The Erdős discrepancy problem

Introduction and statement of problem

The Erdős discrepancy problem is a problem that has been around since the 1930s and has resisted attack ever since. It is extremely easy to state, and a very appealing problem, so perhaps Polymath can succeed where individual mathematicians have failed. If so, then, given the notoriety of the problem, it would be a significant coup for the multiply collaborative approach to mathematics. But even if a full solution is not reached, preliminary investigations have already thrown up some interesting ideas, which could perhaps be developed into publishable results.

To state the problem, we first define a homogeneous arithmetic progression to be a set of positive integers of the form {d,2d,3d,...,nd}. (It is also reasonable to define it to be a set of non-negative integers of the form {0,d,2d,3d,...,nd}: at the time of writing it is not clear what the "official" definition should be.) Suppose that we have a sequence [math]\displaystyle{ x_1,x_2,x_3,\dots }[/math] of elements of the set [math]\displaystyle{ \{-1,1\} }[/math] (henceforth to be referred to as a [math]\displaystyle{ \pm 1 }[/math] sequence). We say that the discrepancy of the sequence [math]\displaystyle{ (x_n) }[/math] with respect to a set A of positive integers is [math]\displaystyle{ |\sum_{n\in A}x_n| }[/math]. The reason for this name is that we can think of the sequence as a red/blue-colouring of the positive integers, and the discrepancy with respect to A is then the difference between the number of red elements of A and the number of blue elements of A. The Erdős discrepancy problem is the following question.

Problem. Let [math]\displaystyle{ (x_n) }[/math] be a [math]\displaystyle{ \pm 1 }[/math] sequence. Must it be the case that for every constant C there exists a homogeneous arithmetic progression P such that the discrepancy of the sequence [math]\displaystyle{ (x_n) }[/math] with respect to P is at least C?

Simple observations

• To answer the problem negatively, it is sufficient to find a completely multiplicative sequence (that is, a sequence [math]\displaystyle{ (x_n) }[/math] such that [math]\displaystyle{ x_{mn}=x_mx_n }[/math] for every m,n) such that its partial sums [math]\displaystyle{ x_1+\dots+x_n }[/math] are bounded. First mentioned in the proof of the theorem in this post.

• The function that takes n to 1 if the last non-zero digit of n in its ternary representation is 1 and -1 if the last non-zero digit is 2 is completely multiplicative and the partial sum up to n is easily shown to be at most [math]\displaystyle{ \log_3n +1 }[/math]. Therefore, the rate at which the worst discrepancy grows, as a function of the length of the homogeneous progression, can be as slow as logarithmic. [1] It turns out that the base can be made significantly higher than 3, so this example is not best possible. Correction

• To answer the problem negatively, it is also sufficient to find a completely multiplicative function f from [math]\displaystyle{ \mathbb{N} }[/math] to a finite Abelian group G (meaning that f(mn)=f(m)+f(n) for every m,n) and a function h from G to {-1,1} such that for each g in G there exists d with f(d)=g and with the partial sums hf(d)+hf(2d)+...+hf(nd) bounded. In that case, one can set [math]\displaystyle{ x_n=hf(n) }[/math]. (Though this observation is simple with the benefit of hindsight, it might not have been made had it not been for the fact that this sort of structure was identified in a long sequence that had been produced experimentally). First mentioned here. Se also this post and this comment.

• It can be shown that if a [math]\displaystyle{ \pm 1 }[/math] sequence [math]\displaystyle{ (x_n) }[/math] starts with 1 and has only finitely many distinct subsequences of the form [math]\displaystyle{ (x_d,x_{2d},x_{3d},\dots) }[/math], then it must arise in the above way.

• The problem for the positive integers is equivalent to the problem for the positive rationals. [2](Sketch proof: if you have a function f from the positive integers to {-1,1} that works, then define [math]\displaystyle{ f_q(x) }[/math] to be f(q!x) whenever q!x is an integer. Then take a pointwise limit of a subsequence of the functions [math]\displaystyle{ f_q }[/math].)

Experimental evidence

A nice feature of the problem is that it lends itself well to investigation on a computer. Although the number of [math]\displaystyle{ \pm 1 }[/math] sequences grows exponentially, the constraints on these sequences are such that depth-first searches can lead quite quickly to interesting examples. For instance, at the time of writing they have yielded a sequence of length 1124 with discrepancy 2. See this page for more details about this and for further links.

Annoted Bibliography

Mathias, A. R. D. On a conjecture of Erdős and Čudakov. Combinatorics, geometry and probability (Cambridge, 1993).

This is a short paper establishing the maximal length of sequence for the case where C=1 is 11, and is the starting point for our experimental studies.