First obtain multiplicative structure and then obtain a contradiction
A great deal of multiplicative structure has been observed in the longest sequences we have of discrepancy 2. (See the experimental results page for more details about this.) For example, if you look at the beginnings of HAP sequences, you tend to find that they can be partitioned into a small number of classes of very similar sequences. And if you look at the values at the elements of a geometric progression, you tend to find that they are periodic, or approximately periodic in the way that a sequence such as [math]\lfloor \alpha(n+1)\rfloor-\lfloor\alpha n\rfloor[/math] is approximately periodic if α is a real number between 0 and 1. The description "multiplicative structure" is appropriate here because the number of distinct HAP subsequences of a completely multiplicative sequence is either one (if the sequence is constant) or 2 (otherwise, and in this case one of the two sequences is minus the other). This suggests the following proof strategy.
1. Prove that if there exists an infinite bounded-discrepancy sequence then there exists an infinite bounded-discrepancy sequence with multiplicative structure.
2. Prove that if there exists an infinite bounded-discrepancy sequence with multiplicative structure, then there exists an infinite completely multiplicative sequence with bounded discrepancy.
3. Prove that there is no infinite completely multiplicative sequence with bounded discrepancy.
There are some grounds for hope that a programme like this could be carried out, at least in part. For instance, the experimental evidence points very strongly towards 1, and we have a sketch proof of 2. It would be extremely interesting if the entire problem could be reduced to the much more specific multiplicative case 3.