涉及p进对偶序列的新同余式
New Congruences Involving $p$-adic dual sequences
AI总结：
本文研究涉及中心二项式系数与p进对偶序列的模$p^2$同余式，证明相关同余关系，建立新公式并证明Z.-W. Sun关于广义中心三项式系数的部分猜想。
AI中文摘要：
设$(a_n)_{n\geqslant 0}$为整数序列，其对偶序列$(a_n^*)_{n\geqslant 0}$定义为$a_n^*:= \sum_{k=0}^{n} \binom{n}{k}(-1)^k a_k$。设$p>3$为素数，本文主要研究涉及中心二项式系数和p进对偶序列的模$p^2$同余式。例如，证明对任意p进整数序列$(a_k)_{k\ge0}$，有$\sum^{(p-1)/2}_{k=0}\binom{2k}{k}^2\frac{a_{2k}}{16^k}\equiv\left( \frac{-1}{p}\right) \sum_{k=0}^{p-1}\frac{\mathcal{P}_{k}}{16 ^{k}}a_{k}^*\pmod{p^2}$，其中$(\mathcal{P}_n)_{n\ge0}$为Catalan-Larcombe-French数，满足$\mathcal{P}_0=1$、$\mathcal{P}_1=8$及递推式$n^2 \mathcal{P}_n = 8(3n^2-3n+1)\mathcal{P}_{n-1}-128(n-1)^2\mathcal{P}_{n-2} \\ (n\ge2)$。还建立了$\sum_{k=0}^{(p-1)/2}\binom{2k}{k}a_{2k}^*/4^k \pmod{p^2}$的新公式，进而证明了Z.-W. Sun关于广义中心三项式系数$T_{2k}(b,c)$（即$(x^2+bx+c)^{2k}$中$x^{2k}$的系数，$b,c$为整数）的部分猜想。
英文摘要：
Let $(a_n)_{n\geqslant 0}$ be a sequence of integers. Its dual sequence $(a_n^*)_{n\geqslant 0}$ is defined by \begin{equation*} a_n^* := \sum_{k=0}^{n} \binom{n}{k}(-1)^k a_k. \end{equation*} Let $p>3$ be a prime. In this paper we mainly investigate congruences modulo $p^2$ involving central binomial coefficients and $p$-adic dual sequences. For example, we prove that for any sequence $(a_k)_{k\ge0}$ of $p$-adic integers, \begin{align*} \sum^{(p-1)/2}_{k=0}\binom{2k}{k}^2\frac{a_{2k}}{16^k}\equiv\left( \frac{-1}{p}\right) \sum_{k=0}^{p-1}\frac{\mathcal{P}_{k}}{16 ^{k}}a_{k}^*\pmod{p^2}, \end{align*} where $(\mathcal{P}_n)_{n\ge0}$ are the Catalan--Larcombe--French numbers given by \begin{equation*} \mathcal{P}_0=1,\quad \mathcal{P}_1=8, \quad n^2 \mathcal{P}_n = 8(3n^2-3n+1)\mathcal{P}_{n-1}-128(n-1)^2\mathcal{P}_{n-2} \quad (n\ge2). \end{equation*} We also establish a new formula for $\sum_{k=0}^{(p-1)/2}\binom{2k}{k}a_{2k}^*/4^k \pmod{p^2}$ and as a consequence we confirm some conjectures of Z.-W. Sun \cite{Sun2014CANT} on the generalized central trinomial coefficients $T_{2k}(b,c)$, i.e., the coefficient of $x^{2k}$ in $(x^2+bx+c)^{2k}$, where $b,c$ are integers.
