomial coefficients; $q$-congruences; cyclotomic polynomials

###### 2010 Mathematics Subject Classification:

Primary 33D15; Secondary 11A07, 11B65

##  1\. Introduction

In 2019, Straub [18] gave the following $q$-supercongruence:

$\genfrac{[}{]}{0.0pt}{0}{an}{bn}\equiv\genfrac{[}{]}{0.0pt}{0}{a}{b}_{q^{n^{2}}}-(a-b)b\dbinom{a}{b}\frac{n^{2}-1}{24}(q^{n}-1)^{2}\pmod{\Phi_{n}(q)^{3}},$
(1.1)

which is a $q$-analogue of the Wolstenholme–Ljunggren congruence: for any
prime $p\geq 5$,

$\dbinom{ap}{bp}\equiv\dbinom{a}{b}\pmod{p^{3}}.$

Here and throughout the paper, $\Phi_{n}(q)$ stands for the $n$-th cyclotomic
polynomial in $q$:

$\displaystyle\Phi_{n}(q)=\prod_{\begin{subarray}{c}1\leqslant k\leqslant n\\\
\gcd(n,k)=1\end{subarray}}(q-\zeta^{k}),$

where $\zeta$ is an $n$-th primitive root of unity. The $q$-binomial
coefficient is defined as

$\displaystyle\genfrac{[}{]}{0.0pt}{0}{n}{k}=\genfrac{[}{]}{0.0pt}{0}{n}{k}_{q}=\begin{cases}\displaystyle\frac{(q;q)_{n}}{(q;q)_{k}(q;q)_{n-k}},&\text{if
$0\leq k\leq n$; }\\\\[11.38109pt] 0,&\text{otherwise},\end{cases}$

where the $q$-shifted factorial is defined as $(a;q)_{0}=1$ and
$(a;q)_{n}=(1-a)(1-aq)\cdots(1-aq^{n-1})$ with $n\in\mathbb{Z}^{+}$.

For $n\in\mathbb{N}$ and integer $j$ with $-n\leq j\leq n$, the trinomial
coefficient is the coefficient of $x^{j}$ in the expansion of
$(1+x+x^{-1})^{n}$. Namely,

$\bigg{(}\\!\\!\dbinom{n}{j}\\!\\!\bigg{)}=[x^{j}](1+x+x^{-1})^{n},$

and it has a simple expression (see[17])

$\bigg{(}\\!\\!\dbinom{n}{j}\\!\\!\bigg{)}=\sum_{k=0}^{n}\dbinom{n}{k}\dbinom{n-k}{k+j}.$

Six different $q$-analogues of the trinomial coefficients, which play
significant roles in hard hexagon model, were introduced by Andrews and Baxter
[1]. One of these $q$-analogues is

$\bigg{(}\\!\\!\dbinom{n}{j}\\!\\!\bigg{)}_{q}=\sum_{k=0}^{n}q^{k(k+j)}\genfrac{[}{]}{0.0pt}{0}{n}{k}\genfrac{[}{]}{0.0pt}{0}{n-k}{k+j}.$
(1.2)

During the past few years, some experts have paid attention to $q$-analogues
of supercongruences. We refer the reader to [14, 15, 9, 5, 6] for some of
their work. Moreover, some congruences for $q$-binomial coefficients and
$q$-trinomial coefficients can be found in [7, 16, 21, 3, 13, 2, 8].

Recently, Liu [12] established the following beautiful $q$-supercongruences:
for any positive integer $n$, modulo $\Phi_{n}(q)^{2}$,

$\displaystyle\bigg{(}\\!\\!\dbinom{n}{0}\\!\\!\bigg{)}_{q}$
$\displaystyle\equiv\begin{cases}(-1)^{m}(1+q^{m})q^{m(3m-1)/2},&\text{if \
$n=3m$};\\\\[11.38109pt] (-1)^{m}q^{m(3m+1)/2},&\text{if \
$n=3m+1$};\\\\[11.38109pt] (-1)^{m}q^{m(3m-1)/2},&\text{if \
$n=3m-1$},\end{cases}$ (1.3)
$\displaystyle\bigg{(}\\!\\!\dbinom{2n}{n}\\!\\!\bigg{)}_{q}$
$\displaystyle\equiv\begin{cases}2(-1)^{m}(1+q^{m})q^{m(3m-1)/2}-3m(1-q^{3m}),&\text{if
\ $n=3m$};\\\\[11.38109pt] 2(-1)^{m}q^{m(3m+1)/2}-(3m+1)(1-q^{3m+1}),&\text{if
\ $n=3m+1$};\\\\[11.38109pt]
2(-1)^{m}q^{m(3m-1)/2}-(3m-1)(1-q^{3m-1}),&\text{if \ $n=3m-1$}.\end{cases}$
(1.4)

Motivated by the work just mentioned, we shall establish the following
$q$-supercongruence similar to (1.3) and (1.4).

######  Theorem 1.1.

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

$\displaystyle\bigg{(}\\!\\!\dbinom{2n}{0}\\!\\!\bigg{)}_{q}\equiv\begin{cases}2(-1)^{m}(1+q^{m})q^{m(3m-1)/2}-9m(1-q^{3m})+1,&\text{if
\ $n=3m$};\\\\[11.38109pt]
2(-1)^{m}q^{m(3m+1)/2}-3(3m+1)(1-q^{3m+1})+1,&\text{if \
$n=3m+1$};\\\\[11.38109pt]
2(-1)^{m}q^{m(3m-1)/2}-3(3m-1)(1-q^{3m-1})+1,&\text{if \
$n=3m-1$}.\end{cases}$ (1.5)

More generally, we shall give the following two new $q$-supercongruences.

######  Theorem 1.2.

For any positive integers $a$ and $n$, modulo ${\Phi_{n}(q)^{2}}$,

$\displaystyle\bigg{(}\\!\\!\dbinom{an}{an-n}\\!\\!\bigg{)}_{q}\equiv\begin{cases}(-1)^{m}a(1+q^{m})q^{m(3m-1)/2}-3m(1-q^{3m}){\binom{a}{2}},&\text{if
\ $n=3m$};\\\\[14.22636pt]
(-1)^{m}aq^{m(3m+1)/2}-(3m+1)(1-q^{3m+1})\binom{a}{2},&\text{if \
$n=3m+1$};\\\\[14.22636pt]
(-1)^{m}aq^{m(3m-1)/2}-(3m-1)(1-q^{3m-1})\binom{a}{2},&\text{if \
$n=3m-1$}.\end{cases}$ (1.6)

######  Theorem 1.3.

For any positive integers $a$ and $n$ with $a\geq 2$, modulo
${\Phi_{n}(q)^{2}}$,

$\displaystyle\bigg{(}\\!\\!\dbinom{an}{an-2n}\\!\\!\bigg{)}_{q}$
$\displaystyle\equiv\begin{cases}(-1)^{m}2\binom{a}{2}(1+q^{m})q^{m(3m-1)/2}-9m(1-q^{3m})\binom{a+1}{3}+\frac{3a-a^{2}}{2},&\text{if
\ $n=3m$};\\\\[11.38109pt]
(-1)^{m}2\binom{a}{2}q^{m(3m+1)/2}-3(3m+1)(1-q^{3m+1})\binom{a+1}{3}+\frac{3a-a^{2}}{2},&\text{if
\ $n=3m+1$};\\\\[11.38109pt]
(-1)^{m}2\binom{a}{2}q^{m(3m-1)/2}-3(3m-1)(1-q^{3m-1})\binom{a+1}{3}+\frac{3a-a^{2}}{2},&\text{if
\ $n=3m-1$}.\end{cases}$ (1.7)

It is easily seen that Liu’s two supercongruences (1.3) and (1.4) are the
special cases of Theorem 1.2 with $a=1$ and $a=2$, respectively, and Theorem
1.1 can be obtained by taking $a=2$ in Theorem 1.3.

As serendipitous discoveries, we obtain the following $q$-congruences which
are the contiguous forms of (1.6) and (1.3) modulo $\Phi_{n}(q)$.

######  Theorem 1.4.

For any integer $a\geq 1$, integer $j$ and positive odd integer $n$ with
$0\leq j\leq n-1$, there holds

$\displaystyle\bigg{(}\\!\\!\dbinom{an-1}{an-n+j}\\!\\!\bigg{)}_{q}$
$\displaystyle\equiv\bigg{(}\frac{n-j}{3}\bigg{)}q^{(n-j-1)(n-j-2)/6-(j^{2}+j)/2}\pmod{\Phi_{n}(q)}.$
(1.8)

Here and in what follows, $\big{(}\frac{.}{p}\big{)}$ denotes the Legendre
symbol.

######  Theorem 1.5.

For any integer $a\geq 2$, integer $j$ and positive odd integer $n$ with
$0\leq j\leq n$, there holds, modulo $\Phi_{n}(q)$,

$\displaystyle\bigg{(}\\!\\!\dbinom{an-1}{an-2n+j}\\!\\!\bigg{)}_{q}$
$\displaystyle\equiv(a-1)\bigg{(}\frac{n-j}{3}\bigg{)}q^{\frac{(n-j-1)(n-j-2)}{6}-\frac{j^{2}+j}{2}}+\bigg{(}\frac{j}{3}\bigg{)}q^{\frac{1-j^{2}}{3}}.$
(1.9)

It should be mentioned that if we take $j=0$ in Theorem 1.4 and $j=n$ in
Theorem 1.5, then the right-hand sides of (1.8) and (1.9) are congruent to
each other modulo $\Phi_{n}(q)$.

######  Theorem 1.6.

For any integer $a\geq 2$, positive integer $b$ satisfying $a\leq b+2$ and
positive odd integer $n$, there holds

$\displaystyle\bigg{(}\\!\\!\dbinom{an-1}{bn-1}\\!\\!\bigg{)}_{q}\equiv\dbinom{a}{b}+\dbinom{a-1}{b}\bigg{(}\frac{n+1}{3}\bigg{)}q^{n(n-1)/6}\pmod{\Phi_{n}(q)}.$
(1.10)

By taking $j=n-1$ in Theorem 1.5 and $b=a-1$ in Theorem 1.6, we can easily
find that they have the same left-hand sides. As a result, the right-hand
sides of (1.9) and (1.10) are congruent to each other modulo $\Phi_{n}(q)$. In
other words, for any integer $a\geq 2$ and positive odd integer $n$, there
holds

$a+\bigg{(}\frac{n+1}{3}\bigg{)}q^{n(n-1)/6}\equiv(a-1)q^{-n(n-1)/2}+\bigg{(}\frac{n-1}{3}\bigg{)}q^{-n(n-2)/3}\pmod{\Phi_{n}(q)}.$

The rest of the paper is arranged as follows. In the next section, we shall
give proofs of Theorems 1.2 and 1.3. The proofs of Theorems 1.4 and 1.5 will
be presented in Sections 3 and 4, respectively. In the last section, a sketch
of the proof of Theorem 1.6 will be provided.

##  2\. Proofs of Theorems 1.2 and 1.3

In order to prove Theorems 1.2 and 1.3, we need the following two lemmas which
have been proved by Liu.

######  Lemma 2.1 (Liu[11]).

For any non-negative integer $n$, there holds

$\displaystyle(1-q^{n})\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}\frac{(-1)^{k}q^{k(k-1)/2}}{1-q^{n-k}}\genfrac{[}{]}{0.0pt}{0}{n-k}{k}=\begin{cases}(-1)^{m}(1+q^{m})q^{m(3m-1)/2},&\text{if
\ $n=3m$};\\\\[5.69054pt] (-1)^{m}q^{m(3m+1)/2},&\text{if \
$n=3m+1$};\\\\[5.69054pt] (-1)^{m}q^{m(3m-1)/2},&\text{if \
$n=3m-1$},\end{cases}$ (2.1)

where $\lfloor x\rfloor$ is the integral part of real $x$.

######  Lemma 2.2 (Liu[12]).

For any positive integer $n$, there holds

$\sum_{k=1}^{\lfloor\frac{n}{2}\rfloor}\frac{q^{-k(k-1)}}{[2k]_{q}}\genfrac{[}{]}{0.0pt}{0}{2k}{k}\equiv\frac{(1-q)(1-R_{n}(q))}{1-q^{n}}\pmod{\Phi_{n}(q)},$

where $R_{n}(q)$ denotes the right-hand side of (2.1).

It is not difficult to see that $R_{n}(q)\equiv 1\pmod{\Phi_{n}(q)}$.

###### Proof of Theorem 1.2.

Firstly, we express the left-hand side of (1.6) by Andrews and Baxter’s
expression (1.2). Note that${an-k\brack k+an-n}$ for any
$k>\lfloor\frac{n}{2}\rfloor$. Therefore, we get

$\displaystyle\bigg{(}\\!\\!\binom{an}{an-n}\\!\\!\bigg{)}_{q}$
$\displaystyle=\sum_{k=0}^{an}q^{k(k+an-n)}\genfrac{[}{]}{0.0pt}{0}{an}{k}\genfrac{[}{]}{0.0pt}{0}{an-k}{k+an-n}$
$\displaystyle=\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}q^{k(k+an-n)}\genfrac{[}{]}{0.0pt}{0}{an}{k}\genfrac{[}{]}{0.0pt}{0}{an-k}{k+an-n}$
$\displaystyle=\genfrac{[}{]}{0.0pt}{0}{an}{an-n}+\sum_{k=1}^{\lfloor\frac{n}{2}\rfloor}q^{k(k+an-n)}\genfrac{[}{]}{0.0pt}{0}{an}{k}\genfrac{[}{]}{0.0pt}{0}{an-k}{k+an-n}.$
(2.2)

For $1\leq k\leq\lfloor\frac{n}{2}\rfloor$, we have

$\displaystyle\genfrac{[}{]}{0.0pt}{0}{an}{k}$
$\displaystyle=\frac{(1-q^{an})(1-q^{an-1})\cdots(1-q^{an+1-k})}{(1-q)(1-q^{2})\cdots(1-q^{k})}$
$\displaystyle\equiv\frac{a(1-q^{n})(1-q^{-1})\cdots(1-q^{1-k})}{(1-q)(1-q^{2})\cdots(1-q^{k})}$
$\displaystyle\equiv\frac{a(1-q^{n})(-1)^{k-1}q^{-k(k-1)/2}}{1-q^{k}}\quad\pmod{\Phi_{n}(q)^{2}},$
(2.3)

where the second relation is due to the fact that $1-q^{an}\equiv
a(1-q^{n})\pmod{\Phi_{n}(q)^{2}}$.

Because the reduced form of ${an\brack k}$ contains the factor $\Phi_{n}(q)$,
it is enough to consider the following $q$-binomial coefficient modulo
$\Phi_{n}(q)$. For $1\leq k\leq\lfloor\frac{n}{2}\rfloor$, we have

$\displaystyle\genfrac{[}{]}{0.0pt}{0}{an-k}{k+an-n}$
$\displaystyle=\genfrac{[}{]}{0.0pt}{0}{an}{an-n}\frac{(1-q^{n})(1-q^{n-1})\cdots(1-q^{n+1-2k})}{(1-q^{an})\cdots(1-q^{an-k+1})(1-q^{(a-1)n+1})\cdots(1-q^{(a-1)n+k})}$
$\displaystyle\equiv\genfrac{[}{]}{0.0pt}{0}{an}{n}\frac{1-q^{n}}{1-q^{an}}\frac{(1-q^{-1})\cdots(1-q^{1-2k})}{(1-q^{-1})\cdots(1-q^{-k+1})(1-q)\cdots(1-q^{k})}$
$\displaystyle\equiv\genfrac{[}{]}{0.0pt}{0}{an}{n}\frac{(-1)^{k}}{a}q^{-(3k-1)k/2}\genfrac{[}{]}{0.0pt}{0}{2k-1}{k}\pmod{\Phi_{n}(q)}.$
(2.4)

Substituting the results (2) and (2) into the right-hand side of (2), we
arrive at

$\bigg{(}\\!\\!\binom{an}{an-n}\\!\\!\bigg{)}_{q}\equiv\genfrac{[}{]}{0.0pt}{0}{an}{n}-(1-q^{n})\genfrac{[}{]}{0.0pt}{0}{an}{n}\sum_{k=1}^{\lfloor\frac{n}{2}\rfloor}\frac{q^{-k(k-1)}}{1-q^{k}}\genfrac{[}{]}{0.0pt}{0}{2k-1}{k}\pmod{\Phi_{n}(q)^{2}}.$
(2.5)

Then by Lemma 2.2, we obtain

$\displaystyle\sum_{k=1}^{\lfloor\frac{n}{2}\rfloor}\frac{q^{-k(k-1)}}{1-q^{k}}\genfrac{[}{]}{0.0pt}{0}{2k-1}{k}=\frac{1}{1-q}\sum_{k=1}^{\lfloor\frac{n}{2}\rfloor}\frac{q^{-k(k-1)}}{[2k]_{q}}\genfrac{[}{]}{0.0pt}{0}{2k}{k}\equiv\frac{1-R_{n}(q)}{1-q^{n}}\pmod{\Phi_{n}(q)}.$
(2.6)

On the other hand, by the weaker version of (1.1), we have

$\begin{split}\genfrac{[}{]}{0.0pt}{0}{an}{n}&\equiv\genfrac{[}{]}{0.0pt}{0}{a}{1}_{q^{n^{2}}}\equiv
1+q^{n^{2}}+q^{2n^{2}}+\cdots+q^{(a-1)n^{2}}\\\ &\equiv
a-((1-q^{n^{2}})+(1-q^{2n^{2}})+\cdots+(1-q^{(a-1)n^{2}}))\\\ &\equiv
a-n(1-q^{n})\frac{a(a-1)}{2}\pmod{\Phi_{n}(q)^{2}},\end{split}$ (2.7)

where we have utilized the fact $q^{n}\equiv 1\pmod{\Phi_{n}(q)}$ and the
identity

$1-q^{kn^{2}}=(1-q^{n})(1+q^{n}+q^{2n}+\cdots+q^{(kn-1)n}).$

Finally, substituting (2.6) and (2.7) into the right-hand side of (2.5), we
are led to

$\bigg{(}\\!\\!\binom{an}{an-n}\\!\\!\bigg{)}_{q}\equiv
aR_{n}(q)-n(1-q^{n})\dbinom{a}{2}\pmod{\Phi_{n}(q)^{2}}$

as desired. ∎

###### Proof of Theorem 1.3.

The proof is quite similar to that of Theorem 1.2. In order to apply Lemma
2.2, we express $\big{(}\binom{an}{an-2n}\big{)}_{q}$ as follows:

$\bigg{(}\\!\\!\binom{an}{an-2n}\\!\\!\bigg{)}_{q}=\genfrac{[}{]}{0.0pt}{0}{an}{an-2n}+q^{(a-1)n^{2}}\genfrac{[}{]}{0.0pt}{0}{an}{n}+\sum_{k=1}^{n-1}q^{k(k+an-2n)}\genfrac{[}{]}{0.0pt}{0}{an}{k}\genfrac{[}{]}{0.0pt}{0}{an-k}{k+an-2n}.$

For $1\leq k\leq n-1$, we have

$\displaystyle\genfrac{[}{]}{0.0pt}{0}{an-k}{k+an-2n}$
$\displaystyle=\genfrac{[}{]}{0.0pt}{0}{an}{2n}\frac{(1-q^{2n})\cdots(1-q^{2n+1-2k})}{(1-q^{an})\cdots(1-q^{an-k+1})(1-q^{(a-2)n+1})\cdots(1-q^{(a-2)n+k})}$
$\displaystyle\equiv\genfrac{[}{]}{0.0pt}{0}{an}{2n}\frac{1-q^{2n}}{1-q^{an}}\frac{(1-q^{-1})\cdots(1-q^{1-2k})}{(1-q^{-1})\cdots(1-q^{-k+1})(1-q)\cdots(1-q^{k})}$
$\displaystyle\equiv\genfrac{[}{]}{0.0pt}{0}{an}{2n}\>\frac{2}{a}\>(-1)^{k}q^{-(3k-1)k/2}\genfrac{[}{]}{0.0pt}{0}{2k-1}{k}\quad\pmod{\Phi_{n}(q)}.$
(2.8)

Applying (2), (2.6) and (2), we obtain

$\displaystyle\bigg{(}\\!\\!\binom{an}{an-2n}\\!\\!\bigg{)}_{q}$
$\displaystyle=\genfrac{[}{]}{0.0pt}{0}{an}{an-2n}+q^{(a-1)n^{2}}\genfrac{[}{]}{0.0pt}{0}{an}{n}+\sum_{k=1}^{n-1}q^{k(k+an-2n)}\genfrac{[}{]}{0.0pt}{0}{an}{k}\genfrac{[}{]}{0.0pt}{0}{an-k}{k+an-2n}$
$\displaystyle\equiv\genfrac{[}{]}{0.0pt}{0}{an}{2n}+q^{(a-1)n^{2}}\genfrac{[}{]}{0.0pt}{0}{an}{n}-2(1-q^{n})\genfrac{[}{]}{0.0pt}{0}{an}{2n}\sum_{k=1}^{n-1}\frac{q^{-k(k-1)}}{1-q^{k}}\genfrac{[}{]}{0.0pt}{0}{2k-1}{k}$
$\displaystyle\equiv\genfrac{[}{]}{0.0pt}{0}{an}{2n}+q^{(a-1)n^{2}}\genfrac{[}{]}{0.0pt}{0}{an}{n}-2(1-q^{n})\genfrac{[}{]}{0.0pt}{0}{an}{2n}\frac{1-R_{n}(q)}{1-q^{n}}\pmod{\Phi_{n}(q)^{2}}.$
(2.9)

The third step holds since ${2k-1\brack k}\equiv 0\pmod{\Phi_{n}(q)}$ for
$\lfloor\frac{n}{2}\rfloor+1\leq k\leq n-1$.

By utilizing the method used in (2.7), we get

$\begin{split}\genfrac{[}{]}{0.0pt}{0}{an}{2n}\equiv\genfrac{[}{]}{0.0pt}{0}{a}{2}_{q^{n^{2}}}\equiv\frac{a(a-1)}{2}-n(1-q^{n})\frac{a(a-1)(a-2)}{2}\pmod{\Phi_{n}(q)^{2}}.\end{split}$
(2.10)

Substituting (2.7) and (2.10) into the right-hand side of (2), we arrive at

$\bigg{(}\\!\\!\binom{an}{an-2n}\\!\\!\bigg{)}_{q}\equiv-\frac{(a+1)a(a-1)}{2}n(1-q^{n})+\frac{3a-a^{2}}{2}+a(a-1)R_{n}(q)\pmod{\Phi_{n}(q)^{2}}.$

This completes the proof. ∎

##  3\. Proof of Theorem 1.4

For the sake of proving Theorem 1.4, we need the following congruence on the
sum of central $q$-binomial coefficients.

######  Theorem 3.1.

For any positive odd integer $n$ and integer $j$ with $0\leq j\leq n-1$, there
holds

$\sum_{k=0}^{n-j-1}q^{-k(k+1+j)}\genfrac{[}{]}{0.0pt}{0}{2k+j}{k}\equiv(-1)^{j}\bigg{(}\frac{n-j}{3}\bigg{)}q^{(n-j-1)(n-j-2)/6}\pmod{\Phi_{n}(q)}.$

Obviously, the special case of Theorem 3.1 with $j=0$ is

$\sum_{k=0}^{n-1}q^{-k(k+1)}\genfrac{[}{]}{0.0pt}{0}{2k}{k}\equiv\bigg{(}\frac{n}{3}\bigg{)}q^{(n-1)(n-2)/6}\pmod{\Phi_{n}(q)},$
(3.1)

which is a $q$-analogue of a congruence by Sun and Tauraso [19] (the modulo
$p$ version): for any prime $p\geq 5$,

$\sum_{k=0}^{p-1}\dbinom{2k}{k}\equiv\bigg{(}\frac{p}{3}\bigg{)}\pmod{p^{2}}.$

In order to prove Theorem 3.1, we recall the following result as the lemma,
which firstly appeared in Ekhad and Zeilberger [4] and Krattenthaler’s note
[10] and latter been proved by Warnaar [20] as a special case of a cubic
summation formula.

######  Lemma 3.2.

For any integer $n\geq 0$, there holds

$\sum_{k=0}^{n}(-1)^{k}q^{k(k-1)/2}\genfrac{[}{]}{0.0pt}{0}{n-k}{k}=(-1)^{n}\bigg{(}\frac{n+1}{3}\bigg{)}q^{n(n-1)/6}.$

###### Proof of Theorem 3.1.

Let $\omega=e^{2\pi mi/n}$ with $\gcd(m,n)=1$. Then we have that
$\omega^{n}=1$ and $\omega^{k}\neq 1$ for $1\leq k\leq n-1$. Observing that

$\displaystyle\omega^{-k(k+j+1)}\genfrac{[}{]}{0.0pt}{0}{2k+j}{k}_{\omega}=\omega^{-k(k+j+1)}\prod_{l=1}^{k}\frac{1-\omega^{2k+j+1-l}}{1-\omega^{l}}=(-1)^{k}\omega^{k(k-1)/2}\begin{bmatrix}n-k-j-1\\\
k\end{bmatrix}_{\omega},$

we immediately get

$\displaystyle\sum_{k=0}^{n-j-1}q^{-k(k+j+1)}\genfrac{[}{]}{0.0pt}{0}{2k+j}{k}$
$\displaystyle\equiv\sum_{k=0}^{n-j-1}(-1)^{k}q^{k(k-1)/2}\genfrac{[}{]}{0.0pt}{0}{n-k-j-1}{k}$
$\displaystyle\equiv(-1)^{j}\bigg{(}\frac{n-j}{3}\bigg{)}q^{(n-j-2)(n-j-1)/6}\pmod{\Phi_{n}(q)},$

where we have utilized Lemma 3.2. ∎

Now we begin to prove Theorem 1.4.

###### Proof of Theorem 1.4.

For $1\leq k\leq n-1$, we have

$\genfrac{[}{]}{0.0pt}{0}{an-1}{k}=\frac{(1-q^{an-1})\cdots(1-q^{an-k})}{(1-q)\cdots(1-q^{k})}\equiv(-1)^{k}q^{-k(k+1)/2}\pmod{\Phi_{n}(q)}.$
(3.2)

We can also get the following congruence with $1\leq
k\leq\lfloor(n-1-j)/2\rfloor$,

$\displaystyle\genfrac{[}{]}{0.0pt}{0}{an-k-1}{an-n+k+j}$
$\displaystyle=\genfrac{[}{]}{0.0pt}{0}{an}{n}\frac{(1-q^{n})\cdots(1-q^{n-2k-j})}{(1-q^{an})\cdots(1-q^{an-k})(1-q^{(a-1)n+1})\cdots(1-q^{(a-1)n+k+j})}$
$\displaystyle\equiv(-1)^{k+j}q^{-(3k+j+1)(k+j)/2}\genfrac{[}{]}{0.0pt}{0}{2k+j}{k}\quad\pmod{\Phi_{n}(q)}.$
(3.3)

By combining the congruence (3.2) with (3), we arrive at the following result

$\displaystyle\bigg{(}\\!\\!\dbinom{an-1}{an-n+j}\\!\\!\bigg{)}_{q}$
$\displaystyle=\sum_{k=0}^{\lfloor\frac{n-j-1}{2}\rfloor}q^{k(k+an-n+j)}\genfrac{[}{]}{0.0pt}{0}{an-1}{k}\genfrac{[}{]}{0.0pt}{0}{an-k-1}{k+an-n+j}$
$\displaystyle\equiv\genfrac{[}{]}{0.0pt}{0}{an-1}{an-n+j}-(-1)^{j}q^{-(j^{2}+j)/2}\sum_{k=1}^{\lfloor\frac{n-j-1}{2}\rfloor}q^{-k^{2}-kj-k}\genfrac{[}{]}{0.0pt}{0}{2k+j}{k}$
$\displaystyle\equiv\bigg{(}\frac{n-j}{3}\bigg{)}q^{(n-j-1)(n-j-2)/6-(j^{2}+j)/2}\pmod{\Phi_{n}(q)},$

where we have applied Theorem 3.1 and ${2k+j\brack k}\equiv
0\pmod{\Phi_{n}(q)}$ for $\lfloor\frac{n-j-1}{2}\rfloor+1\leq k\leq n-j-1$ in
the last step. ∎

##  4\. Proof of Theorem 1.5

###### Proof of Theorem 1.5.

Because of the different range of $k$, the proof of Theorem 1.5 is a little
different from that of Theorem 1.4. We first split the summation of the
following $q$-trinomial coefficient into three parts as

$\displaystyle\bigg{(}\\!\\!\dbinom{an-1}{an-2n+j}\\!\\!\bigg{)}_{q}=$
$\displaystyle\genfrac{[}{]}{0.0pt}{0}{an-1}{an-2n+j}+\sum_{k=1}^{n-j-1}q^{k(k+an-2n+j)}\genfrac{[}{]}{0.0pt}{0}{an-1}{k}\genfrac{[}{]}{0.0pt}{0}{an-k-1}{k+an-2n+j}$
$\displaystyle+\sum_{k=n-j}^{\lfloor\frac{2n-j-1}{2}\rfloor}q^{k(k+an-2n+j)}\genfrac{[}{]}{0.0pt}{0}{an-1}{k}\genfrac{[}{]}{0.0pt}{0}{an-k-1}{k+an-2n+j}.$
(4.1)

The first and the second parts of the right-hand side of (4) can be handled
just as what we have done in the proof of Theorem 1.4. Thus, we can get the
following result with no difficulty,

$\displaystyle\genfrac{[}{]}{0.0pt}{0}{an-1}{an-2n+j}+\sum_{k=1}^{n-j-1}q^{k(k+an-2n+j)}\genfrac{[}{]}{0.0pt}{0}{an-1}{k}\genfrac{[}{]}{0.0pt}{0}{an-k-1}{k+an-2n+j}$
$\displaystyle\quad\equiv(a-1)\bigg{(}\frac{n-j}{3}\bigg{)}q^{(n-j-1)(n-j-2)/6-(j^{2}+j)/2}\pmod{\Phi_{n}(q)}.$
(4.2)

Now we begin to calculate the third part of the right-hand side of (4). Since
the third part will disappear if $j=0$, we only need to consider the case
where $j$ is in the range $1\leq j\leq n$. Noticing that ${an-k-1\brack
k+an-2n+j}$ can be simplified through the method used in (3), and recalling
the result (3.2), we have

$\displaystyle\sum_{k=n-j}^{\lfloor\frac{2n-j-1}{2}\rfloor}q^{k(k+an-2n+j)}\genfrac{[}{]}{0.0pt}{0}{an-1}{k}\genfrac{[}{]}{0.0pt}{0}{an-k-1}{k+an-2n+j}$
$\displaystyle\
\equiv(-1)^{j}q^{-(j^{2}+j)/2}\sum_{k=n-j}^{\lfloor\frac{2n-j-1}{2}\rfloor}q^{-k^{2}-kj-k}\genfrac{[}{]}{0.0pt}{0}{2k+j}{k}\pmod{\Phi_{n}(q)}.$
(4.3)

Performing $k\to n-j+k$ in the summation of the right-hand side of (4), we
have

$\displaystyle\sum_{k=n-j}^{\lfloor\frac{2n-j-1}{2}\rfloor}q^{-k^{2}-kj-k}\genfrac{[}{]}{0.0pt}{0}{2k+j}{k}$
$\displaystyle=\sum_{k=0}^{\lfloor\frac{j-1}{2}\rfloor}q^{-(k+n-j)^{2}-(k+n-j)j-(k+n-j)}\genfrac{[}{]}{0.0pt}{0}{2(k+n-j)+j}{k+n-j}$
$\displaystyle\equiv\sum_{k=0}^{\lfloor\frac{j-1}{2}\rfloor}q^{-k^{2}+kj-k+j}\genfrac{[}{]}{0.0pt}{0}{n+2k-j}{k}$
$\displaystyle\equiv(-1)^{n-j}\bigg{(}\frac{j}{3}\bigg{)}q^{(j-1)(j-2)/6}\pmod{\Phi_{n}(q)},$
(4.4)

where we have utilized Theorem 3.1 with $j\to n-j$ in the last step.

Finally, by combining the formulas from (4) to (4) together, we obtain

$\displaystyle\bigg{(}\\!\\!\dbinom{an-1}{an-2n+j}\\!\\!\bigg{)}_{q}$
$\displaystyle\equiv(a-1)\bigg{(}\frac{n-j}{3}\bigg{)}q^{\frac{(n-j-1)(n-j-2)}{6}-\frac{j^{2}+j}{2}}+\bigg{(}\frac{j}{3}\bigg{)}q^{\frac{1-j^{2}}{3}}\pmod{\Phi_{n}(q)}.$
(4.5)

The right-hand side of (4) equals $0$ if $j=0$, and so (4.5) also holds when
$j=0$. Thus we finish the proof of Theorem 1.5. ∎

##  5\. The Proof of Theorem 1.6

###### Sketch of Proof.

For $1\leq k\leq\lfloor(a-b)n/2\rfloor$, we have

$\displaystyle\genfrac{[}{]}{0.0pt}{0}{an-k-1}{k+bn-1}$
$\displaystyle=\genfrac{[}{]}{0.0pt}{0}{an}{bn}\frac{(1-q^{(a-b)n})\cdots(1-q^{(a-b)n-2k+1})}{(1-q^{bn+1})\cdots(1-q^{bn+k-1})(1-q^{an-k})\cdots(1-q^{an})}$
$\displaystyle\equiv(-1)^{k-1}\dbinom{a-1}{b}q^{-3k(k-1)/2}\genfrac{[}{]}{0.0pt}{0}{2k-1}{k}\pmod{\Phi_{n}(q)}.$

Similarly to the proof of Theorem 1.4, we get

$\displaystyle\bigg{(}\\!\\!\dbinom{an-1}{bn-1}\\!\\!\bigg{)}_{q}$
$\displaystyle=\sum_{k=0}^{\lfloor\frac{(a-b)n}{2}\rfloor}q^{k(k+bn-1)}\genfrac{[}{]}{0.0pt}{0}{an-1}{k}\genfrac{[}{]}{0.0pt}{0}{an-k-1}{k+bn-1}$
$\displaystyle\equiv\genfrac{[}{]}{0.0pt}{0}{an-1}{bn-1}-\sum_{k=1}^{\lfloor\frac{(a-b)n}{2}\rfloor}q^{-k^{2}}\genfrac{[}{]}{0.0pt}{0}{2k-1}{k}$
$\displaystyle\equiv\genfrac{[}{]}{0.0pt}{0}{an-1}{bn-1}-\sum_{k=1}^{\lfloor\frac{(a-b)n}{2}\rfloor}(-1)^{k}q^{k(k-1)/2}\genfrac{[}{]}{0.0pt}{0}{n-k}{k}$
$\displaystyle\equiv\dbinom{a}{b}+\dbinom{a-1}{b}\bigg{(}\frac{n+1}{3}\bigg{)}q^{n(n-1)/6}\pmod{\Phi_{n}(q)},$

where we have applied Lemma 3.2 in the last step. ∎

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