Thanks Kristal! Looks like we are in very good shape now. In particular we now have the situation described in 795; I’ll put that argument on the wiki now.

As this thread is now basically full, I am closing it and we will be moving to the 900 thread at

http://terrytao.wordpress.com/2009/03/04/dhj3-900-999-density-hales-jewett-type-numbers/

Case 2

The center slice is 42 c less that 17
Then we must have c is 12
And we must have on of the 8
(6, 24, 12, 0, 0) sets.
without less of generality
we can assume it to be
In particular the middle slice
Contains the points
2122 21212 21221
23322 23232 23223 and the *1*** and *3*** slices
have a d value of at least three and so have at most
41 points now if the center slice is 42 we have 124 points
and we are done so it must be 43 but then it has value 18 or more
and and by case 1 we have 124 points or less so we are done.

I’ve updated the Moser spreadsheet,

http://spreadsheets.google.com/ccc?key=p5T0SktZY9DuqNcxJ171Bbw&hl=en

to input the 41-point data, and as expected the “score” a+5b/4+5c/3+5d/2+5e never reached 62 (though the anomalous d=3 solution got close, at 61.5). Assuming Kristal’s lemma, this implies that there is a coordinate whose two side slices have one of the following statistics:

* (4,16,23,0,0) [score: 62 1/3]
* (3,16,24,0,0) [score: 63]
* (4,15,24,0,0) [score: 62 3/4]
* (2,16,24,0,0) [score: 62]

and furthermore the total score must add up to at least 125. (For comparison, in the known 124 point example, the side slices have statistics (3,16,24,0,0), (2,16,24,0,0), or (1,16,24,0,0), with scores 63, 62, and 61 respectively.) This looks like quite a bit of information with which one can work with.

The HOC for n=6 seems a lot harder than the previous cases but I’ll see what I can do.

A lot of progress! I will have to start a new thread soon, this one is being used up.

Kristal: it seems the second case (when c(2****) < 17) is missing from your 787. I’ve tentatively updated Lemma 2 on the wiki to reflect this improved bound, but we’ll need to check that second case.

Assuming this improvement, it means that 125-point Moser sets have at most 41 points in the middle slice, which forces them to have at least 41 points in the side slices. This is good news, because we’ve got a computer list of all of these points. In fact I have the following strategy to propose. We know that a 125-point Moser set does not contain 22222, and so all the points in the Moser set are contained in at least one side slice. More precisely, points with four 2s are contained in one side slice, points with three 2s are contained in two side slices, and so forth up to points with no 2s, which are contained in five side slices. Double-counting this fact, we see that

125 = sum_S a(S)/5 + b(S)/4 + c(S)/3 + d(S)/2 + e(S)

where S ranges over the 10 side slices. Dividing the side slices into five pairs (1****,3****), etc., and using the pigeonhole principle, we may thus assume without loss of generality that

a(1****) + 5 b(1****)/4 + 5 c(1****) / 3 + 5 d(1****) / 2 + 5 e(1*****)

and

a(3****) + 5 b(3****)/4 + 5 c(3****) / 3 + 5 d(3****) / 2 + 5 e(3*****)

sum to at least 125.

Now, we know (contingent on Kristal’s lemma) that the 1**** and 3**** slices have at least 41 points. On the spreadsheet

http://spreadsheets.google.com/ccc?key=p5T0SktZY9DuqNcxJ171Bbw&hl=en

I’ve computed a+5b/4+5c/3+5d/2+5e for the 43 and 42-point solutions. Encouragingly, the largest this quantity is is 63, and almost all of these numbers are less than 62; only the (4,16,23,0,0), (3,16,24,0,0), (4,15,24,0,0), (2,16,24,0,0) configurations are 62 or higher. I haven’t entered in the 41-point data but I would assume the quantity is smaller than 62 in all those cases. If so, that pins down the 1**** and 3**** slices quite precisely, and I would hope that this gets us close to finishing off the problem.

Klas: Great! I’ve updated the hyper-optimistic conjecture page and the spreadsheet to reflect your new data. Thus far the HOC is holding up very nicely, we’ve verified it up to n=5. If we get n=6 for either c^\mu_n or \overline{c}^\mu_n then it might be time to send this sequence to the OEIS also. (The latter sequence is an integer sequence, and if HOC is true, the former is also.)

Marc: I’ve added your example to the wiki and spreadsheet too. Looks like Jason’s super-optimistic conjecture didn’t make it, unfortunately.

Well, *that* was a short-lived conjecture. (re: Marc.791 in response to Dyer.790 — I just confirmed, and changed the spreadsheet) Was that done with random fiddling or was there a method?

Also I would like to remind people of the optimist conjecture in Jakobsen.813, which seems quite (well, moreso than other things) reasonable to prove.

028,046,055,064,073,118,172,181,190,208,217,235,262,
316,334,352,361,406,433,442,541,550,604,613,622,
721,730,901,1000

For ,

The solutions for n=3 are here
http://abel.math.umu.se/~klasm/solutions-n=3-k=3-HOC

The solutions for n=2 are here
http://abel.math.umu.se/~klasm/solutions-n=2-k=3-HOC