The discussion is continued over here.

I meant to say that there are 4 x 3^{n-1} of +-1 sequences of length 2n and 2 x 3^n of length 2n+1. The norm brackets were meant to be absolute value brackets…

There are , sequences of length such that and of length , so there are of lengthh . There are sequences in total, so the probability that a random sequence of length survives the sum test is . Looking at subsequence or our sequence, the probability that it survives the test is . (This is where the argument turns faulty, because I don’t know how correlated the events , are. I don't know if it kills the argument. Ignoring that and pressing on…) Continuing with the rest of the subsequences we get that the probability that the sequence has discrepancy is . Multiplying by the total number of sequences we get that there are length sequences of discrepancy 2.

Similar argument suggests that there are i.e. 0 for large enough sequences of discrepancy 1.

Thank you for your reply (January 14, 2010 at 6:17 pm)
and explanation.

I still think we can, and almost certainly should, unify the two concepts. I have a sneaking suspicion that there’s some really deep, but rather well-understood, number-theory stuff lurking beneath the surface — adelic analysis? I don’t know enough about that stuff to be sure — and I’d love to have a clearer view of what it is.

For this to work, it’s quite important that the signs Alec was calculating above (but renamed so that I define to be what he calls ) should be completely multiplicative. To my great delight, they seem to be.

+ + + + – - – + …

With the following shifts:

n = 1: Shift of 0
n = 2: Shift of 1
n = 3: Shift of 1
n = 4: Shift of 2
n = 5: Shift of 2
n = 6: Shift of 2
n = 7: Shift of 0 (?!)
n = 8: Shift of 3
n = 9: Shift of 2
n = 10: Shift of 3
n = 11: Shift of 3
n = 12: Shift of 3
n = 13: Shift of 3

This is of course a very small sample but it does in fact look as though the shifts may be of size log n.

If that’s the case, then each sequence will either be biased towards 1 or towards -1. If the former, then it should never have two -1s in a row. So I’d be interested to know whether the sequence for 16 has this property (that it doesn’t have both two 1s in a row and two -1s in a row).

Come to think of it, I can work this out for myself, so here goes. I get + – + + – -, which means that, disappointingly, it seems to be a serious anomaly rather than a mild one. (However, the anomaly first shows up at 1056, so maybe there are other examples that are OK.)

: +1, -1, -1, -1, +1, +1, -1, +1, -1, -1, +1, +1, +1, -1, +1, -1, -1, +1, +1, +1, +1

: 0, 1, 2, 3, 2, 3, 3, 3, 4, 3, 4, 3, 4, 3, 5, 5, 4, 5, 4, 4, 5

Apart from one anomalous value in the 3 sequence (the one starting at 7) and one in the 16 sequence (the one starting at 33), the agreement is perfect up to here.

I’m not entirely convinced that this is a phenomenon that’s really unique to the 1124-sequence — it just seems to encapsulate some of the multiplicative structure in a rather nice way.

So there’s a new item that’s jumped right to the top of my wishlist. Alec, can you get any super-long sequences this way?

The experiment was very simple. I took the numbers in turn and for each I wrote out the values at . Here is a table of the results. I start with what n is and then give the values. I have deliberately shifted the rows by varying amounts (I hope it will come out nicely when I submit the comment, but if not then perhaps someone like Thomas can produce a nicer table). Added later: sorry, it hasn’t come out all that well, so a proper table would be great.

2: – + – – + – – + – +
3: + – + + – + + – +
4: . + – – + – – + – +
5: . – + + – + + – +
6: . – + + – + + – +
7: . . – + – + + – + -
8: . . – – + – – + – +
9: . + – – + – – + -
10: . . + + – + + – +
11: . . – – + – – + -
12: . . + + – + + – +
13: . . + + – + + – +
14: . . . + – + + – + -
15: . . – – + – – + -
16: . . . – + – – + – +
17: . . . + – + + – + -

What I find remarkable about this table is just how perfect it is. The only anomalous value is (which will come as no surprise to Mark). If you changed that, then everything would be a shift, or minus a shift, of the sequence that you get for 2. (Perhaps I should have started with the row at 1.) Also, the row for 2 looks to me very like a “periodic sequence with irrational period”. Could it be the golden ratio? Also, it is notable that the amount you shift is closely related to the size of the number. So if we count the row I didn’t put (starting with 1) as the basic row, then rows 2 and 3 are shifted by 1, rows 4,5,6,9 are shifted by 2, rows 7,8,10,11,12,13,15 are shifted by 3, and rows 14,16,17 are shifted by 4. (The shifts at powers of 2 obviously have to be what they are.)

In the Polymath spirit, I am mentioning this straight away rather than trying to work out exactly what is going on and reporting back. One question I’d like to answer is whether the sequence + – + – – + – – + – + is compatible with a period that is related to the golden ratio. I very much hope so as I haven’t quite let go of that little fantasy. (That’s the sequence of values at 1,2,4,8,16,32,64,128,256,512,1024.) Another is whether everything continues to work wonderfully well if you look at powers of 3, though the sequences will be a bit short. Here’s what the values are at 1,3,9,27,81,243,729: + + + + – – -. And at twice those numbers: – – – + + +. Hmm, it’s looking pretty good!

The code isn’t optimized, and actually is running in Mathematica, so the timing isn’t interesting. The ratio of time is, however. The COMBINATORIAL EXPLOSION happens between drift=5 and drift=6.

Perhaps interesting is that the longest branches of the trees seem to be of the alternating sign sort. Sometimes two plusses followed by two minuses.

But this seems to contradict your earlier observation, so maybe one of us has made a mistake?