نتایج جستجو برای: whenever r divides n

تعداد نتایج: 1334461  

1998
Jerrold R. Griggs Chih-Chang Ho Martin Gardner

In “Bulgarian Solitaire”, a player divides a deck of n cards into piles. Each move consists of taking a card from each pile to form a single new pile. One is concerned only with how many piles there are of each size. Starting from any division into piles, one always reaches some cycle of partitions of n. Brandt proved that for n = 1+2+ · · ·+k, the cycle is just the single partition into piles ...

Journal: :Journal of Combinatorial Designs 2021

An m × n row-column factorial design is an arrangement of the elements a into rectangular array. Such array used in experimental design, where rows and columns can act as blocking factors. Formally, for any integer q, let [ q ] = { 0 , 1 … − } . The k (full) with replication α multiset consisting occurrences each element ; we denote this by A regular (which say type I ( ) such that row (column)...

2013
Salvatore Tringali

We determine all triples (a, b, n) of positive integers such that a and b are relatively prime and nk divides an+bn (respectively, an bn), when k is the maximum of a and b (in fact, we answer a slightly more general question). As a by-product, we see that, for m,n 2 N+ with n 2, nm divides mn + 1 if and only if (m,n) = (2, 3) or (1, 2), which generalizes problems from the 1990 and 1999 editions...

2016
Hojjat Mostafanasab Ece Yetkin Ünsal Tekir Ahmad Yousefian Darani

All rings are commutative with 1 6= 0, and all modules are unital. The purpose of this paper is to investigate the concept of 2-absorbing primary submodules generalizing 2-absorbing primary ideals of rings. Let M be an R-module. A proper submodule N of an R-module M is called a 2-absorbing primary submodule of M if whenever a, b ∈ R and m ∈M and abm ∈ N , then am ∈M -rad(N) or bm ∈M -rad(N) or ...

Journal: :Filomat 2021

This article introduces the concept of S-2-absorbing primary submodule as a generalization 2-absorbing submodule. Let S be multiplicatively closed subset ring R and M an R-module. A proper N is said to if (N :R M) ? = there exists fixed element s such that whenever abm for some a,b m M, then either sam or sbm sab ?(N M). We give several examples, properties characterizations related concept. Mo...

Journal: :Journal of Graph Theory 2022

Morris and Saxton used the method of containers to bound number n-vertex graphs with m edges containing no ℓ-cycles, hence girth more than ℓ. We consider a generalization r-uniform hypergraphs. The hypergraph H is minimum ℓ ≥ 2 such that there exist distinct vertices v 1 , … hyperedges e i + ∈ for all ≤ Letting N r ( n ) denote hypergraphs larger defining λ = ⌈ − ∕ ⌉ we show which tight when di...

Journal: :Rocky Mountain Journal of Mathematics 2022

Bohr phenomenon for analytic functions f, where f(z)=∑n=0∞anzn, was first introduced by Harald in 1914 and deals with finding the largest radius rf, 0<rf<1, such that inequality ∑n=0∞|an||z|n<1 holds |z|=r≤rf whenever |f(z)|<1 unit disk

Journal: :Filomat 2023

Let R be a commutative ring with non-zero identity, S multiplicatively closed subset of and M unital R-module. In this paper, we define submodule N (N :R M) ? = to weakly S-primary if there exists s such that whenever m 0 , am N, then either sa ??(N or sm N. We present various properties characterizations concept (especially in faithful multiplication modules). Moreover, the behavior structure ...

Journal: :Advances in Mathematics 2021

Let $G$ be an abelian group with identity $e$. $R$ a $G$-graded commutative ring identity, $M$ graded $R$-module and $S\subseteq h(R)$ multiplicatively closed subset of $R$. In this paper, we introduce the concept $S$-prime submodules modules over rings. We investigate some properties class their homogeneous components. $N$ submodule such that $(N:_{R}M)\cap S=\emptyset $. say is \textit{a }$S$...

Journal: :The American Mathematical Monthly 2003
Jack B. Brown George Kozlowski

1 pi − 2 j+1 > 7pr − 4pr > pr . If r is odd, set nk = ∏r1 pi − 2k+1. Then nk ≡ 3(4) (since 32 ≡ 1(4)). But no pi divides nk for 1 ≤ i ≤ r , so the integer nk has some prime factor qk ≡ 3(4) with qk > pr . If j = k, say j > k, the assumption that qk also divides n j leads to the same contradiction as earlier: since nk − n j = 2 j+1 − 2k+1 = 2k+1(2 j−k − 1), we have qk | 2 j−k − 1 and hence qk < ...

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