论文标题
$ Q $ - 与Euler号码有关
$q$-Analogues of some supercongruences related to Euler numbers
论文作者
论文摘要
令$ e_n $为$ n $ -th欧拉号和$（a）_n = a（a+1）\ cdots（a+n-1）$ the Rising Factorial。令$ p> 3 $为素数。在2012年，Sun证明了$$ \ sum^{（p-1）/2} _ {k = 0}（ - 1）^k（4k+1）\ frac {（\ frac {1} {2} {2} {2} {2} {2} {2} _ k^3} _ k^3} k！ \ pmod {p^4}，$$，这是对范·哈姆（Van Hamme）著名超级企业的改进。在2016年，Chen，XIE，他确定了以下结果：$$ \ sum_ {k = 0}^{p-1}（ - 1）^k（3k+1）\ frac {（\ frac {1} {1} {2} {2} {2} {2} {2} {2}）_ k^3} {k！ p（-1）^{（P-1）/2}+p^3e_ {p-3} \ pmod {p^4}，$$最初由Sun猜想。在本文中，我们通过采用$ Q $ -WZ方法给出了上述两个超级企业的$ Q $ -Analogues。作为结论，我们提供了以下太阳的以下超容器的$ q $ -ANALOGUE： e_ {p-3} \ pmod {p^3}。 $$
Let $E_n$ be the $n$-th Euler number and $(a)_n=a(a+1)\cdots (a+n-1)$ the rising factorial. Let $p>3$ be a prime. In 2012, Sun proved the that $$ \sum^{(p-1)/2}_{k=0}(-1)^k(4k+1)\frac{(\frac{1}{2})_k^3}{k!^3} \equiv p(-1)^{(p-1)/2}+p^3E_{p-3} \pmod{p^4}, $$ which is a refinement of a famous supercongruence of Van Hamme. In 2016, Chen, Xie, and He established the following result: $$ \sum_{k=0}^{p-1}(-1)^k (3k+1)\frac{(\frac{1}{2})_k^3}{k!^3} 2^{3k} \equiv p(-1)^{(p-1)/2}+p^3E_{p-3} \pmod{p^4}, $$ which was originally conjectured by Sun. In this paper we give $q$-analogues of the above two supercongruences by employing the $q$-WZ method. As a conclusion, we provide a $q$-analogue of the following supercongruence of Sun: $$ \sum_{k=0}^{(p-1)/2}\frac{(\frac{1}{2})_k^2}{k!^2} \equiv (-1)^{(p-1)/2}+p^2 E_{p-3} \pmod{p^3}. $$
