Two supercongruences related to multiple harmonic sums

TL;DR

This paper establishes two new supercongruences involving truncated series of multiple harmonic sums for prime numbers and p-adic integers, extending the understanding of congruence properties in number theory.

Contribution

It introduces novel supercongruences for specific truncated series involving multiple harmonic sums, expanding the theoretical framework of supercongruences in number theory.

Findings

Proves two supercongruences for series involving multiple harmonic sums.

Extends supercongruence results to series with p-adic parameters.

Provides new tools for analyzing congruences in combinatorial and number-theoretic contexts.

Abstract

Let $p$ be a prime and let $x$ be a $p$-adic integer. We provide two supercongruences for truncated series of the form $$\sum_{k=1}^{p-1} \frac{(x)_k}{(1)_k}\cdot \frac{1}{k}\sum_{1\le j_1\le\cdots\le j_r\le k}\frac{1}{j_1^{}\cdots j_r^{}}\quad\mbox{and}\quad \sum_{k=1}^{p-1} \frac{(x)_k(1-x)_k}{(1)_k^2}\cdot \frac{1}{k}\sum_{1\le j_1\le\cdots\le j_r\le k}\frac{1}{j_1^{2}\cdots j_r^{2}}.$$

Equations

$$\sum _{k=1}^{p-1}\frac{(x{)}_k}{(1{)}_k}\cdot \frac{1}{k}\sum _{1\le j_1\le \cdots \le j_r\le k}\frac{1}{j_1\cdots j_r}\text{mbox}and\sum _{k=1}^{p-1}\frac{(x{)}_k(1-x{)}_k}{(1{)}_k^2}\cdot \frac{1}{k}\sum _{1\le j_1\le \cdots \le j_r\le k}\frac{1}{j_1^2\cdots j_r^2}.$$

$$\sum _{k=1}^{p-1}\frac{\left(\genfrac{}{}{0ex}{}{2k}{k}\right)}{k{4}^k}{\equiv }_{p^3}-H_{(p-1)/2}\text{and}\sum _{k=1}^{p-1}\frac{{\left(\genfrac{}{}{0ex}{}{2k}{k}\right)}^2}{k16^k}{\equiv }_{p^3}-2H_{(p-1)/2}$$

$$S_n(t_1,\dots ,t_r):=\sum _{1\le j_1\le \cdots \le j_r\le n}\frac{1}{j_1^{t_1}\cdots j_r^{t_r}}$$

$$\sum _{k=1}^{p-1}\frac{(x{)}_k}{(1{)}_k}\cdot \frac{S_k(\{1{\}}^r)}{k}$$

$$\sum _{k=1}^{p-1}\frac{(x{)}_k(1-x{)}_k}{(1{)}_k^2}\cdot \frac{S_k(\{2{\}}^r)}{k}$$

$$+\frac{2s(1+3sr+2s{r}^2)}{2r+3}{p}^2{B}_{p-2r-3}(x).$$

$$\sum _{k=1}^{p-1}\frac{\left(\genfrac{}{}{0ex}{}{2k}{k}\right)}{k{4}^k}\cdot S_k(\{1{\}}^r){\equiv }_{p^2}\{\begin{array}{ll}-H_{(p-1)/2}^{(r+1)}& \text{if r≡20},\text{vspace}3mm\\ \frac{2^{r+2}-1}{2(r+2)}p{B}_{p-r-2}& \text{if r≡21,}\end{array}$$

$$\sum _{k=1}^{p-1}\frac{{\left(\genfrac{}{}{0ex}{}{2k}{k}\right)}^2}{k16^k}\cdot S_k(\{2{\}}^r){\equiv }_{p^3}-2H_{(p-1)/2}^{(2r+1)}-\frac{r(2^{2r+3}-1)}{2}{p}^2{B}_{p-2r-3}.$$

$$x\to \frac{(x{)}_k}{(x{)}_n}$$

$$\sum _{k=0}^{n-1}\frac{(x{)}_k}{(1{)}_k}\cdot a_k=(x{)}_n\sum _{j=0}^{n-1}\frac{(-1{)}^j{T}_j}{j!(n-1-j)!}\cdot \frac{1}{x+j}$$

$$T_j:=\sum _{k=0}^j(-1{)}^k(\genfrac{}{}{0ex}{}{j}{k})\cdot a_k.$$

$$\sum _{k=0}^{p-1}\frac{(x{)}_k}{(1{)}_k}\cdot a_k{\equiv }_p{T}_{⟨-x{⟩}_p}.$$

$$G_n^{(r)}(x):=\sum _{k=1}^n\frac{(x{)}_k}{(1{)}_k}\cdot S_k(\{1{\}}^r).$$

$$G_n^{(0)}(x)=\frac{(1+x{)}_n}{(1{)}_n}-1$$

$$G_n^{(r)}(x)=\frac{(1+x{)}_n}{(1{)}_n}\cdot S_n(\{1{\}}^r)-\frac{G_n^{(r-1)}(x)}{x}.$$

$$F_n^{(r)}(x+1)-F_n^{(r)}(x)=\frac{G_n^{(r)}(j)}{x}$$

$$F_n^{(r)}(x):=\sum _{k=1}^n\frac{(x{)}_k}{(1{)}_k}\cdot \frac{S_k(\{1{\}}^r)}{k}.$$

$$F_n^{(r)}(x+m)-F_n^{(r)}(x)=\sum _{j=0}^{m-1}\frac{G_n^{(r)}(x+j)}{x+j}.$$

$$G_n^{(r)}(-u)=\frac{G_n^{(r-1)}(-u)}{u}=\cdots =\frac{G_n^{(0)}(-u)}{u^r}=-\frac{1}{u^r}.$$

$$\sum _{k=1}^n(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k})\frac{S_k(\{1{\}}^r)}{k}=-H_n^{(r+1)}.$$

$$G_{p-1}^{(0)}(x)=\frac{(1+x{)}_{p-1}}{(1{)}_{p-1}}-1{\equiv }_{p^2}\frac{sp}{x}-1.$$

$$G_{p-1}^{(r)}(x){\equiv }_{p^2}-\frac{G_{p-1}^{(r-1)}(x)}{x}{\equiv }_{p^2}\cdots {\equiv }_{p^2}(-1{)}^r\frac{G_{p-1}^{(0)}(x)}{x^r}{\equiv }_{p^2}\frac{(-1{)}^{r}sp}{x^{r+1}}-\frac{(-1{)}^r}{x^r}.$$

$$F_{p-1}^{(r)}(sp)$$

$$=sp{S}_{p-1}(\{1{\}}^r,2)){\equiv }_{p^2}sp{B}_{p-r-2}$$

$$F_{p-1}^{(r)}(x)$$

$${\equiv }_{p^2}-\sum _{j=1}^{⟨-x{⟩}_p}\frac{1}{(j-sp{)}^{r+1}}-sp\sum _{j=1}^{⟨-x{⟩}_p}\frac{1}{j^{r+2}}+sp{B}_{p-r-2}$$

$${\equiv }_{p^2}-H_{⟨-x{⟩}_p}^{(r+1)}-(r+2)sp{H}_{⟨-x{⟩}_p}^{(r+2)}+sp{B}_{p-r-2}$$

$${\equiv }_{p^2}-H_{⟨-x{⟩}_p}^{(r+1)}-(-1{)}^{r}sp{B}_{p-r-2}$$

$$H_{⟨-x{⟩}_p}^{(t)}{\equiv }_p\sum _{j=1}^{⟨-x{⟩}_p}{j}^{p-1-t}=\frac{B_{p-t}(⟨-x{⟩}_p+1)-B_{p-t}}{p-t}{\equiv }_p(-1{)}^t\frac{B_{p-t}(x)-B_{p-t}}{t}$$

$$x\to \frac{(x{)}_k(1-x{)}_k}{(x{)}_n(1-x{)}_n}$$

$$\sum _{k=0}^{n-1}\frac{(x{)}_k(1-x{)}_k}{(1{)}_k^2}\cdot a_k=(x{)}_n(1-x{)}_n\sum _{j=0}^{n-1}\frac{(-1{)}^j{A}_j}{(n+j)!(n-1-j)!}\left(\frac{1}{x+j}+\frac{1}{1-x+j}\right)$$

$$A_j:=\sum _{k=0}^j(-1{)}^k(\genfrac{}{}{0ex}{}{j}{k}\left)\right(\genfrac{}{}{0ex}{}{j+k}{k})\cdot a_k.$$

$$\sum _{k=0}^{\infty }\frac{(x{)}_k(1-x{)}_k}{(1{)}_k^2}\cdot a_k=\frac{\sin (\pi x)}{\pi }\sum _{j=0}^{\infty }(-1{)}^j{A}_j\left(\frac{1}{x+j}+\frac{1}{1-x+j}\right).$$

Full text

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1. Introduction and main results

In [9, Theorem 1.1] and [10, Theorem 7] we showed that for any prime $p\not=2$,

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where $H_{n}^{(t)}=\sum_{j=1}^{n}\frac{1}{j^{t}}$ is the $n$-th harmonic number of order $t\geq 1$. Here we present two extensions of such congruences which involves the (non-strict) multiple harmonic sums

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with $t_{1},t_{2},\dots,t_{r}$ positive integers. For the sake of brevity, if $t_{1}=t_{2}=\dots=t_{r}=t$ we write $S_{n}(\{t\}^{r})$.

Let $(x)_{n}:=x(x+1)\cdots(x+n-1)$ be the Pochhammer symbol, and let $B_{n}(x)$ be the $n$-th Bernoulli polynomial. For any prime $p$, $\mathbb{Z}_{p}$ denotes the ring of all $p$-adic integers and $\langle\cdot\rangle_{p}$ is the least non-negative residue modulo $p$ of the $p$-integral argument.

Theorem 1**.**

Let $p$ be a prime, $x\in\mathbb{Z}_{p}$ and $r\in\mathbb{N}$. Let $s:=(x+\langle-x\rangle_{p})/p$.

i) If $p>r+3$ then

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ii) If $p>2r+3$ then

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Note that, when $r=0$, both (1) and (1) have been established by Zhi-Hong Sun in [7]. Moreover, for the special value $x=1/2$, (1) and (1) yield

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and

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For $r=1$, the congruence (4) proves the conjecture [8, Conjecture 5.3].

In the last section we provide $q$-analogs of two binomial identities related to the congruences (1) and (1).

2. Proof of (1) in Theorem 1

By taking the partial fraction expansion of the rational function

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with $0\leq k<n$, we find

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where $T_{j}$ is the binomial transform of the sequence $a_{k}$,

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It is easy to see from (5) that if $a_{0},\dots,a_{p-1},x\in\mathbb{Z}_{p}$ then

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In order to show (1) we introduce the function

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We have that

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and $S_{k}(\{1\}^{r})=S_{k-1}(\{1\}^{r})+S_{k}(\{1\}^{r-1})/k$ implies

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Moreover

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where

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Then, for any positive integer $m$,

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By (7), for $u=1,\dots,n$

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Hence by letting $x=-n$ and $m=n$ in (8) we obtain the known identity (see [1])

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Thus, for $a_{k}=\frac{S_{k}(\{1\}^{r})}{k}$, we have that $T_{j}=-H_{j}^{(r+1)}$, and by (6), we already have the modulo $p$ version of (1).

Proof of (1) in Theorem 1.

Since $sp=x+\langle-x\rangle_{p}$ it follows that

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By [11, Theorem 1.6], $S_{p-1}(\{1\}^{r})\equiv_{p}0$ and therefore

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Moreover

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where we used $S_{p-1}(\{1\}^{r},2)\equiv_{p}B_{p-r-2}$ (see [2, Theorem 4.5]).

Finally, by (8),

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where the last step uses the following congruence: for $2\leq t<p-1$

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which is an immediate consequence of [5, Lemma 3.2]. ∎

3. Proof of (1) in Theorem 1

We follow a similar strategy as outlined in the previous section. We start by considering the partial fraction decomposition of the rational function

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with $0\leq k<n$. We have that

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where

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For $n\to\infty$, if the series is convergent, the identity (11) becomes

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In many cases the transformed sequence $A_{j}$ has a nice formula. For example if $a_{k}=1/(k+z)$ then

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and for $x=z=1/2$ we recover this series representations the Catalan’s constant $G=\sum_{j=0}\frac{(-1)^{j}}{(2j+1)^{2}}$:

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As regards congruences we have the following result.

Theorem 2**.**

Let $p$ be a prime with $a_{0},\dots,a_{p-1},x\in\mathbb{Z}_{p}$. Then

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For $x=1/2$ and $p>2$ then

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Proof.

Rearranging (11) in a convenient way, we have

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If $0\leq k\leq j\leq p-1$ then $A_{p-1-j}\equiv_{p}A_{j}$ because

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Thus, since $\langle-x\rangle_{p}+\langle-(1-x)\rangle_{p}=p-1$, it follows that

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Therefore

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Finally, by using

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we are done. For $x=1/2$ it suffices to note that

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∎

As an application of the previous theorem, we note that when $a_{k}=1$ then $A_{j}=(-1)^{j}$, and, by (12), it follows that

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which has been established in [6, Corollary 2.1]. Another example worth to be mentioned is $a_{k}=1/k^{r}$ for $k\geq 1$ (and $a_{0}=0$). Then by [4, Theorem 1]

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Now we consider the case $a_{k}=S_{k}(\{2\}^{r})/k$. Let

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We have that

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and $S_{k}(\{2\}^{r})=S_{k-1}(\{2\}^{r})+S_{k}(\{2\}^{r-1})/k^{2}$ implies

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Moreover

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where

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Hence

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The next identity is a variation of (9) and it appears to be new.

Theorem 3**.**

For any integers $n\geq 1$ and $r\geq 0$,

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Proof.

By (13), for $u=1,\dots,n$,

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Hence by letting $x=-n$ and $m=n$ in (14)

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∎

Thus by applying $\eqref{A2}$ we find a modulo $p^{2}$ version of $\eqref{SI2}$. A more refined reasoning will lead us to the $p^{3}$ congruence.

Proof of (1) in Theorem 1.

Since $sp=x+\langle-x\rangle_{p}$,

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By [11, Theorem 1.6], $S_{p-1}(\{2\}^{r})\equiv_{p}0$ and therefore

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It follows that

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By (10)

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Moreover

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where we used

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and the congruences

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which have been established in [11, Theorem 1.6] in [2, Theorem 4.1] respectively. Finally,

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∎

We observe that (4) follows by letting $x=1/2$. Then $\langle-x\rangle_{p}-1=(p-1)/2$, $B_{2n}(1/2)=(2^{1-2n}-1)B_{2n}$ and for $p-4>t>1$

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see [5, Theorem 5.2].

4. Final remarks: $q$-analogs of (9) and (15)

It is interesting to note that identities (9) and (15) have both a $q$-version (the first one appears in [3]).

Theorem 4**.**

For any integers $n\geq 1$ and $r\geq 0$,

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and

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where $\genfrac{[}{]}{0.0pt}{}{m}{k}_{q}$ is the Gaussian binomial coefficient

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and

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Proof.

We show (15) and we leave the proof of other one to the interested reader. The procedure is quite similar to the one given for the corresponding ordinary identity (15). Let

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Then for $u=1,\dots,n$, $G_{n}^{(0)}(u)=-1$ and

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Moreover

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where

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Thus, since $F_{n}^{(0)}(n)=0$,

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and the proof is complete. ∎