OFFSET
1,1
COMMENTS
The numbers x, y and z form a HM(1,3)-amicable triple (HM = harmonic mean). See Dimitrov link. An amicable triple forms a HM(1,3)-amicable triple, so the larger member of an amicable triple A125492 is a term of this sequence.
From David A. Corneth, Jun 20 2025: (Start)
Let sx = sigma(x), sy = sigma(y) and sz = sigma(z).
We may write (1/sx + 1/sy + 1/sz)*(x + y + z) = (1/sx + 1/sy) * (x + y) + 1/sz * (x + y + z) + z * (1/sx + 1/sy). As z > y > x we have 2 * z > x + y so z > (x + y) / 2.
Therefore we have 3 = (1/sx + 1/sy) * (x + y) + 1/sz * (x + y + z) + z * (1/sx + 1/sy) > 1.5*(1/sx + 1/sy) * (x + y) + 1/sz * (x + y + z) > 1.5*(1/sx + 1/sy) * (x + y) and so (1/sx + 1/sy) * (x + y) < 2. Possibly 2 could be tightened due to 1/sz * (x + y + z) which is discarded for now.
If we know (1/sx + 1/sy) * (x + y) < U for some U then similarly x/s(x) can be tightened to 0.5*U and maybe more due to term 1/sy * (x + y).
Furthermore 3 = (1/sx + 1/sy + 1/sz)*(x + y + z) > 1/sx * (x + y + z) > 1/sx * (x + y + y) = 1/sx * (x + 2*y) which constraints y and later on z once 1/sx is known.
For every pair (k, m) in {(x, y), (x, z), (y, z)} from solutions (x, y, z) where z <= 10000 we have (k + m) * (1/sigma(k) + 1/sigma(m)) <= 1.5. Is that the case for every solution? (End).
LINKS
David A. Corneth, PARI program
S. I. Dimitrov, Generalizations of amicable numbers, arXiv:2408.07387 [math.NT], 2024.
EXAMPLE
(840, 1020, 1380) is such a triple because (1/sigma(840) + 1/sigma(1020) + 1/sigma(1380))*(840 + 1020 + 1380) = 3.
PROG
(PARI) \\ See Corneth link
CROSSREFS
KEYWORD
nonn,hard
AUTHOR
S. I. Dimitrov, Jun 19 2025
EXTENSIONS
Corrected and extended by David A. Corneth, Jun 20 2025
STATUS
approved
