User blog:IAmNotATRex/Array-Based Extension to the FGH

After creating my own extension to the FGH, I have decided to create a more powerful extension using arrays. My previous extension can be found here: https://googology.wikia.com/wiki/User_blog:IAmNotATRex/My_Extension_to_the_Fast_Growing_Hierarchy.

Notation for Arrays
For this blog post, the notation for arrays I will use is \([a_{1}, a_{2},\cdots, a_{k-1}, a_{k-2}]\), where \(a\) is the name of the arrays and \(k\) is its length. For the sake of clarity, \([\langlea\rangle]\) will be equal to the array \(a\) instead of \(a\) nested inside a one-element array.

Definition
I will first define a less powerful function, \(f[\langlea\rangle](n)\), and then a more powerful function, \(F_{a}[\langleb\rangle](n)\). If I made a mistake in my logic or a typo anywhere, please let me know.

\(f[a_{1}](n)\) reduces to \(f_{a_{1}}(n)\) in the FGH. When array \(a\) contains more than one element, \(f[0, 0, a_{3},\cdots, a_{k}](n)\) is equal to \(f[f[\cdots[f[0, a_{3},\cdots, a_{k}](n), a_{3},\cdots, a_{k}]\cdots](n), a_{3},\cdots, a_{k}](n)\), where \(f\) is repeated (\n\) number of times. \(f[0, a_{2}\rangle0, a_{3},\cdots, a_{k}](n)\) is equal to \(f[f[\cdots[f[a_{2}, a_{3},\cdots, a_{k}](n), a_{3},\cdots, a_{k}]\cdots](n), a_{3},\cdots, a_{k}](n)\), where \(f\) is repeated \(f^{n}[0, a_{2}-1, a_{3},\cdots, a_{k}](n)\) number of times. For \(a_{1}\rangle0\), \(f[a_{1}, 0, a_{3},\cdots, a_{k}](n)\) is equal to \(f[a_{1}-1, f[\cdots[a_{1}-1, f[a_{1}-1, 0, a_{3},\cdots, a_{k}](n), a_{3},\cdots, a_{k}]\cdots](n), a_{3},\cdots, a_{k}](n)\), where \(f\) is repeated (\n\) number of times. When both \(a_{1}\) and \(a_{2}\) are greater than \(0\), \(f[a_{1}, a_{2}, a_{3},\cdots, a_{k}](n)\) is equal to \(f[a_{1}-1, f[\cdots[a_{1}-1, f[a_{1}-1, a_{2}, a_{3},\cdots, a_{k}](n), a_{3},\cdots, a_{k}]\cdots](n), a_{3},\cdots, a_{k}](n)\), where \(f\) is repeated \(f^{n}[a{1}, a_{2}-1, a_{3},\cdots, a_{k}](n)\) number of times.

This means that \(f[x, y](n)\) is equal to \(Fx+1_{y}(n)\) using my old notation.

To define the \(F_{a}[\langleb\rangle](n)\), I will first need to define a new notation. \([\{n\}]\) is equal to an array, with each element being a base-10 digit of \(n\) plus \(1\), where \(n\) is a nonnegative integer.

Examples: \([\{0\}]\) = (\[1]\) \([\{1\}]\) = (\[2]\) \([\{123\}]\) = (\[2, 3, 4]\) \([\{10^{100}\}]\) = (\[2, (100 \mathup{ones}]\)

When \(a\) is equal to \(0\), \(F_{0}[\langleb\rangle](n)\) is equal to f[\{f[\{f[\cdots[\{f[\langleb\rangle}](n)\}]\cdots](n)\}](n)\}](n), where \(f\) is repeated \(n\) number of times. When \(a>0\), \(F_{a}[\langleb\rangle](n)\) is equal to F_{a-1}[\{F_{a-1}[\{F_{a-1}[\cdots[\{F_{a-1}[\langleb\rangle}](n)\}]\cdots](n)\}](n)\}](n), where \(F_{a-1}\) is repeated \(F_{a-1}^{n}[\langleb\rangle](n)\) number of times.

Examples
\(f[0, 0](3) = f[f[f[0](3)](3)](3) = f[f[4](3)](3) = f_{f_{4}(3)}(3)\) \(f[1, 0](3) = f[0, f[0, f[0, 0](3)](3)](3)\) \(f[1, 0, 0](3) = f[0, f[0, f[0, 0, 0](3), 0](3), 0](3)\) \(f[1, 1](3) = f[0, f[\cdots[f[0, 1](3)]\cdots](3)](3)\) with \(f^{3}[1, 0](3)\) number of layers \(F_{0}[1, 1](3) = f[\{f[\{f[1, 1](3)\}](3)\}](3)\) \(F_{1}[1, 1](3) = F_{0}[\{F_{0}[\cdots[\{F_{0}[1, 1](3)\}]\cdots](3)\}](3)\) with \(F_{0}^{3}[1, 1](3)\) number of layers