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Related papers: Congruences concerning Lucas' law of repetition

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Let p be any prime, and let a and n be nonnegative integers. Let $r\in Z$ and $f(x)\in Z[x]$. We establish the congruence $$p^{\deg f}\sum_{k=r(mod p^a)}\binom{n}{k}(-1)^k f((k-r)/p^a) =0 (mod p^{\sum_{i=a}^{\infty}[n/p^i]})$$ (motivated by…

Number Theory · Mathematics 2007-07-25 Zhi-Wei Sun , Donald M. Davis

Let $p$ be a prime. In 1878 \'{E}. Lucas proved that the congruence $$ {p-1\choose k}\equiv (-1)^k\pmod{p}$$ holds for any nonnegative integer $k\in\{0,1,\ldots,p-1\}$. The converse statement was given in Problem 1494 of {\it Mathematics…

Number Theory · Mathematics 2018-04-10 Romeo Mestrovic

For $A,B\in\mathbb Z$, the Lucas sequence $u_n(A,B)\ (n=0,1,2,\ldots)$ are defined by $u_0(A,B)=0$, $u_1(A,B)=1$, and $u_{n+1}(A,B) = Au_n(A,B)-Bu_{n-1}(A,B)$ $(n=1,2,3,\ldots).$ For any odd prime $p$ and positive integer $n$, we establish…

Number Theory · Mathematics 2020-12-15 Zhi-Wei Sun

Harmonic numbers $H_k=\sum_{0<j\le k}1/j (k=0,1,2,...)$ arise naturally in many fields of mathematics. In this paper we initiate the study of congruences involving both harmonic numbers and Lucas sequences. One of our three theorems is as…

Number Theory · Mathematics 2011-05-24 Zhi-Wei Sun

Let $S= \{ p_1, \ldots, p_s\}$ be a finite, non-empty set of distinct prime numbers and $(U_{n})_{n \geq 0}$ be a linear recurrence sequence of integers of order $r$. For any positive integer $k,$ we define $(U_j^{(k)})_{j\geq 1}$ an…

Number Theory · Mathematics 2020-04-16 S. S. Rout , N. K. Meher

Let $p_n$ denote the $n$-th prime number, $\{q_n\}$ be a sequence of positive numbers and $x\in\mathbb{R}$. In this note we prove that the inequality $$q_n p_{n+1}^{x}-q_{n+1}p_{n}^{x}<p_{n}^{x}p_{n+1}^{x-1}, $$ holds for infinitely many…

Number Theory · Mathematics 2017-12-11 Douglas Azevedo , Tiago Reis

(Below, \Box means "perfect square") Let $P$ and $Q$ be non-zero integers. The Lucas sequence $\{U_n(P,Q)\}$ is defined by $U_0=0$, $U_1=1$, $U_n=P U_{n-1}-Q U_{n-2}$, $(n \geq 2)$. Historically, there has been much interest in when the…

Number Theory · Mathematics 2007-05-23 A. Bremner N. Tzanakis

Let P and Q be non-zero integers. The Lucas sequence U_n(P,Q), n=0,1,2,... is defined by U_0=0, U_1=1, U_n= P U_{n-1}-Q U_{n-2} for n>1. For each positive integer n<8 we describe all Lucas sequences with (P,Q)=1 having the property that…

Number Theory · Mathematics 2007-05-23 Andrew Bremner , Nikos Tzanakis

We say that an arithmetical function $S:\mathbb{N}\rightarrow\mathbb{Z}$ has Lucas property if for any prime $p$, \begin{equation*} S(n)\equiv S(n_{0})S(n_{1})\ldots S(n_{r})\pmod p, \end{equation*} where $n=\sum_{i=0}^{r}n_{i}p^{i}$, with…

Number Theory · Mathematics 2016-12-22 Hao Zhong , Tianxin Cai

We give elementary proofs of some congruence criteria to compute binomial coefficients in modulo a prime. These criteria are analogues to the symmetry property of binomial coefficients. We give extended version of Lucas Theorem by using…

Number Theory · Mathematics 2023-09-04 Zubeyir Cinkir , Aysegul Ozturkalan

In this paper we obtain some congruences involving central binomial coefficients and Lucas sequences. For example, we show that if p>5 is a prime then $\sum_{k=0}^{p-1}F_k*binom(2k,k)/12^k$ is congruent to 0,1,-1 modulo p according as p=1,4…

Number Theory · Mathematics 2009-12-14 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,…

Number Theory · Mathematics 2018-01-08 Nicholas J. Newsome , Maria S. Nogin , Adnan H. Sabuwala

Recently, Amdeberhan, Sellers, and Singh introduced the notion of a generalized cubic partition function $a_c(n)$ and proved two isolated congruences via modular forms, namely, $a_3(7n+4)\equiv 0\pmod{7}$ and $a_5(11n+10)\equiv 0\pmod{11}$.…

Number Theory · Mathematics 2025-03-03 Russelle Guadalupe

Suppose that $p$ is an odd prime and $m$ is an integer not divisible by $p$. Sun and Tauraso [Adv. in Appl. Math., 45(2010), 125--148] gave $\sum_{k=0}^{n-1}\binom{2k}{k+d}/m^k$ and $\sum_{k=0}^{n-1}\binom{2k}{k+d}/(km^k)$ modulo $p$ for…

Number Theory · Mathematics 2021-10-22 He-Xia Ni

Let $(L_n^{(k)})_{n\geq 2-k}$ be the sequence of $k$--generalized Lucas numbers for some fixed integer $k\ge 2$ whose first $k$ terms are $0,\ldots,0,2,1$ and each term afterwards is the sum of the preceding $k$ terms. For an integer $m$,…

Number Theory · Mathematics 2023-11-23 Herbert Batte , Florian Luca

Let P and Q be non-zero integers. The Lucas sequence U_n(P,Q) is defined by U_0=0, U_1=1, U_n= P*U_{n-1}-Q*U_{n-2} for n >1. The question of when U_n(P,Q) can be a perfect square has generated interest in the literature. We show that for…

Number Theory · Mathematics 2007-05-23 A. Bremner , N. Tzanakis

Let q>1 and m>0 be relatively prime integers. We find an explicit period $\nu_m(q)$ such that for any integers n>0 and r we have $[n+\nu_m(q),r]_m(a)=[n,r]_m(a) (mod q)$ whenever a is an integer with $\gcd(1-(-a)^m,q)=1$, or a=-1 (mod q),…

Number Theory · Mathematics 2007-08-06 Zhi-Wei Sun , Roberto Tauraso

In this work we consider the congruence $\sum_{j=1}^{n-1} j^{k(n-1)} \equiv -1 \pmod n$ for each $k \in \mathbb{N}$, thus extending Giuga's ideas for $k=1$. In particular, it is proved that a pair $(n,k)\in \mathbb{N}^2$ satisfies this…

Number Theory · Mathematics 2013-11-15 Antonio M. Oller-Marcén , José María Grau

In this paper, we present several new $q$-congruences on the $q$-trinomial coefficients introduced by Andrews and Baxter. As a conclusion, we obtain the following congruence: \begin{align*}…

Number Theory · Mathematics 2023-05-03 Yifan Chen , Xiaoxia Wang

Let $p$ be a prime. In this paper, we present a detailed $p$-adic analysis to factorials and double factorials and their congruences. We give good bounds for the $p$-adic sizes of the coefficients of the divided universal Bernoulli number…

Number Theory · Mathematics 2013-08-23 Shaofang Hong , Jianrong Zhao , Wei Zhao
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