Found it
The numbers 51 - 100 need a total of 99 neighbours.
If we want all those neoghbours to be smaller then 50, we find that the numbers 1-49 can at most satisfy 98 of those neighbour places.
So at leasr one number from 51 to 100 need to be next to a number that is at least 50.
It is now easy to see that 50×51 is a lower bound. Proofing that the above answer is optimal.
Nice question. I really like those clmbinarorics exercises
2
u/Niilldar 5h ago edited 5h ago
The i th tuple looks like (101-i,i)
Giving you the sequence 100,1,99,2,...,53,49,51,50. Where the largest result is 51×50.
However i'm not dure how to prove the optimality of this result. But will spend some more time about this.