# determination of abundant numbers with specified prime factors

###### Theorem 1.

A number $n$ whose factorization into prime numbers  is $\prod_{i=1}^{k}p_{i}^{m_{1}}$ is abundant if and only if

 $\prod_{i=1}^{k}\left({1-p_{i}^{-m_{i}-1}\over 1-p_{i}^{-1}}\right)>2.$
###### Proof.

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

 $\prod_{i=1}^{k}\left({p_{i}^{m_{i}+1}-1\over p_{i}-1}\right)>2\prod_{i=1}^{k}p% _{i}^{m_{i}}.$

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

 ${1\over p_{i}^{m_{i}}}{p_{i}^{m_{i}+1}-1\over p_{i}-1}={p_{i}^{-m_{i}-1}\over p% _{i}^{-1}}{p_{i}-p_{i}^{-m_{i}}\over p_{i}-1}={1-p_{i}^{-m_{i}-1}\over 1-p_{i}% ^{-1}},$

so the condition becomes

 $\prod_{i=1}^{k}\left({1-p_{i}^{-m_{i}-1}\over 1-p_{i}^{-1}}\right)>2$

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

 ${1\over 1-p_{i}^{-1}}={p_{i}\over p_{i}-1}.$

This means that it is only possible to have an abundant number whose prime factors are $\{p_{i}\mid 1\leq i\leq k\}$ if

 $\prod_{i=1}^{k}\left({p_{i}\over p_{i}-1}\right)>2.$

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

###### Theorem 2.

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

 $\prod_{p\in S}\left({p\over p-1}\right)>2.$
###### Proof.

As described above, if $S$ is a set of prime factors of an abundant number, then we may bound each term in the inequality  of the previous theorem to obtain the inequality in the current theorem. Assume, then that $S$ is a finite set of prime numbers which satisfies said inequality. Then, by continuity, there must exist a real number $\epsilon>0$ such that

 $\prod_{p\in S}\left({p\over p-1}-x\right)>2$

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

 ${p^{m(p)}\over p-1}<\epsilon.$

Hence,

 $\prod_{p\in S}\left({p\over p-1}-{p^{m(p)}\over p-1}\right)=\prod_{p\in S}% \left({1-p^{-m(p)-1}\over 1-p^{-1}}\right)>2$

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

Title determination of abundant numbers with specified prime factors DeterminationOfAbundantNumbersWithSpecifiedPrimeFactors 2013-03-22 16:47:41 2013-03-22 16:47:41 rspuzio (6075) rspuzio (6075) 12 rspuzio (6075) Theorem msc 11A05