Divisor graph sequence

[Edwin Clark]

On Tue, 20 May 2008, David W. Wilson wrote:

> Let G(n) the divisor graph of n whose nodes are the divisors of n and edges
> a->b indicate that a is a (proper/any) divisor of n.
>
>
>
> Clearly the number of vertices of G(n) is d(n) = A000005(n).
>
>
>
> What is the number of edges e(n) of G(n)?
>

It may be of interest to note that if we don't allow loops in G(n) then

e(n) =  number of divisors of n's divisors minus the number of divisors of 
n

That is,

e(n) = A007425(n) - tau(n).