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}

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.

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.

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.