# Under what “natural” combinatorial conditions would tetration or higher hyperoperations appear?

This is quite a soft question and is not the same as this question. I want to know what sort of "natural" problems one might examine in combinatorics whose solutions naturally require higher order hyperoperations than exponentiation.

For example, the Ramsey Theory problem in which Graham's Number appeared was the place where I (and, I believe, most people) first encountered up-arrow notation and the idea of hyperoperations in general.

Are there other problems such as that one which would lead you to naturally formulate this hierarchy concept?

(I would also appreciate an explanation as to why hyperoperations are required at all in the Graham's number problem itself, but that's sufficiently different to perhaps require a different question so it's not the priority)

Here's an example problem, which I'm going to solve in a suboptimal way.

Show that there is an $$n$$ large enough that in any coloring of an $$n \times n$$ grid by $$r$$ colors, there is a rectangle whose corners are monochromatic. (That is, we can choose $$1 \le x_1 < x_2 \le n$$ and $$1 \le y_1 < y_2 \le n$$ such that $$(x_1,y_1), (x_1,y_2), (x_2, y_1), (x_2, y_2)$$ are all the same color.)

Let $$n = r^{r+1}+1$$, and take an $$r+1$$ by $$r^{r+1}+1$$ subgrid of the $$n \times n$$ grid which is colored by $$r$$ colors.

First, in every column (which has height $$r+1$$), there are two identically colored cells.

Second, there are $$r^{r+1}$$ ways to color a column. With $$r^{r+1}+1$$ columns, we know that two columns will be identically colored.

So pick the rectangle whose corners are the two identically colored cells in those two identically colored columns.

(This proof can be made more efficient and it doesn't actually need $$n$$ as big as $$r^{r+1}+1$$, but I can give you similar problems in which this is the best thing we know how to do.)

We could take this proof up to $$3$$ dimensions (and find a "cuboid" whose corners are identically colored). To do that, take an $$r+1$$ by $$r^{r+1}+1$$ by $$r^{(r+1)(r^{r+1}+1)}+1$$ grid. In each of its $$r+1$$ by $$r^{r+1}+1$$ layers, we can find a rectangle whose corners are monochromatic. To go from that to a cuboid, we find two identically colored layers. There are $$(r+1)(r^{r+1}+1)$$ total cells in a layer, so there are $$r^{(r+1)(r^{r+1}+1)}$$ ways to color a layer. We have more layers than that, so two layers have an identical coloring.

The 2D result gave us about $$r^r$$ as a bound, and the 3D result gave us about $$r^{r^r}$$ as a bound. If we go up to $$k$$ dimensions, the bound we get is going to be approximately $$r\uparrow\uparrow k$$.

In this particular example, the only thing that we're applying at every step is the pigeonhole principle. If there are more cells in a column, or columns in a 2D grid, or layers in a 3D grid, than there are ways to color each cell/column/layer, then we win.

In a more complicated problem, instead of the pigeonhole principle, we might be applying another Ramsey-type theorem. In that case, in addition to iterated powers, we're going to be iterating some other function.

If that Ramsey-type theorem itself involved some kind of argument like this one, then it could have a hyperoperation for its upper bound. In that case, by applying that theorem over and over, we will get a higher-order hyperoperation for our final bound.

See some of Harvey Friedman's manuscripts here, especially

Long finite sequences (#17 in "1. Preprints, Drafts, and Abstracts")

and

Enormous integers in real life (#17 in "2. Lectures").

Note that Friedman's $$E^{*}(n)$$ is $${}^{n}2$$ and Friedman's $$A(k,n)$$ is $$2\,{\uparrow}^{k-1}\,n,$$ where $${\uparrow}^{k-1}$$ denotes $$(k-1)$$ many Knuth up-arrows. Friedman tells a funny story regarding the fact that $$n(3) > A(7,184)$$ here and in Section 8 of Enormous integers in real life.

Instead of looking at how arithmetic as in tetration is founded in combinatorics, an alternative is looking at how arithmetic may be founded in combinatorics which in turn is founded in arithmetic. This cycle gives a way to build the hyperoperators. Rota's successor Daniel Loeb work with umbral calculus shows how the polynomial umbral calculus generates a number of sequences or combinatorial structures. Loeb shows that extending the polynomial umbral calculus to the logarithmic umbral calculus creates a number of new sequences or combinatorial structures. My understanding is that the umbral calculus can be extended to log(log) and so on.

Tetration and the higher hyperoperators are defined in terms of iteration which in turned is defined in terms of composition. Iterated functions are also iterated compositions. See Continuous Iteration of Dynamical Maps for the first published paper on placing fractional iteration on a mathematical foundation. Also Tetration: an iterative approach. Consider composition and it's embodiment in Faà di Bruno's formula where both unlabeled and labeled partitions appear. Unlabeled partitions are also known as integer partitions while labeled partitions as also called Bell numbers.

Faà di Bruno's formula $$\begin{eqnarray} D^nf(g(z)) = \sum_{\pi(n)} \frac{n!}{k_1! \cdots k_n!} (D^kf)(g(z)) \left(\frac{Dg(z)}{1!}\right)^{k_1} \cdots \left(\frac{D^ng(z)}{n!}\right)^{k_n}\nonumber \label{eq:FaaDiBruno} \end{eqnarray}$$

Iterated Faà di Bruno's formula

A partition of $$n$$ is $$\pi(n)$$, usually denoted by $$1^{k_1}2^{k_2}\cdots n^{k_n}$$ with $$k_1+2k_2+ \cdots nk_n=k$$; where $$k_i$$ is the number of parts of size $$i$$. The partition function $$p(n)$$ is a decategorized version of $$\pi(n)$$; the function $$\pi(n)$$ enumerates the integer partitions of $$n$$, while $$p(n)$$ is the cardinality of the enumeration of $$\pi(n)$$.

Setting $$g(z) = f^{t-1}(z)$$ results in $$\begin{eqnarray} D^n f^t(z)= & \nonumber\\ & \sum_{\pi(n)} \frac{n!}{k_1! \cdots k_n!} (D^k f)(f^{t-1}(z)) \left(\frac{Df^{t-1}(z)}{1!}\right)^{k_1} \cdots \left(\frac{D^n f^{t-1}(z)}{n!}\right)^{k_n}\nonumber \end{eqnarray}$$

Iterated partitions

The Taylors series of the iterated Faà di Bruno's formula results in the combinatorial structure Schröder's Fourth Problem, also known as total partitions in Stanley Enumerative Combinatorics, volume 2. Philippe Flajolet actually referred to these combinatorial structures as hierarchies. See A Problem in Statistical Classification Theory They are also known as phylogenetic trees. Given n books, how many ways can they be classified in a library? This suggests that the laws of physics are based on a hierarchy of physical laws.

A natural representation of hyperoperators

In a series of conversations with Stephen Wolfram he pointed out that there are two mathematical foundations for physics, PDE's and iterated functions. What happens when you want to put something like psychology on a rigorous mathematical foundations. But psychology is based in biology which is in turn based in chemistry and then quantum mechanics. My point is the Universe is filled with many constructs created from multiple levels, from hierarchies. The Ackermann Function ends up being the simplest mathematical hierarchy.

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