# On the number of Goldbach partitions

http://members.chello.nl/k.ijntema/partitions.html?text1=8&area1=You+entered%3A+6%0D%0ANumber+of+Goldbach+partitions+%3D+1%0D%0A%0D%0AGoldbach+partitions%3A%0D%0A3+%2B+3+++%0D%0AEnd

Here you can find the number of Goldbach partitions of an even number. $r(190)=8$, $r(50)= 5$ and so on.

Now $r(4)=r(6)=r(8)=r(12)=1$. Here is my question. Is there any $n>6$ which yields $r(2n)=1$. I tried all $n$ values for $n<96$ and didn't found any.

And is there any research on this function's behaviour?

• – mathlove Jan 4 '16 at 12:30

Heuristically we don't expect that there are a lot of $n$ such that $r\left(n\right)=1$. Note that from PNT we have that the number of representations of $n$ as sum of two primes is roughly $$\frac{n}{\log^{2}\left(n\right)}$$ so if $n$ grows, we expect that there are a lot of representations. There are a lot of papers studying this function and some very related functions, like $$R_{2}\left(n\right)=\sum_{p_{1}+p_{2}=n}\log\left(p_{1}\right)\log\left(p_{2}\right)$$ or $$R'_{2}\left(n\right)=\sum_{m_{1}+m_{2}=n}\Lambda\left(m_{1}\right)\Lambda\left(m_{2}\right)$$ which are, for technical reasons, more tractable.