A307098 The primitive abundant numbers k (A071395) arranged by the decreasing values of their abundancy index sigma(k)/k.

3465, 15015, 4095, 1430, 19635, 16796, 20, 21945, 5355, 692835, 2584, 5985, 23205, 49742, 20332, 22309287, 26565, 188955, 1870, 216315, 838695, 25935, 3128, 22724, 6084351, 7245, 2090, 60214, 2107575, 937365, 1542773001, 25636, 28129101, 33495, 13066965, 3016174

Offset: 1

Author

Amiram Eldar, Mar 25 2019

Keywords

• nonn

Comments

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.

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.

Examples

			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...
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..91
• Graeme L. Cohen, On primitive abundant numbers, Journal of the Australian Mathematical Society, Vol. 34 No. 1 (1983), pp. 123-137.
• Amiram Eldar, Table of n, a(n), prime factorization of a(n), sigma(a(n))/a(n) (rounded) for n = 1..91

Crossrefs

Cf. A000203, A005100, A005101, A071395.