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In a previous question I asked about a counterexample for an observation I did about the Goldbach's comet: it seems that there is always common prime shared between the Goldbach's prime pairs of the even number $n$, $G(n)$, and the even number $n+6$, $G(n+6)$, when $n\ge8$. And for that reason is seems that it is also shared a pair of sexy primes between $G(n)$ and $G(n+6)$, one of the sexy primes is in $G(n)$ and the counterpart sexy prime is in $G(n+6)$. It seems it only happens with distance $d=6$, for other distances I did find a counterexample quickly.

I still did not find any counterexample for $d=6$ because it might be a very big number (my tests with Python initially did not find it, but it could be an error of my code), so I was thinking that if there is no counterexample, there could be a generalization of Goldbach's conjecture as follows:

$\forall n\ / \ n=2t, t\in \Bbb N,\ \exists\ (a_k,b_k) \in \Bbb P\ /\ n = a_k+b_k+6k, k \in [0..(\lfloor \frac{n}{6} \rfloor-1)]$

Where the case "k=0" would be the basic Goldbach's conjecture and $k>0$ would happen if the above mentioned generalization was true.

Does it make sense? If somebody could help me to find a counterexample would be great. Thank you!

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  • $\begingroup$ When if $n=4$ and $k=1$. You need to ensure that the $n\geq 6k+4$, and then this is trivially equivalent to Goldbach. $\endgroup$ – Thomas Andrews Apr 6 '15 at 1:53
  • $\begingroup$ @Thomas Andrews, yes the cases where $n \lt 12$ will follow the basic Goldbach conjecture, the generalization would make sense for $n$ over or equal $12$. $\endgroup$ – iadvd Apr 6 '15 at 2:07
  • $\begingroup$ But then it is just the same conjecture as Goldbach. There is no difference. $\endgroup$ – Thomas Andrews Apr 6 '15 at 2:10
  • $\begingroup$ @ThomasAndrews, I have updated the text, I forgot to add the restrictions for k. I am not sure, but I think it would not be the same one. Goldbach conjectures that every even $n$ has at least on prime pair $(p,q)$ so $n=p+q$ but if I am not wrong, the conjecture does not say anything about existing others $(p_k,q_k)$ covering $n=p_k+q_k+6k$ for $k\ge1$ with the restrictions I wrote... $\endgroup$ – iadvd Apr 6 '15 at 2:24
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The 'generalization' appears to be: for all even $n>2$ and all $0\le k\le \lfloor n/6\rfloor-1$ there is a pair of primes $p,q$ such that $p+q+6k=n.$ But this is just the standard Goldbach conjecture on $n,n-6,n-12,\ldots k$ where $k\in\{6,8,10\}.$ In particular this is equivalent to the Goldbach conjecture by strong induction.

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  • $\begingroup$ thank you! So it would be equivalent to the conjecture by strong induction. Please may I ask you what is your opinion about the other point I mention? the observation about the existence of a prime that seems to be present always in $G(n)$ and $G(n+6)$ the Goldbach pairs of $n$ and $n+6$, (is not always the same one for every $n$,$n+6$). So there is always a pair of sexy primes coupled in $G(n)$ and $G(n+6)$ for $n \ge 8$? The idea of the generalization for $k=6$ came from that observation. In other 'distances' ($k$) is not possible to find them. Thank you! $\endgroup$ – iadvd Apr 7 '15 at 0:36
  • $\begingroup$ @iadvd: I think you are asking: Given some (large enough) even $n$, is there always a prime $p$ such that both $n-p$ and $n-p+6$ are prime? $\endgroup$ – Charles Apr 7 '15 at 15:25
  • $\begingroup$ 'backwards' is difficult to express, a 'forward' expression: for any $n \ge 8$ exists at least a prime pair $(p,q)\ /\ n=p+q$ in which $p$ or $q$ has a sexy prime counterpart, e.g $q+6$, also prime, and $n+6 = p+(q+6)$. E.g. $n=8=3+5$ and $n+6=14=3+11$, where the 'shared' or 'fixed' prime is $3$ and the sexy primes are $(5,11)$. E.g.: $n=10=3+7$ and $n+6=16=3+13$, so the 'shared' prime is $3$ and the sexy primes are $(7,13)$. I tested up to $n= 18006$ and always happens. E.g.: $n=17890=17837+53$ and $n+6=17896=17837+59$, and so on. I did not find counterexamples. Thank you! $\endgroup$ – iadvd Apr 8 '15 at 0:33
  • $\begingroup$ Just in case, sorry I forgot to write $n \ge 8$ even. $\endgroup$ – iadvd Apr 8 '15 at 1:18
  • $\begingroup$ @iadvd: I don't understand what you're asking. Feel free to post it as a new question (and ping me here, if you like). $\endgroup$ – Charles Apr 8 '15 at 14:54

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