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 multiple 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 $2/(2-1)=2$and, for any $p>2$, we have , it follows that, forevery 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 numbersof 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 satisfyour 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 abundantnumber 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 previousresult, $m\le m_0$; 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 abundantnumbers with one odd prime factor none of whose proper factorsare abundant starting with a table of prime numbers, as hasbeen done below. Note that, in the case where $2^{m_0}p$ isperfect, we have listed $2^{m_0+1}p$ and $2^{m_0}{p}^2$ asthese numbers are abundant.