OFFSET
1,2
COMMENTS
The abundancy index of a number k is sigma(k)/k. When n is prime, (n+1)/n is irreducible and abund(n) = (n+1)/n, so a(n) = n + 1.
A known abundancy index is related to a limit. Terms of the sequence have been built with a limit of 10^30. So when n is composite, the values for a(n) are conjectural. A higher limit could provide smaller values.
If m < k < sigma(m) and k is relatively prime to m, then k/m is an abundancy outlaw. Hence if r/s is an abundancy index with gcd(r, s) = 1, then r >= sigma(s). [Stanton and Holdener page 3]. Since a(n) is coprime to n, this implies that a(n) >= sigma(n).
When a(n)=A214413(n), this means that a(n) is sure to be the least m satisfying the property.
LINKS
Michel Marcus, Table of n, a(n) for n = 1..1000
Michel Marcus, Computations for k such that sigma(k)/k = a(n)/n
W. Nissen Augmentation of Table of Abundancies
William G. Stanton and Judy A. Holdener, Abundancy "Outlaws" of the Form (sigma(N) + t)/N, Journal of Integer Sequences , Vol 10 (2007) , Article 07.9.6.
EXAMPLE
For n = 5, a(5) = 6 because 6/5 is irreducible, 6/5 is a known abundancy (namely of 5), and no number below 6 can be found with the same properties.
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Marcus, Jul 16 2012
STATUS
approved