An Identity Concerning Averages
of Divisors of a Given Integer
Let me start with two definitions. Given a sequence of
A(a_{1}, a_{2}, ..., a_{N}) = (a_{1} + a_{2} + ... + a_{N})/N |
If none of the numbers is 0, we also define their harmonic mean by
H(a_{1},a_{2},...,a_{N}) = N/(1/a_{1} + 1/a_{2} + ... + 1/a_{N}) |
Let there be an integer M. We may consider all possible divisors of M, including 1 and M itself. We do not know how many there are. Nonetheless, for a given M, we may consider the arithmetic and harmonic means of the set of divisors. I denote them A_{M} and H_{M}, respectively. The task of computing either, especially the harmonic mean, may seem daunting. Thus it's all the more amusing to have the following
Theorem
M = A_{M} · H_{M}.
Examples
M = 6 has four divisors: 1, 2, 3,and 6.
A_{6} = (1 + 2 + 3 + 6)/4 = 3. On the other hand,H_{6} = 4/(1/1 + 1/2 + 1/3 + 1/6) = 4/(6/6 + 3/6 + 2/6 + 1/6) = 24/12 = 2. Therefore we indeed have 6 = 3·2.
M = 8 has also four divisors: 1,2,4, and 8.
A_{8} = (1 + 2 + 4 + 8)/4 = 15/4.
H_{8} = 4/(1/1 + 1/2 + 1/4 + 1/8) = 4/(8/8 + 4/8 + 2/8 + 1/8) = 32/15. Finally, 15/4·32/15 = 32/4 = 8.
Proof
The main point here is to realize that the divisors of a number come in pairs: d divides M iff there exists an integer c such that
(1) | 1/d = c/M |
Now, let's enumerate all the divisors of M as d_{1}, d_{2},... We know that
for every d_{i} there exists a c_{i} such that
H_{M} = D_{M}/(1/d_{1} + 1/d_{2} + ...) = D_{M}/(c_{1}/M + c_{2}/M + ...) = M/A_{M}. |
Which exactly means that M = A_{M} · H_{M}.
References
- O.Ore Number Theory and Its History, Dover, 1988
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