Of course, with enough patience this can be fixed by adding more cases or, as Kristal did when he fixed my last post, by considering longer partial sums. I tried to get away with partial sums of length 3 to keep checking managable since the proof was essentially generated by computer.

Let me recap the proof up to this point:

“I assume f(2)=-1, f(3)=-1 and f(5)=-1. Then

f[242,245]=-3+f(61) implies f(61)=1.

For the sake of contradiction I assume furthermore f(7) = 1. Then f[1,10]=2 and

thus f(11)=-1.

We have:

f[18,21]=-3+f(19) implies f(19)=1,

f[168,171]=3+f(17) implies f(17)=-1,

f[22,25]=3+f(23) implies f(23)=-1,

f[60,63]=3-f(31) implies f(31)=1,

f[60,65]=3-f(13) implies f(13)=1,

f[114,117]=3+f(29) implies f(29)=-1,

f[184,187]=3-f(37) implies f(37)=1,”

from this we can get f(85,91) = 5 + f(89) -f(43) which implies that
f(43)=1

however further on there was this: f[868,871]=3-f(79) implies f(79)=1.

So there remains things to do to reach 246 as the maximal length.

I assume f(2)=-1, f(3)=-1 and f(5)=-1. Then

f[242,245]=-3+f(61) implies f(61)=1.

For the sake of contradiction I assume furthermore f(7) = -1.

We have:

f[14,17]=3+f(17) implies f(17)=-1 and

f[34,37]=3+f(37) implies f(37)=-1.

Sub-case 1: f(11)=-1.

0: 1 -1 -1 1 -1 1 -1 -1 1 1|0

10: -1 -1 f(13) 1 1 1 -1 -1 f(19) -1|-2+f(13)+f(19)

Then f[1,12]=-2 and thus f(13)=1. Now

f[54,57]=3-f(19) implies f(19)=1.

10: -1 -1 1 1 1 1 -1 -1 1 -1|0

20: 1 1 f(23) 1 1 -1 -1 -1 f(29) -1|f(23)+f(29)

f[1,22]=2 implies f(23)=-1 and then f[1,25]=3 is a contradiction. Thus f(11)=1.

Sub-case 2: f(11)=1.

0: 1 -1 -1 1 -1 1 -1 -1 1 1|0

10: 1 -1 f(13) 1 1 1 -1 -1 f(19) -1|f(13)+f(19)

We have f(9)=f(10)=f(11)=f(14)=f(15)=f(16)=1 and f(12)=-1. This implies f(13)=-1.

f[34,39]=3-f(19) implies f(19)=1,

f[30,33]=-3+f(31) implies f(31)=1,

f[28,33]=-3+f(29) implies f(29)=1 and

f[242,247]=-3+f(41) implies f(41)=1.

10: 1 -1 -1 1 1 1 -1 -1 1 -1|0

20: 1 -1 f(23) 1 1 1 -1 -1 1 -1|1+f(23)

30: 1 -1 -1 1 1 1 -1 -1 1 1|3+f(23)

40: 1 -1 f(43) 1 -1 -f(23) f(47) -1 1 -1|2+f(43)+f(47)

Since f[1,41]=4 +f(23) we get a contradiction and thus f(11) is not 1.

Therefore we get a contradiction in the main case (the case f(7)=-1).

Since f(7)=1 is treated in an earlier post, that completes the proof (by hand) that there is no completely multiplicative sequence with discrepancy 2.

Maximum absolute value attained by partial sums up to 10000 terms for various :

0.1     731
0.2    1111
0.3     844
0.4     639
0.5    1009
0.6    1271
0.7    1324
0.8     371
0.9     480
0.99    181
0.999   138
0.9999  140

http://www.obtext.com/erdos/some_alpha.png

These are plots up to 5000 for 20 values of between 0.5 and 1. I’ll see if I can home in on the ‘good’ values…

I assume f(2)=-1, f(3)=-1 and f(5)=-1. Then

f[242,245]=-3+f(61) implies f(61)=1.

For the sake of contradiction I assume furthermore f(7) = 1. Then f[1,10]=2 and thus f(11)=-1.

We have:

f[18,21]=-3+f(19) implies f(19)=1,

f[168,171]=3+f(17) implies f(17)=-1,

f[22,25]=3+f(23) implies f(23)=-1,

f[60,63]=3-f(31) implies f(31)=1,

f[60,65]=3-f(13) implies f(13)=1,

f[114,117]=3+f(29) implies f(29)=-1,

f[184,187]=3-f(37) implies f(37)=1,

f[558,561]=-3+f(43) implies f(43)=1,

f[40,43]=3+f(41) implies f(41)=-1,

f[50,55]=3+f(53) implies f(53)=-1,

f[58,61]=3+f(59) implies f(59)=-1,

f[92,95]=-3-f(47) implies f(47)=-1 and

f[132,135]=3-f(67) implies f(67)=1.

0: 1 -1 -1 1 -1 1 1 -1 1 1|2

10: -1 -1 1 -1 1 1 -1 -1 1 -1|0

20: -1 1 -1 1 1 -1 -1 1 -1 -1|-2

30: 1 -1 1 1 -1 1 1 -1 -1 1|0

40: -1 1 1 -1 -1 1 -1 -1 1 -1|-2

50: 1 1 -1 1 1 -1 -1 1 -1 1|0

60: 1 -1 1 1 -1 -1 1 -1 1 1|2

70: f(71) -1 f(73) -1 -1 1 -1 1 f(79) -1|-1+f(71)+f(73)+f(79)

Now f[1,70]=2 and thus f(71)=-1. We have

f[72,75]=-3+f(73) implies f(73)=1,

f[88,91]=3+f(89) implies f(89)=-1 and

f[868,871]=3-f(79) implies f(79)=1.

70: -1 -1 1 -1 -1 1 -1 1 1 -1|0

80: 1 1 f(83) -1 1 -1 1 1 -1 1|3+f(83)

90: 1 -1 -1 1 -1 1 f(97) -1 -1 1|2+f(83)+f(97)

Now f[1,91]=4+f(83) is a contradiction and thus f(7)=-1.

It remains to show (by hand) that there is no completely multiplicative sequence with discrepancy 2 and f(2)=f(3)=f(5)=f(7)=-1. If my computer is right, this is much easier.

(i) Whatever you choose, you get big growth, but affects its frequency and amplitude to some extent.

(ii) Most have very little effect, but if you get the right you almost completely kill the growth (with subcases that you make it square-root-like or you make it power-of-log-like).

(iii) There are a few values for , notably 1, that are particularly bad, but in general pretty well any is a big help.

It occurs to me now that the apparent linear behaviour may be an illusion, and that the truth may be growth that is very slightly sublinear. My reasoning is as follows. If you define a multiplicative function by defining to be raised to the power , then you get some very nice behaviour that points in the direction of linear discrepancy. For instance, if and are primes and it so happens that is only just smaller than , then it is very likely that . So there is a tendency for the s to behave very like the primes, at least in the run-up to a power of 2.

This kind of argument works for products of more than two primes as well, but the more primes you take, the worse it gets. Now a typical integer near has about prime factors (despite anything I might have written to the contrary), and if is less than a million, then looks pretty much like a constant. So I think that the apparent linearity that we see in the examples that blow up may well be a function more like that we are just not fully seeing yet. Looking back at some of the plots, there are some signs, though far from conclusive ones, that the gradient is levelling off just a little.

Actually, have you tried the very simple adjustment, designed to avoid this rather extreme behaviour, of setting … wait, I’ve just looked back and seen that you have. I’m talking about having probabilities of 2/3 and 1/3 instead of 1 and 0. You gave a link to some pictures that you claimed were much more square-root-like. I’d be interested to know whether that really is the case — that is, if you go a lot further than 10,000, do the deviations get quite a bit bigger?

Also, one could imagine varying your tanh experiment by sticking a factor of 1/2 in front of the tanh (so that the probabilities vary between 1/4 and 3/4 rather than between 0 and 1). I think that might have the effect of stopping the linear behaviour from ever arising, while maintaining the good control. And if it doesn’t, then perhaps some other constant would.

http://www.obtext.com/erdos/alternate_30000.png

As expected the graphs display a mixture of characteristics. Apart from the green one they do seem to be just about under control.

Now the other method (making -1 when partial sums are positive and +1 when they’re negative) defeats randomness, but seems to give rise to linear behaviour. So what if we combine the two? That is, what if we just alternate the two methods of choosing ? The idea is that one method would keep things at worst random and the other would then kill off the randomness. In short, one method would produce the randomness that the other one needs for the heuristic argument that lay behind it to work.

My heuristic arguments here are obviously pretty vague, and I can well imagine that this won’t work at all, but it definitely seems to be worth a try, and I’d have thought you could patch together some bits of existing code pretty quickly.

On another topic, I am very intrigued by your tanh sequences, and especially by the idea that they may be extremely well controlled but in an unstable way. There are all sorts of phenomena that I’d like to understand better. It feels as though there ought to be differential equations that would predict the behaviour well, but my attempts to do that have so far given obviously wrong predictions.

The program does not move forward in simple steps, as from 1119 to 1120, so it is difficult to estimate the remaining running time.

Define as if and otherwise.

Then is completely multiplicative and up to length has discrepancy at most .

Verifying using the sample multiplicative sequence of length 246 with discrepancy 2 (http://michaelnielsen.org/polymath1/index.php?title=Multiplicative_sequences), I checked to find the largest index where the sequence has value +1 and the partial sum just before that is 0, which is at . Extending that gives a completely multiplicative sequence of length 1935 with discrepancy 3, and it goes to with discrepancy 7.

It's still multiplicative if , which helped improve the original Matryoshka sequence. When I extend that way I get to length 1943 while still at discrepancy 3.

What I'm really hopeful about is iterating this procedure – finding a good constructive way to choose how much to increase the discrepancy and where to pick the length of the next sequence, then constructing "better and better" versions. I thought I had sqrt(log) for a moment, but it didn't work…

which has 66 elements.

which has 729 elements! (Thus all Klas’s sequences satisfy .)