A214409 a(n) is the smallest conjectured m such that the irreducible fraction m/n is a known abundancy index.

1, 3, 4, 7, 6, 13, 8, 15, 13, 21, 12, 31, 14, 31, 26, 31, 18, 49, 20, 51, 32, 45, 24, 65, 31, 49, 40, 57, 30, 91, 32, 63, 52, 63, 48, 91, 38, 75, 56, 93, 42, 127, 44, 93, 88, 93, 48, 127, 57, 93, 80, 105, 54, 121, 72, 127, 80, 105, 60, 217, 62, 127, 104, 127

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

Kate Moore A Geometric Representation of the Abundancy Index

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

Equal to or greater than A214413.

Sequence in context: A324108 A324054 A214413 * A215509 A053480 A258052

Adjacent sequences: A214406 A214407 A214408 * A214410 A214411 A214412

Keyword

nonn

Author

Michel Marcus, Jul 16 2012

Status

approved