# Harmonic function

1. Nov 9, 2008

### AlexHall

Hi

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.

1/H(n)={1/τ(n)}Σ(1/d)

I have found that

H(n)=nτ(n)/σ(n)
H(n)=τ(n)/2

How can I conclude that this is an integer?

Thank you

2. Nov 19, 2008

### yasiru89

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.

3. Nov 19, 2008

### CRGreathouse

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).

4. Nov 19, 2008

### yasiru89

Oh yeah, Euler's result. There we are then.