$p$-adic analogues of hypergeometric identities
We prove some supercongruences for the truncated hypergeometric series.
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arXiv subjects
Guo-Shuai Mao
Publications and source records attributed to Guo-Shuai Mao.
On the divisibility of some truncated hypergeometric series
Let $p$ be an odd prime and $r\geq 1$. Suppose that $α$ is a $p$-adic integer with $α\equiv2a\pmod p$ for some $1\leq a<(p+r)/(2r+1)$. We confirm a conjecture of Sun and prove that $${}_{2r+1}F_{2r}\bigg[\begin{matrix}α&α&\ldots&α\\ &1&\ldots&1\end{matrix}\bigg|\,1\bigg]_{p-1}\equiv0\pmod{p^2},$$ where the truncated hypergeometric series $$ {}_{q+1}F_{q}\bigg[\begin{matrix}x_0&x_1&\ldots&x_{q}\\ &y_1&\ldots&y_q\end{matrix}\bigg|\,z\bigg]_{n}:=\sum_{k=0}^n\frac{(x_0)_k(x_1)_k\cdots(x_q)_k}{(y_1)_k\cdot (y_q)_k}\cdot\frac{z^k}{k!}. $$
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Supercongruences on some binomial sums involving Lucas sequences
In this paper, we confirm several conjectured congruences of Sun concerning the divisibility of binomial sums. For example, with help of a quadratic hypergeometric transformation, we prove that $$ \sum_{k=0}^{p-1}\binom{p-1}k\binom{2k}k^2\frac{P_k}{8^k}\equiv0\pmod{p^2} $$ for any prime $p\equiv 7\mod{8}$, where $P_k$ is the $k$-th Pell number. Further, we also propose three new congruences of the same type.
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Two congruences involving harmonic numbers with applications
The harmonic numbers $H_n=\sum_{0 3$ be a prime. With helps of some combinatorial identities, we establish the following two new congruences: $$\sum_{k=1}^{p-1}\frac{\binom{2k}k}kH_k\equiv\frac13\left(\frac p3\right)B_{p-2}\left(\frac13\right)\pmod{p}$$ and $$\sum_{k=1}^{p-1}\frac{\binom{2k}k}kH_{2k}\equiv\frac7{12}\left(\frac p3\right)B_{p-2}\left(\frac13\right)\pmod{p},$$ where $B_n(x)$ denotes the Bernoulli polynomial of degree $n$. As an application, we determine $\sum_{n=1}^{p-1}g_n$ and $\sum_{n=1}^{p-1}h_n$ modulo $p^3$, where $$g_n=\sum_{k=0}^n\binom nk^2\binom{2k}k\quad\mbox{and}\quad h_n=\sum_{k=0}^n\binom nk^2C_k$$ with $C_k=\binom{2k}k/(k+1)$.
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New congruences involving products of two binomial coefficients
Let $p>3$ be a prime and let $a$ be a positive integer. We show that if $p\equiv1\pmod 4$ or $a>1$ then $$\sum_{k=0}^{\lfloor\frac34p^a\rfloor}\frac{\binom{2k}k^2}{16^k}\equivł(\frac{-1}{p^a}\r)\pmod{p^3}$$ with $(-)$ the Jacobi symbol, which confirms a conjecture of Z.-W. Sun. We also establish the following new congruences: \begin{align*}\sum_{k=0}^{(p-1)/2}\frac{\binom{2k}k\binom{3k}k}{27^k}\equiv&ł(\frac p3\r)\frac{2^p+1}3\pmod{p^2}, \\\sum_{k=0}^{(p-1)/2}\frac{\binom{6k}{3k}\binom{3k}k}{(2k+1)432^k}\equiv&ł(\frac p3\r)\frac{3^p+1}4\pmod{p^2}, \\\sum_{k=0}^{(p-1)/2}\frac{\binom{4k}{2k}\binom{2k}k}{(2k+1)64^k}\equiv&ł(\frac{-1}p\r)2^{p-1}\pmod{p^2}. \end{align*} Note that in 2003 Rodriguez-Villeguez posed conjectures on $$\sum_{k=0}^{p-1}\frac{\binom{2k}k^2}{16^k},\ \sum_{k=0}^{p-1}\frac{\binom{2k}k\binom{3k}k}{27^k},\ \sum_{k=1}^{p-1}\frac{\binom{4k}{2k}\binom{2k}k}{64^k},\ \sum_{k=1}^{p-1}\frac{\binom{6k}{3k}\binom{3k}k}{432^k}$$ modulo $p^2$ which were later proved.
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Proof of a conjecture involving Sun polynomials
The Sun polynomials $g_n(x)$ are defined by \begin{align*} g_n(x)=\sum_{k=0}^n{n\choose k}^2{2k\choose k}x^k. \end{align*} We prove that, for any positive integer $n$, there hold \begin{align*} &\frac{1}{n}\sum_{k=0}^{n-1}(4k+3)g_k(x) \in\mathbb{Z}[x],\quad\text{and}\\ &\sum_{k=0}^{n-1}(8k^2+12k+5)g_k(-1)\equiv 0\pmod{n}. \end{align*} The first one confirms a recent conjecture of Z.-W. Sun, while the second one partially answers another conjecture of Z.-W. Sun. We give three different proofs of the former. One of them depends on the following congruence: $$ {m+n-2\choose m-1}{n\choose m}{2n\choose n}\equiv 0\pmod{m+n}\quad\text{for $m,n\geqslant 1$.} $$
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Proof of two supercongruences conjectured by Z.-W.Sun involving Catalan-Larcombe-French numbers
The harmonic numbers $H_n=\sum_{0 3$, the second one was conjectured by Z.-W. Sun in 2012. These two congruences are very important to prove the following conjectures of Z.W.Sun: For any old prime $p$, we have $$\sum_{k=0}^{p-1}\frac{P_k}{8^k}\equiv1+2\big(\frac{-1}p\big)p^2E_{p-3}\pmod{p^3}$$ and $$\sum_{k=0}^{p-1}\frac{P_k}{{16}^k}\equiv\big(\frac{-1}p\big)-p^2E_{p-3}\pmod{p^3},$$ where $P_n=\sum_{k=0}^n\frac{\binom{2k}k^2\binom{2(n-k)}{n-k}^2}{\binom nk}$ is the n-th Catalan-Larcombe-French number.
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