Does $n = 2p$ where $p$ is prime, have fewer prime pairs than $n \neq 2p?$

Question

Does an even integer $n = 2p$ where $p$ is prime, have relatively fewer prime pairs $(p_1, p_2)$ such that $p_1 + p_2 = n$, than an even integer $n \neq 2p$?

Answer 1

This is just an attempt to understand the question which I think could be an interesting one but seems a little ill defined.

Define a set $S_m=\{ (p_1,p_2): p_1,p_2 \in \mathbb{P} \text{ such that } 2m=p_1+p_2\}$. Where $\mathbb{P}$ is the set of primes.

Define a function $f(m)=|S_m|$.

So now we have some machinery in place. $f$ counts the number of ways we can write the double of a number as the sum of two primes. For example $f(7)=3$ because we can write $14$ in $3$ different ways. $14=11+3=7+7=3+11$.

Note that you may decide not to count them in this way but if that's the case you should really specify how you want to count this in your question...

$f(6)=2$, because $12=5+7=7+5$.

$f(8)=4$, because $16= 5+11=11+5= 3+13=13+3$.

So we have $f(6)<f(7)<f(8)$.

Now after we build up some machinery... what's a good way to frame the question?

Something like this?

Maybe let $\operatorname{avg}(n) = \frac{1}{n}\sum^n_{j=1} f(j)$

and let $\operatorname{avg_p}(n) = \frac{1}{\pi(n)}\sum^n_{p \in \mathbb{P}} f(p)$

Where $\pi(n)$ is the number of primes less than $n$.

Then we can ask how these functions compare as $n$ grows?

Anyway... I think that this is one reasonable interpretation of this question but I am just trying to make the point that there are probably many possible interpretations you could ask. You might have to do some thinking about how to formulate this question so there is something to really investigate and not leave so much interpretation up to others...