Good morning, team, thanks for turning out this morning – the Hilbert Corporation appreciates your dedication to providing top-notch, limitless service to our infinitely many loyal Guests.

As you know, last night was unusually challenging – special shout-out to Dante on the front desk who improvised a solution to the unexpectedly large number… yes, Kurt, I know it wasn’t technically a number … of infinitely-large buses full of guests arriving one after the other. Moving everyone in room $n$ into room $2n$ was innovative and successfully made sure we had space for all of the incoming Guests, great work.

Unfortunately, it also generated an infinite number of complaints – the couple originally in room 1 were especially upset about having to move an infinite number… sorry, Kurt, infinitely many times.

Tonight, then, we’re going to do things a little differently. The zeroth coach – we put every Guest into a square-numbered room, starting at 0. If the first coach then shows up, we put each Guest into a room with a number one more than a square, starting at 2.

That’s right, Georg, then the second coachload goes into rooms with numbers two more than a square, starting at 3. Then we would have a problem, if I hadn’t thought it through: because 4 is already a square, the Guests in bus #3 need to go into rooms with numbers three more than a square, starting at seven. In general, the Guests on bus number $n$ need to go into a room $n$ more than a square, starting at room $\frac{1}{4}\left(n^2 + 6n + 1\right)$ for odd $n$ and $\frac{1}{2}\left(n^2 + 2n\right)$ for even $n$.

Yes, Bertrand, of course there’s a reason for it. See if you can work it out. Now, the next item on the agenda, Don and Glenn have pointed out the problem with having an electronic reservations system and infinitely long corridors, which is that Guests can check out any time they like, but they can never leave.