determination of abundant numbers with specified prime factors

The formula for sums of factors may be used tofind all abundant numbers with a specified setof prime factors or that no such numbersexist. To accomplish this, we first do alittle algebraic manipulation to our formula.

Theorem 1.

A number $n$ whose factorizationinto prime numbers is ${\mathrm{\prod }}_{i\mathrm{=}\mathrm{1}}^k{p}_i^{m_{\mathrm{1}}}$is abundant if and only if

$$\prod _{i=1}^k\left(\frac{1-p_i^{-m_i-1}}{1-p_i^{-1}}\right)>2.$$

Proof.

By definition $n$ is abundant, if the sum ofthe proper divisors of $n$ is greater than $n$.Using our formula, this is equivalent to the condition

$$\prod _{i=1}^k\left(\frac{p_i^{m_i+1}-1}{p_i-1}\right)>2\prod _{i=1}^k{p}_i^{m_i}.$$

Dividing the $k$-th term in the product onthe left-hand side by the $k$-th term on theright-hand side,

$$\frac{1}{p_i^{m_i}}\frac{p_i^{m_i+1}-1}{p_i-1}=\frac{p_i^{-m_i-1}}{p_i^{-1}}\frac{p_i-p_i^{-m_i}}{p_i-1}=\frac{1-p_i^{-m_i-1}}{1-p_i^{-1}},$$

so the condition becomes

$$\prod _{i=1}^k\left(\frac{1-p_i^{-m_i-1}}{1-p_i^{-1}}\right)>2$$

∎

Note that each of the terms in the product is bigger than 1.Furthemore, the $k$-th term is bounded by

$$\frac{1}{1-p_i^{-1}}=\frac{p_i}{p_i-1}.$$

This means that it is only possible to have an abundantnumber whose prime factors are $\{p_i\mid 1\le i\le k\}$if

$$\prod _{i=1}^k\left(\frac{p_i}{p_i-1}\right)>2.$$

As it turns out, the convers also holds, so we have anice criterion for determining when a set of primenumbers happens to be the set of prime divisors ofan abundant number.

Theorem 2.

A finite set $S$ of prime numbers is the set ofprime divisors of an abundant number if and only if

$$\prod _{p\in S}\left(\frac{p}{p-1}\right)>2.$$

Proof.

As described above, if $S$ is a set of prime factorsof an abundant number, then we may bound each termin the inequality of the previous theorem to obtainthe inequality in the current theorem. Assume, thenthat $S$ is a finite set of prime numbers whichsatisfies said inequality. Then, by continuity,there must exist a real number $ϵ>0$ suchthat

$$\prod _{p\in S}\left(\frac{p}{p-1}-x\right)>2$$

whenver . Since ${lim}_{k\to \mathrm{\infty }}{n}^{-k}=0$ when $n>1$, we can, for every $p\in S$,find an $m(p)$ such that

Hence,

$$\prod _{p\in S}\left(\frac{p}{p-1}-\frac{p^{m(p)}}{p-1}\right)=\prod _{p\in S}\left(\frac{1-p^{-m(p)-1}}{1-p^{-1}}\right)>2$$

so, by the previous theorem, ${\prod }_{p\in S}{p}^{m(p)}$is an abundant number.∎