determination of even abundant numbers with one odd prime factor

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

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

$$\frac{(1-2^{-m-1})(1-p^{-n-1})}{\frac{1}{2}(1-p^{-1})}>2$$

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 numbers of the form $2^m{p}^n$.

Theorem 1.

Let $p\mathrm{>}\mathrm{2}$ be prime. Then, for all $n\mathrm{\ge }\mathrm{0}$, we find that ${\mathrm{2}}^m\mathit{}{p}^n$ is abundant for $m$ sufficiently large.

Proof.

When $n\ge 0$, we have

$$1-p^{-n-1}>1-p^{-1}.$$

Since ${lim}_{m\to \mathrm{\infty }}{2}^{-m-1}=0$, it follows that

$$(1-2^{-m-1})(1-p^{-n-1})>1-p^{-1}.$$

for $m$ sufficiently large, hence $2^m{p}^n$ will be abundant for $m$ sufficiently large. ∎

Theorem 2.

Let $p\mathrm{>}\mathrm{2}$ be prime. Then there exists $m_{\mathrm{0}}$ such that:

1. If , then $2^m{p}^n$ is not abundant for any $n$.

2. if $m\ge m_0$, then $2^m{p}^n$ is abundant for $n$ sufficiently large.

Proof.

Set $m_0$ to be the smallest integer such that $2^{m_0+1}>p$. This inequality is equivalent to

$$1-2^{-m_0-1}>1-p^{-1}.$$

On the one hand, if , then it will be impossible to satisfy our criterion for any choice of $n$. On the other hand, if $m\ge m_0$, then, by ther same sort of continuity argument employed previously, the criterion will be satisfied for $m$ sufficiently large. ∎

Theorem 3.

Let $p\mathrm{>}\mathrm{2}$ be prime and let $m_{\mathrm{0}}$ be the unique integer such that

$$2^{m_0+1}>p>2^{m_0}.$$

Then ${\mathrm{2}}^{m_{\mathrm{0}}}\mathit{}p$ is either perfect or abundant and every abundant number of the form ${\mathrm{2}}^m\mathit{}{p}^n$ is a multiple of ${\mathrm{2}}^{m_{\mathrm{0}}}\mathit{}p$.

Proof.

We begin with the equation

$$p+1\le 2^{m_0+1}.$$

Making some algebraic manipulations, we obtain the following:

$p$ $\le {\displaystyle \frac{2^{m_0}+1}{1+p^{-1}}}$

$p^{-1}$ $\ge 2^{-m_0-1}(1+p^{-1})$

$p^{-1}-2^{-m_0-1}(1+p^{-1})$ $\ge 0$

$1+p^{-1}-2^{-m_0-1}(1+p^{-1})$ $\ge 1$

$(1-2^{-m_0-1})(1+p^{-1})$ $\ge 1$

$(1-2^{-m_0-1})(1+p^{-2})$ $\ge 1-p^{-1}$

According to our earlier inequality, this means that $2^{m_0}p$ is either perfect or abundant.

Suppose that $2^m{p}^n$ is abundant. Then, by the previous result, $m\le m_0$; since powers of $2$ are deficient, $n>0$, so $2^{m_0}p\mid 2^m{p}^n$. ∎

This result makes it rather easy to draw up a list of abundant numbers with one odd prime factor none of whose proper factors are abundant starting with a table of prime numbers, as has been done below. Note that, in the case where $2^{m_0}p$ is perfect, we have listed $2^{m_0+1}p$ and $2^{m_0}{p}^2$ as these numbers are abundant.

$3$	$12$ $(2^2\cdot 3)$, $18$ $(2\cdot 3^2)$
$5$	$20$ $(2^2\cdot 5)$
$7$	$56$ $(2^3\cdot 7)$, $196$ $(2^2\cdot 7^2)$
$11$	$88$ $(2^3\cdot 11)$
$13$	$104$ $(2^3\cdot 13)$
$17$	$272$ $(2^4\cdot 17)$
$19$	$304$ $(2^4\cdot 19)$
$23$	$368$ $(2^4\cdot 23)$
$29$	$464$ $(2^4\cdot 29)$
$31$	$992$ $(2^5\cdot 31)$, $7688$ $(2^5\cdot {31}^2)$
$37$	$1184$ $(2^5\cdot 37)$
$41$	$1312$ $(2^5\cdot 41)$
$43$	$1376$ $(2^5\cdot 43)$
$47$	$1504$ $(2^5\cdot 47)$
$53$	$1696$ $(2^5\cdot 53)$
$59$	$1888$ $(2^5\cdot 59)$
$61$	$1952$ $(2^5\cdot 61)$
$67$	$4208$ $(2^6\cdot 67)$
$71$	$4544$ $(2^6\cdot 71)$
$73$	$4672$ $(2^6\cdot 73)$
$79$	$5056$ $(2^6\cdot 79)$
$83$	$5312$ $(2^6\cdot 83)$
$89$	$5696$ $(2^6\cdot 89)$
$97$	$6208$ $(2^6\cdot 97)$
$101$	$6464$ $(2^6\cdot 101)$
$103$	$6592$ $(2^6\cdot 103)$
$107$	$6848$ $(2^6\cdot 107)$
$109$	$6976$ $(2^6\cdot 109)$
$113$	$7232$ $(2^6\cdot 113)$
$127$	$16256$ $(2^7\cdot 127)$, $1032256$ $(2^6\cdot {127}^2)$
$131$	$16768$ $(2^7\cdot 131)$
$137$	$17536$ $(2^7\cdot 137)$
$139$	$17792$ $(2^7\cdot 139)$
$149$	$19072$ $(2^7\cdot 149)$
$151$	$19328$ $(2^7\cdot 151)$
$157$	$20096$ $(2^7\cdot 157)$
$163$	$20864$ $(2^7\cdot 163)$
$167$	$21376$ $(2^7\cdot 167)$
$173$	$22144$ $(2^7\cdot 173)$
$179$	$22912$ $(2^7\cdot 179)$
$181$	$23168$ $(2^7\cdot 181)$
$191$	$24448$ $(2^7\cdot 191)$
$193$	$24704$ $(2^7\cdot 193)$
$197$	$25216$ $(2^7\cdot 197)$
$199$	$25472$ $(2^7\cdot 199)$
$211$	$27008$ $(2^7\cdot 211)$
$223$	$28544$ $(2^7\cdot 223)$
$227$	$29056$ $(2^7\cdot 227)$
$229$	$29312$ $(2^7\cdot 229)$
$233$	$29824$ $(2^7\cdot 233)$
$239$	$30592$ $(2^7\cdot 239)$
$241$	$30848$ $(2^7\cdot 241)$
$251$	$32128$ $(2^7\cdot 251)$

Title	determination of even abundant numbers with one odd prime factor
Canonical name	DeterminationOfEvenAbundantNumbersWithOneOddPrimeFactor
Date of creation	2013-03-22 16:47:51
Last modified on	2013-03-22 16:47:51
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	24
Author	rspuzio (6075)
Entry type	Definition
Classification	msc 11A05