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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 $n \neq 2p$?

Robert Soupe
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temp watts
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    Presumably you are comparing with $n$ twice a composite number – Henry Nov 19 '18 at 17:51
  • It's more a matter of how large $n$ is. Suppose $p = 7$, then $n = 14$ and you just have 3 + 11 = 7 + 7. Suppose instead $n = 720$, which is twice 360, clearly not a prime. You have 709 + 11 = 701 + 19 = 691 + 29 = 683 + 37 = 677 + 43 = etc. – Robert Soupe Nov 20 '18 at 03:30

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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...

Mason
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  • Thank you for your reply. Your interpretation of what I wanted to say is quite accurate. I apologize that my math and writing skills need work. Anyway, what I wanted to know is, if you graph f(x) where x is not prime, and you graph f(p) where p is prime, will the curve for f(x) always lie above f(p)? – temp watts Nov 19 '18 at 19:52