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On two congruence conjectures of Z.-W. Sun involving Franel numbers

Part of: Combinatorics Elementary classical functions Elementary number theory

Published online by Cambridge University Press:  16 May 2023

Guo-Shuai Mao  and
Yan Liu
Guo-Shuai Mao
Affiliation:
Department of Mathematics, Nanjing University of Information Science and Technology, Nanjing 210044, People's Republic of China (maogsmath@163.com, 1325507759@qq.com)
Yan Liu
Affiliation:
Department of Mathematics, Nanjing University of Information Science and Technology, Nanjing 210044, People's Republic of China (maogsmath@163.com, 1325507759@qq.com)
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Abstract

In this paper, we mainly prove the following conjectures of Z.-W. Sun (J. Number Theory 133 (2013), 2914–2928): let $p>2$ be a prime. If $p=x^2+3y^2$ with $x,y\in \mathbb {Z}$ and $x\equiv 1\ ({\rm {mod}}\ 3)$, then

\[ x\equiv\frac14\sum_{k=0}^{p-1}(3k+4)\frac{f_k}{2^k}\equiv\frac12\sum_{k=0}^{p-1}(3k+2)\frac{f_k}{({-}4)^k}\ ({\rm{mod}}\ p^2), \]
and if $p\equiv 1\pmod 3$, then
\[ \sum_{k=0}^{p-1}\frac{f_k}{2^k}\equiv\sum_{k=0}^{p-1}\frac{f_k}{({-}4)^k}\ ({\rm{mod}}\ p^3), \]
where $f_n=\sum _{k=0}^n\binom {n}k^3$ stands for the $n$th Franel number.

Keywords

CongruencesFranel numbersp-adic gamma functiongamma function

MSC classification

Primary: 11A07: Congruences; primitive roots; residue systems
Secondary: 05A10: Factorials, binomial coefficients, combinatorial functions 05A19: Combinatorial identities, bijective combinatorics 33B15: Gamma, beta and polygamma functions
Type
Research Article
Information
Proceedings of the Royal Society of Edinburgh Section A: Mathematics , Volume 154 , Issue 3 , June 2024 , pp. 887 - 905
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Copyright
Copyright © The Author(s), 2023. Published by Cambridge University Press on behalf of The Royal Society of Edinburgh

1. Introduction

In 1894, Franel [Reference Franel2] found that the numbers

\[ f_n=\sum_{k=0}^n\binom{n}k^3\quad (n=0,1,2,\ldots) \]

satisfy the recurrence relation (cf. [Reference Sloane14, A000172]):

\[ (n+1)^2f_{n+1}=(7n^2+7n+2)f_n+8n^2f_{n-1}\quad (n=1,2,3,\ldots). \]

These numbers are now called Franel numbers. Callan [Reference Callan1] found a combinatorial interpretation of the Franel numbers. The Franel numbers play important roles in combinatorics and number theory. The sequence $\{f_n\}_{n \geqslant 0}$ is one of the five sporadic sequences (cf. [Reference Zagier23, § 4]) which are integral solutions of certain Apéry-like recurrence equations and closely related to the theory of modular forms. In 2013, Sun [Reference Sun19] revealed some unexpected connections between the numbers $f_n$ and representations of primes $p \equiv 1 (\text{mod} 3)$ in the form $x^2+3y^2$ with $x,y\in \mathbb {Z}$, for example, Sun [Reference Sun19, (1.2)] showed that

(1.1)\begin{equation} \sum_{k=0}^{p-1}\frac{f_k}{2^k}\equiv\sum_{k=0}^{p-1}\frac{f_k}{({-}4)^k}\equiv 2x-\frac{p}{2x}\ ({\rm{mod}}\ p^2), \end{equation}

and in the same paper, Sun proposed some conjectures involving Franel numbers, one of which is

Conjecture 1.1 Let $p>2$ be a prime. If $p=x^2+3y^2$ with $x,y\in \mathbb {Z}$ and $x\equiv 1 ({\rm {mod}}\ 3)$, then

\[ x\equiv\frac14\sum_{k=0}^{p-1}(3k+4)\frac{f_k}{2^k}\equiv\frac12\sum_{k=0}^{p-1}(3k+2)\frac{f_k}{({-}4)^k}\ ({\rm{mod}}\ p^2). \]

For more details on Franel numbers, we refer the readers to [Reference Guo3, Reference Guo4, Reference Liu6, Reference Mao8, Reference Mao9, Reference Sun18, Reference Sun20] and so on.

In this paper, our first goal is to prove the above conjecture.

Theorem 1.1 Conjecture 1.1 is true.

Combining (1.1) and theorem 1.1, we immediately obtain the following result.

Corollary 1.1 For any prime $p\equiv 1\pmod 3$, we have

\[ \sum_{k=0}^{p-1}\frac{kf_k}{2^k}\equiv2\sum_{k=0}^{p-1}\frac{kf_k}{({-}4)^k}\ ({\rm{mod}}\ p^2). \]

Sun [Reference Sun19] also gave the following conjecture.

Conjecture 1.2 Let $p>2$ be a prime. If $p\equiv 1\pmod 3$, then

\[ \sum_{k=0}^{p-1}\frac{f_k}{2^k}\equiv\sum_{k=0}^{p-1}\frac{f_k}{({-}4)^k}\ ({\rm{mod}}\ p^3). \]

Our last goal is to prove this conjecture.

Theorem 1.2 Conjecture 1.2 is true.

We are going to prove theorem 1.1 in §2. Section 3 is devoted to proving theorem 1.2. Our proofs make use of some combinatorial identities which were found by the package Sigma [Reference Schneider13] via software Mathematica and the $p$-adic gamma function. The proof of theorem 1.2 is somewhat difficult and complex because it is rather convoluted. Throughout this paper, prime $p$ always $\equiv 1\pmod 3$, so in the following lemmas $p>5$ or $p>3$ or $p>2$ is the same, we mention it here first.

2. Proof of theorem 1.1

For a prime $p$, let $\mathbb {Z}_p$ denote the ring of all $p$-adic integers and let $\mathbb {Z}_p^{\times }:=\{a\in \mathbb {Z}_p:\,a\text { is prime to }p\}.$ For each $\alpha \in \mathbb {Z}_p$, define the $p$-adic order $\nu _p(\alpha ):=\max \{n\in \mathbb {N}:\, p^n\mid \alpha \}$ and the $p$-adic norm $|\alpha |_p:=p^{-\nu _p(\alpha )}$. Define the $p$-adic gamma function $\Gamma _p(\cdot )$ by

\[ \Gamma_p(n)=({-}1)^n\prod_{\substack{1\leq k< n\\ (k,p)=1}}k,\quad n=1,2,3,\ldots, \]

and

\[ \Gamma_p(\alpha)=\lim_{\substack{|\alpha-n|_p\to 0\\ n\in\mathbb{N}}}\Gamma_p(n),\quad \alpha\in\mathbb{Z}_p. \]

In particular, we set $\Gamma _p(0)=1$. In the following, we need to use the most basic properties of $\Gamma _p$, and all of them can be found in [Reference Murty11, Reference Robert12]. For example, we know that

(2.1)\begin{gather} \frac{\Gamma_p(x+1)}{\Gamma_p(x)}=\begin{cases}-x, & \text{if }|x|_p=1,\\ - 1, & \text{if }|x|_p<1. \end{cases} \end{gather}
(2.2)\begin{gather} \Gamma_p(1-x)\Gamma_p(x)=({-}1)^{a_0(x)}, \end{gather}

where $a_0(x)\in \{1,2,\ldots,p\}$ such that $x\equiv a_0(x)\ ({\rm {mod}}\ p)$. And a property we need here is the fact that for any positive integer $n$,

(2.3)\begin{equation} z_1\equiv z_2\ ({\rm{mod}}\ p^n)\quad \text{implies}\quad \Gamma_p(z_1)\equiv\Gamma_p(z_2)\ ({\rm{mod}}\ p^n). \end{equation}

Lemma 2.1 ([Reference Sun19, lemma 2.2]) For any $n\in \mathbb {N}$ we have

(2.4)\begin{equation} \sum_{k=0}^n\binom{n}k^3z^k=\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n+k}{3k}\binom{2k}k\binom{3k}kz^k(1+z)^{n-2k} \end{equation}

and

(2.5)\begin{equation} f_n=\sum_{k=0}^n\binom{n+2k}{3k}\binom{2k}k\binom{3k}k({-}4)^{n-k}. \end{equation}

For $n,m\in \{1,2,3,\ldots \}$, define

\[ H_n^{(m)}:=\sum_{1\leq k\leq n}\frac1{k^m},\ \ H_0^{(m)}:=0, \]

these numbers with $m=1$ are often called the classic harmonic numbers. Recall that the Bernoulli polynomials are given by

\[ B_n(x)=\sum_{k=0}^n\binom nkB_kx^{n-k}\quad (n=0,1,2,\ldots). \]

Lemma 2.2 ([Reference Sun15, Reference Sun16]) Let $p>5$ be a prime. Then

\begin{align*} & H_{p-1}^{(2)}\equiv0\ ({\rm{mod}}\ p),\quad H_{{(p-1)}/2}^{(2)}\equiv0\ ({\rm{mod}}\ p),\quad H_{p-1}\equiv0\ ({\rm{mod}}\ p^2),\\ & \frac15H_{\lfloor{p}/6\rfloor}^{(2)}\equiv H_{\lfloor{p}/3\rfloor}^{(2)}\equiv\frac12\left(\frac{p}3\right)B_{p-2}\left(\frac13\right)\ ({\rm{mod}}\ p),\\ & H_{\lfloor{p}/3\rfloor}\equiv{-}\frac32q_p(3)+\frac{3p}4q^2_p(3)-\frac{p}6\left(\frac{p}3\right)B_{p-2}\left(\frac13\right)\ ({\rm{mod}}\ p^2),\\ & H_{{(p-1)}/2}\equiv{-}2q_p(2)+pq^2_p(2)\ ({\rm{mod}}\ p^2),\ H_{\lfloor{p}/4\rfloor}^{(2)}\equiv({-}1)^{{(p-1)}/2}4E_{p-3}\ ({\rm{mod}}\ p),\\ & H_{\lfloor{p}/6\rfloor}\equiv H_{\lfloor{p}/3\rfloor}+H_{{(p-1)}/2}-\frac{p}{4}\left(\frac{p}3\right)B_{p-2}\left(\frac13\right)\ ({\rm{mod}}\ p^2),\\ & H_{\lfloor{2p}/3\rfloor}\equiv{-}\frac32q_p(3)+\frac{3p}4q^2_p(3)+\frac{p}3\left(\frac{p}3\right)B_{p-2}\left(\frac13\right)\ ({\rm{mod}}\ p^2), \end{align*}

where $q_p(a)=(a^{p-1}-1)/p$ stands for the Fermat quotient.

Lemma 2.3 Let $p>5$ be a prime. If $0\leq j\leq (p-1)/2$, then we have

\[ \binom{3j}j\binom{p+j}{3j+1}\equiv \frac{p}{3j+1}(1-pH_{2j}+pH_j)\ ({\rm{mod}}\ p^3). \]

Proof. If $0\leq j\leq (p-1)/2$ and $j\neq (p-1)/3$, then we have

\begin{align*} \binom{3j}j\binom{p+j}{3j+1}& =\frac{(p+j)\cdots(p+1)p(p-1)\cdots(p-2j)}{j!(2j)!(3j+1)}\\ & \equiv\frac{pj!(1+pH_j)({-}1)^{2j}(2j)!(1-pH_{2j})}{j!(2j)!(3j+1)}\\ & \equiv\frac{p}{3j+1}(1-pH_{2j}+pH_j)\ ({\rm{mod}}\ p^3). \end{align*}

If $j=(p-1)/3$, then by lemma 2.2, we have

\begin{align*} & \binom{p-1}{\frac{p-1}3}\binom{p+\frac{p-1}3}{\frac{p-1}3}\\ & \quad\equiv\left(1-pH_{{(p-1)}/3}+\frac{p^2}2(H_{{(p-1)}/3}^2-H_{{(p-1)}/3}^{(2)})\right)\\ & \qquad\left(1+pH_{{(p-1)}/3}+\frac{p^2}2(H_{{(p-1)}/3}^2-H_{{(p-1)}/3}^{(2)})\right)\\ & \quad\equiv 1-p^2H_{{(p-1)}/3}^{(2)}\equiv1-\frac{p^2}2\left(\frac{p}3\right)B_{p-2}\left(\frac13\right)\ ({\rm{mod}}\ p^3) \end{align*}

and

\[ 1-pH_{{(2p-2)}/3}+pH_{{(p-1)}/3}\equiv1-\frac{p^2}2\left(\frac{p}3\right)B_{p-2}\left(\frac13\right)\ ({\rm{mod}}\ p^3). \]

Now the proof of lemma 2.3 is complete.

Proof of theorem 1.1 With the help of (2.4), we have

(2.6)\begin{align} \sum_{k=0}^{p-1}(3k+4)\frac{f_k}{2^k}& =\sum_{k=0}^{p-1}\frac{3k+4}{2^k}\sum_{j=0}^{\lfloor k/2\rfloor}\binom{k+j}{3j}\binom{2j}j\binom{3j}j2^{k-2j}\nonumber\\ & =\sum_{j=0}^{(p-1)/2}\frac{\binom{2j}j\binom{3j}j}{4^j}\sum_{k=2j}^{p-1}(3k+4)\binom{k+j}{3j}. \end{align}

By loading the package Sigma in software Mathematica, we find the following identity:

\[ \sum_{k=2j}^{n-1}(3k+4)\binom{k+j}{3j}=\frac{9nj+3n+9j+5}{3j+2}\binom{n+j}{3j+1}. \]

Thus, replacing $n$ by $p$ in the above identity and then substitute it into (2.6), we have

\[ \sum_{k=0}^{p-1}(3k+4)\frac{f_k}{2^k}=\sum_{j=0}^{(p-1)/2}\frac{\binom{2j}j\binom{3j}j}{4^j}\frac{9pj+3p+9j+5}{3j+2}\binom{p+j}{3j+1}. \]

Hence, we immediately obtain the following result by lemma 2.3,

(2.7)\begin{equation} \sum_{k=0}^{p-1}(3k+4)\frac{f_k}{2^k}\equiv p\sum_{j=0}^{(p-1)/2}\frac{\binom{2j}j}{4^j}\frac{9j+5}{(3j+1)(3j+2)}\ ({\rm{mod}}\ p^2). \end{equation}

It is easy to verify that

(2.8)\begin{align} & p\sum_{j=0}^{{(p-1)}/2}\frac{\binom{2j}j}{4^j}\frac{9j+5}{(3j+1)(3j+2)}\nonumber\\ & \quad=p\sum_{\substack{j=0\\j\neq(p-1)/3}}^{{(p-1)}/2}\frac{\binom{2j}j}{4^j}\frac{9j+5}{(3j+1)(3j+2)}+\frac{3p+2}{p+1}\binom{(2p-2)/3}{(p-1)/3}4^{{(1-p)}/3}\nonumber\\ & \quad\equiv p\sum_{\substack{j=0\\j\neq(p-1)/3}}^{{(p-1)}/2}\binom{\frac{p-1}2}j({-}1)^j\frac{9j+5}{(3j+1)(3j+2)}+\frac{3p+2}{p+1}\binom{(2p-2)/3}{(p-1)/3}4^{{(1-p)}/3}\nonumber\\ & \quad\equiv S_1+S_2\ ({\rm{mod}}\ p^2), \end{align}

where

(2.9)\begin{equation} S_1= p\sum_{j=0}^{(p-1)/2}\binom{(p-1)/2}j({-}1)^j\left(\frac2{3j+1}+\frac1{3j+2}\right) \end{equation}

and

\[ S_2=\frac{3p+2}{p+1}\left(\binom{(2p-2)/3}{(p-1)/3}4^{(1-p)/3}-\binom{(p-1)/2}{(p-1)/3}\right). \]

Applying the famous partial fraction identity

(2.10)\begin{equation} \sum_{k=0}^n\binom{n}k\frac{({-}1)^k}{k+x}=\frac{n!}{x(x+1)\cdots(x+n)} \end{equation}

with $x=1/3, n=(p-1)/2$ and $x=2/3,n=(p-1)/2$, we may simplify (2.9) as

\[ S_1=\frac{4p}{3p-1}\frac{(1)_{(p-1)/2}}{(1/3)_{(p-1)/2}}+\frac{2p}{3p+1}\frac{(1)_{(p-1)/2}}{(2/3)_{(p-1)/2}}, \]

where $(a)_n=a(a+1)\cdots (a+n-1)$ is the rising factorial or the Pochhammer symbol.

In view of (2.2), we have

\begin{align*} & \frac{4p}{3p-1}\frac{(1)_{{(p-1)}/2}}{(\frac13)_{{(p-1)}/2}}=\frac{4p}{3p-1}\frac{\Gamma(\frac{p+1}2)\Gamma(\frac13)}{\Gamma(\frac13+\frac{p-1}2)}=\frac{4p}{3p-1}\frac{({-}1)^{{(p+1)}/2}\Gamma_p(\frac{p+1}2)\Gamma_p(\frac13)}{({-}1)^{{(p-1)}/2}\frac{p}3\Gamma_p(\frac13+\frac{p-1}2)}\\ & \quad= \frac{12}{1-3p}\frac{\Gamma_p(\frac{p+1}2)\Gamma_p(\frac13)}{\Gamma_p(\frac{p}2-\frac16)}=\frac{12({-}1)^{{(p-1)}/6}}{1-3p}\Gamma_p\left(\frac{p+1}2\right)\Gamma_p\left(\frac13\right)\Gamma_p\left(\frac76-\frac{p}2\right), \end{align*}

where $\Gamma (\cdot )$ is the gamma function. In view of [Reference Long and Ramakrishna7, theorem 14] and [Reference Liu5, (2.4)] (or [Reference Mao and Pan10, (3.2)]), for $\alpha,s\in \mathbb {Z}_p$, we have

(2.11)\begin{equation} \Gamma_p(\alpha+ps)\equiv\Gamma_p(\alpha)+ps\Gamma^{'}_p(\alpha)\ ({\rm{mod}}\ p^2)\end{equation}

and

(2.12)\begin{equation} \frac{\Gamma^{'}_p(\alpha)}{\Gamma_p(\alpha)}\equiv1+H_{p-\langle-\alpha\rangle_p-1}\ ({\rm{mod}}\ p), \end{equation}

where $\Gamma _p'(x)$ denotes the $p$-adic derivative of $\Gamma _p(x)$, $\langle \alpha \rangle _n$ denotes the least non-negative residue of $\alpha$ modulo $n$, i.e. the integer lying in $\{0,1,\ldots,n-1\}$ such that $\langle \alpha \rangle _n\equiv \alpha \ ({\rm {mod}}\ n)$.

Therefore,

\begin{align*} & \frac{4p}{3p-1}\frac{(1)_{{(p-1)}/2}}{(\frac13)_{{(p-1)}/2}}\\ & \quad\equiv\frac{12({-}1)^{{(p-1)}/6}\Gamma_p\left(\frac{1}2\right)\Gamma_p\left(\frac13\right)\Gamma_p\left(\frac76\right)}{1-3p}\left(1+\frac{p}2(H_{{(p-1)}/2}-H_{{(p-7)}/6})\right)\ ({\rm{mod}}\ p^2). \end{align*}

In view of (2.1) and (2.2), we have

\begin{align*} & \frac{4p}{3p-1}\frac{(1)_{{(p-1)}/2}}{(\frac13)_{{(p-1)}/2}}\\ & \quad\equiv\frac{2(1+3p)\Gamma_p\left(\frac{1}2\right)\Gamma_p\left(\frac13\right)}{\Gamma_p\left(\frac56\right)}\left(1+\frac{p}2(H_{{(p-1)}/2}-H_{{(p-7)}/6})\right)\ ({\rm{mod}}\ p^2). \end{align*}

In view of [Reference Yeung22, proposition 4.1], we have

\[ \frac{\Gamma_p\left(\frac{1}2\right)\Gamma_p\left(\frac13\right)}{\Gamma_p\left(\frac56\right)}\equiv\frac{\binom{(5p-5)/6}{(p-1)/3}}{\left(1+\frac{p}6(5H_{(5p-5)/6}-3H_{(p-1)/2}-2H_{(p-1)/3})\right)}\ ({\rm{mod}}\ p^2). \]

Then with the help of [Reference Yeung22, theorem 4.12] and lemma 2.2, we have

(2.13)\begin{equation} \frac{4p}{3p-1}\frac{(1)_{{(p-1)}/2}}{(1/3)_{{(p-1)}/2}}\equiv4x+3pxq_p(3)-\frac{p}{x}\ ({\rm{mod}}\ p^2) \end{equation}

and

(2.14)\begin{equation} \frac{2p}{3p+1}\frac{(1)_{(p-1)/2}}{(2/3)_{(p-1)/2}}\equiv\frac{p}{x}\ ({\rm{mod}}\ p^2). \end{equation}

Hence,

(2.15)\begin{equation} S_1\equiv 4x+3pxq_p(3)\ ({\rm{mod}}\ p^2). \end{equation}

Lemma 2.4 Let $p>3$ be a prime. For any $p$-adic integer $t$, we have

(2.16)\begin{equation} \binom{\frac{p-1}2+pt}{\frac{p-1}3}\equiv\binom{\frac{p-1}2}{\frac{p-1}3}\left(1+pt\left(H_{{(p-1)}/2}-H_{{(p-1)}/6}\right)\right)\ ({\rm{mod}}\ p^2). \end{equation}

Proof. Set $m=(p-1)/2$. It is easy to check that

\begin{align*} \binom{m+pt}{(p-1)/3}& =\frac{(m+pt)\cdots(m+pt-(p-1)/3+1)}{((p-1)/3)!}\\ & \equiv\frac{m\cdots(m-(p-1)/3+1)}{((p-1)/3)!}(1+pt(H_m-H_{m-(p-1)/3}))\\ & =\binom{m}{(p-1)/3}(1+pt(H_m-H_{m-(p-1)/3}))\ ({\rm{mod}}\ p^2). \end{align*}

So lemma 2.4 is finished.

Now we evaluate $S_2$ modulo $p^2$. It is easy to obtain that

(2.17)\begin{align} S_2& \equiv2\left(\binom{-\frac12}{\frac{p-1}3}-\binom{\frac{p-1}2}{\frac{p-1}3}\right)\equiv{-}p\binom{\frac{p-1}2}{\frac{p-1}3}\left(H_{{(p-1)}/2}-H_{{(p-1)}/6}\right)\nonumber\\ & \equiv{-}3pxq_p(3)\ ({\rm{mod}}\ p^2) \end{align}

with the help of lemmas 2.2, 2.4 and [Reference Yeung22, theorem 4.12].

Therefore, in view of (2.7), (2.8), (2.15) and (2.17), we immediately get the desired result

\[ \frac14\sum_{k=0}^{p-1}(3k+4)\frac{f_k}{2^k}\equiv x\ ({\rm{mod}}\ p^2). \]

On the contrary, we use equation (2.5) to obtain

\begin{align*} \sum_{k=0}^{p-1}(3k+2)\frac{f_k}{({-}4)^k}& =\sum_{k=0}^{p-1}\frac{3k+2}{({-}4)^k}\sum_{j=0}^k\binom{k+2j}{3j}\binom{2j}j\binom{3j}j({-}4)^{k-j}\\ & =\sum_{j=0}^{p-1}\frac{\binom{2j}j\binom{3j}j}{({-}4)^j}\sum_{k=j}^{p-1}(3k+2)\binom{k+2j}{3j}. \end{align*}

By using the package Sigma again, we find the following identity:

\[ \sum_{k=j}^{n-1}(3k+2)\binom{k+2j}{3j}=\frac{9nj+3n+1}{3j+2}\binom{n+2j}{3j+1}. \]

Thus,

(2.18)\begin{equation} \sum_{k=0}^{p-1}(3k+2)\frac{f_k}{({-}4)^k}=\sum_{j=0}^{p-1}\frac{\binom{2j}j\binom{3j}j\binom{p+2j}{3j+1}}{({-}4)^j}\frac{9pj+3p+1}{3j+2}. \end{equation}

Lemma 2.5 Let $p>5$ be a prime. If $0\leq j\leq (p-1)/2$ and $j\neq (p-1)/3$, then

\[ \binom{3j}j\binom{p+2j}{3j+1}\equiv \frac{p({-}1)^j}{3j+1}(1+pH_{2j}-pH_j)\ ({\rm{mod}}\ p^3). \]

If $(p+1)/2\leq j\leq p-1$, then

\[ \binom{3j}j\binom{p+2j}{3j+1}\equiv \frac{2p({-}1)^j}{3j+1}\ ({\rm{mod}}\ p^2). \]

Proof. If $0\leq j\leq (p-1)/2$ and $j\neq (p-1)/3$, then we have

\begin{align*} \binom{3j}j\binom{p+2j}{3j+1}& =\frac{(p+2j)\cdots(p+1)p(p-1)\cdots(p-j)}{j!(2j)!(3j+1)}\\ & \quad\equiv\frac{p(2j)!(1+pH_{2j})({-}1)^{j}(j)!(1-pH_j)}{j!(2j)!(3j+1)}\\ & \quad\equiv\frac{p({-}1)^j}{3j+1}(1+pH_{2j}-pH_j)\ ({\rm{mod}}\ p^3). \end{align*}

If $(p+1)/2\leq j\leq p-1$, then

\begin{align*} & \binom{3j}j\binom{p+2j}{3j+1}\\ & \quad=\frac{(p+2j)\cdots(2p+1)(2p)(2p-1)\cdots(p+1)p(p-1)\cdots(p-j)}{j!(2j)!(3j+1)}\\ & \quad\equiv\frac{2p^2(2j)\cdots(p+1)(p-1)!({-}1)^{j}(j)!}{j!(2j)!(3j+1)}=\frac{2p({-}1)^j}{3j+1}\ ({\rm{mod}}\ p^2). \end{align*}

Now the proof of lemma 2.5 is complete.

It is known that $\binom {2k}k\equiv 0\ ({\rm {mod}}\ p)$ for each $(p+1)/2\leq k\leq p-1$, and it is easy to check that for each $0\leq j\leq (p-1)/2$:

\[ \binom{3j}j\binom{p+2j}{3j+1}\equiv \frac{p({-}1)^j}{3j+1}\ ({\rm{mod}}\ p^2). \]

These, with (2.18) yield

(2.19)\begin{align} & \sum_{k=0}^{p-1}(3k+2)\frac{f_k}{({-}4)^k}\equiv\sum_{j=0}^{{(p-1)}/2}\frac{\binom{2j}j}{({-}4)^j}\frac{p({-}1)^j}{3j+1}\frac{9pj+3p+1}{3j+2}\nonumber\\ & \quad+\sum_{j={(p+1)}/2}^{p-1}\frac{\binom{2j}j}{({-}4)^j}\frac{2p({-}1)^j}{3j+1}\frac{1}{3j+2}\equiv\sum_{j=0}^{{(p-1)}/2}\binom{\frac{p-1}2}j\frac{p({-}1)^j}{3j+1}\frac{1}{3j+2}+S_3\nonumber\\ & =p\sum_{j=0}^{{(p-1)}/2}\binom{\frac{p-1}2}j({-}1)^j\left(\frac{1}{3j+1}-\frac{1}{3j+2}\right)+S_3\ ({\rm{mod}}\ p^2), \end{align}

where

\begin{align*} S_3& =\binom{\frac{2p-2}3}{\frac{p-1}3}\frac1{(p+1)4^{{(p-1)}/3}}-\binom{\frac{p-1}2}{\frac{p-1}3}\frac1{p+1}-\binom{\frac{4p-4}3}{\frac{2p-2}3}\frac1{4^{{(2p-2)}/3}}\\ & =\frac1{p+1}\left(\binom{-1/2}{(p-1)/3}-\binom{\frac{p-1}2}{\frac{p-1}3}\right)-\binom{-1/2}{(2p-2)/3}. \end{align*}

As above, with (2.10), (2.13), (2.14), lemma 2.2 and [Reference Yeung22, theorem 4.12], we have the following congruence modulo $p^2$:

(2.20)\begin{equation} p\sum_{j=0}^{{(p-1)}/2}\binom{\frac{p-1}2}j({-}1)^j\left(\frac{1}{3j+1}-\frac{1}{3j+2}\right)\equiv2x+\frac{3px}2q_p(3)-\frac{3p}{2x}.\end{equation}

Now we evaluate $S_3$. It is easy to see that

\begin{align*} \binom{-1/2}{(2p-2)/3}& =\frac{(-\frac12)(-\frac12-1)\cdots(-\frac12-\frac{2p-2}3+1)}{(\frac{2p-2}3)!}\\ & =\frac{(\frac12)(\frac32)\cdots(\frac{p}2-1)\frac{p}2(\frac{p}2+1)\cdots(\frac{p}2+\frac{p-7}6)}{(\frac{2p-2}3)!}\\ & =\frac{(\frac{p}2-\frac{p-1}2)\cdots(\frac{p}2-1)\frac{p}2(\frac{p}2+1)\cdots(\frac{p}2+\frac{p-7}6)}{(\frac{2p-2}3)!}\\ & \equiv\frac{({-}1)^{{(p-1)}/2}\frac{p}2(\frac{p-1}2)!(\frac{p-7}6)!}{(\frac{2p-2}3)!}=\frac{({-}1)^{{(p-1)}/2}3p}{p-1}\frac1{\binom{\frac{2p-2}3}{\frac{p-1}2}}\\ & \equiv\frac{-3p({-}1)^{(p-1)/2}}{\binom{\frac{2p-2}3}{\frac{p-1}2}}\ ({\rm{mod}}\ p^2). \end{align*}

In view of (2.17) and [Reference Yeung22, theorem 4.12], we immediately obtain

\[ S_3\equiv{-}\frac{3px}2q_p(3)+\frac{3p}{2x}\ ({\rm{mod}}\ p^2). \]

This, with (2.19) and (2.20) yields

\[ \frac12\sum_{k=0}^{p-1}(3k+2)\frac{f_k}{({-}4)^k}\equiv x\ ({\rm{mod}}\ p^2) \]

Now the proof of theorem 1.1 is complete.

3. Proof of theorem 1.2

Proof of theorem 1.2 With the help of (2.4), we have

(3.1)\begin{align} \sum_{k=0}^{p-1}\frac{f_k}{2^k}& =\sum_{k=0}^{p-1}\frac{1}{2^k}\sum_{j=0}^{\lfloor k/2\rfloor}\binom{k+j}{3j}\binom{2j}j\binom{3j}j2^{k-2j}\nonumber\\ & =\sum_{j=0}^{(p-1)/2}\frac{\binom{2j}j\binom{3j}j}{4^j}\sum_{k=2j}^{p-1}\binom{k+j}{3j}. \end{align}

By loading the package Sigma in software Mathematica, we have the following identity:

\[ \sum_{k=2j}^{n-1}\binom{k+j}{3j}=\binom{n+j}{3j+1}. \]

Thus, replace $n$ by $p$ in the above identity and then substitute it into (3.1), we have

\[ \sum_{k=0}^{p-1}\frac{f_k}{2^k}=\sum_{j=0}^{(p-1)/2}\frac{\binom{2j}j\binom{3j}j}{4^j}\binom{p+j}{3j+1}. \]

Hence, we immediately obtain the following result by lemma 2.3:

(3.2)\begin{equation} \sum_{k=0}^{p-1}\frac{f_k}{2^k}\equiv p\sum_{j=0, j\neq{(p-1)}/3}^{(p-1)/2}\frac{\binom{2j}j}{4^j}\frac{1-pH_{2j}+pH_j}{(3j+1)}+S_1\ ({\rm{mod}}\ p^3), \end{equation}

where

\[ S_1=\frac{\binom{\frac{2p-2}3}{\frac{p-1}3}\binom{p-1}{\frac{p-1}3}\binom{p+\frac{p-1}3}{p}}{4^{{(p-1)}/3}}=\binom{-\frac12}{\frac{p-1}3}\binom{p-1}{\frac{p-1}3}\binom{p+\frac{p-1}3}{p}. \]

It is easy to verify that

\begin{align*} & p\sum_{j=0, j\neq{(p-1)}/3}^{(p-1)/2}\frac{\binom{2j}j}{4^j}\frac{1-pH_{2j}+pH_j}{(3j+1)}\\ & \equiv p\sum_{j=0, j\neq{(p-1)}/3}^{(p-1)/2}\frac{\binom{\frac{p-1}2}j({-}1)^j(1-pH_{2j}+pH_j)}{(3j+1)\left(1-p\sum_{r=1}^j\frac1{2r-1}\right)}\\ & \equiv p\sum_{j=0}^{(p-1)/2}\frac{\binom{\frac{p-1}2}j({-}1)^j\left(1+\frac{p}2H_j\right)}{(3j+1)}-S_2\ ({\rm{mod}}\ p^3), \end{align*}

where

\[ S_2= \binom{\frac{p-1}2}{\frac{p-1}3}\left(1+\frac{p}2H_{{(p-1)}/3}\right). \]

So,

(3.3)\begin{equation} \sum_{k=0}^{p-1}\frac{f_k}{2^k}\equiv p\sum_{j=0}^{(p-1)/2}\frac{\binom{\frac{p-1}2}j({-}1)^j\left(1+\frac{p}2H_j\right)}{(3j+1)}+S_1-S_2\ ({\rm{mod}}\ p^3). \end{equation}

It is easy to see that

(3.4)\begin{equation} \frac{2p}{3p-1}\frac{(1)_{{(p-1)}/2}}{(\frac13)_{{(p-1)}/2}}=\frac{(\frac{p-1}2)!}{\frac13\cdots(\frac{p}3-1)(\frac{p}3+1)\cdots(\frac{p}3+\frac{p-1}6)}\equiv\binom{\frac{p-1}2}{\frac{p-1}3}\ ({\rm{mod}}\ p). \end{equation}

On the other hand, we have

\begin{align*} \sum_{k=0}^{p-1}\frac{f_k}{({-}4)^k}& =\sum_{k=0}^{p-1}\frac1{({-}4)^k}\sum_{j=0}^k\binom{k+2j}{3j}\binom{2j}j\binom{3j}j({-}4)^{k-j}\\ & =\sum_{j=0}^{p-1}\frac{\binom{2j}{j}\binom{3j}j}{({-}4)^{j}}\sum_{k=j}^{p-1}\binom{k+2j}{3j}=\sum_{j=0}^{p-1}\frac{\binom{2j}{j}\binom{3j}j}{({-}4)^{j}}\binom{p+2j}{3j+1}. \end{align*}

So by lemma 2.5 and the fact that for each $0\leq k\leq (p-1)/2$,

\[ \frac{\binom{2k}k}{({-}4)^k}\equiv\frac{\binom{\frac{p-1}2}k}{(1-p\sum_{j=1}^k\frac1{2j-1})}\ ({\rm{mod}}\ p^2), \]

and for each $(p+1)/2\leq j\leq p-1$,

\[ j\binom{2j}j\binom{2p-2j}{p-j}\equiv2p\ ({\rm{mod}}\ p^2), \]

we have the following congruence modulo $p^3$:

\begin{align*} & \sum_{k=0}^{p-1}\frac{f_k}{({-}4)^k}-S_3\equiv p\sum_{j=0\atop j\neq{(p-1)}/3}^{{(p-1)}/2}\frac{\binom{2j}{j}(1+pH_{2j}-pH_j)}{(3j+1)4^{j}}+2p\sum_{j={(p+1)}/2}^{p-1}\frac{\binom{2j}{j}}{(3j+1)4^{j}}\\ & \quad\equiv\sum_{j=0\atop j\neq{(p-1)}/3}^{{(p-1)}/2}\frac{p({-}1)^j\binom{\frac{p-1}2}{j}\left(1+2pH_{2j}-\frac32pH_j\right)}{3j+1}+\sum_{j={(p+1)}/2}^{p-1}\frac{4p^2 }{4^{j}(3j+1)j\binom{2p-2j}{p-j}}\\ & \quad\equiv\sum_{j=0}^{{(p-1)}/2}\frac{p({-}1)^j\binom{\frac{p-1}2}{j}\left(1+2pH_{2j}-\frac32pH_j\right)}{3j+1}+\sum_{j=1}^{{(p-1)}/2}\frac{p^24^j }{(3j-1)j\binom{2j}{j}}-S_4, \end{align*}

where

\begin{align*} & S_3=\frac{\binom{\frac{2p-2}3}{\frac{p-1}3}\binom{p-1}{\frac{p-1}3}\binom{p+\frac{2p-2}{3}}{p}}{({-}4)^{\frac{p-1}3}}=\binom{-\frac12}{\frac{p-1}3}\binom{p-1}{\frac{p-1}3}\binom{p+\frac{2p-2}{3}}{p},\\ & S_4=\binom{\frac{p-1}2}{\frac{p-1}3}\left(1+2pH_{{(2p-2)}/3}-\frac32pH_{{(p-1)}/3}\right). \end{align*}

Hence, we have

(3.5)\begin{align} & \sum_{k=0}^{p-1}\frac{f_k}{({-}4)^k}-\sum_{k=0}^{p-1}\frac{f_k}{2^k}\nonumber\\ & \quad\equiv2p^2\sum_{j=0}^{{(p-1)}/2}\frac{\binom{\frac{p-1}2}{j}({-}1)^j(H_{2j}-H_j)}{3j+1}+S_5+\sum_{j=1}^{{(p-1)}/2}\frac{p^24^j }{(3j-1)j\binom{2j}{j}}\ ({\rm{mod}}\ p^3), \end{align}

where

\[ S_5=S_3-S_4+S_2-S_1. \]

By Sigma, we can find and prove the following identity:

(3.6)\begin{align} & \sum_{j=0}^{n}\frac{2\binom{n}{j}({-}1)^j(H_{2j}-H_j)}{3j+1}\nonumber\\ & \quad=\frac1{3n+1}\prod_{k=1}^n\frac{3k}{3k-2}\left(\sum_{k=1}^n\frac1k\prod_{j=1}^k\frac{3j-2}{3j}-\sum_{k=1}^n\frac1k\prod_{j=1}^k\frac{2(3j-2)}{3(2j-1)}\right)\nonumber\\ & \quad=\frac{(1)_{n}}{(3n+1)\left(\frac13\right)_n}\left(\sum_{k=1}^n\frac{\left(\frac13\right)_k}{k(1)_k}-\sum_{k=1}^n\frac{\left(\frac13\right)_k}{k\left(\frac12\right)_k}\right). \end{align}

In view of [Reference Sun17, lemma 3.1] and lemma 2.2, we have

(3.7)\begin{equation} \sum_{k=1}^{{(p-1)}/2}\frac{\left(\frac13\right)_k}{k(1)_k}=\sum_{k=1}^{{(p-1)}/2}\frac{\binom{-1/3}{k}}{k\binom{-1}k}\equiv\frac32q_p(3)-\frac{3p}4q^2_p(3) -\frac{p}3\sum_{k=1}^{{(p-1)}/3}\frac{4^k}{k^2\binom{2k}k}\ ({\rm{mod}}\ p^2).\end{equation}
(3.8)\begin{align} \sum_{k=1}^{{(p-1)}/2}\frac{\left(\frac13\right)_k}{k\left(\frac12\right)_k}& =\sum_{k=1}^{{(p-1)}/2}\frac{\binom{-1/3}{k}}{k\binom{-1/2}k}\equiv\frac{4p}3({-}1)^{{(p-1)}/2}E_{p-3}+\frac32q_p(3)-\frac{3p}4q^2_p(3)\nonumber\\ & \quad-\frac{2p}3({-}1)^{{(p-1)}/2}\sum_{k=1}^{{(p-1)}/3}\frac{4^k}{(2k-1)k\binom{2k}k}\ ({\rm{mod}}\ p^2). \end{align}

It is easy to check that

(3.9)\begin{equation} \sum_{k=1}^{{(p-1)}/3}\frac{4^k}{(2k-1)k\binom{2k}k}=2\sum_{k=1}^{{(p-1)}/3}\frac{4^k}{(2k-1)\binom{2k}k}-\sum_{k=1}^{{(p-1)}/3}\frac{4^k}{k\binom{2k}k}. \end{equation}

And by [Reference Sury, Wang and Zhao21, (6)], we have

(3.10)\begin{gather} \frac1{\binom{n+1+k}k}=(n+1)\sum_{r=0}^n\binom{n}r({-}1)^r\frac{1}{k+r+1}. \end{gather}
(3.11)\begin{align} & 2\sum_{k=1}^{{(p-1)}/3}\frac{4^k}{(2k-1)\binom{2k}k}\equiv2\sum_{k=1}^{{(p-1)}/3}\frac{({-}1)^k}{(2k-1)\binom{\frac{p-1}2}k}\nonumber\\& \quad\equiv({-}1)^{{(p+1)}/2}\sum_{k={(p-1)}/6}^{{(p-3)}/2}\frac{({-}1)^k}{(k+1)\binom{\frac{p-1}2}{k}}\nonumber\\ & \quad=({-}1)^{{(p+1)}/2}\left(\sum_{k=0}^{{(p-3)}/2}\frac{({-}1)^k}{(k+1)\binom{\frac{p-1}2}{k}}-\sum_{k=0}^{{(p-7)}/6}\frac{({-}1)^k}{(k+1)\binom{\frac{p-1}2}{k}}\right)\ ({\rm{mod}}\ p). \end{align}

By Sigma, we find the following identity which can be proved by induction on $n$:

\[ \sum_{k=0}^{n}\frac{({-}1)^k}{(k+1)\binom{n}{k}}=\frac{2({-}1)^n-1}{n+1}-(n+1)H_n^{(2)}-2(n+1)\sum_{k=1}^n\frac{({-}1)^k}{k^2}. \]

So by setting $n=(p-1)/2$ in the above identity and with lemma 2.2, we have

(3.12)\begin{equation} \sum_{k=0}^{{(p-3)}/2}\frac{({-}1)^k}{(k+1)\binom{\frac{p-1}2}{k}}\equiv2\left(({-}1)^{{(p-1)}/2}-1\right)-({-}1)^{{(p-1)}/2}2E_{p-3}\ ({\rm{mod}}\ p). \end{equation}

And by (3.10), we have

\begin{align*} & \sum_{k=0}^{{(p-7)}/6}\frac{({-}1)^k}{(k+1)\binom{\frac{p-1}2}{k}}\equiv\sum_{k=0}^{{(p-7)}/6}\frac{1}{(k+1)\binom{\frac{p-1}2+k}{k}}\\ & \quad=\sum_{k=0}^{{(p-7)}/6}\frac{1}{k+1}\frac{p-1}2\sum_{r=0}^{{(p-3)}/2}\binom{\frac{p-3}2}{r}({-}1)^r\frac1{k+r+1}\\ & \quad\equiv{-}\frac12\sum_{k=1}^{{(p-1)}/6}\frac{1}{k}\sum_{r=0}^{{(p-3)}/2}\binom{\frac{p-3}2}{r}({-}1)^r\frac1{k+r}\\ & \quad={-}\frac12H_{{(p-1)}/6}^{(2)}-\frac12\sum_{r=1}^{{(p-3)}/2}\frac{({-}1)^r}r\binom{\frac{p-3}2}r\sum_{k=1}^{{(p-1)}/6}\left(\frac1k-\frac1{k+r}\right)\ ({\rm{mod}}\ p). \end{align*}

It is easy to check that

\[ H_{{(p-1)}/6}-\sum_{k=1}^{{(p-1)}/6}\frac1{k+r}\equiv{-}\sum_{k=1}^r\frac1{k(6k-1)}\ ({\rm{mod}}\ p). \]

By Sigma again, we have

\[ \sum_{r=1}^{n}\frac{({-}1)^r}r\binom{n}r\sum_{k=1}^r\frac1{k(6k-1)}=H_n^{(2)}-\sum_{k=1}^n\frac{(1)_k}{k^2\left(\frac56\right)_k}. \]

So in view of lemma 2.2 and [Reference Yeung22], we have

\begin{align*} & \sum_{k=0}^{{(p-7)}/6}\frac{({-}1)^k}{(k+1)\binom{\frac{p-1}2}{k}}\\ & \quad\equiv\frac{({-}1)^{{(p-1)}/2}}{x}-2-\frac54\left(\frac{p}3\right)B_{p-2}\left(\frac13\right)-\frac12\sum_{k=1}^{{(p-1)}/2}\frac{({-}1)^k}{k^2\binom{-\frac56}{k}}\ ({\rm{mod}}\ p). \end{align*}

Thus, by (3.10), we have

\begin{align*} & \sum_{k=1}^{{(p-1)}/2}\frac{({-}1)^k}{k^2\binom{-\frac56}{k}}={-}\frac65\sum_{k=1}^{{(p-1)}/2}\frac{({-}1)^k}{k\binom{-\frac{11}6}{k-1}}\equiv\frac65\sum_{k=0}^{{(p-3)}/2}\frac{({-}1)^k}{(k+1)\binom{\frac{5p-11}6}{k}}\\ & \quad=\frac65\sum_{k=0}^{{(p-3)}/2}\frac{1}{(k+1)\binom{\frac{p+5}6+k}{k}}=\frac65\sum_{k=0}^{{(p-3)}/2}\frac1{k+1}\frac{p+5}6\sum_{r=0}^{{(p-1)}/6}({-}1)^r\binom{\frac{p-1}6}r\frac1{k\!+\!1\!+\!r}\\ & \quad\equiv\sum_{k=1}^{{(p-1)}/2}\frac1{k}\sum_{r=0}^{{(p-1)}/6}({-}1)^r\binom{\frac{p-1}6}r\frac1{k+r}\\ & \quad=H_{{(p-1)}/2}^{(2)}+\sum_{r=1}^{{(p-1)}/6}\frac{({-}1)^r}r\binom{\frac{p-1}6}r\sum_{k=1}^{{(p-1)}/2}\left(\frac1k-\frac1{k+r}\right)\ ({\rm{mod}}\ p). \end{align*}

Also it is easy to see that

\[ H_{{(p-1)}/2}-\sum_{k=1}^{{(p-1)}/2}\frac1{k+r}\equiv{-}\sum_{k=1}^r\frac1{k(2k-1)}\ ({\rm{mod}}\ p). \]

And by Sigma, we have

\[ \sum_{r=1}^{n}\frac{({-}1)^r}r\binom{n}r\sum_{k=1}^r\frac1{k(2k-1)}=H_n^{(2)}-\sum_{k=1}^n\frac{4^k}{k^2\binom{2k}k}. \]

So in view of lemma 2.2, we have

\[ \sum_{k=1}^{{(p-1)}/2}\frac{({-}1)^k}{k^2\binom{-\frac56}{k}}\equiv\sum_{k=1}^{{(p-1)}/6}\frac{4^k}{k^2\binom{2k}k}-\frac52\left(\frac{p}3\right)B_{p-2}\left(\frac13\right)\ ({\rm{mod}}\ p). \]

Hence,

\[ \sum_{k=0}^{{(p-7)}/6}\frac{({-}1)^k}{(k+1)\binom{\frac{p-1}2}{k}}\equiv\frac{({-}1)^{{(p-1)}/2}}{x}-2-\frac12\sum_{k=1}^{{(p-1)}/6}\frac{4^k}{k^2\binom{2k}k}\ ({\rm{mod}}\ p). \]

This, with (3.11) and (3.12) yields

(3.13)\begin{align} & 2\sum_{k=1}^{{(p-1)}/3}\frac{4^k}{(2k-1)\binom{2k}k}\nonumber\\ & \quad\equiv{-}2+\frac1x+2E_{p-3}-\frac12({-}1)^{{(p-1)}/2}\sum_{k=1}^{{(p-1)}/6}\frac{4^k}{k^2\binom{2k}k}\ ({\rm{mod}}\ p). \end{align}

By Sigma, we find the following identity which can be proved by induction on $n$:

(3.14)\begin{equation} \sum_{k=1}^n\frac{4^k}{k\binom{2k}k}={-}2+2\frac{4^n}{\binom{2n}n}. \end{equation}

So in view of [Reference Yeung22], we have

\[ \sum_{k=1}^{{(p-1)}/3}\frac{4^k}{k\binom{2k}k}\equiv{-}2+\frac{2}{\binom{\frac{p-1}2}{\frac{p-1}3}}\equiv{-}2+\frac1x\ ({\rm{mod}}\ p). \]

This, with (3.9) and (3.13) yields

\[ \sum_{k=1}^{{(p-1)}/3}\frac{4^k}{(2k-1)k\binom{2k}k}\equiv2E_{p-3}-\frac12({-}1)^{{(p-1)}/2}\sum_{k=1}^{{(p-1)}/6}\frac{4^k}{k^2\binom{2k}k}\ ({\rm{mod}}\ p). \]

Thus, with (3.8) we have

(3.15)\begin{equation} \sum_{k=1}^{{(p-1)}/2}\frac{\left(\frac13\right)_k}{k\left(\frac12\right)_k}\equiv\frac32q_p(3)-\frac{3p}4q^2_p(3)+\frac{p}3\sum_{k=1}^{{(p-1)}/6}\frac{4^k}{k^2\binom{2k}k}\ ({\rm{mod}}\ p^2).\end{equation}

So by (3.7), we have

\[ \sum_{k=1}^{{(p-1)}/2}\frac{\left(\frac13\right)_k}{k(1)_k}-\sum_{k=1}^{{(p-1)}/2}\frac{\left(\frac13\right)_k}{k\left(\frac12\right)_k}\equiv{-}\frac{p}3\left(\sum_{k=1}^{{(p-1)}/3}\frac{4^k}{k^2\binom{2k}k}+\sum_{k=1}^{{(p-1)}/6}\frac{4^k}{k^2\binom{2k}k}\right)\ ({\rm{mod}}\ p^2). \]

Therefore, by (3.6) and (3.4), we deduce

(3.16)\begin{align} & 2p^2\sum_{j=0}^{{(p-1)}/2}\frac{\binom{\frac{p-1}2}{j}({-}1)^j(H_{2j}-H_j)}{3j+1}\nonumber\\ & \quad\equiv{-}\frac{p^2}3\binom{\frac{p-1}2}{\frac{p-1}3}\left(\sum_{k=1}^{{(p-1)}/3}\frac{4^k}{k^2\binom{2k}k}+\sum_{k=1}^{{(p-1)}/6}\frac{4^k}{k^2\binom{2k}k}\right)\ ({\rm{mod}}\ p^3). \end{align}

Now, we evaluate the second sum on the right-hand side of (3.5). It is easy to see

(3.17)\begin{equation} \sum_{j=1}^{{(p-1)}/2}\frac{4^j }{(3j-1)j\binom{2j}{j}}=3\sum_{j=1}^{{(p-1)}/2}\frac{4^j }{(3j-1)\binom{2j}{j}}-\sum_{j=1}^{{(p-1)}/2}\frac{4^j }{j\binom{2j}{j}}. \end{equation}

By (3.14), we have

(3.18)\begin{equation} \sum_{j=1}^{{(p-1)}/2}\frac{4^j }{j\binom{2j}{j}}\equiv{-}2+2({-}1)^{{(p-1)}/2}\ ({\rm{mod}}\ p). \end{equation}

Now we consider the first sum of the right-hand side in (3.17):

\[ \sum_{j=1}^{{(p-1)}/2}\frac{4^j }{(3j-1)\binom{2j}{j}}=\sum_{j=1}^{{(p-1)}/3}\frac{4^j }{(3j-1)\binom{2j}{j}}+\sum_{j={(p+2)}/3}^{{(p-1)}/2}\frac{4^j }{(3j-1)\binom{2j}{j}}. \]

The following identity is very important to us:

(3.19)\begin{equation} \sum_{k=1}^n\frac{4^k}{(k+n)\binom{2k}k}={-}2+2\frac{4^n}{\binom{2n}n}-\frac{n\binom{2n}n}{4^n}\sum_{k=1}^n\frac{4^k}{k^2\binom{2k}k}. \end{equation}

This, with [Reference Yeung22] yields

(3.20)\begin{align} & 3\sum_{j=1}^{{(p-1)}/3}\frac{4^j }{(3j-1)\binom{2j}{j}}\equiv\sum_{j=1}^{{(p-1)}/3}\frac{4^j }{\left(j+\frac{p-1}3\right)\binom{2j}{j}}\nonumber\\ & \quad\equiv{-}2+\frac{2}{\binom{-1/2}{\frac{p-1}3}}+\frac13\binom{-1/2}{\frac{p-1}3}\sum_{k=1}^{{(p-1)}/3}\frac{4^k}{k^2\binom{2k}k}\nonumber\\ & \quad\equiv{-}2+\frac1x+\frac13\binom{\frac{p-1}2}{\frac{p-1}3}\sum_{k=1}^{{(p-1)}/3}\frac{4^k}{k^2\binom{2k}k}\ ({\rm{mod}}\ p). \end{align}

And by (3.19), we have

(3.21)\begin{align} & 3\sum_{j={(p+2)}/3}^{{(p-1)}/2}\frac{4^j}{(3j-1)\binom{2j}{j}}\equiv3\sum_{j=0}^{{(p-7)}/6}\frac{({-}1)^{{(p-1)}/2-j}}{(3(\frac{p-1}2-j)-1)\binom{\frac{p-1}2}{j}}\nonumber\\ & \quad\equiv6({-}1)^{{(p+1)}/2}\sum_{j=0}^{{(p-7)}/6}\frac{4^j }{(6j+5)\binom{2j}{j}}\equiv({-}1)^{{(p+1)}/2}\sum_{j=0}^{{(p-7)}/6}\frac{({-}1)^j }{(j+\frac{p+5}6)\binom{\frac{p-1}2}{j}}\nonumber\\ & \quad\equiv\frac65({-}1)^{{(p+1)}/2}+({-}1)^{{(p+1)}/2}\sum_{j=1}^{{(p+5)}/6}\frac{4^j }{(j+\frac{p+5}6)\binom{2j}{j}}+\frac3{\binom{\frac{p-1}2}{\frac{p-1}3}}\ ({\rm{mod}}\ p). \end{align}

In view of (3.19) and [Reference Yeung22], we have

\begin{align*} & \sum_{j=1}^{{(p+5)}/6}\frac{4^j }{(j+\frac{p+5}6)\binom{2j}{j}}\equiv{-}\frac{16}5+\frac{5({-}1)^{{(p-1)}/6}}{2x}\\ & \quad -\frac{({-}1)^{{(p-1)}/6}}{3}\binom{\frac{p-1}2}{\frac{p-1}3}\sum_{k=1}^{{(p-1)}/6}\frac{4^k}{k^2\binom{2k}k}\ ({\rm{mod}}\ p). \end{align*}

This, with (3.21) yields

\[ 3\sum_{j={(p+2)}/3}^{{(p-1)}/2}\frac{4^j}{(3j-1)\binom{2j}{j}}\equiv2({-}1)^{{(p-1)}/2}-\frac1x+\frac13\binom{\frac{p-1}2}{\frac{p-1}3}\sum_{k=1}^{{(p-1)}/6}\frac{4^k}{k^2\binom{2k}k}\ ({\rm{mod}}\ p). \]

Combining this with (3.20), we have

\begin{align*} & 3\sum_{j=1}^{{(p-1)}/2}\frac{4^j }{(3j-1)\binom{2j}{j}}\\ & \quad\equiv{-}2+2({-}1)^{{(p-1)}/2}+\frac13\binom{\frac{p-1}2}{\frac{p-1}3}\left(\sum_{k=1}^{{(p-1)}/3}\frac{4^k}{k^2\binom{2k}k}+\sum_{k=1}^{{(p-1)}/6}\frac{4^k}{k^2\binom{2k}k}\right)\ ({\rm{mod}}\ p). \end{align*}

Thus, by (3.17) and (3.18), we have

\[ \sum_{j=1}^{{(p-1)}/2}\frac{4^j }{(3j-1)j\binom{2j}{j}}\equiv\frac13\binom{\frac{p-1}2}{\frac{p-1}3}\left(\sum_{k=1}^{{(p-1)}/3}\frac{4^k}{k^2\binom{2k}k}+\sum_{k=1}^{{(p-1)}/6}\frac{4^k}{k^2\binom{2k}k}\right)\ ({\rm{mod}}\ p). \]

This, with (3.5) and (3.16) yields

(3.22)\begin{equation} \sum_{k=0}^{p-1}\frac{f_k}{({-}4)^k}-\sum_{k=0}^{p-1}\frac{f_k}{2^k}\equiv S_5\ ({\rm{mod}}\ p^3). \end{equation}

While

\begin{align*} S_5& =\binom{-\frac12}{\frac{p-1}3}\binom{p-1}{\frac{p-1}3}\left(\binom{p+\frac{2p-2}3}{\frac{2p-2}3}-\binom{p+\frac{p-1}3}{\frac{p-1}3}\right)\\ & \quad+2p\binom{\frac{p-1}2}{\frac{p-1}3}\left(H_{{(p-1)}/3}-H_{{(2p-2)}/3}\right). \end{align*}

It is easy to check that

\[ \binom{p+\frac{2p-2}3}{\frac{2p-2}3}\equiv1+pH_{{(2p-2)}/3}+\frac{p^2}2\left(H_{{(2p-2)}/3}^2-H_{{(2p-2)}/3}^{(2)}\right)\ ({\rm{mod}}\ p^3) \]

and

\[ \binom{p+\frac{p-1}3}{\frac{p-1}3}\equiv1+pH_{{(p-1)}/3}+\frac{p^2}2\left(H_{{(p-1)}/3}^2-H_{{(p-1)}/3}^{(2)}\right)\ ({\rm{mod}}\ p^3). \]

So by lemma 2.2 and the fact that $H_{p-1-k}^{(2)}\equiv -H_k^{(2)}\ ({\rm {mod}}\ p)$ for each $0\leq k\leq p-1$, we have

\begin{align*} \binom{p+\frac{2p-2}3}{\frac{2p-2}3}-\binom{p+\frac{p-1}3}{\frac{p-1}3}& \equiv p(H_{{(2p-2)}/3}-H_{{(p-1)}/3})+\frac{p^2}2(H_{{(p-1)}/3}^{(2)}-H_{{(2p-2)}/3}^{(2)})\\ & \equiv p^2\left(\frac{p}3\right)B_{p-2}\left(\frac13\right)\ ({\rm{mod}}\ p^3) \end{align*}

and

\[ 2p\left(H_{{(p-1)}/3}-H_{{(2p-2)}/3}\right)\equiv{-}p^2\left(\frac{p}3\right)B_{p-2}\left(\frac13\right)\ ({\rm{mod}}\ p^3). \]

So by $\binom {-\frac 12}{\frac {p-1}3}\equiv \binom {\frac {p-1}2}{\frac {p-1}3}\ ({\rm {mod}}\ p)$ and $\binom {p-1}{\frac {p-1}3}\equiv (-1)^{\frac {p-1}3}=1\ ({\rm {mod}}\ p)$, we can immediately obtain that

\[ S_5\equiv0\ ({\rm{mod}}\ p^3). \]

This, with (3.22) yields

\[ \sum_{k=0}^{p-1}\frac{f_k}{({-}4)^k}\equiv\sum_{k=0}^{p-1}\frac{f_k}{2^k}\ ({\rm{mod}}\ p^3). \]

Now the proof of theorem 1.2 is complete.

Acknowledgements

The authors would like to thank the anonymous referee for careful reading of this manuscript and valuable comments, which make the paper more readable. This research was supported by the National Natural Science Foundation of China (Grant 12001288).

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