论文标题
在Z.-W。的两个猜想超级方面太阳
On two conjectural supercongruences of Z.-W. Sun
论文作者
论文摘要
在本文中，我们主要通过使用以下身份$$证明了太阳的两个猜想\ sum_ {k = 0}^n \ binom {2k} {k}^2 \ binom {2n-2k} {n-k}^2 = 16^n \ sum_ {k = 0}^n \ frac {\ binom {n+k} {k} \ binom {n} {k} {k} {k} \ binom {2k} {k} {k}^2} {（ - 16）^k} $$来自$ {} _ 4f_3 $超几何变换。对于任何prime $ p> 3 $，我们证明\ begin {chater*} \ sum_ {n = 0}^{p-1} \ frac {n+1} {8^n} \ sum_ {k = 0}^n \ binom {2k} {2k} {k}^2 \ bin om {2n-2k} {n-k}^2 \ equiv（-1）^{（p-1）/2} p+5p^3e_ {p-3} \ pmod {p^4}，\\ \ sum_ {n = 0}^{p-1} \ frac {2n+1} {（ - 16）^n} \ sum_ {k = 0}^n \ binom {2k} {k} {k}^2 \ binom {2n-2k} {n-k}^2 \ equiv（-1）^{（p-1）/2} p+3p^3e_ {p-3} \ pmod {p pmod {p^4}， \ end {chater*}其中$ e_ {p-3} $是$（p-3）$ th euler编号。
In this paper, we mainly prove two conjectural supercongruences of Sun by using the following identity $$ \sum_{k=0}^n\binom{2k}{k}^2\binom{2n-2k}{n-k}^2=16^n\sum_{k=0}^n\frac{\binom{n+k}{k}\binom{n}{k}\binom{2k}{k}^2}{(-16)^k} $$ which arises from a ${}_4F_3$ hypergeometric transformation. For any prime $p>3$, we prove that \begin{gather*} \sum_{n=0}^{p-1}\frac{n+1}{8^n}\sum_{k=0}^n\binom{2k}{k}^2\binom{2n-2k}{n-k}^2\equiv(-1)^{(p-1)/2}p+5p^3E_{p-3}\pmod{p^4},\\ \sum_{n=0}^{p-1}\frac{2n+1}{(-16)^n}\sum_{k=0}^n\binom{2k}{k}^2\binom{2n-2k}{n-k}^2\equiv(-1)^{(p-1)/2}p+3p^3E_{p-3}\pmod{p^4}, \end{gather*} where $E_{p-3}$ is the $(p-3)$th Euler number.
