# determination of abundant numbers with specified prime factors

The formula^{} for sums of factors may be used to
find all abundant numbers with a specified set
of prime factors^{} or that no such numbers
exist. To accomplish this, we first do a
little algebraic manipulation to our formula.

###### Theorem 1.

A number $n$ whose factorization
into 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 of
the 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^{} on
the left-hand side by the $k$-th term on the
right-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 abundant number 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 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

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

###### 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 $\u03f5>0$ such
that

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

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. ∎

Title | determination of abundant numbers with specified prime factors |
---|---|

Canonical name | DeterminationOfAbundantNumbersWithSpecifiedPrimeFactors |

Date of creation | 2013-03-22 16:47:41 |

Last modified on | 2013-03-22 16:47:41 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 12 |

Author | rspuzio (6075) |

Entry type | Theorem |

Classification | msc 11A05 |