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A376713 a(n) is the minimum number with exactly n triples (u,m,v) of divisors such that u < v and m^2 is the harmonic mean of u^2 and v^2.

%I A376713 #21 Nov 04 2024 12:14:15
%S A376713 35,70,140,210,560,420,2240,840,1260,1680,35840,2100,143360,6720,5040,
%T A376713 4200,1709435,6300,645575,8400,20160,77350,36728125,12600,45360,
%U A376713 430080,31500,33600,1117484375,25200,24171875,29400,154700,3418870,181440,44100,31633175,1291150
%N A376713 a(n) is the minimum number with exactly n triples (u,m,v) of divisors such that u < v and m^2 is the harmonic mean of u^2 and v^2.
%C A376713 For two integers u, v, by definition the harmonic mean m is given by 2/m = 1/u + 1/v.
%C A376713 We observe that a(n) is divisible by 35.
%C A376713 Using the observed (checked through n=200) relation a(n) <= 35*A005179(n), further terms from a(29) to a(52) are (<=9395240960, 25200, <=37580963840, 29400, 154700, 3418870, 181440, 44100, <=2405181685760, 1291150, 1290240, 58800, <=38482906972160, 100800, <=153931627888640, 232050, 126000, <=440401920, <=2462906046218240, 88200, 1632960, 226800, 6837740, 2150400). - _Hugo Pfoertner_, Oct 05 2024
%H A376713 Lucas A. Brown, <a href="https://github.com/lucasaugustus/oeis/blob/main/A376713.py">Python program</a>.
%e A376713 a(1) = 35 is because the D2(35) = {1, 5^2, 7^2, 35^2} with the unique pair of squares of divisors (5^2, 35^2) we obtain 1/5^2+1/35^2 = 2/7^2. Hence m = 7^2 is in D2(35). There is no other solution.
%e A376713 a(2) = 70 because D2(70) = {1, 2^2, 5^2, 7^2, 10^2, 14^2, 35^2, 70^2} and we find two pairs of squares of divisors: (5^2, 35^2) and (10^2, 70^2) giving respectively:
%e A376713 First solution: 1/5^2+1/35^2 = 2/7^2. Hence m = 7 ^2 is in D2(70);
%e A376713 Second solution: 1/10^2+1/70^2 = 2/14^2. Hence m = 14^2 is in D2(70).
%p A376713 with(numtheory):nn:=5*10^8:
%p A376713 for n from 1 to 25 do:
%p A376713  ii:=0:
%p A376713 for k from 4 to nn while (ii=0)do:
%p A376713 d:=divisors(k):n0:=nops(d):it:=0:
%p A376713   for i from 1 to n0-1 do:
%p A376713    for j from i+1 to n0 do:
%p A376713      s:=1/d[i]^2+ 1/d[j]^2:
%p A376713       for u from 1 to n0 do:
%p A376713        if s=2/d[u]^2 then it:=it+1:else
%p A376713        fi:
%p A376713       od:
%p A376713     od:
%p A376713    od:
%p A376713    if it=n then ii:=1:printf(`%d %d \n`,n,k):else fi:
%p A376713 od:od:
%Y A376713 Cf. A001599, A001600, A005179.
%K A376713 nonn,hard
%O A376713 1,1
%A A376713 _Michel Lagneau_, Oct 02 2024
%E A376713 a(26)-a(28) from _Hugo Pfoertner_, Oct 03 2024
%E A376713 a(17), a(19), a(23), and a(29)-a(38) from _Lucas A. Brown_, Nov 04 2024