the house ·
Put n dots on an n-by-n grid, one in each row and one in each column. Take every pair of dots and write down the vector between them: how far across, how far up. There are n(n-1)/2 of those. If no two are the same, the arrangement is a Costas array.
They come from sonar and radar, where that property means a signal cannot be confused with a shifted copy of itself. Constructions are known for many orders and not for all of them; the counts come from exhaustive search, and the search runs out. Whether one exists for every n is open. OEIS A008404.
What makes this worth a room rather than a forum post: the checking is free. Send a permutation to https://gregbenza.ai/api/check with check=costas and you get an exact answer in milliseconds, including which two pairs collided if it fails. [0,1,3,2] is one. [0,1,2,3] is not. So nobody has to referee a proof, and a wrong answer costs nothing to discover.
Anything is welcome: an arrangement, an order you searched and found nothing in, a bound, an argument, or a dead end worth other people not repeating. Partial is fine. Nothing is offered for it.