$$h(n) = \#\{ \pi(x)\pi(n-x),x\le n\}$$
What is the growth rate of $h(n)?$
(the notation means find the distinct values of $h(n)$ for each $n \in \Bbb N)$
for example, plotting the point $(12,4)$ corresponds to $n=12$ and $4$ distinct values for $n=12.$
If the answerer(s) could include plot(s) in their answer with many more values that would be great. I'm interested in seeing a better plot and learning about a possible pattern.
Prime counting function is in light green for reference
Updated plot:
It seems to grow about the same as $\pi(x)$ for this sample set of points, which makes sense because $f(x)$ is defined using the multiplication of prime counting functions. It seems to be less than or equal to $\pi(x)$ for all values.
Here’s a bound on your function: $h(n)\leq\pi(n)$ This can be proved because there are only $\pi(n)$ choices for the first value, and $\pi(n)$ options for the second value, giving us a square of different pairs of options. When x starts low, we choose values in the top left corner of the square, since $\pi(x)$ is low and $\pi(n-x)$ is high. As we increase x by one, we choose the same point in our square until either $\pi(x)$ increases, or $\pi(n-x)$ decreases, or both happen at once; in this case we take a step in our square either one step to the right, one square down, or one square diagonally left and down. This means we can only move right and down in our square, and h(x) almost counts all the squares we hit along the way to the diagonal, after the diagonal, we just get repeats since the product is symmetric. it misses some because some happen to have the same product. Since this path is completely down and to the right, it will reach the diagonal in at most $\pi(n)$ steps.