Let be the number of triplets such that , note that: because indeed: ( intepret the RHS as the number of triplets such that and at least one of those factors is negative- therefore we must have 2 negatives since )

Note that:

Thus: that is:

By the definition of the floor function:

That is:

Now consider that:

Then we get:

Let's see what happens with that logarithm on the LHS:

Using the definition of the floor function again:

Thus:

That is:

Thus:

Now we have: (use integrals to approximate the second sum)

Thus:

So this would yield:

PS: When you put , be careful, that represents the number ofpositivedivisors of n