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Has anyone found which $n$-digit prime (or primes) is the midpoint for the maximum number of pairs of other $n$-digit primes? For four pairs $47$ is the midpoint for $(5,89), (11,83), (23,71)$, and $(41,53)$. I searched "balanced primes" with no success.

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J. M. Bergot
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    Your question is unclear to me. What is fixed and what varies? – Dzoooks Feb 21 '19 at 18:27
  • I think that already for, lets say , $n=9$ , we can have plenty of pairs. I do not think we can find the "best" prime for, lets say, $n=50$ and neither the possible number of pairs. – Peter Feb 21 '19 at 18:32
  • You are forgetting the point about the number of digits for the midpoint and for ALL other pairs of primes with the SAME number of digits. Can we have n=9 for a mere 2-digit prime?? We may have that many for a 3-digit prime, and perhaps three different midpoints each being a midpoint for the SAME number of pairs of other 3-digit primes. The only thing that is fixed is the number of digits for the primes being considered. – J. M. Bergot Feb 21 '19 at 18:53
  • @J.M.Bergot $n$ is the number of digits, and the number of possible pairs will soon "explode" with increasing $n$ – Peter Feb 21 '19 at 18:54
  • Granted that for n=9 there will be primes as midpoints for many pairs of primes. Has anyone bothered to find out what is the maximum number of pairs with nine digits? If they explode, perhaps this has frighten anyone from searching. – J. M. Bergot Feb 21 '19 at 19:07
  • @J.M.Bergot $n=9$ is surely feasible, $n=15$ with massive computation as well. Perhaps, we can even get to $n=20$, but I doubt that we can count the solutions for $n=50$ , even for a single prime number , let alone for all prime numbers to find the champion. – Peter Feb 21 '19 at 19:12
  • So, so true, and I guess that's why no one has gotten the petaflop machine ready for hot action. File this question under "a threat to national security." – J. M. Bergot Feb 21 '19 at 19:17
  • It seems to me that one could do a search by using even semiprimes that will get one at the median of the n-digit range. So, for 3-digit numbers, start at 2271=542 then go backward to 2269, 2263..and forward to 2277, 2*281..to find the maximum number of pairs. – J. M. Bergot Feb 22 '19 at 19:14

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