Alright! I haven’t done this post for a while, and I wanted to do this. Time for more confusion…

We’re going to be using the same base as the first Fish number.

S_{1}(0, y) = y + 1

S_{z}(0, y) = S_{z-1}(y, y) for z ≥ 1

S_{z}(x, 0) = S_{z}(x – 1, 1)

S_{z}(x, y) = S_{z}(x – 1, S_{z}(x, y – 1))

Now let’s redefine SS map just a little bit. SS:[m, f(x), S] → [n, g(x), S2]. Remember, g(x) is:

B(0, n) = f(n)

B(m + 1, 0) = B(m, 1)

B(m + 1, n + 1) = B(m, B(m + 1, n))

g(x) = B(x, x)

Anyways, the SS map’s definition where SS:[m, f(x), S] → [n, g(x), S2]:

S2 = S^{f(m)}

S2:[m, f(x)] → [n, p(x)]

S2^{x}:[m,f(x)] → [q(x),g(x)]

Now, just apply the new SS map 63 times to [3, x+1, S] and you get Fish number 2 or the Fish function F_{2}(x).

Now, how larger is Fish number 2 than Fish number 1?

In the Fast-Growing Hierarchy, Fish number 1 is f_{ω2+1}(63), and Fish number 2 is f_{ω3}(63). That might seem small, but keep in mind: f_{ω}(n) is the diagonal of the grid in this post. Remember, it starts with 2, then 8, then a number with 121 million digits. Imagine just the f_{ω} function with 63. That’s f_{63}(63), then we have… whatever that is. The difference between the two numbers is huge.

In the 4-variable Ackermann function, the Fish function for Fish number 2 is comparable to A(63, 0, 0, n), and therefore, Fish number 2 is comparable to A(63, 0, 0, 63).

In conclusion, Fish number 2 is huge.

If you have any suggestion for what I should write next, make sure to comment down below!