Here's a simple-sounding unsolved problem:

Starting with the number 8, and repeatedly adding half of the number to itself, rounding down to get a whole number, will there eventually be a point where you have seen more than twice as many odd numbers as even numbers? The sequence starts like this:

8 → 12 → 18 → 27 → 40 → 60 → 90 → 135 → 202 → …

The answer is probably 'no', but settling this question will help people determine BB(6), the 6th busy beaver number. This is the largest number of steps it takes for a 6-state Turing machine to halt, among those that actually halt.

The function BB is uncomputable, in general, but people have computed BB(1), BB(2), BB(3), BB(4) and BB(5). So, we know

BB(5) = 47,176,870

But BB(6) is a lot bigger! Since June we've known

BB(6) ≥ 2 ↑↑ (2 ↑↑ (2 ↑↑ (2 ↑↑ 2)))

What the hell does that mean? Well, this ↑↑ thing works like this: 2 ↑↑ 2 is just 2^2, and 2 ↑↑ 3 = 2^(2^2), and so on. So

2 ↑↑ (2 ↑↑ 2) = 2 ↑↑ 4 = 2^(2^(2^2)) = 2^(2^4) = 2^16 = 65536

so 2 ↑↑ (2 ↑↑ (2 ↑↑ 2)) is 2 raised to itself 2^65536 times, and then 2 ↑↑ (2 ↑↑ (2 ↑↑ (2 ↑↑ 2))) is 2 raised to itself *that* meany times.

So BB(6) is quite big. But nobody knows how big... because there are lots of 6-state Turing machines that experts are still scratching their heads over. Figuring out which ones halt requires solving some hard math problems, including the one I mentioned at the start.

This problem has a name: it's called the Antihydra problem. You can learn more about it here:

https://wiki.bbchallenge.org/wiki/Antihydra

Working on this isn't very useful. But it's fairly harmless, unlike what most rich people do for fun, and if you solve it you'll win this plaque: