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Harmonic function

  1. Nov 9, 2008 #1

    I have the harmonic mean H(n) of the divisors of a positive integer n. I need to show that if n is perfect number, then H(n) must be an integer.


    I have found that


    How can I conclude that this is an integer?

    Thank you
  2. jcsd
  3. Nov 19, 2008 #2
    The sum of reciprocals of divisors of a perfect number is always equal to 2 if the number itself is considered as a divisor. If, in your problem it is not, then we have that the sum of reciprocals of proper divisors is equal to 1 and the result is trivial.

    If it is the case that the standard divisor function is considered, one needs to show that the number of divisors of a perfect number (including itself) is even, which I haven't been able to prove yet.
  4. Nov 19, 2008 #3


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    An even perfect number is of the form 2^(p-1) * (2^p - 1) where 2^p - 1 is prime, so it has 2p divisors.

    An odd perfect number is of the form p^(4a+1) * n^2 with gcd(p, n) = 1, so it has an even number of divisors (in fact, the number of divisors is divisible by 2 and not by 4).
  5. Nov 19, 2008 #4
    Oh yeah, Euler's result. There we are then.
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