Pair up $\{1..2n\}$ that the sums of each pair are different primes.
I found 9 examples:
$\{(1,2)\},$
$\{(1,2),(3,4)\},$
$\{(1,2),(3,4),(5,6)\},$
$\{(1,4),(2,5),(3,8),(6,7)\},\{(2,3),(1,6),(4,7),(5,8)\},$
$\{(1,4),(2,5),(3,8),(6,7),(9,10)\},\{(2,3),(1,6),(4,7),(5,8),(9,10)\},$
$\{(1,4),(2,5),(3,8),(6,7),(9,10),(11,12)\},\{(2,3),(1,6),(4,7),(5,8),(9,10),(11,12)\}.$
Is there another example like this?
Best Answer
There are no more. There are none for $n=7$. We would need the sums to be seven distinct odd primes below $27$. The most the sum of these sums can be is $5+7+11+13+17+19+23=95$ but the sum of all the numbers up to $14$ is $105$.
For $n=8$ we would need the sum of eight primes below $31$ to be $120$. We can only do this with $3,5,11,13,17,19,23,29$. We need $1+2$ to get $3$ but cannot get $5$.
For $n=9$ we need all the primes up to $31$ except $2,5$, but then $1+2=3,3+4=7,5+5=11$ and we are stuck for $13$.
For $n=10$ we need all the primes up to $37$ except $2,5$ and the $n=9$ proof works.
For $n=11$ we need all the primes up to $43$ except $2,7,11$ or $2,5,13$. We can do $21+22=43$ but cannot get $41$.
For $n=12$ the sum of all the primes up to $43$ is $271$ while the sum of the numbers up to $24$ is $276$
Now going from $n$ to $n+1$ the sum of the numbers increases by $4n+3$ while the sum of primes increases by $4n+1, 4n+3,$ or $8n+4$. As less than $\frac {2\cdot 3 \cdot 5}{3 \cdot 5 \cdot 7} \lt \frac 12$ are primes, the sum of primes will never catch up. $n=13$ doesn't add any new primes because the odd numbers added are $49,51$ and again we miss at $n=19$. Each intervening $n$ has only added one prime.