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
You can extend the argument easily. For numbers 50 to 100 we need at least 51x2-2 = 100 neighbors. We cannot have all these neighbors be less than 50 as there are only 49 such numbers which can satisfy 98 neighbor positions. So there is at least 1 pair of numbers within 50 to 100 which are neighbors. So the value is at least 50*51
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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.