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Jul 25, 2016 at 14:57 comment added Bernardo Recamán Santos If n = 12, then {(1,4), (2,5), (3,8), (6,7), (9,10), (11,12)} is the last time that such a pairing can be achieved in which pairs sum to distinct primes.
Jul 20, 2016 at 15:03 comment added Gerhard Paseman One can note that for each odd number, there are something like $C\pi(2n)$ possible choices of even number to pair with it, where $C$ is some constant not much bigger than 1 (and probably less). Even for $C=2$ this gives an improvement ($(C\pi(n))^n$) on $n!$ for $n$ not too big. Gerhard "Can't Prove C Is Small" Paseman, 2016.07.20.
Jul 20, 2016 at 10:40 comment added Moritz Firsching Perhaps an useful upper bound could be obtained by using the formulation as permanent together with the Bregman–Minc inequality. (Together with a useful estimate of the row sums from the prime number theorem). At least this should give something better than the trivial upper bound of $n!$.
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Jul 19, 2016 at 19:10 answer added Gerhard Paseman timeline score: 1
Jul 19, 2016 at 18:59 comment added Gerhard Paseman This observation is nice enough that I will partly spoil it: Use Bertrand to show that there is a prime p between 2n and 4n, and use that for one of the sums. The rest of the proof suggests a partial enumeration: say such a pairing is intervallic if the pairs that form the same sum q are an interval for any q. You might get a good asymptotic on the number of intervallic pairings. Gerhard "If Goldbach Were Easy" Paseman, 2016.07.19.
Jul 19, 2016 at 18:30 comment added Robert Israel See OEIS sequence A000341. However, asymptotics are not given there.
Jul 19, 2016 at 18:24 review First posts
Jul 19, 2016 at 18:25
Jul 19, 2016 at 18:21 history asked MT_ CC BY-SA 3.0