Avoiding Sum Cubes

One of the most interesting and challenging open problems in combinatorics on words is to decide whether there exists an infinite word over a finite subset of N, the non-negative integers, with the property that it contains no two consecutive blocks of the same length and the same sum (a "sum square").

For example, 01231301020103102310313231301020131013230 is a word of length 41 with this property, but if you append any one of {0,1,2,3} to it, it no longer does. Appending 0 gives the sum square 00; appending 1 gives the sum square (3231301020)(1310132301); appending 2 gives the sum square (103132313010)(201310132302); appending 3 gives the sum square (132)(303). So this word cannot be extended to an infinite word avoiding sum squares; the longest such is of length 50. Of course, there could be an infinite word avoiding sum squares over some other subset of N; no one currently knows.

This problem was originally stated by Pirillo and Varricchio in 1994, and independently by Halbeisen and Hungerbühler in 2000.

Today we posted a preprint in the arxiv that solves a related unsolved problem. Instead of avoiding sum squares, we show that we can avoid sum cubes: three consecutive blocks of the same length and same sum.

The construction is actually quite simple: the infinite word in question is the fixed point of the morphism
0 → 03
1 → 43
3 → 1
4 → 01,
and can be obtained by repeatedly applying this morphism starting with 0. Here are the first 50 terms:
03143011034343031011011031430343430343430314301103 .
I found this morphism several years ago.

However, proving that this word has the desired property is not simple. The proof was recently achieved by Luke Schaeffer, using ideas of James Currie at the University of Winnipeg and Julien Cassaigne at the Institute de Mathématiques at Luminy in France.