Generalized recursive operators

[bo198214]

I just found the asymptotes of pentation, hexation, heptation, octation, and beyond! And they're fascinating:
\(
\begin{align}
\lim_{b \rightarrow -2}(a \begin{tabular}{|c|}\hline 4 \\\hline\end{tabular} b) & = -\infty \\
\lim_{b \rightarrow -\infty}(a \begin{tabular}{|c|}\hline 5 \\\hline\end{tabular} b) & = -2 \\
\lim_{b \rightarrow -4}(a \begin{tabular}{|c|}\hline 6 \\\hline\end{tabular} b) & = -\infty \\
\lim_{b \rightarrow -\infty}(a \begin{tabular}{|c|}\hline 7 \\\hline\end{tabular} b) & = -4 \\
\lim_{b \rightarrow -6}(a \begin{tabular}{|c|}\hline 8 \\\hline\end{tabular} b) & = -\infty \\
\lim_{b \rightarrow -\infty}(a \begin{tabular}{|c|}\hline 9 \\\hline\end{tabular} b) & = -6
\end{blign}
\)
...

I suppose you could see this from the integer versions of these operators, but I think the continuous (or if not continuous, mostly real-valued) versions make it easier to see.

First, Andrew, these are really fascinating findings.

\( \lim_{N\rightarrow\infty} (a \begin{tabular}{|c|}\hline N \\\hline\end{tabular} b) = b+1 \) for all \( a>1, b<0 \)

meaning, in the limit, all hyper-operators return to the successor operation, like the circle of life...

I dare a proof by induction which only needs the integer operations.

Proposition:
If we have a sequence of operations [n] on the natural numbers (>0) that satisfy b[n+1]1=b, b[n+1](x+1)=b[n](b[n+1]x) for n\( \ge \) 1, then we can extend the domain of the right operand of [n] to integer k with k\( \ge \) 3-n and the only way to do so still satisfying the above conditions and injectivity of the functions f(x)=b[n]x is by b[n](-k)=-k+1 for 0\( \le \) k\( \le \) n-3.

Proof:
We prove by induction over k that b[n](-k)=-k+1 for all n\( \ge \) k+3.
Induction Start k=0:
b[n]1=b=b[n+1]1=b[n](b[n+1]0), for n\( \ge \)2,
by injectivity follows
1=b[n+1]0 for n+1\( \ge \)3=0+3

Induction Step k=k+1:
by induction assumption for n\( \ge \)k+3 :
b[n](-k)=-k+1=b[n+1](-k)=b[n](b[n+1]-(k+1))
by injectivity:
-k = b[n+1]-(k+1)
which is the induction assertion:
-(k+1)+1=b[n+1]-(k+1) for n+1\( \ge \)k+1+3