Tetration is pentation. This deserve more thinking.

[marraco]

Possibly iterated left bracket tetration has a way to solve real exponents of right bracket tetration.

Here the bracket shows the place where, the result of the last iteration, is placed on the next iteration.
Common tetration is right bracket, and the weak alternative is left bracket:

\( a\^\^{(m)}=a^{a^{a^{...}}}=\,^ma \)
\( (a)\^\^n={(((a)^a)^a ...)^a}=a^{a^{(n-1)}}=N_a\,^{{a^n}} \)
where \( \\[15pt]

{N_a=a^{\frac{1}{a}}} \)

Note that \( \\[15pt]

{(a)\^\^{(a+1)}=a^{{a}}=\,^2a} \)

Let be \( \\[15pt]

{g(x)=N_a\,^{{a^{x+1}}}=a^{a^{x}}} \)
And in general, n iterations: \( \\[15pt]

{f^{\circ n}\{g(a)\}=\,^{1+2n}a}\,\,\,\leftarrow\,\,\, \) this is actually a form of weak pentation, because is an iteration of a weak form of tetration.

In some sense, tetration is a form of pentation, and we are trying to solve a pentation without understanding his base tetration.

Note that if we solve this iteration for \( \\[15pt]

{n \in R} \), we also solve tetration to any real exponents.

[marraco]

An alternative form is
\( \\[15pt]

{g(x)=ln_{N_a}((a)\^\^{x})=a^{{x}}} \)

then
\( \\[15pt]

{f^{\circ n}\{g(a)\}=\,^{n}a} \)

also:
\( \\[25pt]

{
ln_{N_m}(
{(m)\^\^{\,^nm }}
) \,=\, {^{n+2}m} \)

[marraco]

Observe this incredible sequence

\( \\[15pt]

{(a)\^\^a = a^{a-1} = N_a\,^{a^a} \,=\, N_a\,^{N_a\,^{a^2}} \,=\,
N_a\,^{N_a\,^{N_a\,^{2a}}} \,=\,
N_a\,^{N_a\,^{N_a\,^{N_a\,^{ln_{a}(2^a)+a}}}} \,=\, ...
} \)

This hints that the zeration \( \\[15pt]

{a\circ a=ln_{N_a}(2)+a } \), which is even another argument in favor of \( \\[15pt]

{a\circ b=ln_{?}(?^a+?^b) } \), but also hints about what is below zeration, and maybe it can explored over tetration.

-Is trivial that the former sequence of towers converge to \( \\[15pt]

{ a^{a-1} } \).
-Is puzzling that the exponent (2+a) is missing.