If $n=2^{31}\cdot 3^{19}$ then how many divisors of $n^2$ are smaller than $n$ but they aren't divisor of $n$?

Question

If $n=2^{31}\cdot 3^{19}$ then how many divisors of $n^2$ are smaller than $n$ but they aren't divisor of $n$?

It is clear that the power of $2$ or $3$ should be bigger than how much there are in $n$.For example I found some cases:

Power of $2=32\Rightarrow$there are $19$ cases for the power of $3$

Power of $2=33\Rightarrow$there are $18$ cases for the power of $3$

Power of $2=34\Rightarrow$there are $18$ cases for the power of $3$

But it is hard to see how many numbers the power of $3$ or $2$ can be by increasing the power of the other.Any hints?

Answer 1

Notice that if $d|n^2$ and $dn$.

We conclude the number of divisors of $n^2$ less than $n$ are $\frac{\tau(n^2)-1}{2}$

So the answer is $\frac{\tau(n^2)-1}{2}-\tau(n)+1$