# Tetration Forum

Full Version: There is a non recursive formula for T(x,k)?
You're currently viewing a stripped down version of our content. View the full version with proper formatting.
Consider the tetration of the function $\vspace{15}{e^x}$

$^n(e^x)=(e^x)_1^{(e^x)_2^{(e^x)_2^{_..^{.e^x_n}}}}$

For a natural number n, the taylor series of that function is

$^n(e^x)=\sum_{k=0}^{\infty}\frac{1}{k!}*T(n,k)*x^k$

where $\vspace{15}{T(n,k)}$ is the OEIS A210725; When k<n, $\vspace{15}{T(n,k)=T(k,k)}$

then, for $\vspace{15}{n\in\mathbb{N}}$:

$^n(x)=\sum_{k=0}^{\infty}\frac{1}{k!}*T(n,k)*(ln(x))^k$

On the limit for $\vspace{15}{n\to\infty}$, the T(n,k)=T(k,k) are the coefficients of the Lambert w function.

Now, I switch the variables names, because we are interested on constant base, and variable exponent. x=b, and n=x

$^xb=\sum_{k=0}^{\infty}\frac{1}{k!}*T(x,k)*(ln(b))^k$

The function T(x,k),according to OEIS is

$T(x,k)=\sum_{j=1}^{x+1}{\binom{x-1}{j-1}*j*T({j-1},{k-1})*T({x-j},{k}) }$

(Note that T(x,k)=1 if x=0 or k=0)

So, the question is, there is an explicit, non recursive, formula for T(x,k)?

We can easily replace the binomial with the Gamma function, and the recursive formula probably has a fractal structure. A fractal structure means that it can be extended to non integer values of x by using the self similarity.
An explicit formula, probably has  $\vspace{15}{x^k}$ as factor, because it looks like the Taylor series of  $\vspace{15}{^xb}$, which is higly probable to have  $\vspace{15}{ln(b)^k}$ as factor on the k derivative

Since the Taylor series is

$^xb=\sum_{k=0}^{\infty} \frac {1}{k!}*\frac{d^k (^xb)}{dx^k}|_0 *x^{k}$

T(x,k) has to be

$T(x,k)=\frac{d^k (^xb)}{dx^k}|_0 *\frac{x^{k}}{ln(b)^k}$
Because T(x,k) counts "number of forests of labeled rooted trees with x nodes and height at most k", to extend tetration to rational exponents, we need a working definition of "(rooted labeled) tree with non integer height x".

Here is an example of a rooted labeled tree:

The height x=4, because from the root the most distant node is 4 nodes away.
(the root is also labeled 4, but that's an accident and irrelevant)

So, one way we can define what is a tree with non integer height, is to allow the duplication of labels.
This tree has 9 nodes wit labels {1,2,3,4,5,6,7,8,9}. If we allow duplication of labels, we will have 18 nodes labeled {1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,9}

We can agree that n copies of a label g cause the label to hold a weight of 1/n

For example, if we allow 2 copies of each label, the former tree would have a height 2, composed of half of the nodes labeled {3,8,5,2} (I'm not sure if the root 4 should be counted)

Because now we have 18 nodes, the number of forests is much larger, but we need an isomorphism (a bijection) between trees with duplicated labels, and trees with integer labels, so the count T(x,k) returns the same number of forests, for 9 integer nodes, and isomorphic trees with 18 half nodes.

What we need are additional restrictions on the way trees are allowed to be made with fractional nodes, or additional rules on how to count fractional trees and forests.
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.
(12/22/2020, 12:42 AM)Gottfried Wrote: [ -> ]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.
(12/22/2020, 12:42 AM)Gottfried Wrote: [ -> ]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 $\vspace{15}{h \in \mathbb{Q}}$.

That's the most important problem about tetration, and we only need to figure it for $\vspace{15}{0 , or $\vspace{15}{1

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 $\vspace{15}{k^{th}}$ derivative of $\vspace{15}{^he^x}$ 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 $\vspace{15}{^1(e^{x})}$ gives us the formula for counting forests of half nodes $\vspace{15}{k=0.5}$

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.
(12/23/2020, 03:53 PM)marraco Wrote: [ -> ]
(12/22/2020, 12:42 AM)Gottfried Wrote: [ -> ]Hi Marraco -

I'm currently unable to step in, and I do not yet understood your formulae.
Where is the problem?

Above my shoulders... ;-) Don't have momentarily(?) the capacity to really step in and think through the material, might be depend on my age.
Just thought you might be curious to see my proposals of earlier (and more active) years.

Merry Xmas to all :-)

Gottfried