在研究对偶序列的过程中，Sun引入了多项式\[ D_n(x,y)=\sum_{k=0}^{n}{n\choose k}{x\choose k}y^k\text{ and } S_n(x,y)=\sum_{k=0}^{n}\binom{n}{k}\binom{x}{k}\binom{-1-x}{k} y^k. \]，并建立和证明了许多相关的同余式。在这里，我们通过确定\[ \sum_{k=0}^{p-1}D_k(x_1,y_1)D_k(x_2,y_2)\pmod p \text{ and } \sum_{k=0}^{p-1}S_k(x_1,y_1)S_k(x_2,y_2)\pmod p \]对任何奇素数$p$和$p$-adic整数$x_i,\ y_i$与$i\in\{1,2\}$ .考虑到二项式系数和Catalan数之间的直接联系，我们还刻画了\[ \sum_{n=0}^{p-1}\left(\sum_{k=0}^n {n \choose k} \frac{C_k}{a^k}\right)^2 \pmod {p}, \]，其中$C_k$表示$k$第卡塔兰数，$a\in\mathbb{Z}\setminus \{0\}$和$\gcd(a,p)=1$ .这些证实和概括了孙的一些论点。
英文摘要：
During the study of dual sequences, Sun introduced the polynomials \[ D_n(x,y)=\sum_{k=0}^{n}{n\choose k}{x\choose k}y^k\text{ and } S_n(x,y)=\sum_{k=0}^{n}\binom{n}{k}\binom{x}{k}\binom{-1-x}{k} y^k. \] Many related congruences have been established and conjectured by Sun. Here we generalize some of them by determining \[ \sum_{k=0}^{p-1}D_k(x_1,y_1)D_k(x_2,y_2)\pmod p \text{ and } \sum_{k=0}^{p-1}S_k(x_1,y_1)S_k(x_2,y_2)\pmod p \] for any odd prime $p$ and $p$-adic integers $x_i,\ y_i$ with $i\in\{1,2\}$. Considering the immediate connection between binomial coefficients and Catalan numbers, we also characterize \[ \sum_{n=0}^{p-1}\left(\sum_{k=0}^n {n \choose k} \frac{C_k}{a^k}\right)^2 \pmod {p}, \] where $C_k$ denotes the $k$th Catalan number, $a\in\mathbb{Z}\setminus \{0\}$ with $\gcd(a,p)=1$. These confirm and generalise some of Sun's conjectures.
Heaps of rhombic dodecahedra, catalan congruences on alternating sign matrices, and bases of the Temperley-Lieb algebra
LISN (UMR CNRS 9015), Université Paris Saclay, INRIA, CNRS（巴黎萨克雷大学）；Universitat de Barcelona & Centre de Recerca Matemàtica（巴塞罗那大学）；LIGM, Université Gustave Eiffel（巴黎东马恩拉瓦莱大学）
本文建立了中心二项式系数和Catalan数的新同余式。设$p$是素数，$a$是任意正整数。我们确定了$d=0,...,p^a$的$\sum_{k=0}^{p^a-1}\binom{2k}{k+d}$ mod $p^2$和$δ=0,1$的$\sum_{k=0}^{p^a-1}\binom{2k}{k+δ}$ mod $p^3$。我们还表明，$$C_n^{-1}\sum_{k=0}^{p^a-1}C_{p^an+k}=1-3(n+1)((p^a-1)/3) (mod p^2)$$对每个n= 0，1，2，.，其中$C_m$是加泰罗尼亚数字$\binom{2m}{m}/(m+1)$，（-）是勒让德符号。
让$p>5$是一个素数。受已知公式$\sum_{k=1}^\infty(-1)^k/(k^3\binom{2k}{k})=-2ζ(3)/5$和$\sum_{k=0}^\infty \binom{2k}{k}^2/((2k+1)16^k)=4G/π$的启发，$ (where $ G= \sum_{k=0}^\ infty（-1）^k/（2k+1）^2 $ is the Catalan constant), we show that $ $\sum_{k=1}^{(p-1)/2}\frac{(-1)^k}{k^3\binom{2k}{k}}\equiv-2B_{p-3}\pmod{p},$ $ $ $\sum_{k=(p+1)/2}^{p-1}\frac{\binom{2k}{k}^2}{(2k+1)16^k}\equiv-\frac 7{4}p^2B_{p-3}\pmod{p^3}$ $, and $ $\sum_{k=0}^{(p-3)/2}\frac{\binom{2k}{k}^2}{(2k+1)16^k} \equiv-2q_p(2)-pq_p(2)^2+\frac{5}{12}p^2B_{p-3}\pmod{p^3},$ $ where $ B_0，B_1，\ldots $ are Bernoulli numbers and $ q_p（2）$ is the Fermat q…
