2
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.