Related papers: Some congruences for the second-order Catalan numb…

Let $p$ be an odd prime. In the paper we collect the author's various conjectures on congruences modulo $p$ or $p^2$, which are concerned with sums of binomial coefficients, Lucas sequences, power residues and special binary quadratic…

The Catalan number $C_n$ enumerates parenthesizations of $x_0*\dotsb*x_n$ where $*$ is a binary operation. We introduce the modular Catalan number $C_{k,n}$ to count equivalence classes of parenthesizations of $x_0*\dotsb*x_n$ when $*$…

We present a method for obtaining congruences modulo powers of 3 for sequences given by recurrences of finite depth with polynomial coefficients. We apply this method to Catalan numbers, Motzkin numbers, Riordan numbers, Schr\"oder numbers,…

Let $p>3$ be a prime, and let $a$ be a rational p-adic integer. Let $\{B_n(x)\}$ and $\{E_n(x)\}$ denote the Bernoulli polynomials and Euler polynomials, respectively. In this paper we show that $$\sum_{k=0}^{p-1}\binom…

For each positive integer $m$, the $m$th order harmonic numbers are given by $$H_n^{(m)}=\sum_{0<k\le n}\frac1{k^m}\ \ (n=0,1,2,\ldots).$$ We discover exact values of some series involving harmonic numbers of order not exceeding four. For…

Acquaah and Konyagin showed that if $N$ is an odd perfect number where $N= p_1^{a_1}p_2^{a_2} \cdots p_k^{a_k}$ where $p_1 < p_2 \cdots < p_k$ then one must have $p_k < 3^{1/3}N^{1/3}$. Using methods similar to theirs, we show that…

We prove some polynomial identities from which we deduce congruences modulo $p^2$ for the Fermat quotient $\frac{2^p-2}{p}$ for any odd prime $p$ (Proposition 1 and Theorem 1). These congruences are simpler than the one obtained by…

We give a $q$-congruence whose specializations $q=-1$ and $q=1$ correspond to supercongruences (B.2) and (H.2) on Van Hamme's 1997 list: $$ \sum_{k=0}^{(p-1)/2}(-1)^k(4k+1)A_k\equiv p(-1)^{(p-1)/2}\pmod{p^3} \quad\text{and}\quad…

For n=1,2,3,... define S(n) as the smallest integer m>1 such that those 2k(k-1) mod m for k=1,...,n are pairwise distinct; we show that S(n) is the least prime greater than 2n-2 and hence the value set of the function S(n) is exactly the…

We investigate the divisibility properties of \sigma(C_n), the sum-of-divisors function applied to Catalan numbers, in relation to other number-theoretic functions. We establish conditions under which C_n has prime factors of the form 6k-1,…

Let $p$ be an odd prime. Define the Gaussian power sum \[ G_n(p)=\sum_{a=1}^{p-1}\sum_{b=1}^{p-1}(a+bi)^n\in\mathbb Z[i]. \] We determine $G_p(p)$ modulo high powers of $p$: if $p\equiv 1\pmod 4$ then $$G_p(p)\equiv p^2(1+i)\pmod{p^3},$$…

Let $p$ be an odd prime and let $n$ be a positive integer. For any positive integer $\alpha$ and $m\in\{1,2,3\}$, we have \begin{align*}…

We prove that if $q$ is a power of a prime $p$ and $p^k$ divides $a$, with $k\ge 0$, then \[ 1+(q-1)\sum_{0\le b(q-1)<a} \binom{a}{b(q-1)}\equiv 0\pmod{p^{k+1}}. \] The special case of this congruence where $q=p$ was proved by Carlitz in…

Let $p_{k,3}(n)$ enumerate the number of 2-color partition triples of $n$ where one of the colors appears only in parts that are multiples of $k$. In this paper, we prove several infinite families of congruences modulo powers of 3 for…

Recently, using modular forms F. Beukers posed a unified method that can deal with a large number of supercongruences involving binomial coefficients and Ap\'ery-like numbers. In this paper, we use Beukers' method to prove some conjectures…

In 2003, Zhao discovered a curious congruence involving harmonic series and Bernoulli numbers: for any odd prime $p$, $$\sum_{\substack{i,j,k\ge 1\\\gcd(ijk,p)=1\\i+j+k=p}}\frac{1}{ijk}\equiv -2B_{p-3} \pmod{p},$$ where $B_n$ is the $n$-th…

In this paper, we pose lots of challenging conjectures on congruences for the sums involving binomial coefficients and Ap\'ery-like numbers modulo $p^3$, where $p$ is an odd prime.

With help of $q$-congruence, we prove the divisibility of some binomial sums. For example, for any integers $\rho,n\geq 2$, $$\sum_{k=0}^{n-1}(4k+1) \binom{2k}{k}^\rho \cdot (-4)^{\rho(n-1-k)} \equiv 0\pmod{2^{\rho-2}n\binom{2n}{n}}.$$

Nicholas Pippenger and Kristin Schleich have recently given a combinatorial interpretation for the second-order super-Catalan numbers (u_{n})_{n>=0}=(3,2,3,6,14,36,...): they count "aligned cubic trees" on n internal vertices. Here we give…

In this paper, we prove some supercongruences concerning truncated hypergeometric series. For example, we show that for any prime $p>3$ and positive integer $r$, $$ \sum_{k=0}^{p^r-1}(3k+1)\frac{(\frac12)_k^3}{(1)_k^3}4^k\equiv…