Create the slowest growing function you can in under 100 bytes

Brachylog, 100 bytes

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Try it online!

This is probably nowhere near the slowness of some other fancy answers, but I couldn't believe noone had tried this simple and beautiful approach.

Simply, we compute the length of the input number, then the length of this result, then the length of this other result... 100 times in total.

This grows as fast as log(log(log...log(x), with 100 base-10 logs.

If you input your number as a string, this will run extremely fast on any input you could try, but don't expect to ever see a result higher than 1 :D

Haskell, 98 bytes, score=f_ε0⁻¹(n)

_#[]=0
k#(0:a)=k#a
k#(a:b)=1+(k#(([1..k]>>fst(span(>=a)b)++[a-1])++b))
f n=[k|k<-[0..],k#[k]>n]!!0

How it works

This computes the inverse of a very fast growing function related to Beklemishev’s worm game. Its growth rate is comparable to f_ε0, where f_α is the fast-growing hierarchy and ε₀ is the first epsilon number.

For comparison with other answers, note that

• exponentiation is comparable to f₂;
• iterated exponentiation (tetration or ↑↑) is comparable to f₃;
• ↑↑⋯↑↑ with m arrows is comparable to f_{m + 1};
• The Ackermann function is comparable to f_ω;
• repeated iterations of the Ackermann function (constructions like Graham’s number) are still dominated by f_{ω + 1};
• and ε₀ is the limit of all towers ω^ωω⋰ω.

JavaScript (ES6), Inverse Ackermann Function*, 97 bytes

*if I did it right

A=(m,n)=>m?A(m-1,n?A(m,n-1):1):n+1
a=(m,n=m,i=1)=>{while(A(i,m/n|0)<=Math.log2(n))i++;return i-1}

Function A is the Ackermann function. Function a is supposed to be the Inverse Ackermann function. If I implemented it correctly, Wikipedia says that it will not hit 5 until m equals 2^2^2^2^16. I get a StackOverflow around 1000.

Usage:

console.log(a(1000))

Explanations:

Ackermann Function

A=(m,n)=>                           Function A with parameters m and n
         m?                   :n+1  If m == 0, return n + 1; else,
           A(m-1,n?        :1)       If n == 0, return A(m-1,1); else,
                   A(m,n-1)          return A(m-1,A(m,n-1))

Inverse Ackermann Function

a=(m,n=m,i=1)=>{                                                Function a with parameter m, with n preinitialized to m and i preinitialized to 1
                while(A(i,m/n|0)<=Math.log2(n))                 While the result of A(i, floor(m/n)) is less than log₂ n,
                                               i++;             Increment i
                                                   return i-1}  Return i-1