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相关论文: A note on primes dividing alternating sums

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Let n be a positive odd integer and let p>n+1 be a prime. We mainly derive the following congruence: $$\sum_{0<i_1<...<i_n<p}(i_1/3)(-1)^{i_1}/(i_1...i_n)=0 (mod p).$$

数论 · 数学 2010-02-25 Li-Lu Zhao , Zhi-Wei Sun

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.

数论 · 数学 2009-12-25 Roberto Tauraso

We show that for every $0 < \epsilon \leq 1$ and integer $k\geq 1$, there exists an integer $n = n(\epsilon,k)$ so that for all primes $p$, and integers $0 \leq a \leq p-1$, there exist integers $1 \leq x_1 < ... < x_n \leq p^\epsilon$ such…

数论 · 数学 2007-05-23 Ernie Croot

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…

数论 · 数学 2011-04-14 Zhi-Wei Sun

The following congruence for power sums, $S_n(p)$, is well known and has many applications: $1^n+2^n +\dots +p^n \equiv\begin{cases} -1 \text{ mod } p, & \text{ if } \ p-1 \ | \ n; 0 \text{ mod } p, & \text{ if } \ p-1 \ \not| \ n,…

数论 · 数学 2018-01-08 Nicholas J. Newsome , Maria S. Nogin , Adnan H. Sabuwala

We show that if p is an odd prime then $$\sum_{k=0}^{p-1}E_kE_{p-1-k}=1 (mod p)$$ and $$\sum_{k=0}^{p-3}E_kE_{p-3-k}=(-1)^{(p-1)/2}2E_{p-3} (mod p),$$ where E_0,E_1,E_2,... are Euler numbers. Moreover, we prove that for any positive integer…

数论 · 数学 2010-12-22 Zhi-Wei Sun

The purpose of this note is to report on the discovery of the primes of the form $p=1+n!\sum n$, for some natural numbers $n>0$. The number of digits in the prime p are approximately equal to $\lfloor log_{10}(1+n!\sum n)\rceil+1$.

综合数学 · 数学 2018-04-02 Maheswara Rao Valluri

We will prove several congruences modulo a power of a prime such as $$ \sum_{0<k_1<...<k_{n}<p}\leg{p-k_{n}}{3} {(-1)^{k_{n}}\over k_1... k_{n}}\equiv {lll} -{2^{n+1}+2\over 6^{n+1}} p B_{p-n-1}({1\over 3}) &\pmod{p^2} &{if $n$ is odd}…

数论 · 数学 2009-11-06 Roberto Tauraso

Let $\{\cdot\}$ denote the fractional part and $n \geq 1$ be a fixed integer. In this short note, we show for any prime $p$ the one-to-one correspondence $$\sum_{\nu \geq 1} \left\{\frac{n}{p^\nu}\right\} > 1 \quad \iff \quad p \mid…

数论 · 数学 2017-08-24 Bernd C. Kellner

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…

数论 · 数学 2014-04-29 Zhi-Wei Sun

For n=1,2,3,... let p_n be the n-th prime. We mainly show that p_n>n+sum_{k=1}^n p_k/k for all n>124, and sum_{k=1}^n kp_k<n^2p_n/3 for all n>30.

数论 · 数学 2012-09-20 Zhi-Wei Sun

Quite recently, in [8] the authoor of this paper considered the distribution of primes in the sequence $(S_n)$ whose $n$th term is defined as $S_n=\sum_{k=1}^{2n}p_k$, where $p_k$ is the $k$th prime. Some heuristic arguments and the…

数论 · 数学 2018-05-31 Romeo Meštrović

Let $p$ be an odd prime, and let $m$ be an integer with $p\nmid m$. In this paper show that $$\sum_{k=0}^{p-1}\frac{\binom{2k}k\binom ak\binom{-1-a}k}{m^k} \equiv 0\pmod p \quad\hbox{implies}\quad\sum_{k=0}^{p-1}\frac{\binom{2k}k\binom ak…

数论 · 数学 2015-03-12 Zhi-Hong Sun

Let $p$ be a prime and ${\mathcal{P}_{p}}$ the set of positive integers which are prime to $p$. We establish the following interesting congruence \[\sum\limits_{\begin{smallmatrix} i+j+k={{p}^{r}} i,j,k\in {\mathcal{P}_{p}}…

数论 · 数学 2014-07-23 Liuquan Wang

Let $\{A'_n\}$ be the Ap\'ery numbers given by $A'_n=\sum_{k=0}^n\binom nk^2\binom{n+k}k.$ For any prime $p\equiv 3\pmod 4$ we show that $A'_{\frac{p-1}2}\equiv \frac{p^2}3\binom{\frac{p-3}2}{\frac{p-3}4}^{-2}\pmod {p^3}$. Let $\{t_n\}$ be…

数论 · 数学 2024-10-16 Zhi-Hong Sun

Let $p_n$ be $n$th prime, and let $(S_n)_{n=1}^\infty:=(S_n)$ be the sequence of the sums of the first $2n$ consecutive primes, that is, $S_n=\sum_{k=1}^{2n}p_k$ with $n=1,2,\ldots$. Heuristic arguments supported by the corresponding…

数论 · 数学 2018-04-13 Romeo Meštrović

In 2014, Wang and Cai established the following harmonic congruence for any odd prime $p$ and positive integer $r$, \begin{equation*} \sum\limits_{i+j+k=p^{r}\atop{i,j,k\in \mathcal{P}_{p}}}\frac{1}{ijk}\equiv-2p^{r-1}B_{p-3} (\bmod p^{r}),…

数论 · 数学 2015-03-12 Zhongyan Shen , Tianxin Cai

Let p1, p2,..., pn be distinct prime numbers, and let Nn be their product. We prove that, for any positive integer L that is divisible by the least common multiple of p1 minus one, p2 minus one, and so on, and for integers a1, a2,..., an…

数论 · 数学 2025-10-14 Shao-Yuan Huang , Hsiu-Yu Wu

In this paper we establish some new supercongruences motivated by the well-known fact $\lim_{n\to\infty}(1+1/n)^n=e$. Let $p>3$ be a prime. We prove that $$\sum_{k=0}^{p-1}\binom{-1/(p+1)}k^{p+1}\equiv 0\ \pmod{p^5}\ \ \ \mbox{and}\ \ \…

数论 · 数学 2015-02-27 Zhi-Wei Sun

Write $T(n)$ as the sum of the reciprocals of the primes which divide $n$. Write $H(n) = \prod_{p|n}p/(p-1)$ where the product is over the prime divisors of $n$. We prove new bounds for $T(n)$ and $H(n)$ in terms of the smallest prime…

数论 · 数学 2025-02-11 Joshua Zelinsky
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