I've seen this video about BB(5) = 47,176,870, but this statement is actually wrong and was confused by many people. Here's my comment on this video.
The function that he's talking about is the Maximum shifts function (or frantic frog function). S(n) or FF(n) is defined as the maximum number of state transitions made by an n-state, 2-color Turing machine before halting, given blank input, while BB(n) indicates the maximum number of non-blank symbols that can be written (in the finished tape) with an n-state, 2-color halting Turing machine starting from a blank tape before halting. According to the Googology Wiki, it shows that when the news broke out about the confirmation of the fifth busy beaver, the value of BB(5) refers to the frantic frog function. In this piece of news, it shows that they use BB(n) = S(n), which is historically ambiguous, and I think it's ambiguous up to now, since they are 2 different functions. And according to these pieces of news (https://scottaaronson.blog/?p=8088, https://www.quantamagazine.org/amateur-mathematicians-find-fifth-busy-beaver-turing-machine-20240702/, and https://bbchallenge.org/story), it shows that 47,176,870 is the number of steps required for a 5-state, 2-color Turing machine to halt. Clearly, it refers to the maximum shifts function, or the frantic frog function. Therefore, S(5) or FF(5) = 47,176,870, but not BB(5). Here, BB(5) is actually proven to be exactly equal to 4,098, which is exactly equal to the original lower bound of BB(5), because it follows from the fact that this amount of ones is written by the 5-state Turing machine which takes the longest to halt, and any machine which takes less time to halt also writes less ones.