Perfect Numbers and their reciprocals.

  • Thread starter Slats18
  • Start date
  • #1
47
0

Main Question or Discussion Point

For those who don't know, here's the definition:

A Perfect Number is some number n such that the sum of its divisors is equal to 2n (or just n if you don't count n|n =P )
Ex: Let n=6, and the divisors of 6 are 1,2,3 and 6

1+2+3+6 = 12 = 2*6.

Moving on, there's also the neat fact about them that states that the reciprocals of the divisors will always add up to 2.

Ex: n=6, 1/6 + 1/3 + 1/2 + 1/1 = 2

Now, my question/wonderment: is there any possible relation between this fact, and the summation of (1/2)^n, which converges to two as n approaches infinity?
 

Answers and Replies

  • #2
2,967
5
Let

[tex]1 < p_{1} < p_{2} < \ldots < p_{2m} < N[/tex]

be all the divisors of [itex]N[/itex] excluding 1 and itself.

Notice that the number of such divisors must be even and that:

[tex]
p_{1} p_{2 m} = p_{2} p_{2 m - 1} = \ldots p_{m - 1} p_{m + 1} = N
[/tex]

If [itex]N[/itex] is perfect then we also have:

[tex]
N = 1 + p_{1} + \ldots + p_{2m}
[/tex]

Divide this equality by [itex]N[/itex] and you get:

[tex]
1 = \frac{1}{N} + \frac{p_{1}}{N} + \ldots + \frac{p_{2m}}{N}
[/tex]

But, according to the upper equalities:

[tex]
\frac{p_{1}}{N} = \frac{1}{p_{2m}}, \frac{p_{2m}}{N} = \frac{1}{p_{1}}, \ldots
[/tex]

and we get:

[tex]
1 = \frac{1}{N} + \frac{1}{p_{2m}} + \ldots + \frac{1}{p_{1}}
[/tex]

Q.E.D.
 
Last edited:
  • #3
47
0
Thank you for the speedy reply, Dickfore, and thank you on the proof of why the fact holds. But, I'm more interested in the discussion of whether they are related in any way (excluding the trivial), not just proof-wise =)
 

Related Threads on Perfect Numbers and their reciprocals.

  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
6
Views
5K
  • Last Post
Replies
18
Views
8K
Replies
10
Views
4K
  • Last Post
Replies
9
Views
10K
  • Last Post
Replies
9
Views
3K
Replies
1
Views
2K
Replies
6
Views
10K
Replies
3
Views
3K
  • Last Post
Replies
1
Views
1K
Top