OFFSET
1,1
COMMENTS
It is not known whether sigma(k)/k = 5/3 for any k, so conjecturally, these are all numbers k such that sigma(k)/k > 5/3. However, if such a k exists, then it must by necessity be an odd square that is a multiple of 3, but not a multiple of 5 (and thus in A388016). Furthermore, 5*k is then an odd perfect number. See theorem 3 in Weiner paper.
Note that abundancy ratio 127/63 has similar properties, as it is an abundancy ratio of an odd square divisible by 3, but here we know that such an odd square exists, being 173369889 = (3*3*7*11*19)^2. - Peter Munn and Antti Karttunen, Sep 25 2025
The numbers of terms that do not exceed 10^k, for k = 1, 2, ..., are 4, 39, 426, 4196, 42043, 419385, 4199351, 41981198, 419755905, 4197920239, ... . Apparently, the asymptotic density of this sequence exists and equals 0.419... . - Amiram Eldar, Nov 10 2025
LINKS
Antti Karttunen, Table of n, a(n) for n = 1..10000
C. A. Holdener and J. A. Holdener, Characterizing Quasi-Friendly Divisors, Journal of Integer Sequences, Vol. 23 (2020), Article 20.8.4.
Paul A. Weiner, The abundancy ratio, a measure of perfection, Math. Mag. 73 (4) (2000) 307-310. (See Theorem 3)
MATHEMATICA
A388012Q[k_] := 3*DivisorSigma[1, k] >= 5*k;
Select[Range[200], A388012Q] (* Paolo Xausa, Sep 13 2025 *)
PROG
(PARI) is_A388012(n) = ((sigma(n)/n)>=(5/3));
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Antti Karttunen, Sep 13 2025
STATUS
approved
