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I am looking at the Goldbach Conjecture because I think it's interesting. One thing I've noticed using brute force programming methods is this:

It seems up, at least up to numbers with 6 or 7 digits, that given two primes are a "Goldbach Pair", the average relative distance of these two numbers from the number N/2 appears to be N/4.

Let me rephrase this a few ways for clarity, and provide some examples.

Consider the Goldbach pairs for the number (N=8). They are (3,5), and that's it. There's only one pair for (N=8). The numbers "3" and "5" are both (N/2)+1/4*(N/2) and (N/2)-1/4*(N/2). Consider the Goldbach pairs for the number (N=10). They are (3,7) and (5,5). For the purposes of what I am saying here, discount all instances where the Goldbach pairs consist of two of the same numbers. So exclude (5,5) from what I am saying. The numbers "3" and "7" are both (N/2)+2/5*(N/2) and (N/2)-2/5*(N/2).

All Goldbach pairs will be (N/2)+C*(N/2) and (N/2)-C*(N/2)

What I am finding using my program is the results for C seem to tend toward .5 or 1/2 "on average".

Is there a proof that C must be this way? Is there a good explanation for this?

Basically what it seems is that given a random Goldbach pair, it has the same probability of being 1 away from a given N/2 as it does of being any other distance from N/2, which also seems relevant to the twin primes conjecture.

Mr. AM
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  • you can connect the twin prime conjecture to goldbach as a statement of things like 12m = (6m+1) +(6m-1) infinitely often. also I thin this might have to do with expected value. etc. we do know that the distance has to be equal to both parts of any pair. –  Sep 09 '17 at 18:21
  • The average $C$ you have found is what you would also expect if your experiment was, "take two random odd numbers (no matter whether they are prime or not) that sum to $N$". So what you have observed is that restricting to primes does not appear to cause a new pattern to appear -- or at least not one that can be observed by your methods. This absence of a pattern does not sound like something one would expect to be able to prove at our current state of understanding. – hmakholm left over Monica Sep 09 '17 at 18:27
  • The random model for the primes is : The sequence of random variables $X_n = 1_{n \text{ is prime}}$ is independent and $P[X_n = 1] = \frac{1}{\log n},\ P[X_n = 1]+P[X_n=0]=1$. Thus on average you'll find a "large prime pair" $n-m,n+m$ with $m \sim \frac{c}{\log^2 n}$ ($c$ some constant) and another "small prime pair" $m_2,2n-m_2$ with $m_2 \sim \frac{c_2}{\log n}$. To make this probabilistically rigorous, we should also look at the variance. – reuns Sep 09 '17 at 18:30
  • In fact, much sharper versions of the Goldbach conjecture (essentially stating that we can choose a sum with one of the primes "small") are believed to be true, but a proof that such a version actually is true would imply the Goldbach-conjecture itself and it would not be open anymore. Statistical heuristics however support the sharper conjectures which is the reason why almost all (or all?) mathematicians believe that Goldbach's conjecture is true. – Peter Sep 09 '17 at 18:30
  • Plotting the distribution that follows from @reuns's model, it looks like $C$ ought to skew slightly towards $1$, but less so the larger $N$ becomes. – hmakholm left over Monica Sep 09 '17 at 18:40
  • (for the large pair) I had in mind approximating $m$ with a geometric distribution of parameter $\frac{1}{\log^2 n}$ hence of mean $\frac{1}{\log^2 n}$. But maybe the approximation is un-perfect and $\log \log n$ comes into play. Also the variance should be taken in account to predict if the parameters converge. @HenningMakholm – reuns Sep 09 '17 at 18:49
  • @reuns: I plotted simply, $P(m=?)\sim\frac{1}{\log(n-m)}\frac{1}{\log(n+m)}$. The second factor of course doesn't vary much over $m\in[0,n]$ when $n$ is large, so the main source of non-uniformity is $\frac1{\log(n-m)}$ blowing up when $m$ gets close to $n$. – hmakholm left over Monica Sep 09 '17 at 18:56
  • Related question I asked~ https://math.stackexchange.com/questions/1338059/why-are-there-palindromic-subsequences-at-random-among-this-sequence – temporary_user_name May 08 '18 at 13:15

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