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Is there a functon that counts the number of ways in which an even number can be expressed as a sum of two primes?

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5  
there is one, you just defined it... –  Xoff Aug 7 '12 at 12:55

2 Answers 2

up vote 6 down vote accepted

See Goldbach's comet at Wikipedia.

EDIT: To expand on this a little, let $g(n)$ be the number of ways of expressing the even number $n$ as a sum of two primes. Wikipedia gives a heuristic argument for $g(n)$ to be approximately $2n/(\log n)^2$ for large $n$. Then it points out a flaw with the heuristic, and explains how Hardy and Littlewood repaired the flaw to come up with a better conjecture. The better conjecture states that, for large $n$, $g(n)$ is approximately $cn/(\log n)^2$, where $c>0$ depends on the primes dividing $n$. In all cases, $c>1.32$.

I stress that this is all conjectural, as no one has been able to prove even that $g(n)>0$ for all even $n\ge4$.

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Yes, there is such a function and it has been studied for at least a century. See Sloane's A002375. But it doesn't have a set letter specified for it. Here I will use $g$. So, for example, $g(36) = 4$, $g(38) = 2$. (If you're looking in Sloane's, be sure to divide by 2 before looking up). There is also a function which requires the primes to be distinct, so $31 + 7 = 38$ counts but $19 + 19$ does not; that has also been studied for at least a century.

Now, is there a formula that you can plug in an even number $2n$ and have it give you an answer without knowing the primes up to $n$? I think that if the Riemann hypothesis is proven, it could lead to such a formula. As a quick and dirty estimate, I suggest $g(2n) \approx \frac{n}{8}$; your mileage may vary.

EDIT: Gerry Myerson rightly pointed out that $g(2n) < \pi(n)$, and that my quick and dirty estimate is quite inadequate for large numbers. From his comment, I revise my quick and dirty estimate to $g(2n) \approx \frac{n}{4 \log n}$. The point that I was getting at is that Bertrand's postulate tells us there is always at least one prime between $n$ and $2n$, and it seems unlikely to me that each and every prime in that interval would fail to "match" to a prime between 1 and $n$.

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Your $g(n)$ is clearly no bigger than the number of primes up to $2n$. For $n$ large, the number of primes less than $2n$ is roughly $2n/\log n$, which is much smaller than $n/8$. So the "quick and dirty" estimate can't be much good, once $n$ is large enough. Also, to the best of my knowledge, the Riemann Hypothesis has nothing to do with it. –  Gerry Myerson Dec 21 '14 at 19:49
    
Mostly good points. As for the Riemann hypothesis, that's just speculation on my part. –  Robert Soupe Dec 21 '14 at 21:10
    
The Prime Number Theorem tells us much more than Bertrand's Postulate; instead of one prime between $n$ and $2n$, there are asymptotically $n/\log n$ of them. And a careful consideration of the likelihood of matchups with primes between $1$ and $n$ leads to an estimate of $cn/(\log n)^2$ for $g(n)$, for a certain constant $c$. Have a look at the Wikipedia essay on the Goldbach conjecture. –  Gerry Myerson Dec 22 '14 at 8:15
    
I'll take your word for it. I'm at that delicate halfway point between mathematical ignorance and expertise that makes me very susceptible to information poisoning from Wikipedia. –  Robert Soupe Dec 22 '14 at 13:04
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You guys, you're basically in agreement on the essential point here: as $n$ gets larger, so does the number of potential Goldbach sums. A quick and dirty estimate is just that, and not a tight upper or lower bound. But Gerry, I think you'd do well to post your own answer. Another way of saying the same thing is that a Goldbach failure number is probably less likely to exist than an odd perfect number. –  Lisa Dec 22 '14 at 17:31

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