Polynomial Reduction and Super Congruences

TL;DR

This paper introduces a polynomial reduction method for hypergeometric terms, revealing symmetry properties that lead to new super-congruences, advancing understanding in hypergeometric and number theory.

Contribution

It presents a novel polynomial reduction technique for hypergeometric terms and derives new infinite families of super-congruences based on symmetry properties.

Findings

Reduced hypergeometric terms contain only odd or even powers under symmetry

Derived two infinite families of super-congruences

Enhanced understanding of hypergeometric term reductions

Abstract

Based on a reduction processing, we rewrite a hypergeometric term as the sum of the difference of a hypergeometric term and a reduced hypergeometric term (the reduced part, in short). We show that when the initial hypergeometric term has a certain kind of symmetry, the reduced part contains only odd or even powers. As applications, we derived two infinite families of super-congruences.

Equations

$$\sum _{k=0}^{\frac{p-1}{2}}(-1{)}^k(4k+1){\left(\frac{(1/2{)}_k}{(1{)}_k}\right)}^3\equiv (-1{)}^{\frac{p-1}{2}}p(\mathrm{mod}{p}^3),$$

$$\sum _{k=0}^{\frac{p-1}{2}}(-1{)}^k(4k+1){\left(\frac{(1/2{)}_k}{(1{)}_k}\right)}^3\equiv (-1{)}^{\frac{p-1}{2}}p+p^3{E}_{p-3}(\mathrm{mod}{p}^4),$$

$$\frac{2}{e^x+e^{-x}}=\sum _{n=0}^{\infty }{E}_n\frac{x^n}{n!}.$$

$$\sum _{k=0}^{\frac{p-1}{2}}(4k+1){\left(\frac{(1/2{)}_k}{(1{)}_k}\right)}^4\equiv p(\mathrm{mod}{p}^3).$$

$$\sum _{k=0}^{\frac{p^r-1}{2}}(-1{)}^k(4k+1{)}^m{\left(\frac{(1/2{)}_k}{(1{)}_k}\right)}^3\equiv a_{m,p}{p}^r(-1{)}^{\frac{(p-1)r}{2}}(\mathrm{mod}{p}^{r+2}).$$

$$\sum _{k=0}^{\frac{p^r-1}{2}}(4k+1{)}^m{\left(\frac{(1/2{)}_k}{(1{)}_k}\right)}^4\equiv b_{m,p}{p}^r(\mathrm{mod}{p}^{r+3}).$$

$$f(k)t_k=\Delta (g(k)t_k)+h(k)t_k=(g(k+1)t_{k+1}-g(k)t_k)+h(k)t_k,$$

$$f(k)t_k-\Delta x(k)t_k=(f(k)-u(k))t_k$$

$$\frac{t_{k+1}}{t_k}=\frac{a(k)}{b(k)},$$

$${\Delta }_k\left(b(k-1)x(k)t_k\right)=(a(k)x(k+1)-b(k-1)x(k))t_k.$$

$$S_{a,b}=\{a(k)x(k+1)-b(k-1)x(k):x(k)\in K[k]\}.$$

$$u(k)=a(k)-b(k-1),$$

$$d=\max \{\mathrm{deg}u(k),\mathrm{deg}a(k)-1\},$$

$$m_0=-\mathrm{lc}u(k)/\mathrm{lc}a(k),$$

$$\mathrm{deg}u(k)=\mathrm{deg}a(k)-1\text{mbox}and{m}_0\in \mathbb{N},$$

$$p(k)=a(k)x(k+1)-b(k-1)x(k).$$

$$p(k)=u(k)x(k)+a(k)(x(k+1)-x(k)).$$

$$\mathrm{lc}u(k)+\mathrm{lc}a(k)\cdot \mathrm{deg}x(k)=0,$$

$$K[k]/S_{a,b}=\begin{cases}\langle[k^{0}],[k^{1}],\ldots,[k^{d-1}],[k^{d+m_{0}}]\rangle,&\mbox{if $(a(k),b(k))$ is degenerated,}\\[5.0pt] \langle[k^{0}],[k^{1}],\ldots,[k^{d-1}]\rangle,&\mbox{otherwise}.\end{cases}$$

$$p_s(k)=a(k)(k+1{)}^s-b(k-1)k^s.$$

$$\mathrm{deg}{p}_s(k)=d+s,\forall s\ge 0.$$

$$p^{\prime }(k)=p(k)-\frac{\mathrm{lc}p(k)}{\mathrm{lc}{p}_{m-d}(k)}{p}_{m-d}(k)$$

$$K[k]/S_{a,b}=⟨[k^0],[k^1],\dots ,[k^{d-1}]⟩.$$

$$\mathrm{deg}{p}_s(k)=d+s,\forall s\ne m_0\text{mbox}and\mathrm{deg}{p}_{m_0}(k)<d+m_0.$$

$$p(k)-\mathrm{lc}p(k)\cdot k^{d+m_0}$$

$$K[k]/S_{a,b}=⟨[k^0],[k^1],\dots ,[k^{d-1}],[k^{d+m_0}]⟩,$$

$$t_k=(-n{)}_k/k!,$$

$$a(k)=k-n,b(k)=k+1,$$

$$S_{a,b}=\{(k-n)\cdot x(k+1)-k\cdot x(k):x(k)\in K[k]\}.$$

$$K[k]/S_{a,b}=⟨[k^n]⟩$$

Full text

1 Introduction

In recent years, many super congruences involving combinatorial sequences are discovered, see for example, Sun [16]. The standard methods for proving these congruences include combinatorial identities [18], Gauss sums [5], symbolic computation [14] et al.

We are interested in the following super congruence conjectured by van Hamme [19]

[TABLE]

where $p$ is a odd prime and $(a)_{k}=a(a+1)\cdots(a+k-1)$ is the rising factorial. This congruence was proved by Mortenson [13] Zudilin [21] and Long [12] by different methods. Sun [17] proved a stronger version for prime $p\geq 5$

[TABLE]

where $E_{n}$ is the $n$-th Euler number defined by

[TABLE]

A similar congruence was given by van Hamme [19] for $p\equiv 1\pmod{4}$:

[TABLE]

Long [12] showed that in fact the above congruence holds for arbitrary odd prime modulo $p^{4}$. Motivated by these two congruences, Guo [8] proposed the following conjectures (corrected version).

Conjecture 1.1

• •

For any odd prime $p$, positive integer $r$ and odd integer $m$, there exists an integer $a_{m.p}$ such that

[TABLE]

• •

For any odd prime $p>(m-1)/2$, positive integer $r$ and odd integer $m$, there exists an integer $b_{m,p}$ such that

[TABLE]

Liu [11] and Wang [20] confirmed the conjectures for $r=1$ and some initial values $m$. Jana and Kalita [10] and Guo [9] confirmed (1.1) for $m=3$ and $r\geq 1$. We will prove a stronger version of (1.1) for the case of $r=1$ and arbitrary odd $m$ and a weaker version of (1.2) for the case of $r=1$ and arbitrary odd $m$ by a reduction process.

Recall that a hypergeometric term $t_{k}$ is a function of $k$ such that $t_{k+1}/t_{k}$ is a rational function of $k$. Our basic idea is to rewrite the product of a polynomial $f(k)$ in $k$ and a hypergeometric term $t_{k}$ as

[TABLE]

where $g(k),h(k)$ are polynomials in $k$ such that the degree of $h(k)$ is bounded. To this aim, we construct $x(k)$ such that $\Delta x(k)t_{k}$ equals the product of $t_{k}$ and a polynomial $u(k)$ and that $f(k)$ and $u(k)$ has the same leading term. Then we have

[TABLE]

is the product of $t_{k}$ and a polynomial of degree less than $f(k)$. We call such a reduction process one reduction step. Continuing this reduction process, we finally obtain a polynomial $h(k)$ with bounded degree. We will show that for $t_{k}=\left(\frac{(1/2)_{k}}{(1)_{k}}\right)^{r}$, $r=3,4$ and an arbitrary polynomial of form $(4k+1)^{m}$ with $m$ odd, the reduced polynomial $h(k)$ can be taken as $(4k+1)$. This enables us to reduce the congruences (1.1) and (1.2) to the special case of $m=1$, which is known for $r=1$.

We notice that Pirastu-Strehl [15] and Abramov [1, 2] gave the minimal decomposition when $t_{k}$ is a rational function, Abramov-Petkovšek [3, 4] gave the minimal decomposition when $t_{k}$ is a hypergeometric term, and Chen-Huang-Kauers-Li [6] applied the reduction to give an efficient creative telescoping algorithm. These algorithms concern a general hypergeometric term. While we focus on a kind of special hypergeometric term so that the reduced part $h(k)t_{k}$ has a nice form.

The paper is organized as follows. In Section 2, we consider the reduction process for a general hypergeometric term $t_{k}$. Then in Section 3 we consider those $t_{k}$ with the property $a(k)$ is a shift of $-b(k)$, where $t_{k+1}/t_{k}=a(k)/b(k)$. As an application, we prove a stronger version of (1.1) for the case $r=1$. Finally, we consider the case of $a(k)$ is a shift of $b(k)$, which corresponds to (1.2). In this case, we show that there is a rational number $b_{m}$ instead of an integer such that (1.2) holds when $r=1$.

2 The Difference Space and Polynomial Reduction

Let $K$ be a field and $K[k]$ be the ring of polynomials in $k$ with coefficients in $K$. Let $t_{k}$ be a hypergeometric term. Suppose that

[TABLE]

where $a(k),b(k)\in K[k]$. It is straightforward to verify that

[TABLE]

We thus define the difference space corresponding to $a(k)$ and $b(k)$ to be

[TABLE]

We see that for $f(k)\in S_{a,b}$, we have $f(k)t_{k}=\Delta_{k}(p(k)t_{k})$ for a certain polynomial $p(k)\in K[k]$.

Let $\mathbb{N},\mathbb{Z}$ denote the set of nonnegative integers and the set of integers, respectively. Given $a(k),b(k)\in K[k]$, we denote

[TABLE]

[TABLE]

and

[TABLE]

where $\mathop{\rm lc}p(k)$ denotes the leading coefficient of $p(k)$.

We first introduce the concept of degeneration.

Definition 2.1

Let $a(k),b(k)\in K[k]$ and $u(k),m_{0}$ be given by (2.2) and (2.4). If

[TABLE]

we say that the pair $(a(k),b(k))$ is degenerated.

We will see that the degeneration is closely related to the degrees of the elements in $S_{a,b}$.

Lemma 2.2

Let $a(k),b(k)\in K[k]$ and $d,m_{0}$ be given by (2.3) and (2.4). For any polynomial $x(k)\in K[k]$, let

[TABLE]

If $(a(k),b(k))$ is degenerated and $\deg x(k)=m_{0}$, then $\deg p(k)<d+m_{0}$; Otherwise, $\deg p(k)=d+\deg x(k)$.

Proof. Notice that

[TABLE]

If the leading terms of $u(k)x(k)$ and $a(k)(x(k+1)-x(k))$ do not cancel, the degree of $p(k)$ is $d+\deg x(k)$. Otherwise, we have $\deg u(k)=\deg a(k)-1$ and

[TABLE]

i.e., $\deg x(k)=m_{0}$.

It is clear that $S_{a,b}$ is a subspace of $K[k]$, but is not a sub-ring of $K[k]$ in general. Let $[p(k)]=p(k)+S_{a,b}$ denote the coset of a polynomial $p(k)$. We see that the quotient space $K[k]/S_{a,b}$ is finite dimensional.

Theorem 2.3

Let $a(k),b(k)\in K[k]$ and $d,m_{0}$ be given by (2.3) and (2.4). We have

[TABLE]

Proof. For any nonnegative integer $s$, let

[TABLE]

We first consider the case when the pair $(a(k),b(k))$ is not degenerated. By Lemma 2.2, we have

[TABLE]

Suppose that $p(k)$ is a polynomial of degree $m\geq d$. Then

[TABLE]

is a polynomial of degree less than $m$ and $p(k)\in[p^{\prime}(k)]$. By induction on $m$, we derive that for any polynomial $p(k)$ of degree $\geq d$, there exists a polynomial $\tilde{p}(k)$ of degree $<d$ such that $p(k)\in[\tilde{p}(k)]$. Therefore,

[TABLE]

Now assume that $(a(k),b(k))$ is degenerated. By Lemma 2.2,

[TABLE]

The above reduction process (2.5) works well except for the polynomials $p(k)$ of degree $d+m_{0}$. But in this case,

[TABLE]

is a polynomial of degree less than $d+m_{0}$. Then the reduction process continues until the degree is less than $d$. We thus derive that

[TABLE]

completing the proof.

Example 2.1

Let $n$ be a positive integer and

[TABLE]

where $(\alpha)_{k}=\alpha(\alpha+1)\cdots(\alpha+k-1)$ is the raising factorial. Then

[TABLE]

and

[TABLE]

We have

[TABLE]

is of dimension one.

3 The case when $a(k)=-b(k+\alpha)$

In this section, we consider the case when $a(k)=-b(k+\alpha)$ and $b(k)$ has a symmetric property. We will show that in this case, the reduction process maintains the symmetric property. Notice that in this case

[TABLE]

has the same degree as $a(k)$, the pair $(a(k),b(k))$ is not degenerated.

We first consider the relation between the symmetric property and the expansion of a polynomial.

Lemma 3.1

Let $p(k)\in K[k]$ and $\beta\in K$. Then the following two statements are equivalent.

• (1)

$p(\beta+k)=p(\beta-k)$* ($p(\beta+k)=-p(\beta-k)$, respectively).*

• (2)

$p(k)$* is the linear combination of $(k-\beta)^{2i},\,i=0,1,\ldots$ ($(k-\beta)^{2i+1},\,i=0,1,\ldots$, respectively).*

Proof. Suppose that

[TABLE]

Then

[TABLE]

Therefore,

[TABLE]

The case of $p(\beta+k)=-p(\beta-k)$ can be proved in a similar way.

Now we are ready to state the main theorem.

Theorem 3.2

Let $a(k),b(k)\in K[k]$ such that

[TABLE]

for some $\alpha,\beta\in K$. Then for any non-negative integer $m$, we have

[TABLE]

and

[TABLE]

where

[TABLE]

Proof. We only prove the case of $b(\beta+k)=b(\beta-k)$. The case of $b(\beta+k)=-b(\beta-k)$ can be proved in a similar way. By Lemma 3.1, we may assume that

[TABLE]

where $r=\deg a(k)=\deg b(k)$ is even and $b_{r},b_{r-2},\ldots,b_{0}\in K$ are the coefficients.

Since $(a(k),b(k))$ is not degenerated, taking

[TABLE]

in Lemma 2.2, we derive that

[TABLE]

is a polynomial of degree $s+r$. More explicitly, we have

[TABLE]

is a polynomial with leading term $b_{r}k^{s+r}$.

Notice that

[TABLE]

and

[TABLE]

Since $b(\beta+k)=b(\beta-k)$, i.e., $b(k)=b(2\beta-k)$, we deduce that

[TABLE]

By the relation (3.1), we derive that

[TABLE]

Suppose that $p(k)$ is a linear combination of the even powers of $(k+\gamma)$ and $\deg p(k)\geq r$. By Lemma 3.1, we have $p(-\gamma-k)=p(-\gamma+k)$ and thus

[TABLE]

also satisfies $p^{\prime}(-\gamma-k)=p^{\prime}(-\gamma+k)$ since $\deg p(k)$ and $r$ are both even. It is clear that $p(k)\in[p^{\prime}(k)]$ and the degree of $p^{\prime}(k)$ is less than the degree of $p(k)$. Continuing this reduction process, we finally derive that $p(k)\in[\tilde{p}(k)]$ for some polynomial $\tilde{p}(k)$ with degree $<r$ and satisfying $\tilde{p}(-\gamma-k)=\tilde{p}(-\gamma+k)$. Therefore,

[TABLE]

Suppose that $p(k)$ is a linear combination of the odd powers of $(k+\gamma)$ and $\deg p(k)\geq r$. Then we have $p(-\gamma-k)=-p(-\gamma+k)$ and thus

[TABLE]

also satisfies $p^{\prime}(-\gamma-k)=-p^{\prime}(-\gamma+k)$. Continuing this reduction process, we finally derive that

[TABLE]

This completes the proof.

We may further require to express $[(k+\gamma)^{m}]$ as an integral linear combination of $[(k+\gamma)^{i}],\ 0\leq i<r$ when $b(k)=(k+1)^{r}$.

Theorem 3.3

Let

[TABLE]

where $r$ is a positive integer and $\alpha$ is a rational number with denominator $D$. Then for any positive integer $m$, there exist integers $a_{0},\ldots,a_{r-1}$ and a polynomial $x(k)\in\mathbb{Z}[k]$ such that

[TABLE]

Moreover, $a_{i}=0$ if $i\not\equiv m\pmod{2}$.

Proof. We have

[TABLE]

Let

[TABLE]

We see that it is the case of $\beta=-1$ and $\gamma=\alpha/2$ of Theorem 3.2. From (2.1), we derive that

[TABLE]

where $x_{s}(k)$ and $p_{s}(k)$ are given by (3.2) and (3.3) respectively. Multiplying $(2D)^{s+r}$ on both sides, we obtain

[TABLE]

where $k^{\prime}=2Dk+D\alpha$,

[TABLE]

and

[TABLE]

Notice that $\tilde{x}_{s}(k),\tilde{p}_{s}(k)\in\mathbb{Z}[k]$ and $\tilde{p}_{s}(k)$ is a monic polynomial of degree $s+r$. Moreover, $\tilde{p}_{s}(k)$ contains only even powers of $k$ or only odd powers of $k$. Using $\tilde{p}_{s}(k)$ to do the reduction (2.5), we derive that there exists integers $c_{m},c_{m-2},\ldots$ such that

[TABLE]

becomes a polynomial of degree less than $r$. Clearly, $p(k)\in\mathbb{Z}[k]$. Replacing $k$ by $k^{\prime}$ and multiplying $t_{k}$, we derive that

[TABLE]

completing the proof.

As an application, we confirm Conjecture 6 of [11].

Theorem 3.4

Let

[TABLE]

For any positive odd integer $m$, there exist integers $a_{m}$ and $c_{m}$ such that

[TABLE]

holds for any prime $p\geq 5$.

Proof. Taking $r=3$ and $\alpha=1/2$ in Theorem 3.3, there exist an integer $a_{m}$ and a polynomial $q_{m}(k)\in\mathbb{Z}[k]$ such that

[TABLE]

where $t_{k}=(-1)^{k}(\frac{1}{2})_{k}^{3}/(1)_{k}^{3}$. Summing over $k$ from [math] to $\frac{p-1}{2}$, we derive that

[TABLE]

where $\omega=\frac{p+1}{2}$. Noting that

[TABLE]

and

[TABLE]

we have

[TABLE]

Hence

[TABLE]

Let $c_{m}=-4q_{m}(2)$. We then have

[TABLE]

Sun [17] proved that for any prime $p\geq 5$,

[TABLE]

Therefore,

[TABLE]

Remark 1. The coefficient $a_{m}$ and the polynomial $q_{m}(k)$ can be computed by the extended Zeilberger’s algorithm [7].

4 The case when $a(k)=b(k+\alpha)$

We first give a criterion on the degeneration of $(a(k),b(k))$.

Lemma 4.1

Let $a(k),b(k)\in K[k]$ such that $a(k)=b(k+\alpha)$. Suppose that $-(\alpha+1)\deg a(k)\not\in\mathbb{N}$. Then $(a(k),b(k))$ is not degenerated.

Proof. Let $r=\deg a(k)=\deg b(k)$ and

[TABLE]

It is clear that the coefficient of $k^{r}$ in $u(k)$ is [math] and the coefficient of $k^{r-1}$ in $u(k)$ is $\mathop{\rm lc}b(k)\cdot(\alpha+1)r$. Since $(\alpha+1)r\not=0$, we derive that $\deg u(k)=r-1$. Thus,

[TABLE]

Since $-(\alpha+1)r\not\in\mathbb{N}$, the pair $(a(k),b(k))$ is not degenerated.

When $a(k)$ is a shift of $b(k)$, we have a result similar to Theorem 3.2.

Theorem 4.2

Let $a(k),b(k)\in K[k]$ such that

[TABLE]

for some $\alpha,\beta\in K$. Assume further that $-(\alpha+1)\deg a(k)\not\in\mathbb{N}$. Then for any non-negative integer $m$, we have

[TABLE]

and

[TABLE]

where

[TABLE]

Proof. The proof is parallel to the proof of Theorem 3.2. Instead of (3.2), we take

[TABLE]

in Lemma 2.2. By Lemma 4.1, $(a(k),b(k))$ is not degenerated and

[TABLE]

Hence the polynomial

[TABLE]

satisfies

[TABLE]

Moreover, we have

[TABLE]

so that the reduction process maintains the symmetric property. Therefore, the reduction process continues until the degree is less than $\deg a(k)-1$.

Similar to Theorem 3.3, we have the following result.

Theorem 4.3

Let

[TABLE]

where $r$ is a positive integer and $\alpha$ is a rational number with denominator $D$. Suppose that $-\alpha r\not\in\mathbb{N}$. Then for any positive integer $m$, there exist integers $a_{0},\ldots,a_{r-2}$ and a polynomial $x(k)\in\mathbb{Z}[k]$ such that

[TABLE]

where

[TABLE]

Moreover, $a_{i}=0$ if $i\not\equiv m\pmod{2}$.

Proof. The proof is parallel to the proof of Theorem 3.3. Instead of (3.6) and (3.7), we take

[TABLE]

and

[TABLE]

so that (3.5) still holds. It is clear that $\tilde{x}_{s}(k),\tilde{p}_{s}(k)\in\mathbb{Z}[k]$. But in this case, $\tilde{p}_{s}(k)$ is not monic. The leading term of $\tilde{p}_{s}(k)$ is

[TABLE]

Now let us consider the reduction process. Let $p(k)\in\mathbb{Z}[k]$ be a polynomial of degree $\ell\geq r-1$. Assume further that $p(k)$ contains only even powers of $k$ or only odd powers of $k$. Setting

[TABLE]

we see that $p^{\prime}(k)\in\mathbb{Z}[k]$ and $\deg p^{\prime}(k)<\ell$. Since $\tilde{p}_{\ell-r+1}(k)$ contains only even powers of $k$ or only odd powers of $k$, so does $p^{\prime}(k)$. Therefore, $\deg p^{\prime}(k)\leq\ell-2$.

Continuing this reduction process until $\ell<r-1$, we finally obtain that there exist integers $c_{m},c_{m-2},\ldots$ such that

[TABLE]

is a polynomial of degree less than $r-1$ and with integral coefficients, where $C_{m}$ is the product of the leading coefficient of $\tilde{p}_{m-r+1}(k),\tilde{p}_{m-r-1}(k),\ldots$

[TABLE]

as desired.

For the special case of $t_{k}=(1/2)_{k}^{4}/(1)_{k}^{4}$, we may further reduce the factor $C_{m}$.

Lemma 4.4

Let $m$ be a positive integer and

[TABLE]

• •

If $m$ is odd, then there exist an integer $c$ and a polynomial $x(k)\in\mathbb{Z}[k]$ such that

[TABLE]

where $C^{\prime}_{m}=(\frac{m-1}{2})!$.

• •

If $m$ is even, then there exist integers $c,c^{\prime}$ and a polynomial $x(k)\in\mathbb{Z}[k]$ such that

[TABLE]

where $C^{\prime}_{m}=(m-1)!!$.

Proof. This is the special case of Theorem 4.3 for $\alpha=1/2$ and $r=4$. Therefore, $D=2$ and $\alpha r-r+1=-1$.

We need only to show that the coefficients of $\tilde{p}_{s}(k)$ given by (4.2) is divisible by $2$ when $s$ is odd and is divisible by $4$ when $s$ is even. Then we may replace $\tilde{x}_{s}(k)$ given by (4.1) by $\tilde{x}_{s}(k)/4$ and $\tilde{x}_{s}(k)/2$ so that the leading coefficient of $\tilde{p}_{s}(k)$ is reduced. Correspondingly, the product $C_{m}$ of the leading coefficients becomes

[TABLE]

and

[TABLE]

Notice that

[TABLE]

The coefficient of $k^{j}$ is

[TABLE]

If $j-\ell<s$, the corresponding summand is divisible by $2$. If $j-\ell=s$ and $\ell$ is even, then $(-1)^{s-j}=1$ and the coefficient is [math]. Otherwise, $\ell=1$ or $\ell=3$, and thus $4\mid{4\choose\ell}$. Therefore, the coefficient must be divisible by $2$.

Now consider the case of $s$ being even. If $j-\ell<s-1$, the corresponding summand is divisible by $4$. Otherwise $j-\ell=s$ or $j-\ell=s-1$. We have seen that if $j-\ell=s$, then the coefficient is divisible by $4$. If $j-\ell=s-1$. Then

[TABLE]

Thus the summand is also divisible by $4$.

Example 4.2

Consider the case of $m=11$. We have

[TABLE]

where

[TABLE]

As an application, we obtain the following congruences.

Theorem 4.5

Let $m$ be a positive odd integer and $\mu=(m-1)/2$. Denote

[TABLE]

Then there exists an integer $a_{m}$ such that for each prime $p>\mu$,

[TABLE]

Proof. By Lemma 4.4, there exist an integer $a_{m}$ and a polynomial $q_{m}(k)\in\mathbb{Z}[k]$ such that

[TABLE]

where $t_{k}=\left(\frac{(1/2)_{k}}{(1)_{k}}\right)^{4}$. Summing over $k$ from [math] to $(p-1)/2$, we obtain

[TABLE]

where $\omega=(p+1)/2$. When $p>\mu$, $1/\mu!$ is a $p$-adic integer and

[TABLE]

Therefore,

[TABLE]

It is shown by Long [12] that

[TABLE]

completing the proof.

The integer $a_{m}$ and the polynomial $q_{m}(k)$ can be computed by the extended Zeilberger’s algorithm.

By checking the initial values, we propose the following conjecture.

Conjecture 4.6

For any positive odd integer $m$, the coefficient $a_{m}/(\frac{m-1}{2})!$ is an integer.

Acknowledgement. The work was supported by the National Natural Science Foundation of China (grants 11471244, 11771330 and 11701420).