In mathematics, a harmonic divisor number or Ore number is a positive integer whose divisors have a harmonic mean that is an integer. The first few harmonic divisor numbers are
1, 6, 28, 140, 270, 496, 672, 1638, 2970, 6200, 8128, 8190 (sequence A001599 in the OEIS). Harmonic divisor numbers were introduced by Øystein Ore, who showed that every perfect number is a harmonic divisor number and conjectured that there are no odd harmonic divisor numbers other than 1.
Examples The number 6 has four divisors: 1, 2, 3, and 6. Their harmonic mean is an integer:
4 1 1 + 1 2 + 1 3 + 1 6 = 2. {\displaystyle {\frac {4}{{\frac {1}{1}}+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{6}}}}=2.}
Thus 6 is a harmonic divisor number. Similarly, the number 140 has divisors 1, 2, 4, 5, 7, 10, 14, 20, 28, 35, 70, and 140. Their harmonic mean is
12 1 1 + 1 2 + 1 4 + 1 5 + 1 7 + 1 10 + 1 14 + 1 20 + 1 28 + 1 35 + 1 70 + 1 140 = 5. {\displaystyle {\frac {12}{{\frac {1}{1}}+{\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{5}}+{\frac {1}{7}}+{\frac {1}{10}}+{\frac {1}{14}}+{\frac {1}{20}}+{\frac {1}{28}}+{\frac {1}{35}}+{\frac {1}{70}}+{\frac {1}{140}}}}=5.}
Since 5 is an integer, 140 is a harmonic divisor number.
Factorization of the harmonic mean The harmonic mean H(n) of the divisors of any number n can be expressed as the formula
H ( n ) = n σ 0 ( n ) σ 1 ( n ) {\displaystyle H(n)={\frac {n\sigma _{0}(n)}{\sigma _{1}(n)}}}
where σi (n) is the sum of ith powers of the divisors of n: σ0 is the number of divisors, and σ1 is the sum of divisors (Cohen 1997). All of the terms in this formula are multiplicative, so that the harmonic mean H(n) is also multiplicative. It follows that, for any positive integer n, the harmonic mean H(n) can be expressed as the product of the harmonic means of the prime powers in the factorization of n. For instance, we have
H ( 4 ) = 3 1 + 1 2 + 1 4 = 12 7 , {\displaystyle H(4)={\frac {3}{1+{\frac {1}{2}}+{\frac {1}{4}}}}={\frac {12}{7}},}
H ( 5 ) = 2 1 + 1 5 = 5 3 , {\displaystyle H(5)={\frac {2}{1+{\frac {1}{5}}}}={\frac {5}{3}},}
H ( 7 ) = 2 1 + 1 7 = 7 4 , {\displaystyle H(7)={\frac {2}{1+{\frac {1}{7}}}}={\frac {7}{4}},}
and
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