## 305 – A lower bound for n(3)

February 14, 2012

This continues the previous post on Friedman’s theorem.

I want to give a brief idea of the size of Friedman’s $n(3)$, as defined on homework 2. The reference for what follows is

• Harvey Friedman, Long finite sequences,  J. Combin. Theory Ser. A, 95 (1) (2001), 102–144.

Recall that $n(3)$ is the largest possible size $n$ of a sequence $x_1 x_2 x_3\dots x_{n-1} x_n$ with property $*$, all of whose terms are 1, 2, or 3. Here, to have property $*$ means that if $i, then the sequence $x_i x_{i+1}\dots x_{2i}$ is not a subsequence of the sequence $x_j x_{j+1}\dots x_{2j}$.

In lecture we gave an example of such a sequence, of length 216, namely

$12^2131^73^21313^813^513^{20}1^23^{53}13^{108},$

where we use exponential notation to indicate the number of times a symbol appears consecutively, so the sequence begins

$12213111111133131333333331\dots$

Actually, $n(3)$ is much larger than 216. We know it is finite but, as Friedman mentions in his paper, it is an incomprehensibly large number. To explain this, we should mention the Ackermann hierarchy of functions. These are functions $A_1,A_2,\dots$, with each $A_i:{\mathbb Z}^+\to{\mathbb Z}^+$ a strictly increasing function.

Define $A_1(n)=2n$. Also, set

$A_{k+1}(n)=A_k(A_k(\dots(A_k(1))\dots))$,

where there are $n$ occurrences of $A_k$.

This is perhaps best understood by computing a few examples. We have $A_1(1)=2$, $A_1(A_1(1))=A_1(2)=4=2^2$, $A_1(A_1(A_1(1)))=A_1(4)=8=2^3$ and, by an easy induction, $A_2(n)=2^n$.

Now, $A_2(1)=2$, $A_2(A_2(1))=A_2(2)=2^2$, $A_2(A_2(A_2(1)))=A_2(2^2)=2^{2^2}$ and, in general, $A_3(n)$ is an exponential tower of 2s of height $n$.

The descriptions become more complicated now. For example, $A_4(n)$ is an exponential tower of 2s, but its height is itself an exponential tower of 2s (of height $n$). For example,

$\displaystyle A_4(4)=2^{\vdots ^2}$,

where the tower has height $\displaystyle 2^{2^{2^2}}=65536$.

Let me now quote from Friedman’s paper:

I submit that $A_4 ( 4)$ is a ridiculously large number, but it is not an incomprehensibly large number. One can imagine a tower of 2’s of a large height, where that height is 65,536, and 65,536 is not ridiculously large.

However, if we go much further, then a profound level of incomprehensibility emerges. The definitions are not incomprehensible, but the largeness is incomprehensible. These higher levels of largeness blur, where one is unable to sense one level of largeness from another.

For instance, $A_4( 5 )$ is an exponential tower of 2’s of height $A_4( 4 )$.

It seems safe to assert that, say, $A_5( 5 )$ is incomprehensibly large. We propose this number as a sort of benchmark. In Section 4 we prove that $n( 3 ) >A_7( 184 )$, which is considerably larger.

Friedman actually goes much further than this. Using computer data provided by Randall Dougherty, Friedman proves in the last section of the paper that

$n(3)>A_{7198}(158386)$.

For all I know, this could still be very far from the actual value of $n(3)$. And I do not know a reasonable description of an upper bound.

In technical terms, Friedman proves that the function $n(k)$ is not primitive recursive, and actually grows much faster than the usual examples of recursive functions that are not primitive recursive. He also describes a function $H$ (a member of the so-called Hardy hierarchy) that grows faster than $n(k)$. This means that $n(k) for all sufficiently large $k$, but the argument does not give a specific bound when $k=3$.

## 515 – Homework 2

February 14, 2012

This set is due Feb. 29 at the beginning of lecture. Let me know if more time is needed or anything like that. Problem 4 was incorrect as stated; I have fixed it now. Thanks to Tara Sheehan for bringing the problem to my attention.