OFFSET
1,1
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
LINKS
Lucas A. Brown, Python program.
EXAMPLE
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).
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:
CROSSREFS
KEYWORD
nonn,hard
AUTHOR
Michel Lagneau, Oct 02 2024
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
STATUS
approved