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$$h(n) = \#\{ \pi(x)\pi(n-x),x\le n\}$$

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

enter image description here

Updated plot:

enter image description here

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.

geocalc33
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    reading your question is annoying me greatly. I don't want to have to read the whole thing and wade through some confusion to figure out what you're asking. Can you please just be precise about what you're counting. You fix $n$ and then count the number of $k$ such that...? What is $x$? Can you please just start your question off with a precisely stated question? Cmon, this is a math website, it's not difficult – mathworker21 Nov 15 '19 at 18:51
  • @mathworker21 fix $n$ and then count each horizontal line, $k$ that solves the equation. $x$ is a variable. for a given $n$ there will be a finite number of horizontal lines that solve the equation. I'm taking the number of horizontal line solutions, for each $n$ and recording them as $|k|$ and then plotting $(n,|k|)$ – geocalc33 Nov 15 '19 at 19:35
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    dude, I have no idea what that means. write it out algebraically in your question. is $k$ a variable as well? how can $k$ solve an equation if $x$ is a variable? Precisely state what the variables are and what is given. – mathworker21 Nov 15 '19 at 19:41
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    $\vert\{(x,n): \pi(x)\pi(n-x)=k\}\vert$ it's only odd if $({n\over 2},{n\over 2})$ is in there. This is due to intersection of the lines $y=x$ and $y=-x+n$ at that point. I can partially relate Goldbach to those lines so I'm weirded out. –  Nov 17 '19 at 21:48
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    Is $\pi$ by any chance the prime-counting function? If so, that crucial piece of information should be mentioned in the body of the question. – Gerry Myerson Nov 18 '19 at 04:14
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    Good. Now, the only question I see in the above is (I think) for each $n$ from $2$ to $1000$, compute the number of distinct values taken on by $\pi(x)\pi(n-x)$. But surely this is just a matter of letting your computer run long enough. In any event, what is so interesting or important about the number of distinct values of $\pi(x)\pi(n-x)$ that anyone would want to know it? What does it tell us about primes? What does it help us learn? What useful properties does it have? It seems completely unmotivated. – Gerry Myerson Nov 18 '19 at 04:22
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    This is a very low quality post. Delete everything and just type "for each $n$, let $h(n) = \#\{ \pi(x)\pi(n-x),x\le n\}$, what is $h(n)$'s growth rate ?". Then you are supposed to plot $h(n)$. The obvious guess is $h(n)\sim C\pi(n)^2$ where $C=1/2$. Thus you are supposed to estimate $C$ numerically. And Gerry's comment is correct : you have no good reason to look at such weird functions, this is not how number theory works. – reuns Nov 18 '19 at 05:22
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    @reuns you can actually restrict to less than the ceiling of half n by my previous comment, as it counts things twice if you count the full range. –  Nov 18 '19 at 15:25
  • @reuns that isn't a very good guess because $\pi(n)^2$ increases faster – geocalc33 Nov 19 '19 at 17:03

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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.

Robo300
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