OFFSET
1,1
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.
Write the abundancy index of m as abun(m). If a primitive abundant number k does not divide A137825(n), abun(k) < 2 * abun(A137825(n+1)) / abun(A137825(n)). This relationship is useful for constructing the sequence. - Peter Munn, May 08 2026
The first 3 terms of A071395 that do not appear in the first 10000 terms here are 104 = a(244360), 368 = a(72502) and 464 = a(M), where extrapolation of a plot of log(abun(a(n))/2) against 1/log(n) suggests M has about 25 decimal digits. - Peter Munn, Jun 04 2026
LINKS
Peter Munn, Table of n, a(n) for n = 1..10000 (Terms 1..91 from Amiram Eldar)
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
Peter Munn, PARI program.
EXAMPLE
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...
PROG
(PARI) \\ See Links section.
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Amiram Eldar, Mar 25 2019
STATUS
approved