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Has there been any research into the structure of a non-Goldbach number? This seems like it would be a profitable area for proof by contradiction, so I assume that someone has already done it. (i.e. either the structure proves impossible, or it would allow one to calculate a non-goldbach number)

If so, I'd be interested in sampling some of it. I have googled and turned up nothing, hence this question.

(Apologies if this question is at the wrong level for this site).

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This is far too vague for MO. – Andy Putman May 14 '13 at 20:16
I don't think any condition is known to imply in a nontrivial way that a number is a sum of two primes. So I would be delighted if I was wrong, but I guess there are no results of the type you are looking for. – Johan Wästlund May 14 '13 at 20:18
Questions about whether anything is known about a very specific topic are fine, but the topic in question has to be much more specific and better thought out than your question. – Andy Putman May 14 '13 at 20:25
@AndyPutnam My question seems pretty specific to me, although I'll admit that it's not very deeply thought out - the point of looking for research is to shortcut repeating unnecessarily elementary thought. What would be a sufficiently specific question of this type? – user33996 May 14 '13 at 20:32
What's the difference between studying non-Goldbach integers and the Goldbach conjecture? (to me--none). – Włodzimierz Holsztyński May 14 '13 at 23:52

The number of even integers that are not sums of two primes (non-Goldbach integers) is small in the sense that for $n \leq X$ at most $O(X^{1 - \delta})$ integers are non-Goldbach. This can be thought of as a stronger form of Vingoradov's three-prime theorem that every large enough odd number is a sum of three primes (since the former implies the later). For an old very old survey paper see . For a more recent article I would suggest which is a paper of Heath-Brown.

It might be possible to go through Heath-Brown's proof and see if it gives you any idea about properties of the non-Goldbach numbers. Note however that you will be looking at integers that are not accounted for by Heath-Brown's method (the non-H-B numbers) rather than the non-Goldbach number since those most likely do not exist! The situation is a bit similar when we are working with zeros of the Riemann zeta-function lying off the half-line. We can't use any properties of those fictional zeros, and instead we have to assume them to be just a generic point.

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Ah, thank you! I will take a look at those. – user33996 May 15 '13 at 13:33

Two questions can be raised regarding Goldbach's Conjecture: for any given even number $v > 6$, does exist there a procedure in theory to find a pair of primes such that $v$ is the sum of those two primes? and what happens to the even number $v$ if it is not a sum of any two primes? For the first question we will do the following: first, we pick a prime $p < v - 2$ that does not divide $v$. If $v-p$ is a prime, then we can stop here. Otherwise, $v - p$ is a product of at least 2 primes and none of them divides $v$. More specifically, $p \nmid v$ and \begin{equation} v - p = \prod_{i = 1}^n p_i \end{equation} is a product of some primes $p_1, p_2, \cdots, p_n$, and $p_i \nmid v$ for $i = 1, 2, \cdots, n$. Let $v_i = v - p_i$ for $i = 1, 2, \cdots, n$. Next, if any $v_i$ is prime, we stop here. Otherwise we assume each $v_i$ is a product of at least two primes not dividing the even number $v$: \begin{equation} v - p_i = \prod_{j = 1}^{n_i} q_{ij} \end{equation} Thus, we can do the same thing for each $q_{ij}$ as we did for $p_i$. Now assume we keep doing this procedure for a long time, then eventually either we find a pair of primes that $v$ is the sum of them or we have a finite set of primes $R$ such that for each $p\in R$, $v-p$ is a product of primes in $R$. Now we are ready to answer the second question. We assume that $v$ is not a sum of any two distinct primes. Then it is clear that the set $R$ is finite since there are only finitely many primes $p$ such that $p < v$ and $p\nmid v$ for any given $v$. Mathematically, we can describe the procedure as follows. Given an even number $v$ and a finite set of primes $R$, we have $p\nmid v$ for any $p\in R$ and $v - p$ is a product of at least 2 primes in $R$. If we specify $R=\{p_1,p_2,\cdots,p_n\}$ and introduce a non-negative integer matrix $A=(a_{ij})$ of order $n$, then we are able to describe the related numbers in the following formula $$v - p_i = \prod_{j=1}^n p_j^{a_{ij}} \mbox{ or } v =p_i + \prod_{j=1}^n p_j^{a_{ij}}$$ with condition $$\sum_{j=1}^n a_{ij} \geq 2$$ for $i = 1, 2, \cdots, n$. Our aim is to name this even number $v$ along with the finite set of primes $R$ with algebraic equations a numerical pipe or pipe for short.

Readers may notice that the definition of a pipe requires an assumption that there exist an even number being not a sum of two distinct primes. In fact it is easy to get rid of this unrealistic assumption if we know what a pipe should look like. We can start with any even number $v > 6$ and a prime $p < v - 2$ not dividing $v$. If $q = v - p$ is a prime, then we are done with this procedure. If not, then $q$ is a product of at least two primes not dividing $v$. Now we pick a new prime $p$ from the set of prime divisors in $q$, and repeat the procedure as we just have done for the old $p$. As we know, this procedure will end up with two cases: we find a pair of primes that $v$ is the sum of them or the second exceptional case: we get a pipe!

We do have pipes. The example is a pipe with $v = 2200$ and $R=\{3, 13\}$:
$$2200 = 3 +13^3 = 13 + 3^7$$

We call it Pipe Euler for its elegance and simplicity. It is a surprise that there are only six pipes of this kind for the even number $v < 500,000,000$ and they are listed in the chapter for basic pipes. Each of them is named in order of its value after a great mathematician: Archimedes, Bernoulli, Cauchy, Dirichlet, Euler and Fermat. If there is one pipe after the six pipes, we reserve the name Gauss for it. We believe that there are only finite number of pipes of this kind.

It seems that we are ready to present a formal definition on pipe. But we can not do any work if we only have six pipes on the pipe world. We just need more. We need a definition of pipe board enough that we have more pipes to work on instead of just six pipes. For this reason, we allow that $R$ could be a finite set of any positive integers larger than 1 and the number $v$ could be odd too. Under this new conditions, we are ready to present two examples as follows before entering the next section.

This seems a trivial example of a pipe of order 2: $$11 = 2 + 3^2 = 3 + 2^3$$ Only 8 pipes of order 2 of this kind are found so far. They are listed in section for simple pipe of order 2.

This example is a pipe of order 4 with even number $v$: $$22118 = 2 + 6 \cdot 3686 = 6 + 2^4 \cdot 1382 = 1382 + 2^4\cdot 6^4 = 3686 + 2^9\cdot 6^2$$ Reader should be able to verify the equations easily with or without a calculator.

For more info,

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