请输入您要查询的字词:

 

单词 DeterminationOfEvenAbundantNumbersWithOneOddPrimeFactor
释义

determination of even abundant numbers with one odd prime factor


In this entry, we will use the criterion of the parent entry to determinethe first few even abundant numbers. To keep things more managable,we shall take advantage of the fact that a multipleMathworldPlanetmath of an abundant numberis abundant and only look for abundant numbers none of whose proper divisorsare abundant. Once we know these numbers, it becomes a rather easy matterto find the rest of the abundant numbers by taking multiples.

To begin, we look at the criterion of the second thorem. Since 2/(2-1)=2and, for any p>2, we have 1<p/(p-1)<2, it follows that, forevery prime p, there will exist abundant numbers of the form 2mpn.By the first theorem, for such a number to be abundant, we must have

(1-2-m-1)(1-p-n-1)12(1-p-1)>2

or, after a little algebraic simplification,

(1-2-m-1)(1-p-n-1)>1-p-1.

From this inequality, we can deduce a description of abundant numbersof the form 2mpn.

Theorem 1.

Let p>2 be prime. Then, for all n0, we find that 2mpnis abundant for m sufficiently large.

Proof.

When n0, we have

1-p-n-1>1-p-1.

Since limm2-m-1=0, it follows that

(1-2-m-1)(1-p-n-1)>1-p-1.

for m sufficiently large, hence 2mpnwill be abundant for m sufficiently large.∎

Theorem 2.

Let p>2 be prime. Then there exists m0 such that:

  1. 1.

    If m<m0, then 2mpn is not abundant for any n.

  2. 2.

    if mm0, then 2mpn is abundant for nsufficiently large.

Proof.

Set m0 to be the smallest integer such that 2m0+1>p.This inequality is equivalentMathworldPlanetmathPlanetmathPlanetmathPlanetmath to

1-2-m0-1>1-p-1.

On the one hand, if m<m0, then it will be impossible to satisfyour criterion for any choice of n. On the other hand, if mm0,then, by ther same sort of continuity argument employed previously,the criterion will be satisfied for m sufficiently large.∎

Theorem 3.

Let p>2 be prime and let m0 be the unique integer such that

2m0+1>p>2m0.

Then 2m0p is either perfect or abundant and every abundantnumber of the form 2mpn is a multiple of 2m0p.

Proof.

We begin with the equation

p+12m0+1.

Making some algebraic manipulations, we obtain the following:

p2m0+11+p-1
p-12-m0-1(1+p-1)
p-1-2-m0-1(1+p-1)0
1+p-1-2-m0-1(1+p-1)1
(1-2-m0-1)(1+p-1)1
(1-2-m0-1)(1+p-2)1-p-1

According to our earlier inequality, this means that2m0p is either perfect or abundant.

Suppose that 2mpn is abundant. Then, by the previousresult, mm0; since powers of 2 are deficient, n>0,so 2m0p2mpn.∎

This result makes it rather easy to draw up a list of abundantnumbers with one odd prime factor none of whose proper factorsare abundant starting with a table of prime numbersMathworldPlanetmath, as hasbeen done below. Note that, in the case where 2m0p isperfect, we have listed 2m0+1p and 2m0p2 asthese numbers are abundant.

随便看

 

数学辞典收录了18232条数学词条,基本涵盖了常用数学知识及数学英语单词词组的翻译及用法,是数学学习的有利工具。

 

Copyright © 2000-2023 Newdu.com.com All Rights Reserved
更新时间:2025/5/4 5:49:24