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Some new curious congruences involving multiple harmonic sums

Rong Ma
School of Mathematics and Statistics,
Northwestern Polytechnical University
Xi’an, Shaanxi, 710072,
People’s Republic of China
E-mail: [email protected]
   Li Ni
School of Mathematics and Statistics,
Northwestern Polytechnical University
Xi’an, Shaanxi, 710072,
People’s Republic of China
E-mail: [email protected]
Abstract

It is significant to study congruences involving multiple harmonic sums. Let pp be an odd prime, in recent years, the following curious congruence

∑i+j+k=pi,j,k>01i​j​k≡−2​Bp−3(modp)\sum_{\begin{subarray}{c}i+j+k=p\\ i,j,k>0\end{subarray}}\frac{1}{ijk}\equiv-2B_{p-3}\pmod{p}

has been generalized along different directions, where BnB_{n} denote the nnth Bernoulli number. In this paper, we obtain several new generalizations of the above congruence by applying congruences involving multiple harmonic sums. For example, we have

∑k1+k2+⋯+kn=pki>0,1≤i≤n(−1)k1​(k13)k1​⋯​kn≡(n−1)!n​2n−1+13⋅6n−1​Bp−n​(13)(modp),\sum_{\begin{subarray}{c}k_{1}+k_{2}+\cdots+k_{n}=p\\ k_{i}>0,1\leq i\leq n\end{subarray}}\dfrac{(-1)^{k_{1}}\left(\dfrac{k_{1}}{3}\right)}{k_{1}\cdots k_{n}}\equiv\dfrac{(n-1)!}{n}\dfrac{2^{n-1}+1}{3\cdot 6^{n-1}}B_{p-n}\left(\dfrac{1}{3}\right)\pmod{p},

where nn is even, Bn​(x)B_{n}(x) denote the Bernoulli polynomials.

††2020 Mathematics Subject Classification: Primary 11A07; Secondary 11B68.††Key words and phrases: congruences, multiple harmonic sums, Bernoulli polynomials, Bernoulli numbers.

1 Introduction

Let ℕ\mathbb{N} be the set of positive integers. For any nn, N∈ℕN\in\mathbb{N} and s =(s1,⋯,sn)∈ℕn=(s_{1},\cdots,s_{n})\in\mathbb{N}^{n}, we define the multiple harmonic sums (MHS) by

HN​(s):=∑0<k1<⋯<kn≤N1k1s1​⋯​knsn.H_{N}(\textbf{s}):=\sum\limits_{0<k_{1}<\cdots<k_{n}\leq N}\frac{1}{k_{1}^{s_{1}}\cdots k_{n}^{s_{n}}}.

Since the middle of 1980s, multiple harmonic sums have played an important role in the study of mathematics and theoretical physics. From 2008 onwards, JianQiang Zhao in [14, 15] and Hoffman in [4] independently obtained many congruences modulo prime and prime powers for MHS. Subsequently, Roberto Tauraso in [11] began to consider congruence properties of alternating multiple harmonic sums which are defined as follows. Let nn be any positive integer and (r1,⋯,rn)∈(ℤ∗)n(r_{1},\cdots,r_{n})\in(\mathbb{Z^{*}})^{n}. For any N≥nN\geq n, we define the alternating multiple harmonic sums as

H​(r1,r2,⋯,rn;N):=∑0<k1<⋯<kn≤N∏i=1nsgnki​(ri)ki∣ri∣.H(r_{1},r_{2},\cdots,r_{n};N):=\sum\limits_{0<k_{1}<\cdots<k_{n}\leq N}\prod_{i=1}^{n}\frac{\text{sgn}^{k_{i}}(r_{i})}{k_{i}^{\mid r_{i}\mid}}.

Besides, many number theorists also investigated congruences of MHS with coefficients involving invariant sequences (see [9]). For example, in 2010, LiLu Zhao and ZhiWei Sun (see [16, Theorem 1.1]) showed that for any positive odd integer nn, prime pp such that p>n+1p>n+1

(1.1) ∑0<k1<⋯<kn<p(−1)k1​(k13)k1​⋯​kn≡0(modp).\sum\limits_{0<k_{1}<\cdots<k_{n}<p}\frac{(-1)^{k_{1}}\left(\dfrac{k_{1}}{3}\right)}{k_{1}\cdots k_{n}}\equiv 0\pmod{p}.

Sandro Mattarei and Tauraso (see [6, Theorem 4.1]) proved that for any positive integer nn and prime pp with p>max⁡(n+1,3)p>\max(n+1,3)

(1.2) ∑0<k1<⋯<kn<p(p−kn3)​(−1)knk1​⋯​kn≡{−2n+1+26n+1​p​Bp−n−1​(1/3)(modp2), if ​n​ is odd ,−2n+1+4n​6n​Bp−n​(1/3)(modp), if ​n​ is even ,\sum_{0<k_{1}<\cdots<k_{n}<p}\left(\frac{p-k_{n}}{3}\right)\frac{(-1)^{k_{n}}}{k_{1}\cdots k_{n}}\equiv\begin{cases}-\dfrac{2^{n+1}+2}{6^{n+1}}pB_{p-n-1}(1/3)\pmod{p^{2}},\text{ if }n\text{ is odd },\\ -\dfrac{2^{n+1}+4}{n6^{n}}B_{p-n}(1/3)\pmod{p},\quad\text{ if }n\text{ is even },\end{cases}

where Bn​(x)B_{n}(x) denote the Bernoulli polynomials defined by

z​ex​zez−1=∑n=0∞Bn​(x)n!​zn​(∣z∣<2​π).\dfrac{ze^{xz}}{e^{z}-1}=\sum\limits_{n=0}^{\infty}\dfrac{B_{n}(x)}{n!}z^{n}\left(\mid z\mid<2\pi\right).

As an application of multiple harmonic sums’ congruent properties, Zhao (see [14, Corollary 4.2]) first proved the following curious congruence

(1.3) ∑i+j+k=pi,j,k>01i​j​k≡−2​Bp−3(modp)\sum_{\begin{subarray}{c}i+j+k=p\\ i,j,k>0\end{subarray}}\frac{1}{ijk}\equiv-2B_{p-3}\pmod{p}

for any prime p≥3p\geq 3, where BnB_{n} denotes the nnth Bernoulli number given by B0=1B_{0}=1 and Bn=∑j=0nBj​(nj)​(n≥2)B_{n}=\sum\limits_{j=0}^{n}B_{j}\left(\begin{array}[]{c}n\\ j\end{array}\right)\left(n\geq 2\right). Later on, this congruence has been generalized along different directions. For example, Zhou and Cai (see [17]) generalized (1.3) by increasing the number of indices. Wang, Cai (see [13]) and Cai, Shen, Jia (see [1]) generalized the bound from pp to pp-powers, and a product of two odd prime powers, respectively. In addtion, some scholars established and generalized congruences for alternating version of (1.3) (see [7], [2]).

In view of congruences (1.1) and (1.2). In this paper, we shall prove the following theorems.

Theorem 1.1.

Let nn be a positive integer and pp a prime with p>max⁡(n+1,3)p>\max(n+1,3), then

(1.1) ∑0<k1<⋯<kn≤p−1(−1)(p−kn)​(p−kn3)k1​⋯​kn2≡{−2n+13​(n+1)​6n​Bp−n−1​(13)(modp)if 2∤n,∑r=0p−1−n(p−nr)​(2n+r+1+4)​Br​Bp−n−r​(1/3)n​(n+r)​6n+r−(p3)​Bp−n−1(modp)if 2∣n.\begin{aligned} &\sum\limits_{0<k_{1}<\cdots<k_{n}\leq p-1}\dfrac{(-1)^{(p-k_{n})}\left(\dfrac{p-k_{n}}{3}\right)}{k_{1}\cdots k_{n}^{2}}\\ &\equiv\begin{cases}-\dfrac{2^{n}+1}{3(n+1)6^{n}}B_{p-n-1}\left(\dfrac{1}{3}\right)\pmod{p}&\text{if $2\nmid n$},\\ \sum\limits_{r=0}^{p-1-n}\left(\begin{array}[]{c}{p-n}\\ r\end{array}\right)\dfrac{(2^{n+r+1}+4)B_{r}B_{p-n-r}(1/3)}{n(n+r)6^{n+r}}-\left(\dfrac{p}{3}\right)B_{p-n-1}\pmod{p}&\text{if $2\mid n$}.\end{cases}\end{aligned}
Theorem 1.2.

Let n≥2n\geq 2 be a positive integer and pp a prime with p>n+1p>n+1, then

(1.2) ∑k1+k2+⋯+kn=pki>0,1≤i≤n(−1)k1​(k13)k1​⋯​kn≡{(n−1)!n​2n−1+13⋅6n−1​Bp−n​(13)(modp)if 2∣n,−(n−2)!​∑r=0p−n(p−n+1r)​(2n+r+4)​Br​Bp−n−r+1​(1/3)(n+r−1)​6n+r−1+(n−1)!​(p3)​Bp−1−n(modp)if 2∤n.\begin{aligned} &\sum_{\begin{subarray}{c}k_{1}+k_{2}+\cdots+k_{n}=p\\ k_{i}>0,1\leq i\leq n\end{subarray}}\dfrac{(-1)^{k_{1}}\left(\dfrac{k_{1}}{3}\right)}{k_{1}\cdots k_{n}}\\ &\equiv\begin{cases}\dfrac{(n-1)!}{n}\dfrac{2^{n-1}+1}{3\cdot 6^{n-1}}B_{p-n}\left(\dfrac{1}{3}\right)\pmod{p}&\text{if $2\mid n$},\\ -(n-2)!\sum\limits_{r=0}^{p-n}\left(\begin{array}[]{c}{p-n+1}\\ r\end{array}\right)\dfrac{(2^{n+r}+4)B_{r}B_{p-n-r+1}(1/3)}{(n+r-1)6^{n+r-1}}\\ \quad{}+(n-1)!\left(\dfrac{p}{3}\right)B_{p-1-n}\pmod{p}&\text{if $2\nmid n$}.\end{cases}\end{aligned}
Theorem 1.3.

Let n≥2n\geq 2 be a positive integer and pp a prime with p>n+1p>{n+1}, then

(1.3) ∑k1+k2+⋯+kn=pki>0,1≤i≤n(−1)k1​((k1+13)−(k1−13))k1​⋯​kn≡{−(n−2)!​∑r=0p−n(p−n+1r)​(3n+r−2−1)​(2n+r−2−1)​Br​Bp−n−r+1(n+r−1)​6n+r−2if 2∣n,−(n−1)!​(1−(3n−1−1)​(2n−1−1)2​n​6n−1)​Bp−nif 2∤n.(modp)\begin{aligned} &\sum_{\begin{subarray}{c}k_{1}+k_{2}+\cdots+k_{n}=p\\ k_{i}>0,1\leq i\leq n\end{subarray}}\dfrac{(-1)^{k_{1}}\left(\left(\dfrac{k_{1}+1}{3}\right)-\left(\dfrac{k_{1}-1}{3}\right)\right)}{k_{1}\cdots k_{n}}\\ &\equiv\begin{cases}-(n-2)!\sum\limits_{r=0}^{p-n}\left(\begin{array}[]{c}{p-n+1}\\ r\end{array}\right)\dfrac{(3^{n+r-2}-1)(2^{n+r-2}-1)B_{r}B_{p-n-r+1}}{(n+r-1)6^{n+r-2}}&\text{if $2\mid n$},\\ -(n-1)!\left(1-\dfrac{(3^{n-1}-1)(2^{n-1}-1)}{2n6^{n-1}}\right)B_{p-n}&\text{if $2\nmid n$}.\end{cases}\pmod{p}\end{aligned}

2 Basic lemmas

In this section, we introduce the following lemmas that will be used later to prove the theorems.

Lemma 2.1 (See [10]).

Suppose that kk, p∈ℕp\in\mathbb{N} with p>1p>1. If xx, y∈ℤpy\in\mathbb{Z}_{p}, then p​Bk​(x)∈ℤppB_{k}(x)\in\mathbb{Z}_{p} and (Bk(x)−\left(B_{k}(x)-\right. Bk(y))/k∈ℤp\left.B_{k}(y)\right)/k\in\mathbb{Z}_{p}. If pp is an odd prime such that p−1∤kp-1\nmid k, then Bk​(x)/k∈ℤpB_{k}(x)/k\in\mathbb{Z}_{p}.

Lemma 2.2 (See [10]).

Let p,m∈ℕp,m\in\mathbb{N} and k,r∈ℤk,r\in\mathbb{Z} with k⩾0k\geqslant 0. Then

∑x=0x≡r(modm)p−1xk=mkk+1​(Bk+1​(pm+{r−pm})−Bk+1​({rm})).\sum_{\begin{subarray}{c}x=0\\ x\equiv r(\bmod m)\end{subarray}}^{p-1}x^{k}=\frac{m^{k}}{k+1}\left(B_{k+1}\left(\frac{p}{m}+\left\{\frac{r-p}{m}\right\}\right)-B_{k+1}\left(\left\{\frac{r}{m}\right\}\right)\right).
Lemma 2.3 (See [5]).

Let n∈ℕn\in\mathbb{N}, then

B2​n​(12)=(21−n−1)​B2​n,B2​n​(14)=B2​n​(34)=2−22​n42​n​B2​n,B_{2n}\left(\frac{1}{2}\right)=\left(2^{1-n}-1\right)B_{2n},\quad B_{2n}\left(\frac{1}{4}\right)=B_{2n}\left(\frac{3}{4}\right)=\frac{2-2^{2n}}{4^{2n}}B_{2n},

and

B2​n​(13)=B2​n​(23)=3−32​n2⋅32​n​B2​n,B_{2n}\left(\frac{1}{3}\right)=B_{2n}\left(\frac{2}{3}\right)=\frac{3-3^{2n}}{2\cdot 3^{2n}}B_{2n},
B2​n​(16)=B2​n​(56)=(2−22​n)​(3−32​n)2⋅62​n​B2​n.\quad B_{2n}\left(\frac{1}{6}\right)=B_{2n}\left(\frac{5}{6}\right)=\frac{\left(2-2^{2n}\right)\left(3-3^{2n}\right)}{2\cdot 6^{2n}}B_{2n}.

Moreover, through the formulas Bn​(1−x)=(−1)n​Bn​(x)B_{n}(1-x)=(-1)^{n}B_{n}(x) and Bn​(a​x)=am−1​∑k=0a−1Bn​(x+ka)B_{n}(ax)=a^{m-1}\sum\limits_{k=0}^{a-1}B_{n}\left(x+\dfrac{k}{a}\right) for mm, a>0a>0, we have

B2​n+1​(12)=0,B2​n+1​(23)=−B2​n+1​(13),\displaystyle B_{2n+1}\left(\frac{1}{2}\right)=0,\quad B_{2n+1}\left(\frac{2}{3}\right)=-B_{2n+1}\left(\frac{1}{3}\right),
B2​n+1​(16)=−B2​n+1​(56)=(1+21−(2​n+1))​B2​n+1​(13).\displaystyle B_{2n+1}\left(\frac{1}{6}\right)=-B_{2n+1}\left(\frac{5}{6}\right)=\left(1+2^{1-(2n+1)}\right)B_{2n+1}\left(\frac{1}{3}\right).
Lemma 2.4 (See [6]).

Let nn be a positive integer and let pp be a prime with p>n+1p>n+1. Then we have the polynomial congruence

∑0<k1<⋯<kn≤p−1xknk1​⋯​kn≡(−1)n−1​∑k=1p−1(1−x)kkn(modp).\sum\limits_{0<k_{1}<\cdots<k_{n}\leq p-1}\frac{x^{k_{n}}}{k_{1}\cdots k_{n}}\equiv(-1)^{n-1}\sum_{k=1}^{p-1}\frac{(1-x)^{k}}{k^{n}}\left(\bmod p\right).
Lemma 2.5 (See [12]).

Let aa, b∈ℕb\in\mathbb{N} and a prime p≥a+b+2p\geq a+b+2. If a+ba+b is odd then we have

∑0<k1≤k2≤p−11k1a​k2b≡H​(a,b;p−1)≡(−1)ba+b​(a+ba)​Bp−a−b(modp).\sum\limits_{0<k_{1}\leq k_{2}\leq p-1}\frac{1}{k_{1}^{a}k_{2}^{b}}\equiv H(a,b;p-1)\equiv\frac{(-1)^{b}}{a+b}\left(\begin{array}[]{c}a+b\\ a\end{array}\right)B_{p-a-b}\pmod{p}.

If a+ba+b is even then we have

∑0<k1≤k2≤p−11k1a​k2b≡H​(a,b;p−1)≡0(modp).\sum\limits_{0<k_{1}\leq k_{2}\leq p-1}\frac{1}{k_{1}^{a}k_{2}^{b}}\equiv H(a,b;p-1)\equiv 0\pmod{p}.
Lemma 2.6.

Let nn be a positive integer and p>n+1p>n+1 a prime, then we get

(2.1) ∑0<k1<⋯<kn≤p−1xknk1​⋯​kn2≡{∑0<i1≤i2≤p−1(1−x)i1i1​i2n(modp)if 2∤n,−∑0<i1≤i2≤p−1(1−x)i1i1​i2n+Bp−1−n(modp)if 2∣n.\sum\limits_{0<k_{1}<\cdots<k_{n}\leq p-1}\frac{x^{k_{n}}}{k_{1}\cdots k_{n}^{2}}\equiv\begin{cases}\sum\limits_{0<i_{1}\leq i_{2}\leq p-1}\dfrac{(1-x)^{i_{1}}}{i_{1}i_{2}^{n}}\pmod{p}&\text{if $2\nmid n$},\\ -\sum\limits_{0<i_{1}\leq i_{2}\leq p-1}\dfrac{(1-x)^{i_{1}}}{i_{1}i_{2}^{n}}+B_{p-1-n}\pmod{p}&\text{if $2\mid n$}.\end{cases}
Proof.

Taking the derivative of the left-hand side above and applying Lemma 2.4, we have

x​dd​x​(∑0<k1<⋯<kn≤p−1xknk1​⋯​kn2)\displaystyle x\frac{d}{dx}\left(\sum\limits_{0<k_{1}<\cdots<k_{n}\leq p-1}\frac{x^{k_{n}}}{k_{1}\cdots k_{n}^{2}}\right)
=∑0<k1<⋯<kn≤p−1xknk1​⋯​kn≡(−1)n−1​∑k=1p−1(1−x)kkn(modp).\displaystyle=\sum\limits_{0<k_{1}<\cdots<k_{n}\leq p-1}\frac{x^{k_{n}}}{k_{1}\cdots k_{n}}\equiv(-1)^{n-1}\sum_{k=1}^{p-1}\frac{(1-x)^{k}}{k^{n}}\pmod{p}.

Observe that

(−1)n−1​∑k=1p−1(1−x)kkn\displaystyle(-1)^{n-1}\sum_{k=1}^{p-1}\frac{(1-x)^{k}}{k^{n}} =(−1)n−1​∑k=1p−11kn​∑r=0k(−x)k​(kr)\displaystyle=(-1)^{n-1}\sum_{k=1}^{p-1}\frac{1}{k^{n}}\sum\limits_{r=0}^{k}(-x)^{k}\left(\begin{array}[]{c}k\\ r\end{array}\right)
=(−1)n−1​∑k=1p−11kn+(−1)n−1​∑k=1p−11kn​∑r=1k(−x)r​(kr)\displaystyle=(-1)^{n-1}\sum_{k=1}^{p-1}\frac{1}{k^{n}}+(-1)^{n-1}\sum_{k=1}^{p-1}\frac{1}{k^{n}}\sum\limits_{r=1}^{k}(-x)^{r}\left(\begin{array}[]{c}k\\ r\end{array}\right)
≡(−1)n−1​∑k=1p−11kn​∑r=1k(−x)r​(kr)(modp),\displaystyle\equiv(-1)^{n-1}\sum_{k=1}^{p-1}\frac{1}{k^{n}}\sum\limits_{r=1}^{k}(-x)^{r}\left(\begin{array}[]{c}k\\ r\end{array}\right)\pmod{p},

where we use the congruence ∑k=1p−11kn≡0(modp)\sum\limits_{k=1}^{p-1}\dfrac{1}{k^{n}}\equiv 0\pmod{p} (see [8]). Hence we can easily get

x​dd​x​(∑0<k1<⋯<kn≤p−1xknk1​⋯​kn2−(−1)n−1​∑k=1p−11kn​∑r=1k(−x)rr​(kr))≡0(modp).x\frac{d}{dx}\left(\sum\limits_{0<k_{1}<\cdots<k_{n}\leq p-1}\frac{x^{k_{n}}}{k_{1}\cdots k_{n}^{2}}-(-1)^{n-1}\sum_{k=1}^{p-1}\frac{1}{k^{n}}\sum\limits_{r=1}^{k}\frac{(-x)^{r}}{r}\left(\begin{array}[]{c}k\\ r\end{array}\right)\right)\equiv 0\pmod{p}.

Thus ∑0<k1<⋯<kn≤p−1xknk1​⋯​kn2−(−1)n−1​∑k=1p−11kn​∑r=1k(−x)rr​(kr)≡c(modp)\sum\limits_{0<k_{1}<\cdots<k_{n}\leq p-1}\dfrac{x^{k_{n}}}{k_{1}\cdots k_{n}^{2}}-(-1)^{n-1}\sum\limits_{k=1}^{p-1}\dfrac{1}{k^{n}}\sum\limits_{r=1}^{k}\dfrac{(-x)^{r}}{r}\left(\begin{array}[]{c}k\\ r\end{array}\right)\equiv c\pmod{p} for some constant cc. Replacing xx with 0 we obtain

(2.2) ∑0<k1<⋯<kn≤p−1xknk1​⋯​kn2≡(−1)n−1​∑k=1p−11kn​∑r=1k(−x)rr​(kr)(modp).\sum\limits_{0<k_{1}<\cdots<k_{n}\leq p-1}\frac{x^{k_{n}}}{k_{1}\cdots k_{n}^{2}}\equiv(-1)^{n-1}\sum_{k=1}^{p-1}\frac{1}{k^{n}}\sum\limits_{r=1}^{k}\frac{(-x)^{r}}{r}\left(\begin{array}[]{c}k\\ r\end{array}\right)\pmod{p}.

Since ∫01ur−1​𝑑u=1r\int_{0}^{1}u^{r-1}du=\dfrac{1}{r}, taking v=1−x​uv=1-xu we see that

∑r=1k(−x)rr​(kr)\displaystyle\sum_{r=1}^{k}\frac{(-x)^{r}}{r}\left(\begin{array}[]{l}k\\ r\end{array}\right) =∫01∑r=1k(−x)r​(kr)​ur−1​d​u=∫01(1−x​u)k−1u​𝑑u\displaystyle=\int_{0}^{1}\sum_{r=1}^{k}(-x)^{r}\left(\begin{array}[]{l}k\\ r\end{array}\right)u^{r-1}du=\int_{0}^{1}\frac{(1-xu)^{k}-1}{u}du
=∫11−x∑i=1kvi−1​d​v=∑i=1k(1−x)i−1i.\displaystyle=\int_{1}^{1-x}\sum_{i=1}^{k}v^{i-1}dv=\sum_{i=1}^{k}\frac{(1-x)^{i}-1}{i}.

Combining the identity above with (2.1) and Lemma 2.5, we complete the proof of Lemma 2.6. ∎

Remark.

If we set n=1,x=2n=1,x=2 in Lemma 2.6, we can easily get

∑0<i≤j≤p−1(−1)ii​j≡H​(−1,1;p−1)≡−qp2​(2)(modp)\sum\limits_{0<i\leq j\leq p-1}\dfrac{(-1)^{i}}{ij}\equiv H(-1,1;p-1)\equiv-{{q_{p}}^{2}(2)}\pmod{p}

by using the congruence ∑k=1p−12kk2≡−qp2​(2)(modp)\sum\limits_{k=1}^{p-1}\dfrac{2^{k}}{k^{2}}\equiv-{{q_{p}}^{2}(2)}\pmod{p} (see [3]), where qp​(2){q_{p}(2)} is Euler’s quotient of 2 with base pp.

Lemma 2.7.

Let nn be a positive integer and pp be a prime with p>max⁡(n+1,3)p>\max(n+1,3), then we have

∑i=1p−1(−1)(p+i)​(p+i3)in≡{−2n+1+4n​6n​Bp−n​(1/3)(modp)if 2∣n,0(modp)if 2∤n.\sum\limits_{i=1}^{p-1}\frac{(-1)^{(p+i)}\left(\frac{p+i}{3}\right)}{i^{n}}\equiv\begin{cases}-\dfrac{2^{n+1}+4}{n6^{n}}B_{p-n}(1/3)\pmod{p}&\text{if $2\mid n$},\\ 0\pmod{p}&\text{if $2\nmid n$}.\end{cases}
Proof.

For prime p>max⁡(n+1,3),p>\max(n+1,3), according to Euler’s theorem and Lemma 2.2 we have

∑i=1i≡r(mod6)p−11in≡∑i=1i≡r(mod6)p−1ip−n−1≡−1n​6n​(Bp−n​(p6+{r−p6})−Bp−n​({r6}))(modp),,\begin{aligned} \sum_{\begin{subarray}{c}i=1\\ i\equiv r(\bmod 6)\end{subarray}}^{p-1}\frac{1}{i^{n}}\equiv&\sum_{\begin{subarray}{c}i=1\\ i\equiv r(\bmod 6)\end{subarray}}^{p-1}i^{p-n-1}\\ \equiv&-\frac{1}{n6^{n}}\left(B_{p-n}\left(\frac{p}{6}+\left\{\frac{r-p}{6}\right\}\right)-B_{p-n}\left(\left\{\frac{r}{6}\right\}\right)\right)\pmod{p},\end{aligned},

where {x}\{x\} is the decimal part of xx. By using the following formula for Bernoulli polynomials

Bn​(x+y)=∑r=0n(nr)​Bn−r​(y)​xrB_{n}(x+y)=\sum_{r=0}^{n}\left(\begin{array}[]{l}n\\ r\end{array}\right)B_{n-r}(y)x^{r}

and Lemma 2.1, we have

(2.3) ∑i=1i≡r(mod6)p−11in≡1n​6n​(Bp−n​({r6})−Bp−n​({r−p6}))(modp).\sum_{\begin{subarray}{c}i=1\\ i\equiv r(\bmod 6)\end{subarray}}^{p-1}\frac{1}{i^{n}}\equiv\frac{1}{n6^{n}}\left(B_{p-n}\left(\left\{\frac{r}{6}\right\}\right)-B_{p-n}\left(\left\{\frac{r-p}{6}\right\}\right)\right)\pmod{p}.

On the other hand, (−1)(p+i)​(p+i3)(-1)^{(p+i)}(\dfrac{p+i}{3}) takes the values 0,−1,−1,0,1,10,-1,-1,0,1,1 according as p+i≡0,1,2,3,4,5(mod6).p+i\equiv 0,1,2,3,4,5\pmod{6}. When nn is even, p≡1(mod6)p\equiv 1\pmod{6}, we have

∑i=1p−1(−1)(p+i)​(p+i3)in=−∑i=1i≡0(mod6)p−11in−∑i=1i≡1(mod6)p−11in+∑i=1i≡3(mod6)p−11in+∑i=1i≡4(mod6)p−11in.\sum\limits_{i=1}^{p-1}\frac{(-1)^{(p+i)}\left(\frac{p+i}{3}\right)}{i^{n}}=-\sum_{\begin{subarray}{c}i=1\\ i\equiv 0\pmod{6}\end{subarray}}^{p-1}\frac{1}{i^{n}}-\sum_{\begin{subarray}{c}i=1\\ i\equiv 1\pmod{6}\end{subarray}}^{p-1}\frac{1}{i^{n}}+\sum_{\begin{subarray}{c}i=1\\ i\equiv 3\pmod{6}\end{subarray}}^{p-1}\frac{1}{i^{n}}+\sum_{\begin{subarray}{c}i=1\\ i\equiv 4\pmod{6}\end{subarray}}^{p-1}\frac{1}{i^{n}}.

Combining the above identity, (2.2) and Lemma 2.3, we obtain

∑i=1p−1(−1)(p+i)​(p+i3)in≡−2n+1+4n​6n​Bp−n​(1/3)(modp).\sum\limits_{i=1}^{p-1}\frac{(-1)^{(p+i)}\left(\frac{p+i}{3}\right)}{i^{n}}\equiv-\dfrac{2^{n+1}+4}{n6^{n}}B_{p-n}(1/3)\pmod{p}.

Similarly, when p≡−1(mod6)p\equiv-1\pmod{6}, we have

∑i=1p−1(−1)(p+i)​(p+i3)in=−∑i=1i≡2(mod6)p−11in−∑i=1i≡3(mod6)p−11in+∑i=1i≡5(mod6)p−11in+∑i=1i≡0(mod6)p−11in.\sum\limits_{i=1}^{p-1}\frac{(-1)^{(p+i)}\left(\frac{p+i}{3}\right)}{i^{n}}=-\sum_{\begin{subarray}{c}i=1\\ i\equiv 2\pmod{6}\end{subarray}}^{p-1}\frac{1}{i^{n}}-\sum_{\begin{subarray}{c}i=1\\ i\equiv 3\pmod{6}\end{subarray}}^{p-1}\frac{1}{i^{n}}+\sum_{\begin{subarray}{c}i=1\\ i\equiv 5\pmod{6}\end{subarray}}^{p-1}\frac{1}{i^{n}}+\sum_{\begin{subarray}{c}i=1\\ i\equiv 0\pmod{6}\end{subarray}}^{p-1}\frac{1}{i^{n}}.

Applying (2.2) and Lemma 2.3 again, we obtain the same congruence as the case of p≡1(mod6)p\equiv 1\pmod{6}. Thus we prove the case where nn is even.

When nn is odd, p−np-n is even, we can prove the second congruence of Lemma 2.7 in the similar way . ∎

3 Proofs of the theorems

Proof of Theorem 1.1 Let ω\omega be a complex primitive cubic root of unity. For any integer ii, we note that (i3)=(ωi−ω−i)/(ω−ω−1)\left(\dfrac{i}{3}\right)=(\omega^{i}-\omega^{-i})/\left(\omega-\omega^{-1}\right) and 1−(−ω)−1=−ω1-(-\omega)^{-1}=-\omega. Hence, applying Lemma 2.6 and Lemma 2.5 we have

(3.1) ∑0<k1<⋯<kn≤p−1(−1)(p−kn)​(p−kn3)k1​⋯​kn2\displaystyle\sum\limits_{0<k_{1}<\cdots<k_{n}\leq p-1}\frac{(-1)^{(p-k_{n})}\left(\frac{p-k_{n}}{3}\right)}{k_{1}\cdots k_{n}^{2}}
=1ω−ω−1​((−ω)p​∑0<k1<⋯<kn≤p−1(−ω)−knk1​⋯​kn2−(−ω)−p​∑0<k1<⋯<kn≤p−1(−ω)knk1​⋯​kn2)\displaystyle=\frac{1}{\omega-\omega^{-1}}\left((-\omega)^{p}\sum\limits_{0<k_{1}<\cdots<k_{n}\leq p-1}\frac{(-\omega)^{-k_{n}}}{k_{1}\cdots k_{n}^{2}}-(-\omega)^{-p}\sum\limits_{0<k_{1}<\cdots<k_{n}\leq p-1}\frac{(-\omega)^{k_{n}}}{k_{1}\cdots k_{n}^{2}}\right)
≡(−1)n−1ω−ω−1​((−ω)p​∑0<i1≤i2≤p−1(−ω)i1−1i1​i2n−(−ω)−p​∑0<i1≤i2≤p−1(−ω)−i1−1i1​i2n)\displaystyle\equiv\frac{(-1)^{n-1}}{\omega-\omega^{-1}}\left((-\omega)^{p}\sum\limits_{0<i_{1}\leq i_{2}\leq p-1}\frac{(-\omega)^{i_{1}}-1}{i_{1}i_{2}^{n}}-(-\omega)^{-p}\sum\limits_{0<i_{1}\leq i_{2}\leq p-1}\frac{(-\omega)^{-i_{1}}-1}{i_{1}i_{2}^{n}}\right)
≡(−1)n−1​(∑0<i1≤i2≤p−1(−1)(p+i1)​(p+i13)i1​i2n+(p3)​∑0<i1≤i2≤p−11i1​i2n)\displaystyle\equiv(-1)^{n-1}\left(\sum\limits_{0<i_{1}\leq i_{2}\leq p-1}\frac{(-1)^{(p+i_{1})}\left(\frac{p+i_{1}}{3}\right)}{i_{1}i_{2}^{n}}+\left(\frac{p}{3}\right)\sum\limits_{0<i_{1}\leq i_{2}\leq p-1}\frac{1}{i_{1}i_{2}^{n}}\right)
≡{∑0<i1≤i2≤p−1(−1)(p+i1)​(p+i13)i1​i2n(modp)if 2∤n,−∑0<i1≤i2≤p−1(−1)(p+i1)​(p+i13)i1​i2n−(p3)​Bp−1−n(modp)if 2∣n.\displaystyle\equiv\begin{cases}\sum\limits_{0<i_{1}\leq i_{2}\leq p-1}\dfrac{(-1)^{(p+i_{1})}\left(\dfrac{p+i_{1}}{3}\right)}{i_{1}i_{2}^{n}}\pmod{p}&\text{if $2\nmid n$},\\ -\sum\limits_{0<i_{1}\leq i_{2}\leq p-1}\dfrac{(-1)^{(p+i_{1})}\left(\dfrac{p+i_{1}}{3}\right)}{i_{1}i_{2}^{n}}-\left(\dfrac{p}{3}\right)B_{p-1-n}\pmod{p}&\text{if $2\mid n$}.\end{cases}

When nn is odd. Under substitutions i1→p−i2i_{1}\rightarrow p-i_{2} and i2→p−i1i_{2}\rightarrow p-i_{1}, we obtain

∑0<i1≤i2≤p−1(−1)(p+i1)​(p+i13)i1​i2n\displaystyle\sum\limits_{0<i_{1}\leq i_{2}\leq p-1}\dfrac{(-1)^{(p+i_{1})}\left(\dfrac{p+i_{1}}{3}\right)}{i_{1}i_{2}^{n}} ≡∑0<i1≤i2≤p−1(−1)(2​p−i2)​(2​p−i23)i1n​i2\displaystyle\equiv\sum\limits_{0<i_{1}\leq i_{2}\leq p-1}\dfrac{(-1)^{(2p-i_{2})}\left(\dfrac{2p-i_{2}}{3}\right)}{i_{1}^{n}i_{2}}
≡∑0<i1≤i2≤p−1(−1)(p+i2)​(p+i23)i1n​i2(modp).\displaystyle\equiv\sum\limits_{0<i_{1}\leq i_{2}\leq p-1}\dfrac{(-1)^{(p+i_{2})}\left(\dfrac{p+i_{2}}{3}\right)}{i_{1}^{n}i_{2}}\pmod{p}.

Thus, by ∑k=1p−11kn≡0(modp)(n<p−1)\sum\limits_{k=1}^{p-1}\dfrac{1}{k^{n}}\equiv 0\pmod{p}(n<p-1) and Lemma 2.7, we have

∑0<i1≤i2≤p−1(−1)(p+i1)​(p+i13)i1​i2n\displaystyle\sum\limits_{0<i_{1}\leq i_{2}\leq p-1}\dfrac{(-1)^{(p+i_{1})}\left(\dfrac{p+i_{1}}{3}\right)}{i_{1}i_{2}^{n}}
≡12​(∑0<i1,i2≤p−1(−1)(p+i1)​(p+i13)i1​i2n+∑i=1p−1(−1)(p+i)​(p+i3)in+1)\displaystyle\equiv\frac{1}{2}\left(\sum\limits_{0<i_{1},i_{2}\leq p-1}\dfrac{(-1)^{(p+i_{1})}\left(\dfrac{p+i_{1}}{3}\right)}{i_{1}i_{2}^{n}}+\sum\limits_{i=1}^{p-1}\dfrac{(-1)^{(p+i)}\left(\dfrac{p+i}{3}\right)}{i^{n+1}}\right)
≡12​∑i=1p−1(−1)(p+i)​(p+i3)in+1\displaystyle\equiv\frac{1}{2}\sum\limits_{i=1}^{p-1}\dfrac{(-1)^{(p+i)}\left(\dfrac{p+i}{3}\right)}{i^{n+1}}
≡−2n+13​(n+1)​6n​Bp−n−1​(13)(modp).\displaystyle\equiv-\frac{2^{n}+1}{3(n+1)6^{n}}B_{p-n-1}\left(\frac{1}{3}\right)\pmod{p}.

Now using (3.1), we complete the proof of the first case of theorem 1.1.

When nn is even. Note that

(3.2) ∑0<i1≤i2≤p−1(−1)(p+i1)​(p+i13)i1​i2n\displaystyle\sum\limits_{0<i_{1}\leq i_{2}\leq p-1}\dfrac{(-1)^{(p+i_{1})}\left(\dfrac{p+i_{1}}{3}\right)}{i_{1}i_{2}^{n}}
=∑0<i1,i2≤p−1(−1)(p+i1)​(p+i13)i1​i2n−∑0<i2<i1≤p−1(−1)(p+i1)​(p+i13)i1​i2n\displaystyle=\sum\limits_{0<i_{1},i_{2}\leq p-1}\dfrac{(-1)^{(p+i_{1})}\left(\dfrac{p+i_{1}}{3}\right)}{i_{1}i_{2}^{n}}-\sum\limits_{0<i_{2}<i_{1}\leq p-1}\dfrac{(-1)^{(p+i_{1})}\left(\dfrac{p+i_{1}}{3}\right)}{i_{1}i_{2}^{n}}
≡−∑0<j<i≤p−1(−1)(p+i)​(p+i3)i​jn\displaystyle\equiv-\sum\limits_{0<j<i\leq p-1}\dfrac{(-1)^{(p+i)}\left(\dfrac{p+i}{3}\right)}{ij^{n}}
≡−∑0<i≤p−1(−1)(p+i)​(p+i3)i​∑0<j≤i−11jn(modp).\displaystyle\equiv-\sum\limits_{0<i\leq p-1}\dfrac{(-1)^{(p+i)}\left(\dfrac{p+i}{3}\right)}{i}\sum\limits_{0<j\leq i-1}\dfrac{1}{j^{n}}\pmod{p}.

From the difference equation of Bernoulli polynomials Bn​(x+1)−Bn​(x)=n​xn−1​(n≥1)B_{n}(x+1)-B_{n}(x)=nx^{n-1}\left(n\geq 1\right), we can obtain the following well-known identity

∑x=1n−1xk=∑r=0k(k+1r)​Brk+1​nk+1−r,∀n,k≥1.\sum_{x=1}^{n-1}x^{k}=\sum_{r=0}^{k}\left(\begin{array}[]{c}k+1\\ r\end{array}\right)\frac{B_{r}}{k+1}n^{k+1-r},\quad\forall n,k\geq 1.

Applying Euler’s theorem and the above identity to the last congruence of (3.2), we get

∑0<i1≤i2≤p−1(−1)(p+i1)​(p+i13)i1​i2n\displaystyle\sum\limits_{0<i_{1}\leq i_{2}\leq p-1}\dfrac{(-1)^{(p+i_{1})}\left(\dfrac{p+i_{1}}{3}\right)}{i_{1}i_{2}^{n}}
≡−∑r=0p−1−n(p−nr)​Brp−n​∑0<i≤p−1(−1)(p+i)​(p+i3)in+r(modp).\displaystyle\equiv-\sum_{r=0}^{p-1-n}\left(\begin{array}[]{c}{p-n}\\ r\end{array}\right)\frac{B_{r}}{{p-n}}\sum\limits_{0<i\leq p-1}\dfrac{(-1)^{(p+i)}\left(\dfrac{p+i}{3}\right)}{i^{n+r}}\pmod{p}.

Then we can prove the second congruence of Theorem 1.1 by using Lemma 2.7 and (3.1), hence Theorem 1.1 is proved. ∎

Proof of Theorem 1.2 Let l=k2+⋯+knl=k_{2}+\cdots+k_{n}. It’s easy to see that

(3.3) ∑k1+k2+⋯+kn=pki>0,1≤i≤n(−1)k1​(k13)k1​⋯​kn=∑k2+⋯+kn=l<p0<ki,2≤i≤n(−1)(p−l)​(p−l3)(p−l)​k2​⋯​kn≡−∑k2+⋯+kn=l<p0<ki,2≤i≤n(−1)p−l​(p−l3)l​k2​⋯​kn(modp)=−(n−1)​∑k2+⋯+kn−1=l1<l<p0<ki,2≤i≤n−1(−1)p−l​(p−l3)l2​k2​⋯​kn−1=−(n−1)​(n−2)​∑k2+⋯+kn−2=l2<l1<l<p0<ki,2≤i≤n−2(−1)p−l​(p−l3)l2​l1​k2​⋯​kn−2=⋯=−(n−1)!​∑0<ln−2<⋯<l1<l<p(−1)p−l​(p−l3)l2​l1​⋯​ln−2.\begin{aligned} \sum_{\begin{subarray}{c}k_{1}+k_{2}+\cdots+k_{n}=p\\ k_{i}>0,1\leq i\leq n\end{subarray}}\dfrac{(-1)^{k_{1}}\left(\dfrac{k_{1}}{3}\right)}{k_{1}\cdots k_{n}}&=\sum_{\begin{subarray}{c}k_{2}+\cdots+k_{n}=l<p\\ 0<k_{i},2\leq i\leq n\end{subarray}}\dfrac{(-1)^{(p-l)}\left(\dfrac{p-l}{3}\right)}{(p-l)k_{2}\cdots k_{n}}\\ &\equiv-\sum_{\begin{subarray}{c}k_{2}+\cdots+k_{n}=l<p\\ 0<k_{i},2\leq i\leq n\end{subarray}}\dfrac{(-1)^{p-l}\left(\dfrac{p-l}{3}\right)}{lk_{2}\cdots k_{n}}\pmod{p}\\ &=-(n-1)\sum_{\begin{subarray}{c}k_{2}+\cdots+k_{n-1}=l_{1}<l<p\\ 0<k_{i},2\leq i\leq{n-1}\end{subarray}}\dfrac{(-1)^{p-l}\left(\dfrac{p-l}{3}\right)}{l^{2}k_{2}\cdots k_{n-1}}\\ &=-(n-1)(n-2)\sum_{\begin{subarray}{c}k_{2}+\cdots+k_{n-2}=l_{2}<l_{1}<l<p\\ 0<k_{i},2\leq i\leq_{n}-2\end{subarray}}\dfrac{(-1)^{p-l}\left(\dfrac{p-l}{3}\right)}{l^{2}l_{1}k_{2}\cdots k_{n-2}}\\ &=\cdots=-(n-1)!\sum_{\begin{subarray}{c}0\end{subarray}<l_{n-2}<\cdots<l_{1}<l<p}\dfrac{(-1)^{p-l}\left(\dfrac{p-l}{3}\right)}{l^{2}l_{1}\cdots l_{n-2}}.\end{aligned}

Then we can immediately obtain Theorem 1.2 by applying Theorem 1.1 .∎

Proof of Theorem 1.3 Similar to Theorem 1.2, for positive integer n≥2n\geq 2 and prime p>n+1p>n+1, we have

(3.4) ∑k1+k2+⋯+kn=pki>0,1≤i≤n(−1)k1​((k1+13)−(k1−13))k1​⋯​kn≡−(n−1)!​∑0<ln−2<⋯<l1<l<p(−1)p−l​((p−l+13)−(p−l−13))l2​l1​⋯​ln−2(modp).\begin{aligned} &\sum_{\begin{subarray}{c}k_{1}+k_{2}+\cdots+k_{n}=p\\ k_{i}>0,1\leq i\leq n\end{subarray}}\dfrac{(-1)^{k_{1}}\left(\left(\dfrac{k_{1}+1}{3}\right)-\left(\dfrac{k_{1}-1}{3}\right)\right)}{k_{1}\cdots k_{n}}\\ &\equiv-(n-1)!\sum_{\begin{subarray}{c}0\end{subarray}<l_{n-2}<\cdots<l_{1}<l<p}\dfrac{(-1)^{p-l}\left(\left(\dfrac{p-l+1}{3}\right)-\left(\dfrac{p-l-1}{3}\right)\right)}{l^{2}l_{1}\cdots l_{n-2}}\pmod{p}.\end{aligned}

Let ω\omega be a complex primitive cubic root of unity. we observe that (i+13)−(i−13)=ωi+ω−i\left(\dfrac{i+1}{3}\right)-\left(\dfrac{i-1}{3}\right)={\omega}^{i}+{\omega}^{-i} for any integer ii. Using the same method as (3.1) and the fact that (p+13)−(p−13)=−1\left(\dfrac{p+1}{3}\right)-\left(\dfrac{p-1}{3}\right)=-1 for prime p>3p>3, we obtain

(3.5) ∑0<ln−2<⋯<l1<l<p(−1)p−l​((p−l+13)−(p−l−13))l2​l1​⋯​ln−2≡{∑0<i1≤i2≤p−1(−1)(p+i1)​((p+i1+13)−(p+i1−13))i1​i2n−1(modp)if 2∣n,−∑0<i1≤i2≤p−1(−1)(p+i1)​((p+i1+13)−(p+i1−13))i1​i2n−1+Bp−n(modp)if 2∤n.\begin{aligned} &\sum_{\begin{subarray}{c}0\end{subarray}<l_{n-2}<\cdots<l_{1}<l<p}\dfrac{(-1)^{p-l}\left(\left(\dfrac{p-l+1}{3}\right)-\left(\dfrac{p-l-1}{3}\right)\right)}{l^{2}l_{1}\cdots l_{n-2}}\\ &\equiv\begin{cases}\sum\limits_{0<i_{1}\leq i_{2}\leq p-1}\dfrac{(-1)^{(p+i_{1})}\left(\left(\dfrac{p+i_{1}+1}{3}\right)-\left(\dfrac{p+i_{1}-1}{3}\right)\right)}{i_{1}i_{2}^{n-1}}\pmod{p}&\text{if $2\mid n$},\\ -\sum\limits_{0<i_{1}\leq i_{2}\leq p-1}\dfrac{(-1)^{(p+i_{1})}\left(\left(\dfrac{p+i_{1}+1}{3}\right)-\left(\dfrac{p+i_{1}-1}{3}\right)\right)}{i_{1}i_{2}^{n-1}}\\ \quad{}+B_{p-n}\pmod{p}&\text{if $2\nmid n$}.\end{cases}\end{aligned}

On the other hand, (−1)(p+i)​((p+i+13)−(p+i−13))(-1)^{(p+i)}\left(\left(\dfrac{p+i+1}{3}\right)-\left(\dfrac{p+i-1}{3}\right)\right) takes the values 22, 11, −1-1, −2-2, −1-1, 11 as p+i≡0,1,2,3,4,5(mod6).p+i\equiv 0,1,2,3,4,5\pmod{6}. Referring to the proof of Lemma 2.7, we can get

(3.6) ∑i=1p−1(−1)(p+i)​((p+i+13)−(p+i−13))in≡{(3n−1−1)​(2n−1−1)n​6n−1​Bp−n(modp)if 2∤n,0(modp)if 2∣n.\begin{aligned} &\sum\limits_{i=1}^{p-1}\frac{(-1)^{(p+i)}\left(\left(\dfrac{p+i+1}{3}\right)-\left(\dfrac{p+i-1}{3}\right)\right)}{i^{n}}\\ \qquad{}&\equiv\begin{cases}\dfrac{(3^{n-1}-1)(2^{n-1}-1)}{n6^{n-1}}B_{p-n}\pmod{p}&\text{if $2\nmid n$},\\ 0\pmod{p}&\text{if $2\mid n$}.\end{cases}\end{aligned}

Applying (3.6) and (3.5) , we can obtain the following congruences in a way similar to Theorem 1.1,

(3.7) ∑0<ln−2<⋯<l1<l<p(−1)p−l​((p−l+13)−(p−l−13))l2​l1​⋯​ln−2≡{∑r=0p−n(p−n+1r)​(3n+r−2−1)​(2n+r−2−1)​Br​Bp−n−r+1(n−1)​(n+r−1)​6n+r−2(modp)if 2∣n,(1−(3n−1−1)​(2n−1−1)2​n​6n−1)​Bp−n(modp)if 2∤n.\begin{aligned} &\sum_{\begin{subarray}{c}0\end{subarray}<l_{n-2}<\cdots<l_{1}<l<p}\dfrac{(-1)^{p-l}\left(\left(\dfrac{p-l+1}{3}\right)-\left(\dfrac{p-l-1}{3}\right)\right)}{l^{2}l_{1}\cdots l_{n-2}}\\ &\equiv\begin{cases}\sum\limits_{r=0}^{p-n}\left(\begin{array}[]{c}{p-n+1}\\ r\end{array}\right)\dfrac{(3^{n+r-2}-1)(2^{n+r-2}-1)B_{r}B_{p-n-r+1}}{(n-1)(n+r-1)6^{n+r-2}}\pmod{p}&\text{if $2\mid n$},\\ \left(1-\dfrac{(3^{n-1}-1)(2^{n-1}-1)}{2n6^{n-1}}\right)B_{p-n}\pmod{p}&\text{if $2\nmid n$}.\end{cases}\end{aligned}

Combining (3.7) and (3.4), we comgplete the proof of Theorem 1.3.∎

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