Number of solutions for $xy1$ Is there a way we can determine number of solutions for equation

$$xy < d$$

where $d$ is a constant and $x$ and $y$ are positive integers greater than $1$?
I am not interested in actual values, but just number of possible solutions.
EDIT: Would it help if $d$ is represented as two integers $d=d_1 d_2$?
 A: In terms of programming a (probably not efficient way) to do it would be to preset two arrays for $x$ and $y$ that run through all of the values from $(2,...,d)$ write a for loop with an if statement, that multplies $x$ and $y$ together, and the if statement seperates them, so if $xy\lt d $ send tjose values to an array, then when the for loop ends, the size of that array would be your answer.
A: I think that if you fix a value for x, then it's easy to know how many y's make xy < d true. it is $\lceil d/x - 1 \rceil$ possible y's.
For example, imagine $d = 10$, and fix $x = 3$, so all the possible value to y is 1,2 and 3, i.e., 1 to $\lceil d/x - 1 \rceil$.
The reason that I think it's $\lceil d/x - 1 \rceil$, is that when d/x is not an integer, the floor of this number $\lfloor d/x \rfloor$ is the answer, but if it is an integer, then xy = d if $\lfloor d/x \rfloor$ is in the set of possible y's, so $\lfloor d/x \rfloor$ can't be in the solution set of the y's.
Finally, we need to find this to each x, so the final formula will be $\sum\limits_{x=1}^{d-1}\lceil d/x - 1 \rceil$. I know that it can be computed in $O(d)$, but maybe you can find a closed formula to this summation.
I think this is right, but not sure. If any doubts, just ask.
