Why does this convolution of the prime counting function $\pi$ look like a parabola?

Question

In this previous question it is shown that the convolution of the prime counting function $\pi$ with itself, is related to the Goldbach conjecture:

$$\pi^*(n):=\sum_{k=0}^n \pi(k) \pi(n-k)$$

The Goldbach conjecture might be written as:

$$\forall n \ge 2: \frac{\pi^*(2n)+\pi^*(2n-2)}{2} > \pi^*(2n-1)$$

I have plotted the function $\pi^*$ and it looks like a parabola:

Question: Is there any explanation for this observation or is this superficial observation maybe wrong?

Here is some SageMath Code to compute parameters $a,b,c,d,e$ for regression:

$$an^4+bn^3+cn^2+dn+e \approx \pi^*(n)$$

Thanks for your help!

Answer 1

It is straightforward to show that $$\pi^*(n)\sim\frac{n^3}{6\log^2 n}.\tag{$\ast$}\label{ast}$$ So the graph of $\pi^*(n)$ does not look like a parabola. Instead, it looks like the graph of $\frac{n^3}{\log^2 n}$.

Indeed, we have $$\sum_{\min(k,n-k)<\frac{n}{\log n}} \pi(k) \pi(n-k)\ll\frac{n}{\log n}\pi(n)^2\ll\frac{n^3}{\log^3 n},$$ and also \begin{align*}\sum_{\min(k,n-k)\geq\frac{n}{\log n}} \pi(k) \pi(n-k)&\sim\frac{1}{\log n^2}\sum_{\min(k,n-k)\geq\frac{n}{\log n}}k(n-k)\\&\sim\frac{1}{\log n^2}\sum_{k=0}^n k(n-k)\\&\sim\frac{n^3}{6\log^2 n}.\end{align*} Here we used that for $\min(k,n-k)\geq\frac{n}{\log n}$, both $\log k$ and $\log(n-k)$ are asymptotically $\log n$.

Added. The asymptotic formula \eqref{ast} converges rather slowly. Here are some numeric data: \begin{align*} \pi^*(10^2)&=16329\\ \pi^*(10^3)&=6311273\\ \pi^*(10^4)&=3119183737\\ \pi^*(10^5)&=1817310193749\\ \pi^*(10^6)&=1181102034701650\\ \pi^*(10^7)&=827525141442938787\\ \pi^*(10^8)&=611768346585852887680 \end{align*} The corresponding ratios of the two sides of \eqref{ast} are: \begin{align*} n=10^2\quad \rightsquigarrow\quad 2.0778\\ n=10^3\quad \rightsquigarrow\quad 1.8069\\ n=10^4\quad \rightsquigarrow\quad 1.5876\\ n=10^5\quad \rightsquigarrow\quad 1.4453\\ n=10^6\quad \rightsquigarrow\quad 1.3526\\ n=10^7\quad \rightsquigarrow\quad 1.2899\\ n=10^8\quad \rightsquigarrow\quad 1.2455 \end{align*}