![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
sniffnoyOK! Let's do this. Or, rather, let me comment on our lack of having done this.
Arias precedes these last four conjectures with the comment, "What follows is more tentative and is based on only a few cases." Which makes them more interesting!
Conjecture 8: For all ordinals β<ξ - well, we know what ξ is, so we can say, for all ordinals β<ωω -
limn→∞ aβω+n=cβ/3 ;
limn→∞ bβω+n=aβ ;
limn→∞ cβω+n=bβ
(Reminder: He's using a_α for the 0 mod 3 sequence, b_α for the 4 mod 3 sequence, c_α for the 2 mod 3 sequence. Also reminder: So far as I am concerened right now, the number 1 is not 4 mod 3, but rather is a special case to be excluded.)
(He notes afterward that this conjecture is the basis for his claim that ωξ should equal ξ - a rather different route than we arrived at it! We observed that it looked like defects less than n should have order type ωn, so you just get ωω as the limit of those.)
Thought: Did he mean ωβ, instead of βω? That would seem to me to make a lot more sense.
What I think he intended conjecture 8 to be: For all β<ωω,
limn→∞ aωβ+n=cβ/3 ;
limn→∞ bωβ+n=aβ ;
limn→∞ cωβ+n=bβ
Now this is interesting! But before I comment on the status of this - even though you can guess what that'll be - let's reformulate it. You'll notice it's not fully symmetric - it's got two 1s and a 1/3. Do I intend to make it symmetric? No. But two 1s and a 1/3 looks off to me - the familiar pattern that turns up in integer complexity is not 1, 1, 1/3, but rather 2/3, 2/3, 3/4. (Or 3/2, 3/2, 4/3, if you prefer.)[0] Now, as I am writing this, I have not actually *done* the reformulation I intend, so I don't know if that's what'll come out of it, but I'm betting it will.
What is the reformulation I intend? Well, like I said earlier, I don't like Arias's choice of denominator. So, allow me to define a'α, b'α, and c'α, which will of course be exactly the same except using a denominator of g(||n||) instead of 3⌊||n||/3⌋. Much more natural, isn't that? So a'_α=a_α, but b'_α=(3/4)b_α and c'_α=c_α/2. So, doing that, we get:
Reformulated version of above: For all β<ωω,
limn→∞ a'ωβ+n=(2/3)c'β ;
limn→∞ b'ωβ+n=(3/4)a'β ;
limn→∞ c'ωβ+n=(2/3)b'β
...yup, that's 2/3, 2/3, 3/4, just as expected.
Hm, now I'm curious - what would we get if we put it in terms of denominator 3||n||/3? Probably nothing nice, but let's do it anyway - after all, we have formulated pretty much everything in these terms so far, so it may make things easier. We'll have a''_α=a_α, b''_α=3-2/3b_α, c''_α=3-1/3c_α. Put this way, the conjecture becomes:
Reformulated (again) conjecture 8: For all β<ωω,
limn→∞ a''ωβ+n=3-2/3c''β ;
limn→∞ b''ωβ+n=3-2/3a''β ;
limn→∞ c''ωβ+n=31/3b''β
...aw, I was hoping it would come out symmetric, all factors of 3-1/3. Evidently not. I suppose I could figure out how to do that, but there's really no way that's worth my time. Interestingly the one here that's different is the 2-4 boundary, rather than the 4-0 boundary. (Well, Arias had the 0-2 boundary as the one that's different, but that's because he formulated it badly. :) )
In an case, what's the status? Well... we haven't worked on this. At all, to my knowledge. But hey, we have some tables - I kept track of ordering up to 16d(2) before I decided that was too tedious - so let's check if it they satisfy this! Hopefully I bothered to write down ||n|| mod 3...
...crap, no, I didn't. Well, that will make this rather tedious. OK, first things first, I'll make a version of the table that includes that, then check it, and then come back and report on the results, though I can, at least, tell you off the top of my head that it's true for β=0 (and I doubt Arias would have conjectured it otherwise...)
-Harry
[0]These correspond to the 0-2 boundary, the 2-4 boundary, and the 0-4 boundary, respectively, but since I guess things really go 0-4-2, I guess that should be 2/3, 3/4, 2/3. Well, it's a cycle, so it doesn't matter.
Arias precedes these last four conjectures with the comment, "What follows is more tentative and is based on only a few cases." Which makes them more interesting!
Conjecture 8: For all ordinals β<ξ - well, we know what ξ is, so we can say, for all ordinals β<ωω -
limn→∞ aβω+n=cβ/3 ;
limn→∞ bβω+n=aβ ;
limn→∞ cβω+n=bβ
(Reminder: He's using a_α for the 0 mod 3 sequence, b_α for the 4 mod 3 sequence, c_α for the 2 mod 3 sequence. Also reminder: So far as I am concerened right now, the number 1 is not 4 mod 3, but rather is a special case to be excluded.)
(He notes afterward that this conjecture is the basis for his claim that ωξ should equal ξ - a rather different route than we arrived at it! We observed that it looked like defects less than n should have order type ωn, so you just get ωω as the limit of those.)
Thought: Did he mean ωβ, instead of βω? That would seem to me to make a lot more sense.
What I think he intended conjecture 8 to be: For all β<ωω,
limn→∞ aωβ+n=cβ/3 ;
limn→∞ bωβ+n=aβ ;
limn→∞ cωβ+n=bβ
Now this is interesting! But before I comment on the status of this - even though you can guess what that'll be - let's reformulate it. You'll notice it's not fully symmetric - it's got two 1s and a 1/3. Do I intend to make it symmetric? No. But two 1s and a 1/3 looks off to me - the familiar pattern that turns up in integer complexity is not 1, 1, 1/3, but rather 2/3, 2/3, 3/4. (Or 3/2, 3/2, 4/3, if you prefer.)[0] Now, as I am writing this, I have not actually *done* the reformulation I intend, so I don't know if that's what'll come out of it, but I'm betting it will.
What is the reformulation I intend? Well, like I said earlier, I don't like Arias's choice of denominator. So, allow me to define a'α, b'α, and c'α, which will of course be exactly the same except using a denominator of g(||n||) instead of 3⌊||n||/3⌋. Much more natural, isn't that? So a'_α=a_α, but b'_α=(3/4)b_α and c'_α=c_α/2. So, doing that, we get:
Reformulated version of above: For all β<ωω,
limn→∞ a'ωβ+n=(2/3)c'β ;
limn→∞ b'ωβ+n=(3/4)a'β ;
limn→∞ c'ωβ+n=(2/3)b'β
...yup, that's 2/3, 2/3, 3/4, just as expected.
Hm, now I'm curious - what would we get if we put it in terms of denominator 3||n||/3? Probably nothing nice, but let's do it anyway - after all, we have formulated pretty much everything in these terms so far, so it may make things easier. We'll have a''_α=a_α, b''_α=3-2/3b_α, c''_α=3-1/3c_α. Put this way, the conjecture becomes:
Reformulated (again) conjecture 8: For all β<ωω,
limn→∞ a''ωβ+n=3-2/3c''β ;
limn→∞ b''ωβ+n=3-2/3a''β ;
limn→∞ c''ωβ+n=31/3b''β
...aw, I was hoping it would come out symmetric, all factors of 3-1/3. Evidently not. I suppose I could figure out how to do that, but there's really no way that's worth my time. Interestingly the one here that's different is the 2-4 boundary, rather than the 4-0 boundary. (Well, Arias had the 0-2 boundary as the one that's different, but that's because he formulated it badly. :) )
In an case, what's the status? Well... we haven't worked on this. At all, to my knowledge. But hey, we have some tables - I kept track of ordering up to 16d(2) before I decided that was too tedious - so let's check if it they satisfy this! Hopefully I bothered to write down ||n|| mod 3...
...crap, no, I didn't. Well, that will make this rather tedious. OK, first things first, I'll make a version of the table that includes that, then check it, and then come back and report on the results, though I can, at least, tell you off the top of my head that it's true for β=0 (and I doubt Arias would have conjectured it otherwise...)
-Harry
[0]These correspond to the 0-2 boundary, the 2-4 boundary, and the 0-4 boundary, respectively, but since I guess things really go 0-4-2, I guess that should be 2/3, 3/4, 2/3. Well, it's a cycle, so it doesn't matter.
