Measuring Abundance with Abundancy Index

@inproceedings{Guha2021MeasuringAW,
  title={Measuring Abundance with Abundancy Index},
  author={Kalpok Guha and Sourangshu Ghosh},
  year={2021}
}
A positive integer n is called perfect if σ(n) = 2n, where σ(n) denote the sum of divisors of n. In this paper we study the ratio σ(n) n . We define the function Abundancy Index I : N → Q with I(n) = σ(n) n . Then we study different properties of Abundancy Index and discuss the set of Abundancy Index. Using this function we define a new class of numbers known as superabundant numbers. Finally we study superabundant numbers and their connection with Riemann Hypothesis. 

References

SHOWING 1-10 OF 52 REFERENCES
Abundancy “Outlaws” of the Form σ(N)+t N
The abundancy index of a positive integer n is defined to be the rational number I(n) = �(n)/n, whereis the sum of divisors function �(n) = P d|n d. An abundancy outlaw is a rational number greaterExpand
Abundant Numbers and the Riemann Hypothesis
  • K. Briggs
  • Mathematics, Computer Science
  • Exp. Math.
  • 2006
TLDR
A computational study of the successive maxima of the relative sum-of-divisors function ρ(n) := σ(n)/n, which occurs at superabundant and colossally abundant numbers and the density of these numbers is studied. Expand
Measuring the Abundancy of Integers
The origins of the study of perfect numbers are lost in antiquity, but the concept was clearly recognized well over 2000 years ago and involves such contributors as Euclid, Fermat, Descartes,Expand
Superabundant Numbers and the Riemann Hypothesis
TLDR
This note proposes a method that will establish explicit upper bounds for σ(n)/e n log log n, and observation is that the least number violating the inequality (2) should be a superabundant number. Expand
Robin's inequality and the Riemann hypothesis
Let f(n) = σ(n)/e γ n log log n, n = 3, 4,..., where a denotes the sum of divisors function. In 1984 Robin proved that the inequality f(n) 5041, is equivalent to the Riemann hypothesis. Here we showExpand
The second largest prime divisor of an odd perfect number exceeds ten thousand
TLDR
The latter bound of the statement in the title of this paper is improved, showing that the largest prime divisor of an odd perfect number must exceed 10 6 , and Hagis showed that the second largest must exceeds 10 3 . Expand
The third largest prime divisor of an odd perfect number exceeds one hundred
TLDR
It is proved that the third largest prime divisor of an odd perfect number must exceed 100. Expand
ON THE NONEXISTENCE OF ODD PERFECT NUMBERS
In this article, we show how to prove that an odd perfect number with eight distinct prime factors is divisible by 5. A perfect number N is equal to twice the sum of its divisors: σ(N) = 2N . TheExpand
The Abundancy Ratio, a Measure of Perfection
Acknowledgment. I am indebted to Hessel Pot from Woerden in the Netherlands who in a personal communication to me in 1997 pointed out the additional properties to Theorems 1 and 2, as well as TheoremExpand
Odd perfect numbers are greater than 101500
TLDR
It is obtained that N has at least 101 not necessarily distinct prime factors and that its largest component (i.e. divisor p^a with p prime) is greater than 10^62. Expand
...
1
2
3
4
5
...