Is there any known relationship between Goldbach's comet G(n) and the prime counting function (${\pi(n)}$)?

Question

The "extended" Goldbach conjecture defines R(n) as the number of representations of an even number n as the sum of two primes, but the approach is not related directly with ${\pi(n)}$, is there any kind of Goldbach-${\pi(n)}$ (I will call it G${\pi(n)}$ for short) function?

http://mathworld.wolfram.com/GoldbachConjecture.html

I have tried an approach to that idea as follows.

1. Calculate ${\pi(n)}$

2. Calculate ${\pi(n/2)}$

3. I defined then G${\pi(n/2)}$ as the subset of primes p from ${\pi(n/2)}$ that are symmetrical on n/2, so they have a counterpart prime pc = n-p in [n/2,n-2] so n=p+pc.

4. I did a test for the first 2000 even numbers (my computer slows down very much after that point) and prepared a graph showing ${\pi(n)}$, ${\pi(n/2)}$, G${\pi(n/2)}$ and the linear interpolation of the value of G${\pi(n/2)}$ (a kind of average value of the growing G${\pi}$ set of primes.

5. Then I played with lower values of the ${\pi}$ function looking for a lower bound of the linear interpolation. Finally ${\pi(n/10)}$ seemed a good value to define a lower bound for G${\pi(n/2)}$ because the linear interpolation is always over that value (at least in the test it seems so).

Here is the graph.

When I checked the results, I wondered if there is a way to work on the conjecture through a relationship between Goldbach's comet value for n, and ${\pi(n)}$, specially if the density of the subset of primes in [2,n/2] that are symmetrical in n/2 is always greater than the density of primes in a lower subset of ${\pi}$ function (e.g. ${\pi(n/10)}$).

So the question is: is there any relationship already known or being researched with ${\pi(n)}$ or only the extended Goldbach conjecture R(n) function is the correct approach to a solution of the Goldbach conjecture?

Answer 1

I don't believe there's any deep relationship between $\pi(n)$ and your function $R(n).$ For one thing, their expected 'bulk' rates of growth are different: $R(n)$ should grow something like $n/\log^2(n),$ while $\pi(n)\sim n/\log(n).$ So dividing $\pi(n)$ by 10 isn't going to be enough -- you'll need to divide by more and more as $n$ grows. For example, $\pi(10^{14})=3204941750802$ but $R(10^{14})=90350630388$ and so their quotient is already 35.

Second, $\pi(n)$ grows smoothly, in the sense that $\pi(n)\le\pi(n+1)\le\pi(n)+1.$ But $R(n)$ grows wildly, more even than your graph so far suggests. Essentially, $R$ is sensitive to the small prime divisors in $n$, while $\pi$ doesn't care. So $R(30n)$ grows differently from $R(30n+1)$, while the same is not true for $\pi(30n)$ and $\pi(30n+1).$

Answer 2

There is indeed a simple relationship between the total number of primes pi(2n) upto 2n and the total number of goldbach partitions R(2n) for that even number.

$$R(2n)=\frac{1}{2n}\sum_{k=0}^{2k-1}a_{k}^{2}-b_{k}^{2}$$

and

$$π[2n]=\frac{1}{2n}\sum_{k=0}^{2k-1}a_{k}^{2}+b_{k}^{2}$$

where a = Real{ F[k] } and b = Imag { F[k] }

and F[k] is the Fourier transform of the prime number function f[x] where f]x] = 1 if x is prime otherwise f[x] = 0 for x =0,1,2....2n-1.

Answer 3

It is interested to note that the zero order harmonic of the Fourier components of function $f(x)$ mentioned in PWMs post are $\operatorname{Re}\{F[0]\} = \pi[2n]$ and $\operatorname{Im}\{F[0]\} = 0$ and so their contribution to the sums $R(2n)$ and $\pi[2n]$ is $\pi^2[2n]/(2n)$ respectively, which is approximately $(2n/\log(2n))^2/(2n) = 2n/(\log^2(2n))$ which is equivalent to the bulk rate mentioned in Charles reply.

My intuitive guess for a lower bound for the number of Goldbach partitions would be more like $\pi^2[2n]/(2n)$.

Answer 4

For those number crunchers out there I have included a number of further interrelated relationships between the number of goldbach partitions R(2n) upto 2n , the number of odd primes π(2n) upto 2n, and the number of twin primes T(2n) upto 2n .

$$R(2n)=\frac{1}{2n}\sum_{k=0}^{k=2n-1}F^{2}(k)$$

$$R(2n-2)=\nabla+\frac{1}{2n}\sum_{k=0}^{k=2n-1}F^{2}(k)e^{-j2π.2k/2n}$$

$$R(2n+2)=\frac{1}{2n}\sum_{k=0}^{k=2n-1}F^{2}(k)e^{j2π.2k/2n}$$

$$π(2n)=\frac{1}{2n}\sum_{k=0}^{k=2n-1}F(k).F^{*}(k)$$

$$T(2n)=\frac{1}{2n}\sum_{k=0}^{k=2n-1}F(k).F^{*}(k)e^{-j2π.2k/2n}$$

where $$\nabla=-1$$ if 2n-1 is an odd prime otherwise $$\nabla=0$$

and where F(k) is the discrete Fourier transform of the prime number function f(x)

$$F(k)=\sum_{p\, odd\, primes=3}^{p<2n}f(x).e^{-j2π pk/2n}\quad where\: k=0,1,2....,2n-1$$

where f(x)=1 if x is an odd prime otherwise f(x)=0 for x=0,1,2...,2n-1.

I have source code for these if any one is interested. Email address [email protected]