New Congruences Involving $p$-adic dual sequences

Abstract

Let $(a_n{)}_{n⩾0}$ be a sequence of integers. Its dual sequence $(a_n^{\ast }{)}_{n⩾0}$ 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 $(a_k{)}_{k\ge 0}$ 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 $({\mathcal{P}}_n{)}_{n\ge 0}$ 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 ${\sum }_{k=0}^{(p-1)/2}\left(\genfrac{}{}{0ex}{}{2k}{k}\right)a_{2k}^{\ast }/4^k(\mathrm{mod}{p}^2)$ 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.