There is a non recursive formula for T(x,k)?

[marraco]

Hi Marraco -

I'm currently unable to step in, and I do not yet understood your formulae.

Where is the problem?

I can make it clearer. I edited the latex.

Hi Marraco -

I'm currently unable to step in, and I do not yet understood your formulae. But there is some resemblance to some investigation I did years ago, and especially the topic "labelled rooted trees" reminded me of my treatize on the tetrated pascalmatrix, and of the links to the resp. entries in OEIS. Perhaps there is something in it for you; see this Pascalmatrix tetrated Perhaps the used procedure contains a method to compute the coefficients of T(x,k)... but might come out that the explicite computation of the matrix routines are eventually recursive; I don't know at the moment.

I don't know what to do with it.

T(x,k) appears again and again. It obviously is a central aspect of tetration.

Probably the trees represent different ways to write the same tetration.

The key thing we need to do is to figure what is a tree of rational height \( \\[15pt]

{h \in \mathbb{Q}} \).

That's the most important problem about tetration, and we only need to figure it for \( \\[15pt]

{0<h<1} \) , or \( \\[15pt]

{1<h<2} \)

Since the forest count of trees of height h is T(h,k)
\( \frac{\mathrm{d^{ k}}(^he^x) }{\mathrm{d} x^{ k}} |_{\textbf 0}=T({h},{} k) \)

Because k correspond to the \( \\[15pt]

{k^{th}} \) derivative of \( \\[15pt]

{^he^x} \) on the Taylor series of that function, and also is the number of nodes k of a tree of height h, maybe the half derivative of \( \\[15pt]

{^1(e^{x})} \) gives us the formula for counting forests of half nodes \( \\[15pt]

{k=0.5} \)

That should give us a tip about what is a half node. What we really need to figure is what is a half height h, but figuring what is a half node should help.

Now, the problem is that there are many different definitions of half derivative, and they include a constant of integration that I have no clue what to make of.
The half derivative of \( e^x \) can also be calculated using the Laplace transform.