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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.

G-pi 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?

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  • $\begingroup$ An interesting consequence of combining the aforementioned mentioned equations for R(2n), R(2n+2), R(2n-2), T(2n) and Pi(2n) and noting the fact that F[0] = pi(2n) we get the inequality R(2n+2) + 2*R(2n) +R(2n-2) + 2*T(2n) - delta > (4*pi(2n)/2n) * (4*pi(2n) - n) where R(2n) is the number of goldbach partitions for the even number 2n T(2n) is the number of twin primes upto the even number 2n pi(2n) it the total number of odd primes upto 2n and delta = -1 if 2n-1 is an odd prime otherwise it is zero. Example let 2n = 1460 computing the LHS of the inequality we get 383 and computing the RHS we g $\endgroup$ – user201944 Dec 29 '14 at 0:28
  • $\begingroup$ @PWM it seems that your comment is unfinished, please could you review it? $\endgroup$ – iadvd Jan 6 '15 at 8:14
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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 $\pi(30n+1).$

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  • $\begingroup$ I appreciate your review, it is interesting that, probably because it is not enough data at all, heuristically seemed to be valid. Just in case, to clarify, the relationship I tried to express is: G(n) = (2*G${\pi(n/2)}$) > 2*(${\pi(n/10)}$) $\endgroup$ – iadvd Nov 13 '14 at 12:19
  • $\begingroup$ @iadvd: Your formula doesn't make sense (typo/TeX problem?), what did you mean by the middle part $2*G\pi(n/2)$? But extracting $G(n)>2\cdot\pi(n/10)$ I get (with $n=10^{14}$) 90350630388 > 2*346065536839 which is false. $\endgroup$ – Charles Nov 13 '14 at 19:01
  • $\begingroup$ the middle part would mean "the subset of primes in [2,n/2] symmetrical on n/2, so their counterparts are in [n/2,n-2] so basically they are all together the partition G(n), so two times the primes under or equal to n/2 are all the couples for G(n), a kind of (existing or not) so named "GoldbachPi" function based on bounds related with the ${\pi}$ function. I just named it "G${\pi(n/2)}$". $\endgroup$ – iadvd Nov 13 '14 at 22:54
  • $\begingroup$ @iadvd: It would be useful if you would collect all the definitions for each of the functions you have defined in your question ($R(n), G(n), G\pi(n), G-\pi(n),$ and any others I may have missed -- some of which may be synonyms). $\endgroup$ – Charles Nov 13 '14 at 23:40
  • $\begingroup$ you are right, I rewrote the text. Basically R(n) is the extended Goldbach conjecture prime counting function, G(n) is a synonym. ${\pi}$ is the prime counting function. The rest of them are the new Goldbach-${\pi}$ function I was trying to catch. It was a name too long, so G-${\pi}$ and G${\pi}$ are synonyms of it. I rewrote the text, so now only G${\pi}$ is shown (except the graph image). Hope it helps! $\endgroup$ – iadvd Nov 14 '14 at 10:47
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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 $\pi^2[2n]/(2n)$.

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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 pmackenzie08@gmail.com

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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.

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  • $\begingroup$ PS. It does not follow from this that R(2n) > 0 for all 2n, it could still be zero for some 2n. A further property on the distribution of primes is needed to prove goldbachs one way or the other. PWM $\endgroup$ – P.W. Mackenzie Dec 13 '14 at 8:22
  • $\begingroup$ Welcome to math.SE! LaTeX works inline here; see the MathJax tutorial for details. You can click the edit button to modify your answer at any time. Be sure to check out the help center as well. $\endgroup$ – aes Dec 13 '14 at 8:25
  • $\begingroup$ @P.W.Mackenzie, thank you for the information! $\endgroup$ – iadvd Dec 14 '14 at 7:18
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There is a relationship between Goldbach's comet G(n) and the prime counting function π(n). This sum gives the exact result for G(n) (A002372) $$G(n) = \sum_{i=2}^{\pi(2n)} \pi(2n-p(i))-\pi(2n-1-p(i))$$ where p(n) is the nth prime.

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