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.

35, 70, 140, 210, 560, 420, 2240, 840, 1260, 1680, 35840, 2100, 143360, 6720, 5040, 4200, 1709435, 6300, 645575, 8400, 20160, 77350, 36728125, 12600, 45360, 430080, 31500, 33600, 1117484375, 25200, 24171875, 29400, 154700, 3418870, 181440, 44100, 31633175, 1291150

Offset: 1

Author

Michel Lagneau, Oct 02 2024

Keywords

• nonn
• hard

Comments

For two integers u, v, by definition the harmonic mean m is given by 2/m = 1/u + 1/v.

We observe that a(n) is divisible by 35.

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

Examples

			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.
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:
First solution: 1/5^2+1/35^2 = 2/7^2. Hence m = 7 ^2 is in D2(70);
Second solution: 1/10^2+1/70^2 = 2/14^2. Hence m = 14^2 is in D2(70).
		

Links

• Lucas A. Brown, Python program.

Crossrefs

Cf. A001599, A001600, A005179.

Programs

Maple

with(numtheory):nn:=5*10^8:
for n from 1 to 25 do:
 ii:=0:
for k from 4 to nn while (ii=0)do:
d:=divisors(k):n0:=nops(d):it:=0:
  for i from 1 to n0-1 do:
   for j from i+1 to n0 do:
     s:=1/d[i]^2+ 1/d[j]^2:
      for u from 1 to n0 do:
       if s=2/d[u]^2 then it:=it+1:else
       fi:
      od:
    od:
   od:
   if it=n then ii:=1:printf(`%d %d \n`,n,k):else fi:
od:od:

Extensions

a(26)-a(28) from Hugo Pfoertner, Oct 03 2024

a(17), a(19), a(23), and a(29)-a(38) from Lucas A. Brown, Nov 04 2024