On the Radical of Multiperfect Numbers and Applications

Nithin Kavi; Xinyi Zhang; Viraj Jayam; and Ajit Kadaveru

arXiv:1812.11239·math.NT·January 1, 2019

On the Radical of Multiperfect Numbers and Applications

Nithin Kavi, Xinyi Zhang, Viraj Jayam, and Ajit Kadaveru

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TL;DR

This paper investigates bounds on the radical of multiperfect numbers, explores their distribution and representation, and establishes finiteness results for certain multiperfect multirepdigit numbers, assuming the ABC conjecture.

Contribution

It provides new bounds on the radical of multiperfect numbers based on their abundancy index and proves finiteness of multiperfect multirepdigit numbers in any base with digit counts as powers of two.

Findings

01

Bounds on the radical of multiperfect numbers depending on their abundancy index.

02

Conditional results on gaps and polynomial representations of multiperfect numbers assuming ABC conjecture.

03

Finiteness of multiperfect multirepdigit numbers in any base with digit counts as powers of two.

Abstract

It is conjectured that for a perfect number $m,$ $rad (m) ≪ m^{\frac{1}{2}} .$ We prove bounds on the radical of multiperfect number $m$ depending on its abundancy index. Assuming the ABC conjecture, we apply this result to study gaps between multiperfect numbers, multiperfect numbers represented by polynomials. Finally, we prove that there are only finitely many multiperfect multirepdigit numbers in any base $g$ where the number of digits in the repdigit is a power of $2.$ This generalizes previous works of several authors including O. Klurman, F. Luca, P. Polack, C. Pomerance and others.

Equations220

rad (m) < m^{\frac{4 n + 1}{4 n + 4}} .

rad (m) < m^{\frac{4 n + 1}{4 n + 4}} .

rad m < m^{\frac{2 n + 2 α + 1}{2 n + 2 α + 2}} .

rad m < m^{\frac{2 n + 2 α + 1}{2 n + 2 α + 2}} .

2^{n} t = (1 + \frac{1}{p _{1}}) (1 + \frac{1}{p _{2}}) (1 + \frac{1}{p _{3}} + \frac{1}{p _{3}^{2}}) (1 + \frac{1}{p _{4}} + \frac{1}{p _{4}^{2}}) < \frac{4}{3} \cdot \frac{6}{5} \cdot \frac{31}{25} \cdot \frac{57}{49} < 4.

2^{n} t = (1 + \frac{1}{p _{1}}) (1 + \frac{1}{p _{2}}) (1 + \frac{1}{p _{3}} + \frac{1}{p _{3}^{2}}) (1 + \frac{1}{p _{4}} + \frac{1}{p _{4}^{2}}) < \frac{4}{3} \cdot \frac{6}{5} \cdot \frac{31}{25} \cdot \frac{57}{49} < 4.

2^{n} \cdot t \cdot p_{1} \cdot p_{2} \cdot p_{3}^{2} \cdot p_{4}^{2} \cdot p_{5}^{2} = (1 + p_{1}) (1 + p_{2}) (1 + p_{3} + p_{3}^{2}) (1 + p_{4} + p_{4}^{2}) (1 + p_{5} + p_{5}^{2}) .

2^{n} \cdot t \cdot p_{1} \cdot p_{2} \cdot p_{3}^{2} \cdot p_{4}^{2} \cdot p_{5}^{2} = (1 + p_{1}) (1 + p_{2}) (1 + p_{3} + p_{3}^{2}) (1 + p_{4} + p_{4}^{2}) (1 + p_{5} + p_{5}^{2}) .

2^{n} t = (1 + \frac{1}{p _{1}}) (1 + \frac{1}{p _{2}}) (1 + \frac{1}{p _{3}} + \frac{1}{p _{3}^{2}}) (1 + \frac{1}{p _{4}} + \frac{1}{p _{4}^{2}}) (1 + \frac{1}{p _{5}} + \frac{1}{p _{5}^{2}})

2^{n} t = (1 + \frac{1}{p _{1}}) (1 + \frac{1}{p _{2}}) (1 + \frac{1}{p _{3}} + \frac{1}{p _{3}^{2}}) (1 + \frac{1}{p _{4}} + \frac{1}{p _{4}^{2}}) (1 + \frac{1}{p _{5}} + \frac{1}{p _{5}^{2}})

< \frac{4}{3} \cdot \frac{6}{5} \cdot \frac{31}{25} \cdot \frac{57}{49} \cdot \frac{133}{121} < 4.

< \frac{4}{3} \cdot \frac{6}{5} \cdot \frac{31}{25} \cdot \frac{57}{49} \cdot \frac{133}{121} < 4.

σ (p^{e}) = 1 + p + p^{2} + \dots + p^{e - 1} + p^{e} \equiv 1 + 1 + \dots + 1 \equiv 1 mod 2,

σ (p^{e}) = 1 + p + p^{2} + \dots + p^{e - 1} + p^{e} \equiv 1 + 1 + \dots + 1 \equiv 1 mod 2,

σ (p^{e}) = \frac{p ^{e + 1} - 1}{p - 1} = \frac{p ^{s} - 1}{p - 1} 1 \leq i \leq r - 1 \prod p^{s \cdot 2^{i}} - 1 .

σ (p^{e}) = \frac{p ^{e + 1} - 1}{p - 1} = \frac{p ^{s} - 1}{p - 1} 1 \leq i \leq r - 1 \prod p^{s \cdot 2^{i}} - 1 .

\frac{p ^{s} - 1}{p - 1} = p^{s - 1} + p^{s - 2} + \dots + p + 1 \equiv_{2} 1 + 1 + \dots + 1 \equiv_{2} 1.

\frac{p ^{s} - 1}{p - 1} = p^{s - 1} + p^{s - 2} + \dots + p + 1 \equiv_{2} 1 + 1 + \dots + 1 \equiv_{2} 1.

p = (2^{k} \cdot t)^{s} + 1 = (2^{j} \cdot t)^{s} - (2^{j} \cdot t)^{s - 1} (2^{j} \cdot t)^{s - 2} + \dots + (2^{j} \cdot t)^{1}

p = (2^{k} \cdot t)^{s} + 1 = (2^{j} \cdot t)^{s} - (2^{j} \cdot t)^{s - 1} (2^{j} \cdot t)^{s - 2} + \dots + (2^{j} \cdot t)^{1}

2^{n} t p_{1}^{α_{1}} \dots p_{r}^{α_{r}} = (1 + p_{1} + \dots + p_{1}^{α_{1}}) (1 + p_{2} + \dots + p_{2}^{α_{2}}) \dots (1 + p_{r} + \dots + p_{r}^{α_{r}}) .

2^{n} t p_{1}^{α_{1}} \dots p_{r}^{α_{r}} = (1 + p_{1} + \dots + p_{1}^{α_{1}}) (1 + p_{2} + \dots + p_{2}^{α_{2}}) \dots (1 + p_{r} + \dots + p_{r}^{α_{r}}) .

2^{n} t = (1 + \frac{1}{p _{1}} + \dots + \frac{1}{p _{1}^{α_{1}}}) (1 + \frac{1}{p _{2}} + \dots + \frac{1}{p _{2}^{α_{2}}}) \dots (1 + \frac{1}{p _{r}} + \dots + \frac{1}{p _{r}^{α_{r}}}) .

2^{n} t = (1 + \frac{1}{p _{1}} + \dots + \frac{1}{p _{1}^{α_{1}}}) (1 + \frac{1}{p _{2}} + \dots + \frac{1}{p _{2}^{α_{2}}}) \dots (1 + \frac{1}{p _{r}} + \dots + \frac{1}{p _{r}^{α_{r}}}) .

(1 + \frac{1}{p _{1}} + \dots + \frac{1}{p _{1}^{α_{1}}}) (1 + \frac{1}{p _{2}} + \dots + \frac{1}{p _{2}^{α_{2}}}) \dots (1 + \frac{1}{p _{r}} + \dots + \frac{1}{p _{r}^{α_{r}}})

(1 + \frac{1}{p _{1}} + \dots + \frac{1}{p _{1}^{α_{1}}}) (1 + \frac{1}{p _{2}} + \dots + \frac{1}{p _{2}^{α_{2}}}) \dots (1 + \frac{1}{p _{r}} + \dots + \frac{1}{p _{r}^{α_{r}}})

< (1 + \frac{1}{p _{1}} + \dots) (1 + \frac{1}{p _{2}} + \dots) \dots (1 + \frac{1}{p _{r}} + \dots)

< (1 + \frac{1}{p _{1}} + \dots) (1 + \frac{1}{p _{2}} + \dots) \dots (1 + \frac{1}{p _{r}} + \dots)

\leq \frac{3}{2} \cdot \frac{5}{4} \cdot \frac{7}{6} \cdot \frac{11}{10} \cdot \frac{5}{4} \cdot \dots \cdot \frac{5}{4} \leq (\frac{5}{4})^{r - 3} \cdot \frac{3}{2} \cdot \frac{7}{6} \cdot \frac{11}{10} < (\frac{5}{4})^{r} .

\leq \frac{3}{2} \cdot \frac{5}{4} \cdot \frac{7}{6} \cdot \frac{11}{10} \cdot \frac{5}{4} \cdot \dots \cdot \frac{5}{4} \leq (\frac{5}{4})^{r - 3} \cdot \frac{3}{2} \cdot \frac{7}{6} \cdot \frac{11}{10} < (\frac{5}{4})^{r} .

\frac{c - 1}{c + 2} \leq \frac{1}{k _{1}} + \frac{1}{k _{1} k _{2}} + \dots + \frac{1}{k _{1} k _{2} \dots k _{e}} \leq \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2 ^{e}} \leq \frac{2 ^{e} - 1}{2 ^{e}} .

\frac{c - 1}{c + 2} \leq \frac{1}{k _{1}} + \frac{1}{k _{1} k _{2}} + \dots + \frac{1}{k _{1} k _{2} \dots k _{e}} \leq \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2 ^{e}} \leq \frac{2 ^{e} - 1}{2 ^{e}} .

rad (m) < m^{\frac{4 n + 1}{4 n + 4}} .

rad (m) < m^{\frac{4 n + 1}{4 n + 4}} .

k m = (1 + p_{1} + \dots + p_{1}^{α_{1}}) \dots (1 + p_{r} + \dots + p_{r}^{α_{r}}) .

k m = (1 + p_{1} + \dots + p_{1}^{α_{1}}) \dots (1 + p_{r} + \dots + p_{r}^{α_{r}}) .

m = p_{1} p_{2} \dots p_{g} p_{g + 1}^{2 α_{g + 1} + 1} \dots p_{s}^{2 α_{s} + 1} p_{s + 1}^{2 α_{s + 1}} \dots p_{r}^{2 α_{r}}

m = p_{1} p_{2} \dots p_{g} p_{g + 1}^{2 α_{g + 1} + 1} \dots p_{s}^{2 α_{s} + 1} p_{s + 1}^{2 α_{s + 1}} \dots p_{r}^{2 α_{r}}

β = \frac{4 n + 1}{4 n + 4} .

β = \frac{4 n + 1}{4 n + 4} .

p_{1} p_{2} \dots p_{r} > p_{1}^{β} p_{2}^{β} \dots p_{g}^{β} p_{g + 1}^{2 α_{g + 1} β + β} \dots p_{s}^{2 α_{s} β + β} p_{s + 1}^{2 α_{s + 1} β} \dots p_{r}^{2 α_{r} β} .

p_{1} p_{2} \dots p_{r} > p_{1}^{β} p_{2}^{β} \dots p_{g}^{β} p_{g + 1}^{2 α_{g + 1} β + β} \dots p_{s}^{2 α_{s} β + β} p_{s + 1}^{2 α_{s + 1} β} \dots p_{r}^{2 α_{r} β} .

p_{1} p_{2} \dots p_{g} > p_{g + 1}^{\frac{2 α _{g + 1} β + β - 1}{1 - β}} \dots p_{s}^{\frac{2 α _{s} β + β - 1}{1 - β}} p_{s + 1}^{\frac{2 α _{s + 1} β - 1}{1 - β}} \dots p_{r}^{\frac{2 α _{r} β - 1}{1 - β}} .

p_{1} p_{2} \dots p_{g} > p_{g + 1}^{\frac{2 α _{g + 1} β + β - 1}{1 - β}} \dots p_{s}^{\frac{2 α _{s} β + β - 1}{1 - β}} p_{s + 1}^{\frac{2 α _{s + 1} β - 1}{1 - β}} \dots p_{r}^{\frac{2 α _{r} β - 1}{1 - β}} .

2^{n} t p_{1} p_{2} \dots p_{g} p_{g + 1}^{2 α_{g + 1} + 1} \dots p_{s}^{2 α_{s} + 1} p_{s + 1}^{2 α_{s + 1}} \dots p_{r}^{2 α_{r}}

2^{n} t p_{1} p_{2} \dots p_{g} p_{g + 1}^{2 α_{g + 1} + 1} \dots p_{s}^{2 α_{s} + 1} p_{s + 1}^{2 α_{s + 1}} \dots p_{r}^{2 α_{r}}

= (1 + p_{1}) \dots (1 + p_{g}) (1 + p_{g + 1} + \dots + p_{g + 1}^{2 α_{g + 1} + 1}) \dots (1 + p_{s} + \dots + p_{s}^{2 α_{s} + 1})

= (1 + p_{1}) \dots (1 + p_{g}) (1 + p_{g + 1} + \dots + p_{g + 1}^{2 α_{g + 1} + 1}) \dots (1 + p_{s} + \dots + p_{s}^{2 α_{s} + 1})

(1 + p_{s + 1} + \dots + p_{s + 1}^{2 α_{s + 1}}) \dots (1 + p_{r} + \dots + p_{r}^{2 α_{r}}) .

(1 + p_{s + 1} + \dots + p_{s + 1}^{2 α_{s + 1}}) \dots (1 + p_{r} + \dots + p_{r}^{2 α_{r}}) .

p_{g + 1}^{(2 α_{g + 1} + 1) g} > p_{g}^{g} > p_{1} p_{2} \dots p_{g}

p_{g + 1}^{(2 α_{g + 1} + 1) g} > p_{g}^{g} > p_{1} p_{2} \dots p_{g}

> p_{g + 1}^{\frac{2 α _{g + 1} β + β - 1}{1 - β}} \dots p_{s}^{\frac{2 α _{s} β + β - 1}{1 - β}} p_{s + 1}^{\frac{2 α _{s + 1} β - 1}{1 - β}} \dots p_{r}^{\frac{2 α _{r} β - 1}{1 - β}} > p_{g + 1}^{\frac{2 α _{g + 1} β + β - 1}{1 - β}} .

> p_{g + 1}^{\frac{2 α _{g + 1} β + β - 1}{1 - β}} \dots p_{s}^{\frac{2 α _{s} β + β - 1}{1 - β}} p_{s + 1}^{\frac{2 α _{s + 1} β - 1}{1 - β}} \dots p_{r}^{\frac{2 α _{r} β - 1}{1 - β}} > p_{g + 1}^{\frac{2 α _{g + 1} β + β - 1}{1 - β}} .

2 n α_{g + 1} + n > \frac{( 8 n + 2 ) α _{g + 1} - 3}{3} ⟹ α_{g + 1} < 1.5 ⟹ α_{g + 1} = 1.

2 n α_{g + 1} + n > \frac{( 8 n + 2 ) α _{g + 1} - 3}{3} ⟹ α_{g + 1} < 1.5 ⟹ α_{g + 1} = 1.

p_{g + 1}^{\frac{n + 1}{3}} > p_{g + 2}^{\frac{( 8 n + 2 ) α _{g + 2} - 3}{3}} \dots p_{s}^{\frac{( 8 n + 2 ) α _{s} - 3}{3}} p_{s + 1}^{\frac{( 8 n + 2 ) α _{s + 1} - ( 4 n + 4 )}{3}} \dots p_{r}^{\frac{( 8 n + 2 ) α _{r} - ( 4 n + 4 )}{3}},

p_{g + 1}^{\frac{n + 1}{3}} > p_{g + 2}^{\frac{( 8 n + 2 ) α _{g + 2} - 3}{3}} \dots p_{s}^{\frac{( 8 n + 2 ) α _{s} - 3}{3}} p_{s + 1}^{\frac{( 8 n + 2 ) α _{s + 1} - ( 4 n + 4 )}{3}} \dots p_{r}^{\frac{( 8 n + 2 ) α _{r} - ( 4 n + 4 )}{3}},

p_{g + 1} > p_{g + 2}^{\frac{( 8 n + 2 ) α _{g + 2} - 3}{n + 1}} \dots p_{s}^{\frac{( 8 n + 2 ) α _{s} - 3}{n + 1}} p_{s + 1}^{\frac{( 8 n + 2 ) α _{s + 1} - ( 4 n + 4 )}{n + 1}} \dots p_{r}^{\frac{( 8 n + 2 ) α _{r} - ( 4 n + 4 )}{n + 1}} .

p_{g + 1} > p_{g + 2}^{\frac{( 8 n + 2 ) α _{g + 2} - 3}{n + 1}} \dots p_{s}^{\frac{( 8 n + 2 ) α _{s} - 3}{n + 1}} p_{s + 1}^{\frac{( 8 n + 2 ) α _{s + 1} - ( 4 n + 4 )}{n + 1}} \dots p_{r}^{\frac{( 8 n + 2 ) α _{r} - ( 4 n + 4 )}{n + 1}} .

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TopicsAnalytic Number Theory Research · Advanced Mathematical Identities · Algebraic Geometry and Number Theory

Full text

On the Radical of Multiperfect Numbers and Applications

Viraj Jayam

Ajit Kadaveru

Nithin Kavi

and Xinyi Zhang

Abstract

It is conjectured that for a perfect number $m,$ $\operatorname{rad}\left(m\right)\ll m^{\frac{1}{2}}.$ We prove bounds on the radical of multiperfect number $m$ depending on its abundancy index. Assuming the ABC conjecture, we apply this result to study gaps between multiperfect numbers, multiperfect numbers represented by polynomials. Finally, we prove that there are only finitely many multiperfect multirepdigit numbers in any base $g$ where the number of digits in the repdigit is a power of $2.$ This generalizes previous works of several authors including O. Klurman, F. Luca, P. Polack, C. Pomerance and others.

1 Introduction
2 Preliminary Lemmas
3 Bounds on the radical
3.1 Odd Multiperfect Numbers
3.2 Even Multiperfect Numbers
4 ABC Conjecture and Multiperfect Numbers
4.1 Multiperfect numbers and polynomials
4.2 Distance Between Perfect and Multiperfect Numbers
4.3 Multirepdigit multiperfect numbers
4.4 Multiperfect numbers and factorials

1 Introduction

A positive integer $n$ is perfect if $\sigma\left(n\right)=2n$ , where $\sigma\left(n\right)$ is the sum of all the positive integer divisors of $n$ . Perfect numbers have been studied for many years since Euclid proved a formation rule whereby ${2}^{p-1}\left(2^{p}-1\right)$ is an even perfect number whenever $2^{p-1}$ is prime. He also showed that if an odd perfect number were to exist, it would be in the form $p^{\alpha}d^{2},$ where $p\equiv\alpha\equiv 1\pmod{4}.$ In the study of perfect numbers, there are still two famous questions that remain wide open. First, whether there are infinitely many perfect numbers, and second, whether there exists an odd perfect number. Though not many positive results have come with these two problems, several authors have shown that there are finitely many perfect numbers with certain properties. For example, Pollack proved this for numbers with all identical digits in [6], and Luca proved this for Fibonacci numbers in [8]. More up-to-date results and notions related to perfect numbers are introduced in [12]. Luca and Pomerance related the ABC-conjecture to the study of perfect numbers in [3]. They first proved that $\operatorname{rad}\left(n\right)<2n^{\frac{17}{26}}$ . This allowed them to apply the ABC-conjecture to make conclusions about the gaps between perfect numbers. Acquaah and Konyagin gave a simple proof of a slightly weaker inequality: $\operatorname{rad}\left(n\right)<2n^{\frac{2}{3}}$ in [1]. Since then, Klurman proved a stronger bound in [2], namely $\operatorname{rad}\left(n\right)<2n^{\frac{9}{14}}$ . Though not proven yet, it is conjectured that $\operatorname{rad}\left(n\right)<2n^{\frac{1}{2}}$ . In fact, Ellia in [9] along with Ochem and Rao in [10] showed that if $\operatorname{rad}\left(n\right)<n^{\frac{1}{2}}$ , then the special prime $p$ with odd exponent must be greater $222$ , and greater than than $148207$ if $3\nmid n$ .

A positive integer $m$ is multiperfect (or multiply perfect) if $\sigma\left(m\right)=km$ , where $k>2$ is referred to as the abundancy of $m$ , and $m$ is called $k$ -perfect. A method for determining up to 1,000,000,000 multiperfect numbers was first introduced by Carmichael [11] in 1907. Website [13] provides the latest database for all known multiperfect numbers.

In this paper, we prove a bound on the radical of multiperfect number through the following two theorems:

Theorem 1. Suppose $m$ is an odd multiperfect number such that $\sigma\left(m\right)=km.$ Then, if $k$ is odd, we have that $\operatorname{rad}\left(m\right)\leq\sqrt{m}.$ If $k\equiv 2\pmod{4},$ we have $\operatorname{rad}\left(m\right)<m^{\frac{9}{14}}.$ Otherwise, suppose $k=2^{n}t$ , where $t$ is odd and $n\geq 2.$ We have

[TABLE]

Theorem 2. Suppose $m$ is an even multiperfect number such that $\sigma\left(m\right)=km.$ Then if $\nu_{2}\left(m\right)=n$ and $\nu_{2}\left(k\right)=\alpha,$ we have

[TABLE]

We then use this radical bound to show that the ABC-conjecture implies that generic polynomials must have finitely many $k$ -perfect numbers in their range of outputs. This generalizes the work of Klurman [2], who established similar result for perfect numbers (the case $k=1$ ). Finally, we prove an extension of Pollack’s result with numbers having identical digits in [7], to multiperfect numbers with multirepdigits.

Acknowledgments

This research was conducted at the 2018 AwesomeMath Summer Program. We would like to thank Dr. Oleksiy Klurman for being our research advisor and Mr. George Catalin Turcas for providing helpful feedback on our work.

2 Preliminary Lemmas

Lemma 1

A number of the form $p_{1}\cdot p_{2}\cdot p_{3}^{2}\cdot p_{4}^{2}$ for odd primes $p_{1},p_{2},p_{3},p_{4}$ is not a $k$ -perfect number if $4\mid k.$

Proof: Since $k$ is even, let $k=2^{n}\cdot t$ , where $t$ is odd. Then we have the equation $2^{n}\cdot t\cdot p_{1}\cdot p_{2}\cdot{p_{3}}^{2}\cdot{p_{4}}^{2}=\left(1+p_{1}\right)\left(1+p_{2}\right)\left(1+p_{3}+p_{3}^{2}\right)\left(1+p_{4}+p_{4}^{2}\right).$ If we divide both sides by $p_{1}p_{2}p_{3}^{2}p_{4}^{2},$ we get:

[TABLE]

However, since $4|k$ , we have that $n\geq 2,$ so this is impossible. $\square$

Lemma 2

A number of the form $p_{1}\cdot p_{2}\cdot p_{3}^{2}\cdot p_{4}^{2}\cdot p_{5}^{2}$ is not a $k$ -perfect number for any $k$ divisible by $4.$

Proof: Since $k$ is even, let $k=2^{n}\cdot t$ , where $t$ is odd. Then we have the equation

[TABLE]

If we divide both sides by $p_{1}p_{2}p_{3}^{2}p_{4}^{2}p_{5}^{2},$ we get:

[TABLE]

However, since $4|k$ , we have that $n\geq 2,$ so this is impossible. $\square$

Lemma 3

For a prime number $p$ and odd number $e$ , we have $\nu_{2}\left(\sigma\left(p^{e}\right)\right)=\nu_{2}\left(e+1\right)+\nu_{2}\left(p+1\right)-1$ .

Proof: Let us consider two separate cases: when $e$ is even and when it is odd. If $e$ is even,

[TABLE]

This clearly contradicts the lemma. Our next case is when $e$ is odd. We may express $e+1=2^{r}\cdot s$ , for some $r$ and odd $s$ . Thus $\sigma\left(p^{e}\right)$ can be expressed as below:

[TABLE]

For every $p^{2^{j}*s}+1$ term, where $i\geq 1$ , $p^{2^{j}\cdot s}\equiv_{4}1$ . Therefore, $2\parallel p^{2^{j}\cdot s}+1$ . We now know that since $s$ is odd, we have

[TABLE]

If we let

[TABLE]

Thus, we have that $j=\nu_{2}\left(p+1\right)$ , and $2\parallel p^{s}+1$ . Finally, we conclude that $\nu_{2}\left(\sigma\left(p^{e}\right)\right)=\nu_{2}\left(p+1\right)+\nu_{2}\left(e+1\right)-1$ . $\square$

Lemma 4

Suppose $m$ is a $k$ -perfect number, where $k$ is even and $m$ is odd. Then if $r$ is the number of distinct prime factors of $m$ , we have that $\nu_{2}\left(k\right)<\frac{1}{3}r$ if $r\geq 4.$

Proof: Letting $k=2^{n}t$ where $t$ is odd, we have the equation:

[TABLE]

We divide both sides by $p_{1}^{\alpha_{1}}\ldots p_{r}^{\alpha_{r}}$ to get:

[TABLE]

Suppose that each of these series were infinite. We have:

[TABLE]

From this, we get $2^{n}\leq 2^{n}t<\left(\frac{5}{4}\right)^{r}.$ Thus, we have that $n<r\log_{2}{1.25}<\frac{1}{3}r$ as desired. $\square$

Lemma 5

Suppose $A=k_{e}k_{e-1}\ldots k_{1}$ and $B=k_{e}k_{e-1}\ldots k_{2}+k_{e}k_{e-1}\ldots k_{3}+\ldots+k_{e}k_{e-1}+k_{e}$ where $k_{i}\geq 2.$ Then if the integer $p_{y}\geq 3\cdot 2^{e},$ it is impossible to have $Ap_{y}-B\mid p_{y}^{2}+p_{y}+1.$

Proof: Suppose $Ap_{y}-B\mid p_{y}^{2}+p_{y}+1.$ Then $Ap_{y}-B\mid Ap_{y}^{2}+Ap_{y}+A-\left(Ap_{y}^{2}-Bp_{y}\right)=Ap_{y}+Bp_{y}+A.$ From this, we have $Ap_{y}-B\mid Bp_{y}+A+B.$ Then $Ap_{y}-B\leq Bp_{y}+A+B,$ so $p_{y}\leq\frac{A+2B}{A-B}.$ Then if $p_{y}\geq c,$ we have $Ac-Bc\leq A+2B,$ so $A\left(c-1\right)\leq B\left(c+2\right).$ Thus $\frac{c-1}{c+2}\leq\frac{B}{A}.$ Plugging in $A$ and $B$ gives us:

[TABLE]

Then $2^{e}c-2^{e}\leq 2^{e}c-c+2\cdot 2^{e}-2,$ which is false for all $c\geq 3\cdot 2^{e}-2.$ Since $p_{y}\geq 3\cdot 2^{e}-2,$ this is impossible as desired. $\square$

3 Bounds on the radical

3.1 Odd Multiperfect Numbers

Theorem 6

Suppose $m$ is an odd multiperfect number such that $\sigma\left(m\right)=km.$ Then, if $k$ is odd, we have that $\operatorname{rad}\left(m\right)\leq\sqrt{m}.$ If $k\equiv 2\pmod{4},$ we have $\operatorname{rad}\left(m\right)<m^{\frac{9}{14}}.$ Otherwise, suppose $k=2^{n}t$ , where $t$ is odd and $n\geq 2.$ We have

[TABLE]

Proof: We begin by considering if $k,m$ are both odd in the equation $\sigma\left(m\right)=km.$ Let $m=p_{1}^{\alpha_{1}}\ldots p_{r}^{\alpha_{r}}.$ We have:

[TABLE]

Since $k,m$ are both odd, so is $km.$ This means that the product on the right in the equation above is also odd, so each sum of the powers of the $p_{i}$ must be odd for all $i$ where $1\leq i\leq r$ . This implies that all of the nonzero $\alpha_{i}$ are even, so they are at least $2.$ Therefore $\operatorname{rad}\left(m\right)\leq\sqrt{m}$ if $k,m$ are both odd. Now we consider what happens when $k$ is even.

If $m$ is squarefull, then its radical is clearly no larger than $\sqrt{m}.$ Therefore, we may assume that some of the prime factors of $m$ are raised to only the first power. Let

[TABLE]

and

[TABLE]

Suppose by contradiction that $\operatorname{rad}\left(m\right)>m^{\beta}.$ This would give us:

[TABLE]

From this, we get:

[TABLE]

Since $\sigma\left(m\right)=km=2^{n}tm,$ we have:

[TABLE]

Without loss of generality, let $p_{1}<p_{2}<\ldots<p_{g}.$ This means $p_{g}$ cannot divide $1+p_{1},1+p_{2},\ldots,1+p_{g}.$ We have 2 cases: either $p_{g}|1+p_{g+1}+\ldots+p_{g+1}^{2\alpha_{g+1}+1}$ or $p_{g}|1+p_{s+1}+\ldots+p_{s+1}^{2\alpha_{s+1}}$ where $p_{g+1}$ and $p_{s+1}$ are arbitrary primes with odd and even exponents, respectively.

Case 1: $p_{g}|1+p_{g+1}+\ldots+p_{g+1}^{2\alpha_{g+1}+1}.$ We have $p_{g}<p_{g+1}^{2\alpha_{g+1}+1}\left(1+\epsilon\right).$ Combining this with our above inequality, we have:

[TABLE]

From $n\geq s\geq g,$ we get that $2n\alpha_{g+1}+n>\frac{2\alpha_{g+1}\beta+\beta-1}{1-\beta}.$ From $\beta=\frac{4n+1}{4n+4}$ we get:

[TABLE]

Then,

[TABLE]

or

[TABLE]

Now consider what $p_{g+1}$ divides. Suppose $p_{g+1}\mid 1+p_{g+2}+\ldots+p_{g+2}^{2\alpha_{g+2}+1}.$ Then $2\alpha_{g+2}+1>\frac{\left(8n+2\right)\alpha_{g+2}-3}{n+1},$ or $6n\alpha_{g+2}<n+4.$ Contradiction. Suppose $p_{g+1}\mid 1+p_{s+1}+\ldots+p_{s+1}^{2\alpha_{s+1}}.$ Then $2\alpha_{s+1}\geq\frac{\left(8n+2\right)\alpha_{s+1}-\left(4n+4\right)}{n+1}.$ This means $6n\alpha_{s+1}\leq 4n+4,$ which we know is false for $n\geq 3.$ If $n=2,$ that forces $\alpha_{s+1}=1$ and means $p_{s+1}$ is the only prime on the right. Then $m$ has exactly $g+2$ prime factors, but by Lemma 5:

[TABLE]

which is false for all $g\geq 1.$ Therefore we have a contradiction for all $n\geq 2.$ The only other possibility is that $p_{g+1}\mid 1+p_{i}$ where $1\leq i\leq g.$ Then $p_{i}\geq 2p_{g+1}-1.$ Since $p_{g+1}$ was too large to divide the series on the right and $p_{i}>p_{g+1},$ the same is true for $p_{i}$ so we must have $p_{i}\mid 1+p_{j}$ and so on. Note

[TABLE]

Writing $p_{j}=k_{1}p_{i}-1$ and so on, we use Lemma 5 to eliminate this case.

Case 2: $p_{g}|1+p_{s+1}+\ldots+p_{s+1}^{2\alpha_{s+1}}$ From $p_{g}|1+p_{s+1}+\ldots+p_{s+1}^{2\alpha_{s+1}},$ we have that

[TABLE]

Combining this with our inequality above, we have:

[TABLE]

From this we get that $2\alpha_{s+1}g>\frac{2\alpha_{s+1}\beta-1}{1-\beta}.$ From $n\geq g$ and $\beta=\frac{4n+1}{4n+4}$ we have:

[TABLE]

This implies that $\alpha_{s+1}\leq 2,$ but if $\alpha_{s+1}=2,$ then $p_{s+1}$ is the only prime on the right in the above inequality, violating Lemma 5. Therefore, assume $\alpha_{s+1}=1.$ Plugging this into the inequality above, we get:

[TABLE]

From $\beta=\frac{4n+1}{4n+4},$ we have that $\frac{2\beta-1}{1-\beta}=\frac{4n-2}{3}.$ We have:

[TABLE]

Raising both sides to the $\frac{3}{2n+2}$ power gives us:

[TABLE]

Now we consider what $p_{s+1}$ divides. Suppose $p_{s+1}^{2}\mid 1+p_{g+1}+\ldots+p_{g+1}^{2\alpha_{g+1}+1}.$ Then

[TABLE]

Then $\alpha_{g+1}+1\geq\frac{\left(8n+2\right)\alpha_{g+1}-3}{2n+2}.$ This is impossible for $n\geq 2$ , contradiction. Now suppose $p_{s+1}\mid 1+p_{g+1}+\ldots+p_{g+1}^{2\alpha_{g+1}+1}$ and $p_{s+1}\mid 1+p_{g+2}+\ldots+p_{g+2}^{2\alpha_{g+2}+1}.$ Then

[TABLE]

Contradiction. Now suppose $p_{s+1}^{2}\mid 1+p_{s+2}+\ldots+p_{s+2}^{2\alpha_{s+2}}.$ Then:

[TABLE]

Contradiction. Now suppose $p_{s+1}\mid 1+p_{s+2}+\ldots+p_{s+2}^{2\alpha_{s+2}}$ and $p_{s+1}\mid 1+p_{s+3}+\ldots+p_{s+3}^{2\alpha_{s+3}}.$ Then:

[TABLE]

Contradiction. Now suppose $p_{s+1}\mid 1+p_{g+1}+\ldots+p_{g+1}^{2\alpha_{g+1}+1}$ and $1+p_{s+2}+\ldots+p_{s+2}^{2\alpha_{s+2}}.$ Then:

[TABLE]

Contradiction. Now suppose that $p_{s+1}\mid 1+p_{i}$ and $p_{s+1}\mid 1+p_{g+1}+\ldots+p_{g+1}^{2\alpha_{g+1}+1}.$ Then:

[TABLE]

From this we have that $2\alpha_{g+1}+1\geq\frac{\left(8n+2\right)\alpha_{g+1}-3}{2n+2},$ or $\alpha_{g+1}\leq\frac{2n+5}{4n-2}.$ For $n\geq 4$ this is clearly impossible. Thus we only need to consider if $n=2$ or $n=3.$

Case 2.1: $n=2$ The inequality above implies $\alpha_{g+1}=1.$ We recall that $1\leq g\leq s\leq n=2.$ If $g=s=2,$ then $p_{3}\mid 1+p_{3}+\ldots+p_{3}^{2\alpha_{3}+1},$ which is absurd. Thus we would have to have $g=1$ and $s=2.$

However, this means that $p_{3}\mid 1+p_{1},$ $p_{3}\mid 1+p_{2}+p_{2}^{2}+p_{2}^{3}$ and $p_{1}\mid 1+p_{3}+p_{3}^{2}.$ This is impossible by Lemma 5.

Case 2.2: $n=3$ As before, we can’t have $g=s.$ Thus we have $\left(g,s\right)=\left(1,2\right),\left(1,3\right)$ or $\left(2,3\right).$ Since $p_{g+1}$ is odd, we note that we have $4\mid 1+p_{g+1}+p_{g+1}^{2}+p_{g+1}^{3}.$ This means $s<n,$ so the only case to consider is $g=1$ and $s=2,$ which is eliminated analogously to case 2.1. Now suppose that $p_{s+1}\mid 1+p_{i}$ and $p_{s+1}\mid 1+p_{s+2}+\ldots+p_{s+2}^{2\alpha_{s+2}}.$ From $p_{s+1}\mid 1+p_{s+2}+\ldots+p_{s+2}^{2\alpha_{s+2}},$ we get that:

[TABLE]

This forces $\alpha_{s+2}=1.$ Now suppose $p_{s+2}\mid 1+p_{s+3}+\ldots+p_{s+3}^{2\alpha_{s+3}}.$ Then, using the implied inequalities, we get that $\alpha_{s+3}\leq\frac{2n+2}{4n-5.}$ This is impossible for all $n\geq 4,$ and $n=2$ and $n=3$ are eliminated as above. Therefore, we must have that $p_{s+3}\mid 1+p_{d}$ and $1+p_{e}$ where $1\leq d,e\leq g.$ Then we have $p_{d}\mid 1+p_{f}$ and so on, which is eliminated by Lemma 5. $\square$

3.2 Even Multiperfect Numbers

Theorem 7

Suppose $m$ is an even multiperfect number such that $\sigma\left(m\right)=km$ . Let $k=2^{n}t$ , where $t$ is odd and $n\geq 0$ . Suppose $m=2^{\alpha}h$ , where $h$ is odd and $\alpha\geq 1.$ We have

[TABLE]

Proof: Let $m=2^{\alpha}p_{1}p_{2}\ldots p_{i}p_{i+1}^{2\alpha_{i+1}+1}\ldots p_{j}^{2\alpha_{j}+1}p_{j+1}^{2\alpha_{j+1}}\ldots p_{k}^{2\alpha_{k}}$ . Assume that $\operatorname{rad}\left(m\right)>m^{\beta}$ , where $\beta=\frac{2n+2\alpha+1}{2n+2\alpha+2}$ . Without loss of generality, suppose that $p_{1}<p_{2}<\ldots<p_{i}$ . Suppose that for large $m$ , we have

[TABLE]

Since $m$ is multiperfect, we have

[TABLE]

Case 1: If $\alpha=1$ , we have

[TABLE]

If $i=0$ , since $2^{n+1}\mid\sigma\left(m\right)$ , $j\neq 0$ . We have $\operatorname{rad}\left(m\right)\gg\sqrt{m}$ and the result follows. So we can assume that $i\geq 1$ . Since we assume that $\operatorname{rad}\left(m\right)\gg m^{\beta}$ ,

[TABLE]

Since

[TABLE]

so $p_{i}$ must divide one of

[TABLE]

or

[TABLE]

If $p_{i}\mid 1+p_{c}+\cdots+p_{c}^{2\alpha_{c}}$ , combining that

[TABLE]

then $p_{i}<\sigma\left({p_{c}}^{2\alpha_{c}}\right)<\frac{3}{2}{p_{c}}^{2\alpha_{c}}$ . From Lemmas 2 and 3, we have that $m$ must have at least 3 distinct prime divisors. From Lemma 4, we have $i\leq v_{2}\left(m\right)=n+\alpha$ . Thus, we have

[TABLE]

We proceed by comparing exponents of primes. For $\alpha_{c}=1,2\left(n+1\right)=\frac{2\beta+\beta-1}{1-\beta}$ , when $\beta=\frac{2n+2\alpha+1}{2n+2\alpha+3}$ . For $\alpha_{c}\geq 2$ ,

[TABLE]

because $\frac{d}{d\alpha_{c}}\left(2\alpha_{c}\left(n+1\right)\right)=2n+2>\frac{d}{d\alpha_{c}}\left(\frac{2\alpha_{c}\beta+\beta-1}{1-\beta}\right)=n+\frac{3}{2}$ . We have ${p_{c}}^{2\alpha\left(n+1\right)}\leq{p_{c}}^{\frac{2\alpha_{c}\beta+\beta-1}{1-\beta}}$ and $\frac{3}{2}<p_{e}$ , for $1<e\leq k$ . Thus,

[TABLE]

which is a contradiction. Else if $p_{i}\mid 1+p_{d}+\cdots+p_{d}^{2\alpha_{d}+1}$ , then $p_{i}<\frac{3}{2}{p_{d}}^{2\alpha_{d}+1}$ . By analogy with the previous proof ,

[TABLE]

and $\left(2\alpha_{d}+1\right)\left(n+1\right)\leq\frac{\left(2\alpha_{d}+1\right)\beta-1}{1-\beta}$ , for $\beta=\frac{2\alpha+2n+1}{2\alpha+2n+2}$ , which is a contradiction.

Case 2: If $\alpha\geq 2$ , we have

[TABLE]

Similar to Case 1, we can assume that $i\geq 1$ . If $p_{i}\mid 1+p_{c}+\cdots+p_{c}^{2\alpha_{c}}$ , where $\left(c\in[i+1,j]\right)$ , we have

[TABLE]

Also,

[TABLE]

for $\beta=\frac{2\alpha+2n+1}{2\alpha+2n+3}$ , which is impossible. Else if $p_{i}\mid 1+p_{d}+\cdots+p_{d}^{2\alpha_{d}+1}$ , we have

[TABLE]

Also, ${p_{d}}^{\left(2\alpha_{d}+1\right)\left(n+\alpha\right)}\leq{p_{d}}^{\frac{2\alpha_{d}\beta+\beta-1}{1-\beta}}$ for $\beta=\frac{2\alpha+2n+1}{2\alpha+2n+2}$ , which is again impossible. $\square$

4 ABC Conjecture and Multiperfect Numbers

We proceed by applying our bounds to study the gaps between multiperfect and perfect numbers. Recall the ABC Conjecture, which states that for a fixed $\epsilon>0$ , there exists a constant $C_{\epsilon}$ (dependent on $\epsilon$ ) such that for all coprime $a,b\in\mathbb{N}_{>0}$ and $a+b=c$ , the following inequality is true:

[TABLE]

4.1 Multiperfect numbers and polynomials

Theorem 8

Assume that the ABC conjecture is true. Suppose that $f\left(x,y\right)\in\mathbb{Z}[x,y]$ is homogeneous, without any repeated linear factors. Fix $\epsilon>0$ . Then, for any coprime integers $m$ , $n$ ,

[TABLE]

.

Theorem 9

Assume that the ABC conjecture is true. Let $f\left(X\right)\in\mathbb{Z}[X]$ be a polynomial of degree $d\geq 1$ without repeated roots. Fix $\epsilon>0$ . Then

[TABLE]

.

Proposition 10

Assume that the ABC conjecture is true. Suppose $P\left(x\right)\in\mathbb{Z}[x]$ of degree $>\lfloor{\frac{4n+4}{3}}\rfloor$ where $n=\nu_{2}\left(k\right)\geq 2$ has no repeated factors. Then, there are only finitely many integers $q$ , such that $P\left(q\right)$ is an odd multiperfect number.

Proof: Suppose $P\left(q\right)$ is an odd multiperfect number with a large value of $q$ , $n=\nu_{2}{k}\geq 2$ , and $\deg{P}=d>\lfloor{\frac{4n+4}{3}}\rfloor$ . Fix $\epsilon>0$ . By Theorems $1$ , and $7$ ,

[TABLE]

Combining the two,

[TABLE]

Since it is possible to choose $\epsilon$ such that $d\leq\frac{4n+4}{3}\left(1+\epsilon\right)<\lfloor{\frac{4n+4}{3}}\rfloor+1$ , the contradiction implies the desired result. $\square$

Proposition 11

Assume that the ABC- conjecture is true. Let $f\left(x,y\right)\in\mathbb{Z}[x]$ be a homogeneous form of degree $>2\lfloor{\frac{4n+4}{3}}\rfloor$ where $n=\nu_{2}\left(k\right)\geq 2$ without repeated linear factors. Then there are only finitely many perfect numbers of the form $f\left(m,n\right)$ for $m,n\in\mathbb{Z}$ .

Proof: By Theorems 6 and 8, we have:

[TABLE]

Choosing $\epsilon$ small enough yields the desired conclusion. $\square$

4.2 Distance Between Perfect and Multiperfect Numbers

Luca and Pomerance [3] showed that under the assumption of the ABC conjecture, the equation

[TABLE]

has only finitely many solutions in perfect numbers $x$ and $y$ when $k$ is odd. They also prove that Equation 1 has finitely many solutions when $x,y$ are perfect and squarefull [8]. We consider similar question: how close can perfect ad multiperfect number come together?

Proposition 12

Assume the ABC Conjecture holds. Then there exists finitely many solutions to Equation 1 when $k$ is odd, $x$ is an even perfect number and $y$ is an odd multiperfect number.

Proposition 13

Assume the ABC Conjecture holds. Then there exists finitely many solutions to Equation 1 when $k$ is odd, $x$ is an even perfect number and $y$ is an even multiperfect number.

The proofs of Propositions 12 and 13 are analogous to those of Luca and Pomerance in [3]. The only difference is inserting the bounds on radical in Theorems 6 and 7 for even and odd multiperfect numbers.

4.3 Multirepdigit multiperfect numbers

Pollack recently proved [7] that there exist finitely many ”repdigit” perfect and multiperfect numbers. Along with Luca, he also showed that there exist only finitely many repdigit multiperfect numbers in any base g [6]. We will prove the following extension:

Theorem 14

There exist only finitely many multiperfect multirepdigit numbers in any base g if the abundancy of the multiperfect number is a power of $2.$

While we define repdigit numbers in the following way: consider the Lucas sequence

[TABLE]

the repdigit number in some base $g$ will be defined as

[TABLE]

On the other hand, a multirepdigit number will be the case when $D\in\mathbb{N}$ , rather than being restricted by its base. We will focus on the bounding of multirep digit numbers when the abundancy is $2^{s}$ for a positive integer $s$ .

We know that $U_{1}\mid U_{2}\mid U_{4}\mid U_{8}\mid\cdots\mid U_{2^{s}}\mid\cdots$ . Thus, we know that $\frac{\sigma\left(D\cdot U_{2^{s}}\right)}{D\cdot U_{2^{s}}}$ will be strictly increasing when $s=1,2,\cdots$ . We also know that $\frac{\sigma\left(n\right)}{n}=\sum_{d\mid n}{\frac{1}{d}}$ . Since $D$ is not bounded (it ranges throughout the natural numbers), we will work on restricting the value of $l$ rather than $D$ , where $l$ is the abundancy of the multiperfect number. We know that

[TABLE]

We can prove that $\frac{\sigma\left(U_{2^{s}}\right)}{U_{2^{s}}}$ is bounded by some constant in the following way.

Lemma 15

[TABLE]

where $\mathcal{P}$ is a finite set of primes and $\mathcal{P}^{*}$ is the set of natural numbers all of whose prime factors belong to $\mathcal{P}$ .

Proof: We will work with each term independently. To begin, we will insert the Von-Mangoldt Function.

[TABLE]

We may now insert Euler Product to obtain

[TABLE]

Now we will deal with $\sum_{d\in\mathcal{P}^{*}}\Lambda\left(d\right)$ . This is just summing over all $d|n$ of $\Lambda\left(d\right)$ , and thus we can substitute $\log p$ , for some $p\in\mathbb{P}$ back into the expression. We need to then reinsert all the prime factors that do not divide $n$ . By infinite geometric series, we obtain

[TABLE]

and we are done. $\square$

Lemma 16

[TABLE]

where $m>1$ and is an integer.

Proof: We know that

[TABLE]

We also know that if $p\mid U_{m}$ , it implies that $z\left(p\right)>1$ where $z\left(p\right)$ is the content of $p$ ). If $p\nmid g-1$ , then $z\left(p\right)\mid p-1$ .

[TABLE]

If $d$ is a divisor of $m$ and $U_{d}<g^{d}$ , there exists a progression $1\ \left(\mathrm{mod}\ d\right)$ . And hence, it is bounded by $\frac{\log\left(g^{d}\right)}{log\left(d\right)}$ . So, we may apply this in the following:

[TABLE]

This implies that

[TABLE]

If we let $m=q_{1}^{\alpha_{1}}\cdot q_{2}^{\alpha_{2}}\cdots q_{k}^{\alpha_{k}}$ (where $k=\omega\left(m\right)$ ) and $d\mid m$ , then $d=q_{1}^{\beta_{1}}\cdot q_{2}^{\alpha_{2}}\cdots q_{k}^{\alpha_{k}},0\leq\beta_{i}\leq\alpha_{i}$ . Note that the map from $x\mapsto\frac{\log\left(ex\right)}{x}$ is decreasing for $x\geq 1$ . This implies that $\log\left(\frac{\log\left(ed\right)}{d}\right)\leq\frac{\log\left(ed^{\prime}\right)}{d^{\prime}},d^{\prime}=p_{1}^{\beta_{1}}\cdot p_{2}^{\beta_{2}}\cdots p_{k}^{\beta_{k}}$ , where $p_{i}$ is the $i^{th}$ prime. We can let $\mathcal{P},$ or a finite set of primes $={p_{1},p_{2},\cdots,p_{k}}$ .

[TABLE]

Note: The conclusion resulted from Merten’s Estimate and Prime Number Theorem [6]. $\square$

By combining everything, we obtain what we had originally wanted. Recall that we have only proven this when $m$ is a $2^{s}$ -perfect number.

4.4 Multiperfect numbers and factorials

In this section, we collect a few simple observations regarding factorials and multiperfect numbers.

Proposition 17

The only non-negative integer value of $n$ such that $n!$ is a perfect number is $3$ .

Proof: Clearly, $n=1$ does not work, so $n!$ must be even. We know that

[TABLE]

where $p$ and $2^{p}-1$ are prime. Also we can assume $n\geq 3$ since $n=2$ also does not work. So $3\mid{n!=2^{p-1}*\left(2^{p}-1\right)}\implies 3\mid{2^{p}-1}\implies 2^{p}-1=3\implies p=2\implies n=3$ $\square$

Lemma 18

For sufficiently large $n$ , $\operatorname{rad}\left(n!\right)<\left(n!\right)^{\epsilon}$ .

Proof: By Stirling’s approximation, $n!$ can be estimated as $\left(\frac{n}{e}\right)^{n}$ . By the prime number theorem, the number of primes less than $n$ can be estimated as $\frac{n}{\log{n}}$ . So,

[TABLE]

It suffices to show that

[TABLE]

Clearly, if $n$ is sufficiently large:

[TABLE]

$\square$

Proposition 19

For a multiperfect number $m$ where $\sigma(m)=km,$ there exists at most one multiperfect number that can also be expressed as $n!$ for some $n$ for any natural number $k\geq 2.$

Proof: Let $n!=m=2^{\alpha_{1}}3^{\alpha_{2}}\dots p_{r}^{\alpha_{r}}$ , where $m$ is the smallest multiperfect number such that $\sigma\left(m\right)=km$ for a given $k$ . Then,

[TABLE]

Assume that there exists $n^{\prime}!=m^{\prime}=2^{\beta_{1}}3^{\beta_{2}}\dots p_{c}^{\beta_{c}}$ , where $n^{\prime}>n$ and $c\geq r$ , such that $m^{\prime}$ is the second smallest k-multiperfect number. We have $\beta_{i}\geq\alpha_{i}$ for every $i\leq c$ , because $n^{\prime}>n$ . Hence,

[TABLE]

When comparing each bracket on both side of the equation, for every $i\leq r$ , since $v_{p_{i}}\left(m\right)\leq v_{p_{i}}\left(m^{\prime}\right),$ or in other words $\alpha_{i}\leq\beta_{i}$ ,

[TABLE]

For every $r<j\leq c$ ,

[TABLE]

Since $n!=m\neq\left(n^{\prime}\right)!=m^{\prime}$ ,

[TABLE]

which is impossible. Hence, for every given natural number $k$ , there exists at most one multiperfect number $m$ such that $\sigma\left(m\right)=km$ as desired. $\square$

Next result concerns shifted factorials, that have been extensively studied by many authors throughout the years.

Proposition 20

Assume the ABC conjecture is true. There exists finitely many $n$ such that $n!+1$ is a multi-perfect number.

Proof: Let $n!+1=m$ , where $m$ is a multi-perfect number. By the ABC conjecture,

[TABLE]

where $t$ is the largest power of $2$ that divides m. Clearly, we can now choose an $\epsilon>0$ such that this does not hold. So, we have a contradiction, implying that there are finitely many multiperfect numbers of the form $n!+1$ . $\square$

Bibliography13

The reference list from the paper itself. Each links out to its DOI / PubMed record.

1[1] Acquaah, Peter, and Sergei Konyagin. On prime factors of odd perfect numbers . International Journal of Number Theory 8.06 (2012): 1537-1540.
2[2] Klurman, Oleksiy. Radical of Perfect Numbers and Perfect Numbers among Polynomial Values . International Journal of Number Theory 12.03 (2016): 585-591.
3[3] Luca, Florian, and Carl Pomerance. On the radical of a perfect number . New York J. Math 16 (2010): 23-30.
4[4] Ellia, Philippe. On the distance between perfect numbers . ar Xiv preprint ar Xiv:1210.0450 (2012).
5[5] Broughan, Kevin A., and Zhou, Qizhi. Odd multiperfect numbers of abundancy 4 . Journal of Number Theory 128.6 (2008): 1566-1575.
6[6] Luca, Florian, and Pollack, Paul. Multiperfect Numbers with Identical Digits . Journal of Number Theory 131.2 (2011): 260-284.
7[7] Pollack, Paul. Perfect Numbers with Identical Digits . Integers 11.4 (2011): 519-529.
8[8] Luca, Florian. Perfect Fibonacci and Lucas numbers . Rendiconti del Circolo Matematico di Palermo 49.2 (2000): 313-318.

TL;DR

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On the Radical of Multiperfect Numbers and Applications

Abstract

Contents

1 Introduction

Acknowledgments

2 Preliminary Lemmas

Lemma 1

Lemma 2

Lemma 3

Lemma 4

Lemma 5

3 Bounds on the radical

3.1 Odd Multiperfect Numbers

Theorem 6

3.2 Even Multiperfect Numbers

Theorem 7

4 ABC Conjecture and Multiperfect Numbers

4.1 Multiperfect numbers and polynomials

Theorem 8

Theorem 9

Proposition 10

Proposition 11

4.2 Distance Between Perfect and Multiperfect Numbers

Proposition 12

Proposition 13

4.3 Multirepdigit multiperfect numbers

Theorem 14

Lemma 15

Lemma 16

4.4 Multiperfect numbers and factorials

Proposition 17

Lemma 18

Proposition 19

Proposition 20