Guo-Shuai Mao

In this paper, we mainly prove a congruence conjecture of Z.-W. Sun \cite{Sjnt}: Let $p>5$ be a prime. Then $$ \sum_{k=(p+1)/2}^{p-1}\frac{\binom{2k}k^2}{k16^k}\equiv-\frac{21}2H_{p-1}\pmod{p^4}, $$ where $H_n$ denotes the $n$-th harmonic number.

In this paper, we prove two conjectures of Z.-W. Sun: $$2n\binom{2n}n\big|\sum_{k=0}^{n-1}(3k+1)\binom{2k}k^3{16}^{n-1-k}\ \mbox{for}\ \mbox{all}\ n=2,3,\cdots,$$ and $$\sum_{k=0}^{(p-1)/2}\frac{3k+1}{16^k}\binom{2k}{k}^3\equiv p+2\left(\frac{-1}{p}\right)p^3E_{p-3}\pmod{p^4},$$ where $p>3$ is a prime and $E_0,E_1,E_2,\cdots$ are Euler numbers.

Let $n$ and $r$ be positive integers. Define the numbers $S_n^{(r)}$ by $S_n^{(r)}=\sum_{k=0}^n\binom{n}{k}^2\binom{2k}{k}(2k+1)^r.$ In this paper we prove some conjectures of Guo and Liu which extend some conjectures of Z.-W. Sun \cite{Su1}, such as: There exist integers $a_{2r-1}$ and $b_r$, independent of $n$, such that $$a_{2r-1}\sum_{k=0}^{n-1}S_k^{(2r-1)}\equiv0\pmod{n^2}\ \mbox{and}\ b_r\sum_{k=0}^{n-1}kS_k^{(r)}\equiv0\pmod{n^2}.$$ By Zeilberger algorithm, we find that for all $0\leq j<n$, $$(2j+1)\binom{2j}j\sum_{k=j}^{n-1}(2k-j+1)\binom kj^2\equiv0\pmod{n^2}.$$

Let $p$ be an odd prime and $r\geq 1$. Suppose that $\alpha$ is a $p$-adic integer with $\alpha\equiv2a\pmod p$ for some $1\leq a<(p+r)/(2r+1)$. We confirm a conjecture of Sun and prove that $${}_{2r+1}F_{2r}\bigg[\begin{matrix}\alpha&\alpha&\ldots&\alpha\\ &1&\ldots&1\end{matrix}\bigg|\,1\bigg]_{p-1}\equiv0\pmod{p^2},$$ where the truncated hypergeometric series $$ {}_{q+1}F_{q}\bigg[\begin{matrix}x_0&x_1&\ldots&x_{q}\\ &y_1&\ldots&y_q\end{matrix}\bigg|\,z\bigg]_{n}:=\sum_{k=0}^n\frac{(x_0)_k(x_1)_k\cdots(x_q)_k}{(y_1)_k\cdot (y_q)_k}\cdot\frac{z^k}{k!}. $$