Related papers: On convolutions of Euler numbers

Using an identity arising in the known Banach probability problem on matchboxes, we prove an unexpected congruence for odd prime $p:$ for $1\leq k\leq \frac{p-1}{2},\enskip \sum_{i=1}^{p-2k-1}2^{i-1}\binom{k-1+i}{k}\equiv 0\pmod p.$

In this paper we study mixed sums of primes and linear recurrences. We show that if m=2(mod 4) and m+1 is a prime then $(m^{2^n-1}-1)/(m-1)\not=m^n+p^a$ for any n=3,4,... and prime power p^a. We also prove that if a>1 is an integer, u_0=0,…

Let p be an odd prime and let a be a positive integer. In this paper we investigate the sum $\sum_{k=0}^{p^a-1}\binom{hp^a-1}{k}\binom{2k}{k}/m^k$ mod p^2, where h,m are p-adic integers with m\not=0 (mod p). For example, we show that if…

Let $p$ be an odd prime and let $d\in\{2,3,7\}$. When $(\frac{-d}p)=1$ we can write $p=x^2+dy^2$ with $x,y\in\mathbb Z$; in this paper we aim at determining $x$ or $y$ modulo $p^2$. For example, when $p=x^2+3y^2$, we show that if $p\equiv…

Let $p$ be any odd prime number. Let $k$ be any positive integer such that $2\leq k\leq [\frac{p+1}3]+1$. Let $S = (a_1,a_2,...,a_{2p-k})$ be any sequence in ${\Bbb Z}_p$ such that there is no subsequence of length $p$ of $S$ whose sum is…

The Franel numbers are defined by $ f_n=\sum_{k=0}^n {n\choose k}^3. $ Motivated by the recent work of Z.-W. Sun on Franel numbers, we prove that \begin{align*} \sum_{k=0}^{n-1}(3k+1)(-16)^{n-k-1} {2k\choose k} f_k &\equiv…

The numbers $R_n$ and $W_n$ are defined as \begin{align*} R_n=\sum_{k=0}^{n}{n+k\choose 2k}{2k\choose k}\frac{1}{2k-1},\ \text{and}\ W_n=\sum_{k=0}^{n}{n+k\choose 2k}{2k\choose k}\frac{3}{2k-3}. \end{align*} We prove that, for any positive…

Let $p$ be a prime and let $a$ be a positive integer. In this paper we investigate $\sum_{k=0}^{p^a-1}\binom[(h+1)k,k+d]/m^k$ modulo a prime $p$, where $d$ and $m$ are integers with $-h<d<=p^a$ and $m\not=0 (mod p)$. We also study…

Let b be an odd integer such that b=+/-1 (mod 8) and let q be a prime with primitive root 2 such that q does not divide b. We show that if (p(k)) is a sequence of odd primes, with 0<=k<=q-2 such that p(k)=2p(k-1)+b for all 1<=k<=q-2, then…

We show that for any prime prime $p\not=2$ $$\sum_{k=1}^{p-1} {(-1)^k\over k}{-{1\over 2} \choose k} \equiv -\sum_{k=1}^{(p-1)/2}{1\over k} \pmod{p^3}$$ by expressing the l.h.s. as a combination of alternating multiple harmonic sums.

For a positive integer $n$ let $H_n=\sum_{k=1}^{n}1/k$ be the $n$th harmonic number. Z. W. Sun conjectured that for any prime $p\ge 5$, $$ \sum_{k=1}^{p-1}\frac{H_k}{k\cdot 2^k} \equiv7/24pB_{p-3}\pmod{p^2}. $$ This conjecture is recently…

In this paper, we confirm several conjectures posed by Sun recently; for example, we prove that for any odd prime $p$ we have $$ \sum_{k=0}^{p-1}A_k\equiv\begin{cases}4x^2-2p\pmod{p^2}\quad&\text{if $p=x^2+2y^2\ (x,y\in\mathbb{Z})$},\\…

The Ap\'ery polynomials are given by $$A_n(x)=\sum_{k=0}^n\binom nk^2\binom{n+k}k^2x^k\ \ (n=0,1,2,\ldots).$$ (Those $A_n=A_n(1)$ are Ap\'ery numbers.) Let $p$ be an odd prime. We show that…

Let $p>3$ be a prime. Euler numbers $E_{p-3}$ first appeared in H. S. Vandiver's work (1940) in connection with the first case of Fermat Last Theorem. Vandiver proved that $x^p+y^p=z^p$ has no solution for integers $x,y,z$ with…

Let $m$ and $n>0$ be integers. Suppose that $p$ is a prime dividing $m-4$ but not dividing $m$. We show that $\nu_p(\sum_{k=0}^{n-1}\frac{\binom{2k}k}{m^k})$ and $\nu_p(\sum_{k=0}^{n-1}\binom{n-1}{k}(-1)^k\frac{\binom{2k}k}{m^k})$ are at…

For various positive integers $k$, the sums of $k$th powers of the first $n$ positive integers, $S_k(n+1)=1^k+2^k+...+n^k$, have got to be some of the most popular sums in all of mathematics. In this note we prove that for each $k\ge 2 $$…

In this paper, we apply the power-partible reduction to study arithmetic properties of sums involving Delannoy numbers $D_k$ and polynomials $D_k(z)$. Let $v\in\bN$ and $p$ be an odd prime. It is proved that, for any…

We present some congruences modulo $p^{6-d}$ for sums of the type $\sum_{k=0}^{(p-3)/2}x^k{2k\choose k}/(2k+1)^d$, for $d=1,2,3$ where $p>5$ is a prime.

We define A_n=\sum_{i=1}^n (-1)^i\frac{1}{i} and we show that, for every prime p, there exists a number n such that A_n\equiv 0 (mod p).

Let $p>3$ be a prime. For any $p$-adic integer $a$, we determine $$\sum_{k=0}^{p-1}\binom{-a}k\binom{a-1}kH_k,\ \ \sum_{k=0}^{p-1}\binom{-a}k\binom{a-1}kH_k^{(2)},\ \ \sum_{k=0}^{p-1}\binom{-a}k\binom{a-1}k\frac{H_k^{(2)}}{2k+1}$$ modulo…