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%I #30 May 12 2019 22:22:14
%S 3465,15015,4095,1430,19635,16796,20,21945,5355,692835,2584,5985,
%T 23205,49742,20332,22309287,26565,188955,1870,216315,838695,25935,
%U 3128,22724,6084351,7245,2090,60214,2107575,937365,1542773001,25636,28129101,33495,13066965,3016174
%N The primitive abundant numbers k (A071395) arranged by the decreasing values of their abundancy index sigma(k)/k.
%C Cohen proved that for any given eps > 0 there are only finitely many primitive abundant numbers k with sigma(k)/k >= 2 + eps. Thus the primitive abundant numbers can be arranged by their decreasing value of their abundancy index. In case of more than one primitive abundant number with the same abundancy index, the terms are ordered by their value.
%C Cohen calculated the first 91 terms of this sequence, all the terms with abundancy index >= 2.05 - see the link for the corresponding values of the abundancy index.
%H Amiram Eldar, <a href="/A307098/b307098.txt">Table of n, a(n) for n = 1..91</a>
%H Graeme L. Cohen, <a href="https://doi.org/10.1017/S1446788700019819">On primitive abundant numbers</a>, Journal of the Australian Mathematical Society, Vol. 34 No. 1 (1983), pp. 123-137.
%H Amiram Eldar, <a href="/A307098/a307098.txt">Table of n, a(n), prime factorization of a(n), sigma(a(n))/a(n) (rounded) for n = 1..91</a>
%e a(1) = 3465 since it is the primitive abundant number (A071395) with the largest possible abundancy index among the primitive abundant numbers: sigma(3465)/3465 = 832/385 = 2.161003...
%Y Cf. A000203, A005100, A005101, A071395.
%K nonn
%O 1,1
%A _Amiram Eldar_, Mar 25 2019
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