Accelerated Hierarchy

The Accelerated HierarchyMumuji, Accelerated Hierarchy, github pages, Retrieved 3:53 pm Monday, 19 July 2021 (UTC) refers to a hierarchy of functions defined by Googology Wiki User Mumuji.
==Definition==
Let μ be a large countable ordinal such to every limit ordinal α < μ there is assigned a fundamental sequence (a strictly increasing sequence of ordinals whose supremum is α).  A accelerated-growing hierarchy of functions hα: N → N, for α < μ, is then defined as follows:
\(h_0(n)= n\rightarrow^nn\) (Using Powerful arrow notation)
\(h_{\alpha+1}(n) = h_{\alpha}\alpha\rightarrow^\alpha\alpha(n\rightarrow^nn)\) until n = 0, in which case, refer to the top rule.(Using Powerful arrow notation and below)
\(h_{a}(n) = h_{a[n\rightarrow^nn]}(n\rightarrow^nn)\) (Using Powerful arrow notation and below)
Although it is not clarified, the third rule is perhaps assumed to be applied only when \(a\) is a limit ordinal so that the partial specialisations do not cause overlapping. As we will explain in #Issues, the notation is ill-defined.
== Issues ==
Since Powerful arrow notation is ill-defined, this notation is also ill-defined even when μ and a system of fundamental sequences for limit ordinals below μ are given. Moreover, the right hand side of the second rule includes an ill-formed expression "\(h_{\alpha}\alpha\rightarrow^\alpha\alpha\)", and hence the intention is unclear. As we will explain in #Uparrow, \(\rightarrow\) might be a typo of \(\uparrow\), but replacing it by \(\uparrow\) also fails to generate a well-formed expression, too.
== Uparrow ==
The creator wrote "Where the \(\uparrow\) is defined as follows:", but there is no occurrence of \(\uparrow\) in the definition of this notation. So, perhaps the definition includes some unintentional error. For example, thefirst occurrence of \(\rightarrow\) in the right hand side of the second rule might be a typo of \(\uparrow\). Anyway, we explain how \(\uparrow\) is intended to work:
\begin{eqnarray*} f_x \uparrow^n b(c) = \left\{ \begin{array}{ll} {f_x}^b(c) & (n = 1) \\ f_x(c) & (n > 1, b = 1) \\ f_x \uparrow^{n-1} (f_x \uparrow^{n-1} (f_x \uparrow^{n-1} (f_x \uparrow^n (b-1)(c))(c) & (n > 1, b > 1) \end{array} \right. \end{eqnarray*}
   Where the chained arrows is as follows: 
   \(a \rightarrow b = a^{b}\) 
   \(f_x \rightarrow b \rightarrow c(d) = f_x\uparrow^cb \) (in which up-arrow notation is used) 
   \(f_x \rightarrow\ldots\rightarrow b \rightarrow 1(d) = f_x \rightarrow\ldots\rightarrow b(d)\) — When last entry is 1, it will be ignored. 
   \(f_x \rightarrow\ldots\rightarrow b \rightarrow 1 \rightarrow c(d) = f_x \rightarrow\ldots\rightarrow b(d)\) 
   \(f_x \rightarrow\ldots\rightarrow b \rightarrow (c + 1) \rightarrow (d + 1)(e) = f_x \rightarrow\ldots\rightarrow b \rightarrow (f_x \rightarrow\ldots\rightarrow b \rightarrow c \rightarrow (d + 1) ) \rightarrow d(e)\) 
   \(a\rightarrow^1 b = a^b\)
   \(f_x\rightarrow^x b(d) = f(a)\rightarrow^{x-1} f(a)\rightarrow^{x-1} f(a) \ldots \rightarrow^{x-1} f(a)(d)\) where there are \(b\rightarrow^{x-1}b\) copies of \(f(a)\), where \(f(a) = a\uparrow^aa\) 
   \(f_x\#\rightarrow^y 1(d) = f_x\#(d)\) (here, \(\#\) denotes an arbitrarily long part of the chain before the relevant terms) 
   \(f_x\#\rightarrow^y 1\rightarrow^z a(d) = f_x\#(d)\) 
   \(f_x\#\rightarrow^n a\rightarrow^1 b(d) = \#\rightarrow^n (\#\rightarrow^n (a-1)\rightarrow^1 b)\rightarrow^1 (b-1)(d)\) 
   \(f_x\#\rightarrow^n a\rightarrow^y b(d) = f_x\#\rightarrow^n a\rightarrow^{y-1} a\rightarrow^{y-1} a \ldots \rightarrow^{y-1} a(d)\) where there are \(b\) copies of \(a\) 
== Sources ==
Category:Numbers and functions by Mumuji
Category:Incomplete
Category:Hierarchies
Category:Functions