For comparison let me reput your formula from an early thread.
I am only focus on the superfunction not so much on the specific constants such that it is \( 0\mapsto 1 \). Your double limit can be split into two limits, first:
\( T_b(x) = x + b^{x-1+b^{x-2 + b^{\dots}}} \) and then
the superexponential to base b:
\( \operatorname{sexp}_b(x) = \lim_{n\to \infty} \log_b^{[n]} ( T_b(x_0+x+n) ) \)
for a suitable \( x_0 \)
\( \operatorname{sexp}_b \) indeed satisfies the required equality:
\( \begin{align*}\operatorname{sexp}_b(x+1)&=\lim_{n\to \infty} \log_b^{[n]} ( T_b(x_0+x+1+n) )\\
&=\lim_{n\to \infty} \exp_b\{\log_b^{[n+1]} ( T_b(x_0+x+(1+n))\} \\
&=\exp_b\{\lim_{n\to \infty}\log_b^{[n+1]} ( T_b(x_0+x+(n+1))\}\\
&=b^{\operatorname{sexp}_b(x)}\end{align*} \)
Interestingly this does not depend that much on \( T_b \).
We can choose any function \( T_b \) as long as the limit for \( \operatorname{sexp}_b \) exists.
I am only focus on the superfunction not so much on the specific constants such that it is \( 0\mapsto 1 \). Your double limit can be split into two limits, first:
\( T_b(x) = x + b^{x-1+b^{x-2 + b^{\dots}}} \) and then
the superexponential to base b:
\( \operatorname{sexp}_b(x) = \lim_{n\to \infty} \log_b^{[n]} ( T_b(x_0+x+n) ) \)
for a suitable \( x_0 \)
\( \operatorname{sexp}_b \) indeed satisfies the required equality:
\( \begin{align*}\operatorname{sexp}_b(x+1)&=\lim_{n\to \infty} \log_b^{[n]} ( T_b(x_0+x+1+n) )\\
&=\lim_{n\to \infty} \exp_b\{\log_b^{[n+1]} ( T_b(x_0+x+(1+n))\} \\
&=\exp_b\{\lim_{n\to \infty}\log_b^{[n+1]} ( T_b(x_0+x+(n+1))\}\\
&=b^{\operatorname{sexp}_b(x)}\end{align*} \)
Interestingly this does not depend that much on \( T_b \).
We can choose any function \( T_b \) as long as the limit for \( \operatorname{sexp}_b \) exists.