Rosie Zhao and I have just uploaded our paper “Arithmetic subsequences in a random ordering of an additive set” to the arXiv. In its simplest form, the problem we consider is the following. Given an ordering of the numbers $1$ through $n$, what is the length of the longest arithmetic subsequence that is embedded in this permutation? For example, in the ordering $(2,7,1,6,3,4,5)$, the longest arithmetic subsequence is $(2,3,4,5)$, which has length $4$. If we store the permutation in an array, the problem of finding the length of the longest arithmetic subsequence is a popular programming interview question that can be solved efficiently using dynamic programming (it is a medium-difficulty problem on LeetCode). But if we take the ordering to be random, with each of the $n!$ possibilities occurring with equal probability, then the length $L_n$ of the longest arithmetic subsequence becomes a random variable. Our paper studies the asymptotic behaviour of $L_n$ as $n$ gets large.

### The integer case

Let $[1,n]$ denote the discrete interval ${1,2,\ldots,n}$. We regard an *arithmetic progression* in a sequence
to be a subsequence

where $a$ is called the *base point*, $r$ is the *step size*, and $k$ is the *length*. Traditionally, arithmetic
progressions are studied as *subsets* of some larger set, and one is interested in finding out if sets contain
long arithmetic progressions as subsets. In our case, we start with the arithmetic progression $[1,n]$,
scramble its elements into a sequence, and look for a long subsequence. For small $n$, using the dynamic
programming techniques mentioned above, one can exhaustively compute the distribution of the longest
arithmetic subsequence $L_n$. Letting $f_n(k)$ denote the number of permutations of $[1,n]$ whose longest
arithmetic subsequence is of length $k$, we have the following table (click to enlarge):

Notice that each row sums to $n!$, since there are $n!$ orderings of $[1,n]$, and we can compute probabilities by simply dividing; for example, the probability that $L_9 = 3$ is $168368/(9!)\approx 46.4\%$. The first main result of our paper is the following theorem:

**Theorem 4.** *Let $L_n$ denote the longest arithmetic sequence in an ordering of $[1,n]$, chosen uniformly
at random. There exists a function $\psi(n)$ with $\psi(n)\sim 2\log n/\log\log n$ such that the probabilility
that $L_n$ does not equal either $\lfloor \psi(n)\rfloor$ or $\lceil\psi(n)\rceil$ tends to $0$ as $n$ approaches
infinity.*

(In fact, we proved a slightly more general form of this theorem in the paper, considering $[1,n]^d\subseteq {\bf Z}^d$ for $d\geq 1$ and finding a function $\psi(n,d)\sim 2d\log n/\log\log n$.) To prove the theorem, we let $N_k$ denote the number of arithmetic subsequences of length $k$ in the random ordering. By the union bound, we have \({\bf P}\{L_n\geq k\}\leq {\bf E}\{N_k\}\), which goes to $0$ if $k\geq \lceil\psi(n)\rceil$. Then by the second moment method, it can be shown that

\[{\bf P}\{L_n\geq k\} = {\bf P}\{N_k\geq 1\} \geq { {\bf E}\{N_k\}^2\over {\bf E}\{ {N_k} ^2\} },\]which approaches $1$ when $k<\lfloor\psi(n)\rfloor$. Since $k$ must be an integer and there are no integers strictly between $\lfloor\psi(n)\rfloor$ and $\lceil\psi(n)\rceil$, this implies that as $n\to\infty$, $L_n$ is either the floor or ceiling of $\psi(n)$ with probability approaching $1$. By a counting argument, one can show that

\[{\bf E}\{N_k\} \sim {n^2\over k!(k-1)},\]which can be inverted to show that $\psi(n)\sim 2\log n/\log\log n$.

### Cyclic groups

If we consider addition in the cyclic group ${\bf Z}/n{\bf Z}$, then $L_n$ is larger, since any arithmetic progression in $[1,n]\subseteq {\bf Z}$ is also a progression in ${\bf Z}/n{\bf Z}$, but not the other way around. For example, the sequence $(0,2,6,1,3,5,4)$ has the $4$-term arithmetic subsequence $(0,6,5,4)$ when addition is taken modulo $7$ (the base point is $0$ and the step size is $6$), but it has no arithmetic subsequence of length $4$ when the addition is ordinary integer addition. The distribution of $L_n$ for small $n$ in the cyclic case looks like this ($g_n(k)$ is the cyclic analogue of $f_n(k)$ from Table 1):

We prove the following analogue of Theorem 4 in the cyclic case:

**Theorem 7.** For a positive integer $n$, let $L_n$ denote the longest arithmetic subsequence in an
ordering of ${\bf Z}/n{\bf Z}$, chosen uniformly at random. There exists a function $\chi(n)$ with
$\chi(n)\sim 2\log n/\log\log n$ such that $L_n$ is either the floor or ceiling of $\chi(n)$ with probability
tending to $1$ as $n\to\infty$.

Note that while both $\psi(n)$ from Theorem 4 and $\chi(n)$ from Theorem 7 both have the same leading asymptotic term, we always have $\psi(n)<\chi(n)$. A closer look at Table 2 also reveals some interesting curiosities regarding $L_n$ in the cyclic case. Firstly, \({\bf E}\{L_n\}\) does not increase monotonically with $n$ as in the integer case. For example, we have \({\bf E}\{L_7\} = 4.25\) and \({\bf E}\{L_8\}\approx 4.136\). This is because there are “more ways” to form arithmetic subsequences in permutations of ${\bf Z}/n{\bf Z}$ when $n$ is prime. We also notice that there are only permutations without arithmetic subsequences of length $3$ when $n$ is a power of $2$. This can be formulated as the following lemma:

**Lemma 8.** The number $g_n(2)$ of orderings of ${\bf Z}/n{\bf Z}$ that do not contain any arithmetic
subsequence of length $3$ equals $2^{n-1}$ if $n=2^m$ for some $m\geq 1$, and is zero otherwise. Any ordering
of ${\bf Z}/2^m{\bf Z}$ that contains no progression of length $3$ consists of $2^{m-1}$ elements of the same
parity, followed by the remaining $2^{m-1}$ elements of the opposite parity.

The proof was by induction in the case that $n$ is a power of $2$, and for $n$ not a power of $2$ we can reduce to the case where $n$ is an odd prime.

### Possible noncommutative generalisations

The first and second moment methods we used did not rely on the fact that the groups we considered (${\bf Z}^d$ and ${\bf Z}/n{\bf Z}$) are abelian. One could extend these results to random orderings of other finite groups, such as the dihedral group $D_n$ or the symmetric group $S_n$. In the noncommutative case, one needs to specify whether the step size is multiplied by the base point on the left or on the right, but if the sequence contains every element of the (finite) group, then the distribution of $L_n$ will be the same for both choices.