Proof of a conjecture of Adamchuk

Abstract

In this paper, we prove a congruence which contains a congruence conjectured by Adamchuk (OEIS A066796 in 2006, http://oeis.org/A066796). For any prime $p\equiv 1(\mathrm{mod}3)$ and $a\in {\mathbb{Z}}^{+}$, we have

$$\sum _{k=1}^{\frac{2}{3}(p^a-1)}\left(\begin{array}{c}2k\\ k\end{array}\right)\equiv 0(\mathrm{mod}{p}^2).$$

Introduction

In 2006, Adamchuk [1] conjectured that for any prime $p\equiv 1(\mathrm{mod}3)$,

$$\sum _{k=1}^{\frac{2}{3}(p-1)}\left(\begin{array}{c}2k\\ k\end{array}\right)\equiv 0(\mathrm{mod}{p}^2).$$

After that, many people studied congruences for sums of binomial coefficients (see, for instance, [3], [5], [8], [13], [14], [15], [16], [17], [21], [25], [26], [27]). For example, Pan and Sun [21] used a combinatorial identity to deduce that if p is prime then

$$\sum _{k=0}^{p-1}\left(\begin{array}{c}2k\\ k+d\end{array}\right)\equiv \left(\frac{p-d}{3}\right)(\mathrm{mod}p)\text{for}d=0,1,\dots ,p,$$

where $\left(\frac{\cdot }{\cdot }\right)$ is the Jacobi symbol. Then in 2011, Sun and Tauraso [27] gave a generalization of the above result: For any odd prime p and $a\in {\mathbb{Z}}^{+}$, $d=0,\dots ,p^a$,

$$\sum _{k=0}^{p^a-1}\left(\begin{array}{c}2k\\ k+d\end{array}\right)\equiv \left(\frac{p^a-d}{3}\right)(\mathrm{mod}{p}^2).$$

Apagodu and Zeilberger [3] proved that for any prime $p\ge 5$ and any positive integer r,

Pan and Sun [21, Corollary 1.3] also showed that for any odd prime p,

$$\sum _{n=0}^{p-1}(3n+1)\left(\begin{array}{c}2n\\ n\end{array}\right)\equiv -\left(\frac{p}{3}\right)(\mathrm{mod}p).$$

In 2018, Apagodu [2] gave the following conjecture: For any odd prime p,

$$\sum _{n=0}^{p-1}(5n+1)\left(\begin{array}{c}4n\\ 2n\end{array}\right)\equiv -\left(\frac{p}{3}\right)(\mathrm{mod}p).$$

Then the author and Cao [14] confirmed this conjecture and showed another congruence: For any odd prime p,

$$\sum _{n=0}^{p-1}(15n+5)\left(\begin{array}{c}4n\\ 2n\end{array}\right)\equiv -\left(\frac{p}{5}\right)(\mathrm{mod}p).$$

Recall that the Bernoulli numbers $\{B_n\}$ and the Bernoulli polynomials $\{B_n(x)\}$ are defined as follows:

$$\frac{x}{e^x-1}=\sum _{n=0}^{\infty }{B}_n\frac{x^n}{n!}(0<|x|<2\pi )\text{and}{B}_n(x)=\sum _{k=0}^n\left(\begin{array}{c}n\\ k\end{array}\right)B_k{x}^{n-k}(n\in \mathbb{N}).$$

Mattarei and Tauraso [18] proved that for any prime $p>3$, we have

$$\sum _{k=0}^{p-1}\left(\begin{array}{c}2k\\ k\end{array}\right)\equiv \left(\frac{p}{3}\right)-\frac{p^2}{3}{B}_{p-2}\left(\frac{1}{3}\right)(\mathrm{mod}{p}^3).$$

Many researchers also studied numerous complicated congruences involving binomial coefficients and q-binomial coefficients (for instance, [6], [7], [9], [11]).

The main objective of this paper is to obtain the following result which contains the conjecture of Adamchuk.

Theorem 1.1

Let p be an odd prime and let $a\in {\mathbb{Z}}^{+}$. If $p\equiv 1(\mathrm{mod}3)$ and $a\in {\mathbb{Z}}^{+}$, then

$$\sum _{k=1}^{\frac{2}{3}(p^a-1)}\left(\begin{array}{c}2k\\ k\end{array}\right)\equiv 0(\mathrm{mod}{p}^2).$$

In order to prove Theorem 1.1, we first show the following interesting congruence.

Theorem 1.2

For any prime $p\equiv 1(\mathrm{mod}3)$, we have

$$\sum _{\begin{array}{c}\hfill k=0\hfill \\ \hfill k\ne (p-1)/3\hfill \end{array}}^{(p-1)/2}\frac{\left(\begin{array}{c}2k\\ k\end{array}\right)}{3k+1}\equiv 0(\mathrm{mod}p).$$

We shall prove Theorem 1.2 in Section 2, Section 3 is devoted to prove Theorem 1.1.