I just realized I never explained how that function is related to the exponential factorial. The exponential factorial is defined as \( EF(x) = x^{EF(x-1)} \) where \( EF(1) = 1 \), which is a univariate function. To make this function from the function above, you can first notice the pattern in its iterates:
\(
\begin{tabular}{rl}
E^{\circ 1}(c, 1) = E(c, 1) & = (1, 2) \\
E^{\circ 2}(c, 1) = E(1, 2) & = (2^1, 3) \\
E^{\circ 3}(c, 1) = E(2, 3) & = (3^{2^1}, 4) \\
E^{\circ 4}(c, 1) = E(9, 4) & = (4^{3^{2^1}}, 5)
\end{tabular}
\)
where c could theoretically be anything, since \( 1^c = 1 \). And since I haven't indicated whether we are using 0-based or 1-based vectors, I'm going to use another contraction to get the top value:
Andrew Robbins
\(
\begin{tabular}{rl}
E^{\circ 1}(c, 1) = E(c, 1) & = (1, 2) \\
E^{\circ 2}(c, 1) = E(1, 2) & = (2^1, 3) \\
E^{\circ 3}(c, 1) = E(2, 3) & = (3^{2^1}, 4) \\
E^{\circ 4}(c, 1) = E(9, 4) & = (4^{3^{2^1}}, 5)
\end{tabular}
\)
where c could theoretically be anything, since \( 1^c = 1 \). And since I haven't indicated whether we are using 0-based or 1-based vectors, I'm going to use another contraction to get the top value:
\(
EF(x) = \left[{1 \atop 0}\right] \cd E^{\circ x}\left[{c \atop 1}\right]
\)
Now it should be obvious from this that \( EF(0) = c \) which begs the question in my Conjectures post: what is EF(0)?EF(x) = \left[{1 \atop 0}\right] \cd E^{\circ x}\left[{c \atop 1}\right]
\)
Andrew Robbins
