Let \(n\ge 3\) be an integer and p be a prime with \(p\equiv 1\pmod {n}\). In this paper, we show that

$$\begin{aligned} {}_nF_{n-1}\bigg [\begin{array}{llll} \frac{n-1}{n}&{}\frac{n-1}{n}&{}\ldots &{}\frac{n-1}{n}\\ &{}1&{}\ldots &{}1\end{array}\bigg | \, 1\bigg ]_{p-1}\equiv -\Gamma _p\bigg (\frac{1}{n}\bigg )^n\pmod {p^3}, \end{aligned}$$

where the truncated hypergeometric series

$$\begin{aligned} {}_nF_{n-1}\bigg [\begin{array}{llll} x_1&{}x_2&{}\ldots &{}x_n\\ &{}y_1&{}\cdots &{}y_{n-1}\end{array}\bigg | \, z\bigg ]_m=\sum _{k=0}^{m}\frac{z^k}{k!}\prod _{j=0}^{k-1}\frac{(x_1+j)\cdots (x_{n}+j)}{(y_1+j)\cdots (y_{n-1}+j)} \end{aligned}$$

and \(\Gamma _p\) denotes the p-adic Gamma function. This confirms a conjecture of Deines et al. (Hypergeometric Series, Truncated Hypergeometric Series, and Gaussian Hypergeometric Functions, Directions in Number Theory, vol. 3, pp. 125–159. Assoc.WomenMath. Ser., Springer, New York, 2016). Furthermore, under the same assumptions, we also prove that

$$\begin{aligned} p^n\cdot {}_{n+1}F_{n}\bigg [\begin{matrix} 1&{}1&{}\ldots &{}1\\ &{}\frac{n+1}{n}&{}\ldots &{}\frac{n+1}{n}\end{matrix}\bigg | \, 1\bigg ]_{p-1} \equiv -\Gamma _p\left( \frac{1}{n}\right) ^n\pmod {p^3}, \end{aligned}$$

which solves another conjecture in [5].

中文翻译：

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关于截断超几何级数的超同余

让\(n\ge 3\)是一个整数，而p是一个质数，其中\(p\equiv 1\pmod {n}\)。在本文中，我们表明

$$\begin{aligned} {}_nF_{n-1}\bigg [\begin{array}{llll} \frac{n-1}{n}&{}\frac{n-1}{n}& {}\ldots &{}\frac{n-1}{n}\\ &{}1&{}\ldots &{}1\end{array}\bigg | \, 1\bigg ]_{p-1}\equiv -\Gamma _p\bigg (\frac{1}{n}\bigg )^n\pmod {p^3}, \end{aligned}$$

其中截断的超几何级数

$$\begin{aligned} {}_nF_{n-1}\bigg [\begin{array}{llll} x_1&{}x_2&{}\ldots &{}x_n\\ &{}y_1&{}\cdots &{ }y_{n-1}\end{array}\bigg | \, z\bigg ]_m=\sum _{k=0}^{m}\frac{z^k}{k!}\prod _{j=0}^{k-1}\frac{(x_1 +j)\cdots (x_{n}+j)}{(y_1+j)\cdots (y_{n-1}+j)} \end{aligned}$$

和\（\伽玛_p \）表示p进制伽玛功能。这证实了 Deines 等人的猜想。（超几何级数、截断超几何级数和高斯超几何函数，数论方向，第 3 卷，第 125-159 页。Assoc.WomenMath. Ser.，Springer，纽约，2016 年）。此外，在相同的假设下，我们还证明

$$\begin{aligned} p^n\cdot {}_{n+1}F_{n}\bigg [\begin{matrix} 1&{}1&{}\ldots &{}1\\ &{}\ frac{n+1}{n}&{}\ldots &{}\frac{n+1}{n}\end{matrix}\bigg | \, 1\bigg ]_{p-1} \equiv -\Gamma _p\left( \frac{1}{n}\right) ^n\pmod {p^3}, \end{aligned}$$

这解决了[5]中的另一个猜想。