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New Congruences Involving pp-adic dual sequences

Published 14 Aug 2026 in math.NT and math.CO | (2608.14453v1)

Abstract: Let (an)<em>n0(a_n)<em>{n\geqslant 0} be a sequence of integers. Its dual sequence (an<sup>)</sup></em>n0(a_n<sup>*)</sup></em>{n\geqslant 0} is defined by \begin{equation*} a_n* := \sum_{k=0}{n} \binom{n}{k}(-1)k a_k. \end{equation*} Let $p&gt;3$ be a prime. In this paper we mainly investigate congruences modulo p<sup>2p<sup>2 involving central binomial coefficients and pp-adic dual sequences. For example, we prove that for any sequence (ak)<em>k0(a_k)<em>{k\ge0} of pp-adic integers, \begin{align*} \sum{(p-1)/2}{k=0}\binom{2k}{k}2\frac{a_{2k}}{16k}\equiv\left( \frac{-1}{p}\right) \sum_{k=0}{p-1}\frac{\mathcal{P}_{k}}{16 {k}}a_{k}*\pmod{p2}, \end{align*} where (P<em>n)</em>n0(\mathcal{P}<em>n)</em>{n\ge0} are the Catalan--Larcombe--French numbers given by \begin{equation*} \mathcal{P}0=1,\quad \mathcal{P}_1=8, \quad n2 \mathcal{P}_n = 8(3n2-3n+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 k=0<sup>(p1)/2(2kk)a2k<sup>/4<sup>k</sup></sup></sup>(modp<sup>2)\sum_{k=0}<sup>{(p-1)/2}\binom{2k}{k}a_{2k}<sup>*/4<sup>k</sup></sup></sup> \pmod{p<sup>2} and as a consequence we confirm some conjectures of Z.-W. Sun \cite{Sun2014CANT} on the generalized central trinomial coefficients T2k(b,c)T_{2k}(b,c), i.e., the coefficient of x<sup>2kx<sup>{2k} in (x<sup>2+bx+c)<sup>2k(x<sup>2+bx+c)<sup>{2k}, where b,cb,c are integers.

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