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a(n) is the smallest conjectured m such that the irreducible fraction m/n is a known abundancy index.
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%I #20 Jan 27 2019 05:42:57

%S 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,

%T 49,40,57,30,91,32,63,52,63,48,91,38,75,56,93,42,127,44,93,88,93,48,

%U 127,57,93,80,105,54,121,72,127,80,105,60,217,62,127,104,127

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

%C 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.

%C 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.

%C 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).

%C When a(n)=A214413(n), this means that a(n) is sure to be the least m satisfying the property.

%H Michel Marcus, <a href="/A214409/b214409.txt">Table of n, a(n) for n = 1..1000</a>

%H Michel Marcus, <a href="/A214409/a214409.txt">Computations for k such that sigma(k)/k = a(n)/n</a>

%H Kate Moore <a href="http://biology.kenyon.edu/HHMI/posters_2011/moorek.pdf">A Geometric Representation of the Abundancy Index</a>

%H W. Nissen <a href="http://upforthecount.com/math/black.html">Augmentation of Table of Abundancies</a>

%H William G. Stanton and Judy A. Holdener, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL10/Holdener/holdener7.html">Abundancy "Outlaws" of the Form (sigma(N) + t)/N</a>, Journal of Integer Sequences , Vol 10 (2007) , Article 07.9.6.

%e 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.

%Y Equal to or greater than A214413.

%K nonn

%O 1,2

%A _Michel Marcus_, Jul 16 2012