Sum of divisors of a square I was wondering if there is a nice formula for the number of divisors $d$ of a perfect square ($n^2$), such that $\frac{n}{2} <d < n$. For example, for $n = 12$, the divisors $d$ of $12^2=144$ such that $12/2=6<d<12$ consist of $8$ and $9$, so $f(12) = 2$. 
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Another way to express this is that $12 = 2^2 * 3^1$, so $144=2^4*3^2$. Now we must find the number of tuples $(e_1, e_2)$, such that $6 \lt 2^{e_1} * 3^{e_2} \lt 12$
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Is there a closed formula for this or at least another way to express this?

What I'm about to show isn't a formula but rules on how to quickly find the divisors ($d$) between $n/2$ and $n$. For each $d$ $\exists a,b\in\Bbb{N}|2a>b>a,gcd(a,b)=1,a|n,b|n$
$d=\frac{n*a}{b}$
I will show three examples below. 
If $n=12$ then $n$'s factors are $1,2,3,4,6, \text{ and } 12$. All $a$ and $b$ must be one of these numbers. All $(a,b)$ pairs that satisfy the inequality ($2a>b>a$) are $(2,3),(3,4),(4,6)$. The gcd of the third pair isn't one, so it is removed. 
$$8=\frac{12*2}{3}\quad 9=\frac{12*3}{4}$$
If $n=45$ then $n$'s factors are $1,3,5,9,15,$ and $45$. The three pairs $(3,5),(5,9),(9,15)$ satisfy the inequality. The third pair isn't coprime so it is removed.
$$27=\frac{3*45}{5}\quad 25=\frac{5*45}{9}$$
If $n=60$ 
Factors of $n$: $1,2,3,4,5,6,10,12,15,20,30,60$
Coprime pairs:
$(2,3),(3,4),(3,5),(4,5),(5,6)$
$$d\in (40,45,36,48,50)$$
While I don't have a general formula for $f(n)$ there is a formula for a specific case. If all factors of $n^2$ less than $n/2$ divides $n$ then the formula for $f(n^2)$ is $$\lceil\frac{\sigma_0(n^2)}{2}\rceil -\sigma_0(n)$$ where $\sigma_0(x)$ is the number of factors of $x$
This formula does work for $12$ and $45$ but not $60$.  
