A214409 -id:A214409 - cp's OEIS frontend

A214413 a(n) is the smallest m such that the irreducible fraction m/n is not an abundancy outlaw.

1, 3, 4, 7, 6, 13, 8, 15, 13, 19, 12, 29, 14, 25, 26, 31, 18, 41, 20, 47, 32, 37, 24, 65, 31, 43, 40, 57, 30, 73, 32, 63, 50, 55, 48, 91, 38, 61, 56, 93, 42, 97, 44, 85, 82, 73, 48, 125, 57, 93, 74, 99, 54, 121, 72, 125, 80, 91, 60, 169, 62, 97, 104, 127, 84

Offset: 1

Michel Marcus, Jul 22 2012

nonn

The theorem on page 7 of Stanton and Holdener gives conditions for a rational to be an abundancy outlaw.

For a given n, these conditions have been checked by starting with m/n=sigma(n)/n and then increasing m until they fail.

			a(3) = 4 because 4/3 is the abundancy index of 3, so 4/3 is not an abundancy outlaw.

Michel Marcus, Table of n, a(n) for n = 1..1000
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.

Smaller than or equal to A214409.

A215509 Numerator of sigma(n)/n when n belongs to A162657.

Original entry on oeis.org

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

Offset: 1

Michel Marcus, Aug 14 2012

nonn

When p is prime, A162657(p)=p so a(p)=p+1.

			A162657(2) = 2, so a(2)=numerator(sigma(2)/2)=numerator(3/2)=3.

Michel Marcus, Table of n, a(n) for n = 1..1000

Equal to or greater than A214409.

cp's OEIS Frontend

A214413 a(n) is the smallest m such that the irreducible fraction m/n is not an abundancy outlaw.

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