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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A385155 Numbers z such that there exist two integers 0 such that (1/sigma(x) + 1/sigma(y) + 1/sigma(z))*(x + y + z) = 3.

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%I A385155 #24 Jun 25 2025 11:03:14
%S A385155 1380,1540,1560,1638,2016,2250,2520,2556,2700,2772,3024,3120,3312,
%T A385155 3360,3408,3480,3640,3654,3780,3816,3828,3876,4200,4320,4440,4452,
%U A385155 4620,4920,4956,5220,5280,5292,5304,5340,5400,5460,5472,5640,5700,5724,5760,5940,6048,6060,6180
%N A385155 Numbers z such that there exist two integers 0<x<y<z such that (1/sigma(x) + 1/sigma(y) + 1/sigma(z))*(x + y + z) = 3.
%C A385155 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.
%C A385155 From _David A. Corneth_, Jun 20 2025: (Start)
%C A385155 Let sx = sigma(x), sy = sigma(y) and sz = sigma(z).
%C A385155 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.
%C A385155 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.
%C A385155 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).
%C A385155 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.
%C A385155 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).
%H A385155 David A. Corneth, <a href="/A385155/a385155.gp.txt">PARI program</a>
%H A385155 S. I. Dimitrov, <a href="https://arxiv.org/abs/2408.07387">Generalizations of amicable numbers</a>, arXiv:2408.07387 [math.NT], 2024.
%e A385155 (840, 1020, 1380) is such a triple because (1/sigma(840) + 1/sigma(1020) + 1/sigma(1380))*(840 + 1020 + 1380) = 3.
%o A385155 (PARI) \\ See Corneth link
%Y A385155 Cf. A000203, A125492, A384814, A384487.
%K A385155 nonn,hard
%O A385155 1,1
%A A385155 _S. I. Dimitrov_, Jun 19 2025
%E A385155 Corrected and extended by _David A. Corneth_, Jun 20 2025