Powerful arrow Notation and Chained variant.
Note that all variables in this function are natural numbers.
\begin{eqnarray*} a \uparrow^n b = \left\{ \begin{array}{ll} {a}^b & (n = 1) \\ a & (n > 1, b = 1) \\ [a \uparrow^{n-1} (]^{b\uparrow^{n-1}b}a \uparrow^n (b-1)) & (n > 1, b > 1) \end{array} \right. \end{eqnarray*}
Example of reduction
\(3\uparrow^23 = [3\uparrow^1(]^{3\uparrow^13}(3\uparrow^22) = 3\uparrow^1(3\uparrow^1(3\uparrow^1(3\uparrow^1(3\uparrow^1(3\uparrow^1(3\uparrow^1(3\uparrow^1(3\uparrow^1(3\uparrow^1(3\uparrow^1(3\uparrow^1(3\uparrow^1(3\uparrow^1(3\uparrow^1(3\uparrow^1(3\uparrow^1(3\uparrow^1(3\uparrow^1(3\uparrow^1(3\uparrow^1(3\uparrow^1(3\uparrow^1(3\uparrow^1(3\uparrow^1(3\uparrow^1(3\uparrow^1([3\uparrow^1]^{2\uparrow^12}(3\uparrow^21) = 3\uparrow^231 \text{in regular uparrow notation.}\)
\(a \rightarrow b = a^{b}\)
Where the chained variant is as follows:
\(a \rightarrow b \rightarrow c = a\uparrow^cb = a\underbrace{\uparrow\ldots\uparrow}_cb\) (in which powerful-arrow notation is used) = a[c+2]b (in which the bracket notation of hyper operator is used)
\(a \rightarrow\ldots\rightarrow b \rightarrow 1 = a \rightarrow\ldots\rightarrow b\) — When last entry is 1, it will be ignored.
\(a \rightarrow\ldots\rightarrow b \rightarrow 1 \rightarrow c = a \rightarrow\ldots\rightarrow b\)
\(a \rightarrow\ldots\rightarrow b \rightarrow (c + 1) \rightarrow (d + 1) = a \rightarrow\ldots\rightarrow b \rightarrow (a \rightarrow\ldots\rightarrow b \rightarrow c \rightarrow (d + 1) ) \rightarrow d\)
\(a\rightarrow^1 b = a^b\)
\(a\rightarrow^x b = f(a)\rightarrow^{x-1} f(a)\rightarrow^{x-1} f(a) \ldots \rightarrow^{x-1} f(a)\) where there are \(b\rightarrow^{x-1}b\) copies of \(f(a)\), where \(f(a) = a\uparrow^aa\) (refer above)
\(\#\rightarrow^x 1 = \#\) (here, \(\#\) denotes an arbitrarily long part of the chain before the relevant terms)
\(\#\rightarrow^x 1\rightarrow^y a = \#\)
\(\#\rightarrow^n a\rightarrow^1 b = \#\rightarrow^n (\#\rightarrow^n (a-1)\rightarrow^1 b)\rightarrow^1 (b-1)\)
\(\#\rightarrow^n a\rightarrow^x b = \#\rightarrow^n a\rightarrow^{x-1} a\rightarrow^{x-1} a \ldots \rightarrow^{x-1} a\) where there are \(b\) copies of \(a\)