First assume that we can talk about the density of 1s along the AP 1,4,7,… Let call this density p. Now we would expect that if we take two numbers =1 (mod 3) the term at the product would be a 1 with probability , and (at least if we worked with rationals) this would give a random number =1 (mod 3). The equation have two solutions p=1 and p=1/2. This suggest that either the sequence is almost identical to the character, or it is not at all correlated with the character. If this argument works, I think it generalize to all other primes.

Of course this doesn’t prove that there exist such sequence not correlated with the character. I’m not sure if we should expect that such sequences exists.

I have to go, but when I get back I want to apply this remark to the parameter I have been thinking about — it seems that it can be cleaned up in a nice way.

This is how we prove that Walters example is unbounded (with p=3). Perhaps this could be used for a general multiplicative sequence? It would explain why the discrepancy is logarithmic, if it is logarithmic. And if the discrepancy is sublogaritmic, this approach might still work, by letting p depend on C (that is: We want to prove that no multiplicative sequence is bounded by C. Now let p=…).

A toy problem might be to try to use this idea with to prove that Walters example has unbounded discrepancy.

Prob( f(x) = 1 ) = Prob( f(2x) = 1 )

and the averaging argument also gives

Prob(f(x)=1) = 1/2

but as f(x) and f(2x) can’t both be -1, we get f(2x) = -f(x) almost surely. But we can always remove probability zero events from the space and get f(2x)=-f(x) everywhere.

That is a very nice question, and one that just begs to be tested experimentally. What is the longest multiplicative sequence anyone can find that satisfies these conditions, has discrepancy 2, and begins + – 0 + – 0 – - 0, say?

Maybe I can do that one by hand. Filling in the forced values up to 24, I get

+ – 0 + – 0 – - 0 + ? 0 ? + 0 + ? 0 ? – 0 ? ? 0

No, there seems to be far too much flexibility there. So does anyone feel like looking into this with a computer search?

But an alternative argument would be to say that E has density 1/2 inside the even numbers, E and 2E are virtually disjoint, and 2E has density 1/2 inside the even integers (because it has density 1/4 and consists just of even integers). Therefore, after passing to a suitable limit to get exact disjointness, if is any even number, we have . At least, I think, without really checking, that this should work if you want to stay in the integers.

Some other thing I’ve noticed is that there are some exact identities between some of these partial sums at the final value 1124 (although since these curves cross before as grows perhaps the final identities are more of an accident than anything else): denoting them by their indices I’ve seen for that , , , , . The data and a new plot containing is here http://thomas1111.wordpress.com/2010/01/25/correlations-of-the-first-1124-sequence/

http://michaelnielsen.org/polymath1/index.php?title=Upper_and_lower_discrepancy

The main trick (which I first found by working with the ergodic theory formulation of the problem) is to observe that a bounded discrepancy sequence must have “mean zero” in a suitable sense, by averaging the bound |f(x)+\ldots+f(nx)| \leq C in x and then setting n to infinity. Thus the set E of x for which f(x)=+1 has density 1/2.

On the other hand, with lower discrepancy 1, f(x) and f(2x) can’t both be -1, thus E and 2E cover the whole space. Since E and 2E both have density 1/2 (using the right sort of notion of density, and working on the rationals rather than integers) we conclude that E and 2E are almost disjoint. Passing to a subsequence one can make E and 2E genuinely disjoint, so f(2x) = -f(x). This allows one to flip lower discrepancy 1 to upper discrepancy 1 as well, and now it is easy to eliminate the sequence (e.g. one can use the non-existence of drift 2 sequences, though it is easier than this.)

The main gap in what I have written up to now is a proper account of how this parameter might be used in a proof. This comment aims to fill that gap. I don’t mean that I will actually use the parameter or actually prove anything, but I will argue that the parameter is set up in such a way as to make it rather plausible that it could be used in an inductive argument.

Recall that we have up to now discussed two potential strategies for making longer sequences out of shorter ones. The obvious strategy is simply to take the shorter sequence and try to extend it. But another way (more “quotient-like” than “subspace-like”) is to expand the original sequence by some factor and attempt to fill in the gaps. That is, given a sequence , one defines to be and then tries to choose values of when is not a multiple of . An easy observation is that this can be used to multiply the length of a sequence by about 9 while adding only 1 to its discrepancy.

Here, then, is something we could hope for. Note that if we have a multiplicative sequence , then the sequence is either the same sequence or minus that sequence. So we can regard as with the gaps filled in. Now it seems to be hard to fill in the gaps without increasing drifts a certain amount. What we don’t seem to be able to do is show that we are forced to increase the discrepancy (given the existence of the sequence of length 246 and discrepancy 2), but what we do know is that the drift of the scale puts pressure on the -sequence at three times that scale, and we now have a new contribution at the bottom that should often have the effect of increasing drifts in intervals. (I’m imagining here a much more sophisticated version of the kind of easy observation I was making here.)

So I think that it is at least vaguely believable that one could prove that if you expand a sequence by a factor of at least (for some fixed ) then you have to increase the average-oscillation parameter by at least 1, which would imply a logarithmic lower bound for multiplicative functions.

Now let’s consider HAPs with common difference d, where d is congruent to 1 mod 3. The sequences themselves will go + – ? + – ? + – ?, where the question marks denote the values at the HAP with common difference 3d. For this HAP to work, we need that the sum along an HAP with common difference 3d should never equal 2. With this extra condition we’ll be OK.

A similar argument shows that if d is congruent to 2, then the sum along an HAP with common difference 3d should never equal -2.

Returning to S, this says that S must have discrepancy 2, but also that for HAPs with common difference equal to 1 mod 3 we need the partial sums between -2 and 1, and for HAPs with common difference equal to 2 mod 3 we need them to be between -1 and 2.

In a small way, this provides more evidence that low-discrepancy sequences are forced to become more “chaotic” as they get longer (since if they don’t then you get stronger discrepancy conditions such as the one just described).

In general, if is the correlation between and , then the sequence will have bounded partial sums, by an averaging argument: if , then the average value of is greater than . If we also had some kind of multiplicative structure, this would be a strong statement. (If the were bounded below, it would be even stronger, but that may be too much to hope for.)

For example, if we have to build it out of the blocks

+-+
+--

then exhaustive search shows that the longest sequence of discrepancy 2 has length 402 (composed of 134 such blocks).

http://michaelnielsen.org/polymath1/index.php?title=Drift

The argument is quite combinatorial (with an initial use of passing to subsequences to improve things a little bit). It was lengthier than I anticipated. It relies quite a bit on the constrictive nature of drift 2; it would require quite a bit of ingenuity to push it up even to drift 3. Still, perhaps it may trigger some ideas in someone else…

Suppose we have a multiplicative sequence. If it is character-like, then it puts big restrictions on the possible subsequences of various lengths. I haven’t done the calculations, but I think the number of distinct subsequences of length k (by which I mean restrictions to intervals of length k) grows very modestly as a function of k — perhaps even polynomially but as I say I’d need to do the calculation.

Perhaps it is possible to prove that a multiplicative sequence of bounded discrepancy cannot be significantly longer than a character-like example without the number of subsequences growing much faster, to the point that we’re eventually forced to have long strings of 1s or -1s.

This is a very vague idea at the moment. I think the point would be to prove that either a sequence has very few distinct restrictions to intervals of length k, in which case it is forced to be character-like, in which case we are done, or it has lots, in which case we are done for a different reason. But this could turn out to fail miserably.

Assume that we have a multiplicative sequence with bounded discrepancy. Is seems reasonable to assume that the set of terms where the partial sum is at its maximum has positive density. We know that there must be a + at this term and a – after it. Can we use this for anything? Probably not, since we already know that the set of +’s followed by a – must have positive density (perhaps this was a bit too trivial to be posted?).

However, I still think it might be interesting to look at the density of the set of terms where the partial sum is S. If the sequence has bounded discrepancy, we would probably expect the density to be positive for all possible partial sums S, and given the densities we would be able to calculate the chance of a + at a term given the partial sum up to that term.