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Under Goldbach's conjecture, let's consider once again the map $r_0\colon n\mapsto r_{0}(n)$ such that $r_{0}(n)$ is the smallest non negative integer $r$ such that both $n-r$ and $n+r$ are prime. Let's now define the notion of primal spread $spr(m)$ of an integer $m$ greater than $3/2$ as follows:

$$spr(m):=\sup\{k\geq 0, \forall 0\leq i\leq k, r_{0}(m+i)=r_{0}(m-i)\}$$

Is the map $m\mapsto spr(m)$ bounded above? More generally, is there a rather natural way to express $spr(m)$ as a smooth function of $m$ up to an error term? If so, does RH give any idea of what this error term should be?
Thanks in advance.

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  • $\begingroup$ From m=1 to $7*10^6$, there are 474 values that have $spr(m)\ge3$. Full list for $\ge3$: $\begin{array}{ccccccc} 3&4&5&6&7&8&12\\ 430&25&4&10&1&1&3 \end{array}$. For $m=113160$ with $r_0(m)=12$, looking at the pairs $r_0(m\pm k)$ beyond 12, it turns out that for $k=1,...,23$ they are equal for not less than 18 of them. Likewise for $m=2580660$, again with $r_0(m)=12$, there is a proportion 21/27. For $m=1150695$, with $r_0(m)=8$, there is still a proportion 19/28. etc. That means that some $m$ have a high "degree of symmetry" in this sense, (cont'd) $\endgroup$
    – Wolfgang
    Jan 29, 2015 at 14:02
  • $\begingroup$ ... at least for quite small $k$. BUT this does not seem to survive as $k$ grows, so it cannot be compared with, say, the number of primes among $n^2+n+41$. I don't know whether heuristically this gives some more reason to think that $spr$ is bounded. $\endgroup$
    – Wolfgang
    Jan 29, 2015 at 14:02
  • $\begingroup$ Even though my not-that-smart phone has a display issue, you may have misunderstood something. There is absolutely no possibility for an even integer $m$ to fulfill $r_{0}(m)=12$, since the only even prime number is $2$. Maybe you meant $spr(m)=12$? $\endgroup$ Jan 29, 2015 at 22:53
  • $\begingroup$ Yes of course, typo. All 3 times it should be $spr(m)$ instead of $r_0(m)$. $\endgroup$
    – Wolfgang
    Jan 30, 2015 at 8:43

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