Chronicles of Harry

So some time ago I decided I no longer liked that old write-up I did of my proof of formulae for the r-th highest number writable with k ones (so long as k is sufficiently large relative to r). I mean, the proof was fine, but the context was out of date; a lot of the things I bothered to point out seemed silly, for instance, because I didn't have the concepts down very well when I wrote it. And I'm not even using that same notation anymore!

So a few days ago I went and edited it, cut out all that stuff so it could serve as sort of an "appendix" to the paper. Then I realized, well, I want to put this up where people can see it, but nobody will be able to read it in that form. So today I went and put back in some introductory stuff and other parts I had edited out. (Just the relevant parts, of course.) So now that's back up on my little website, here.

I didn't rewrite the proof itself at all, so if that was hard to understand before, it still will be. Still, I think this makes the rest of it a lot nicer.

UPDATE 10/29: The errors Josh points out below have been fixed now.

-Harry

OK. Presented without explanation, here is my most recent fix and generalization for the conjecture I previously erroneously stated here and here.

Let ||n|| denote the complexity of n, the smallest number of 1s needed to write n using addition and multiplication. For r≥0, define A_r to be the set of all n such that ||n||-3log_3(n)≤r. (Note: I actually define A_r with strict inequality, but my notation for nonstrict is to put a bar over the A, and I don't want to figure out how to do that right now.) Let A_r(x) denote the number of elements of A_r that are at most x.

Let r=k+s, with k an integer and 0≤s<1. Then I conjecture that A_r(x) is asymptotic to C(k+1)^(k-2)/(k!^2)(log_3 x)^(k+1), where C is determined as follows:

For 0≤s<2-3log_3(2), C=C₀=1.
For 2-3log_3(2)≤s<4-6log_3(2), C=C₁=C₀+k+1.
For 4-6log_3(2)≤s<6-9log_3(2), C=C₂=C₁+(k+2 choose 2).
For 6-9log_3(2)≤s<8-12log_3(2), C=C₃=C₂+(k+3 choose 3).
For 8-12log_3(2)≤s<10-15log_3(2), C=C₄=C₃+(k+4 choose 4).
For 10-15log_3(2)≤s<5-3log_3(5), C=C₅=C₄+(k+5 choose 5).
For 5-3log_3(5)≤s<12-18log_3(2), C=C₆=C₅+k+1.
For 12-18log_3(2)≤s<6-3log_3(7), C=C₇=C₆+(k+6 choose 6).
For 6-3log_3(7)≤s<7-3log_3(10), C=C₈=C₇+k+1.
For 7-3log_3(10)≤s<14-21log_3(2), C=C₉=C₈+(k+1)^2.
For 14-21log_3(2)≤s<8-3log_3(14), C=C₁₀=C₉+(k+7 choose 7).
For 8-3log_3(14)≤s<9-3log_3(20), C=C₁₁=C₁₀+(k+1)^2.
For 9-3log_3(20)≤s<16-24log_3(2), C=C₁₂=C₁₁+(k+1)(k+2 choose 2).
For 16-24log_3(2)≤s<10-3log_3(28), C=C₁₃=C₁₂+(k+8 choose 8).
For 10-3log_3(28)≤s<11-3log_3(40), C=C₁₄=C₁₃+(k+1)(k+2 choose 2).
For 11-3log_3(40)≤s<9-3log_3(19), C=C₁₅=C₁₄+(k+1)(k+3 choose 3).
For 9-3log_3(19)≤s<18-27log_3(2), C=C₁₆=C₁₅+k+1.
For 18-27log_3(2)≤s<8-3log_3(13), C=C₁₇=C₁₆+(k+9 choose 9).
For 8-3log_3(13)≤s<1, C=C₁₆+(N-2)(k+1), where N=⌊-log₃(3^((1-s)/3)-1)⌋.

Now is this version, which should finally be correct, good enough to deduce (if it's true, which we don't know how to prove) that ||n|| is not asymptotic to 3log_3(n)? I don't know, Josh'll figure that out...

(Of course we only need the simple version, when r=k is an integer, to determine that, but finally we have a version we can actually *state* in generality, so I wanted to...)

EDIT: D'oh, it's actually possible to do these sums non-horribly. Oh well, I'm not going to do that here, because that would mean more binomial coefficients, which I don't have a nice way of writing here...

-Harry

1. I still haven't gotten around to sitting down and working out whether "small intersection" makes sense! I gotta do that, like, tomorrow. Or the day after or so. Meanwhile Josh has already written up a draft of all his upper bound stuff. It's a huge amount to check, so of course I don't intend to, mostly. Meanwhile I'm acting as something of a proofreader. :P I'm glad my enormous computations are ones that can be safely left out of the paper itself...

(Which reminds me - it would be nice if we could refer to that ||a^n-1|| result those Lithuanian guys discovered. I have to wonder what the appropriate way of giving them credit is, if we refer to it? I don't want to somehow "scoop" them, if that idea makes sense, and if they had planned to write about it.)

Lagarias is back now; I guess I should go talk to him at some point? I'm hoping he didn't find my earlier emails totally creepy... I'm bad at this...

2. I ran into Ginny on the street! And she's living in Ann Arbor now! Working at the hospital. That was unexpected!

3. Man, having an office with just 4 people is going to be dull. Am I the only one who likes having a big common office? How are we going to get people for Dominion now? Maybe I really should buy Flash Duel. :)

4. I brought the PowerJoy Voyager with me back to Ann Arbor. Let's hope it confuses some people. :)

5. Unrelatedly, allow me to state the obvious and note that here are some good ways to quickly change a subject: "Unrelatedly"; "Tangentially"; "That reminds me". (I guess maybe not everyone does this, but when I say "that reminds me", I typically just mean, "that has triggered my brain to think about this", and not that there's any clear relation between it and what was just said.) Further statement of the obvious: Illusion of transparency is a bitch.

6. It is not so easy to find a good copy of the SSBB soundtrack on the internet. (Though I did find something including a number of songs that were cut from the game, which is quite interesting.) Which reminds me, I'm starting to like Josh's idea that we should call USB drives "data crystals" to make the present sound more science-fictiony.

7. Unrelatedly, here is a silly conversation with Erick from yesterday.
( Cut because it's kind of incongruous )
(After that, the conversation turned to the Film Board of Canada...)

-Harry

OK, guys, time to finally write this. I've written repeatedly about the ridiculous tedious computation I did to classify n with d(n)<21d(2); I've mentioned I started writing a program to do it automatically for d(n)<18d(2), but I never finished it; I've said about how, despite this computation being *almost* algorithmic, I haven't been able to figure out how to automate the process in general. And I've said I would post here a fulls tatement of just what it was that I was trying to automate and ask if anyone has any ideas how this can be fully algorithmized. So... time to actually do that.
( Long explanatory background )
( So what's the problem, anwyay? )
So... we've got a data representation problem. I'll admit I haven't banged my head against this as much as I probably could have, but still, I don't think I can figure this out myself. Any ideas, guys?

(I *think* I've pretty much covered everything, but perhaps not. Do ask if anything's unclear...)

-Harry

This is where things start to look complicated. Let's see if they actually are.

Once again, wherever Arias has written βω, I'm going to assume he meant ωβ. (Do some people use the reverse convention for ordinal multiplication, or something?)

Google Translate seems to be having a little more trouble with these ones, but I think with some common sense I can figure out what's meant.

Note - here I'm just transcribing what Arias wrote (except for the ordinal multiplication switch), so that means that for now we're back to his a,b,c, not the ones I used last time.

Well, no, one more thing stands out as "obviously a mistake", so I'll fix that too - the last line of these conjectures, he talks about a_γ and a_β in all three. I'm going to assume that was supposed to vary appropriately. I'm also going to multiply by 1/3 in the last case, as it doesn't seem to make sense otherwise.

Conjecture 9 (fixed): The numbers of the sequence b_ωβ+n that converge to a_β=b/3^a (with ||b||=3a) are the numbers of the sequences p(q3^j+1)/3^(a+j), where b=pq and ||p(q3^j+1)||=3a+3j+1, and those terms of the sporadic succession 2^(3j+2)/3^(2j+1) which are contained between sup_γ<βa_γ and a_β.

Conjecture 10 (fixed): The numbers of the sequence c_ωβ+n that converge to b_β=b/3^a (with ||b||=3a+1) are the numbers of the sequences p(q3^j+1)/3^(a+j), where b=pq and ||p(q3^j+1)||=3a+3j+2, and those terms of the sporadic succession 2^(3j+1)/3^(2j) which are contained between sup_γ<βb_γ and b_β.

Conjecture 11 (fixed): The numbers of the sequence a_ωβ+n that converge to c_β/3=b/3^a (with ||b||=3a-1) are the numbers of the sequences p(q3^j+1)/3^(a+j), where b=pq and ||p(q3^j+1)||=3a+3j, and those terms of the sporadic succession 2^(3j)/3^(2j) which are contained between (1/3)sup_γ<βc_γ and c_β/3.

...hm. So part of this is just a formalization of what I was saying yesterday, and part of it appears to be, well, wrong. I could reformulate these in terms I would prefer, but I'm not going to, because I get what these are saying so I don't think that's really necessary.

For the rest of this, I'm just going to consider #9, as the rest should be analogous, but I don't want to think about how to modify appropriately.

Let's start by summing up: He's saying that the convergents to b/g(||b||)*appropriatefraction discussed previously discussed are some sporadic stuff, plus precisely what you get from the infinite families p(q*3^j+1), where pq=b, and (additional requirement). Now, we're going to immediately need an additional requirement, that ||b||=||p||+||q||, i.e., that this representation is minimal; otherwise, stuff won't line up. But I think his next requirement might cover this - he requires that ||p(q3^j+1)||=3j+||b||+1. Which looks stronger offhand; is it? A quick check says "I can't tell". Well, whatever. So far this is basically a formalization of the correspondence I mentioned before.

Now before we continue it's worth noting that while I do expect this correspondence to hold up down in the regime of +1s, I see little reason to expect it to continue to hold once +6s and such become relevant. Maybe conjecture 8 is still true, but it's pretty doubtful. I was talking to Josh yesterday and the conclusion seemed to be that this would be a good problem to not work on. If it's true, we don't have the tools to prove it; and if it's false, a disproof would require absurd amounts of computation (barring the automation that I still haven't talked about here), if we even had the tools at all.

Anyway, back to 9. 9 says the relevant numbers are those infinite families, plus some sporadic stuff. What sporadic stuff? Powers of 2. Well... that part just isn't right, except at low defects. *searches through files to find a counterexample* ...OK, let's take 32, with its corresponding sequences 32*3^k+1, etc. ||32||=10, add 1, that's 2 mod 3. Now, the range condition on those sporadic things - if you think about it, and apply conjecture 8, it's just saying, "stuff inbetween this collection of infinite families and the previous one of the same complexity mod 3". In this case, the latter would be 4*3^k+1, etc. So let's look between those two at stuff with complexity congruent to 1 mod 3, and check whether it all either is a power of 2, or actually falls into one of 32*3^k+1, etc.

First up is 320, but that's 32(3*3+1). Next is 35, which is... which is... none of those. It's not even, and it's certainly not 32*3^k+1, but there it is anyway. So that part of conjectures 9, 10, 11, is false. (Technically, this is only a counterexample to one of them; but IINM, 35, 70, and 133 together make counterexamples to all three. And there should be plenty more.)

As for the rest? Well, that's what I described above... although, seeing as the conjectures as written are trying to write down *all* the numbers in that range, removing the second part, well, offhand I'm not sure how you'd formalize the result. I guess just say that those numbers approach it, and everything else is bounded away from it?

So, conclusion:
1,3,4,5,6,7 are resolved (either proved or salvaged).
2 is unresolved.
The parts of 9,10,11 dealing with what the sporadic stuff is, is false. The rest can be folded into 8.
8 is unresolved, but too hard to handle without additional tools.

So the point of all this, is that just leaves #2 as something possibly to work on. Now enough of this.

-Harry

OK! 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<ω^ω -

lim_n→∞ a_βω+n=c_β/3 ;
lim_n→∞ b_βω+n=a_β ;
lim_n→∞ c_βω+n=b_β

(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 ω^ω 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<ω^ω,

lim_n→∞ a_ωβ+n=c_β/3 ;
lim_n→∞ b_ωβ+n=a_β ;
lim_n→∞ c_ωβ+n=b_β

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<ω^ω,

lim_n→∞ a'_ωβ+n=(2/3)c'_β ;
lim_n→∞ b'_ωβ+n=(3/4)a'_β ;
lim_n→∞ 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<ω^ω,

lim_n→∞ a''_ωβ+n=3^-2/3c''_β ;
lim_n→∞ b''_ωβ+n=3^-2/3a''_β ;
lim_n→∞ c''_ωβ+n=3^1/3b''_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.

In truth, I don't think the fact that this is in Spanish would be a real problem, were it not for the fact that he was looking at the problem slightly differently from us.

Since these two conjectures are ones we've salvaged and proved, this is going to basically just going to be a review of stuff we've already done. But it may make explicit some stuff I've left implicit before.

Note on notation: While I'm using Arias's ||n|| notation, I'm still going to use 'g' and 'L' to mean the things I have before... Arias uses these to mean rather different things. (What I call 'L', Arias calls 'g', and what he calls 'L' is something we haven't had any use for. I don't think he gave a name to what we call 'g'?)

...actually, now that I run this through Google Translate (annoyance: accented chars don't copy/paste correctly from the PDF, need to enter those manually), and can read the motivation for the conjectures, the mistakes make a little more sense. ("The mistakes" that I've seen so far being basically that even though he noted that ||3n|| and ||n||+3 aren't always equal, he didn't really take this to heart, and conjectured several things that would require this to be true.)

He was looking at the highest numbers writable with n 1's, and the largest few *do* obey this law, as I showed[0]. Interesting to see the different perspectives we originally came at the problem from. When I was doing my original "enormous tree" approach, I honestly didn't even think to look at the actual data, since I already had the method...

Anyway! Let's get on with this. I'm going to consider these two together. These are coming mostly straight from Google Translate.

Conjecture 3. There are three transfinite sequences of rational numbers (a_α)_(α<ξ), (b_α)_(α<ξ), (c_α)_(α<ξ), such that the (greater) numbers of complexity 3n (respectively 3n+1, 3n+2) are the (first) natural numbers contained in the sequence (3ⁿa_α) (resp. (3ⁿb_α), (3ⁿc_α)).
ξ is an infinite countable ordinal such that ωξ = ξ.

Conjecture 4. The three sequences are decreasing. The denominators of each term of a_α, b_α, or c_α are powers of 3.

Status: OK. So. Before I point out why it's false as written, I'd like to reformulate conjecture 3 a little. As he wrote it, it bugs me a little. First thing I'd like to do is say, we should replace "3n+1" with "3n+4". In this context, {0,2,4} is a much more natural set of representatives for Z/(3) than {0,1,2}. This excludes the number 1 from consideration - which it should. 1 is a special case. Similarly, it bugs me that he used 3^n, 3^n, 3^n for the multipliers, rather than 3^n, 4*3^(n-1) [sticking with his use of 3n+1 instead of 3n+4 here], 2*3^n.

The second thing I'd like to do is, as long as I'm going to point out why this conjecture is false, let's turn it into a bolder conjecture!

Hypothetical bolder version of conjecture 3: There is an transfinite sequence of rational numbers (a_α)_(α<ξ) such that the (greater) numbers of complexity n are the (first) natural numbers contained in the sequence (g(n)a_α). Etc.

Of course, I'm going to guess that the reason that he didn't make this hypothetical bolder version, is that it's false even for the low defects Arias was looking at. (Though back when I was programming this, due to a bug in my program, it took me quite a while to notice this!) Why? Suppose it were true. Then consider the number 82. ||82||=13, and g(13)=4*3^3=108. So the term in the sequence giving you 82 would be 82/108=41/54. But 54=2*3^3=g(11), so 41=g(11)*(41/54) would show up in the sequence for g(11), meaning 41 would be writable with 11 ones. Which it isn't; ||41||=12. The problem is that ||2*41|| isn't equal to ||41||+2. (Note - this is not the minimal counterexample; that would be 2*23=46. But it is the counterexample of minimal defect; 23 and 46 have way higher defect than what Arias was looking at.)

Splitting it into three sequences avoids this problem... but unfortunately there's no way to avoid the problem that ||3n|| isn't always equal to ||n||+3. As written, the conjecture is false. Suppose it were true and consider 321. ||321||=18, 18=3*6, so 321/3^6=107/3^5 would lie in the sequence of a_α's. But then 107=3^5*(107/3^5) would have to be writable with 15 ones - which, again, it isn't; ||107||=16. In general, if true, this conjecture would imply that ||3n||=||n||+3 for all n. As I said above - he may have noted that this point is false, but he didn't really take the lesson to heart, evidently[3].

Before we get to salvaging this, let's actually look at conjecture 4. The second part of it, about the denominators being powers of 3, I don't understand why he put that there, because that's true by definition. (Though it would be modified to "power of 3, two times power of 3, or four times power of 3" if we made the modifications above.) Anything whose denominator wasn't a power of 3 could be safely excluded from the sequence. So, I'm just going to ignore that.

But OK, how can we salvage this? The whole idea of this sequence where you can multiply by a power of 3 (or perhaps a 2*3^n or 4*3^n) through to find what numbers have a given complexity, is simply wrong. Get rid of that and you're just left saying that a countable set can be enumerated by an ordinal, which doesn't say anything at all. Fortunately, we can turn it nontrivial again by combining it with #4: The requirement that the sequence be decreasing. Then what you're saying is, this set (or these sets) are reverse well-ordered. Though, I see more than one possible way to salvage this... let me list three of them.

Salvage of 3 and 4, version 1: The sets {n/g(||n||) : ||n||=0 (mod 3)}, {n/g(||n||) : ||n||=2 (mod 3)}, and {n/g(||n||) : ||n||=1 (mod 3), n≠1}, are reverse well-ordered.

(OK, excluding n=1 doesn't actually do anything there, it works perfectly well with 1 included, I'm just excluding it out of principle. :) )

This is closest to Arias's version, in that it's split in three; note that changing his 3^{⌊||n||/3⌋} to a g(||n||) doesn't affect anything, as it's just a constant factor applied to each of the three sets. Indeed, instead of changing to g(||n||), we could also just get rid of the floors there and use 3^||n||/3. Same statement.

But hey, since ||2n||≠||n||+2 is no longer really a problem for us, why split in three at all? Let's merge it all back into one! The resulting statement will be equivalent - well-ordering is preserved under subsets and under finite unions.

Salvage of 3 and 4, version 2: The set {n/g(||n||) : n∈N} is reverse well-ordered.

But like I said, instead of g(||n||) in the denominator in the first version, we could also put 3^||n||/3... let's merge those sequences, instead. It's equivalent, yes, but let's do it anyway.

Salvage of 3 and 4, version 3: The set {n/3^||n||/3 : n∈N} is reverse well-ordered.

Well, these statements are all equivalent, and as we recently proved, they're all true.

I could include a version 4, based on 3^{⌊||n||/3⌋} like Arias did, but I see no reason to do so. Why include version 2? Because it's the most natural way to state it, I think, considering the problem. But then why include version 3? Ah, well, that would be because that's how we actually proved it. :) We've been talking about the set of defects being well-ordered, but if you haven't noticed, you take d(n)=||n||-3log_3(n), you raise 3^1/3 to that power and you take the reciprocal (reversing the order), and the notions are equivalent. Fundamentally, 3^n/3 is multiplicative, whereas g(n) isn't, and that makes the former a lot easier to work with.

Somewhere along the way here I started ignoring the question of just what the (obviously countable) ordinal in question is, though. Arias called it ξ and conjectured that it satisfied ωξ=ξ. This is true as well - in fact, the order type of all five of the sets I've listed above is exactly equal to ω^ω. (Though equivalence here is not as obvious, and SFAICT relies on actually knowing some facts about these sets.)

Finally, to recover at least some of what Arias originally conjectured, we can note that if we stick to the split-in-three-sets version, this his idea is in fact true if we stick to indices less than ω². (That's not the absolute limit here, but I don't really think being more specific than that is worth my time, and this way it's true regardless of which of the three sets we're considering.)

ADDENDUM next day for clarity: That is to say, using the three-sequence formulation, at indices less than ω², if you multiply through the numbers you get will in fact have the appropriate complexity. And they will be in decreasing order, seeing as how you took a decreasing sequence and multiplied through by a constant.

Later ones will come... later. Then we'll be getting into ones we really haven't considered. I think you can see why I'm stopping here for now.

-Harry

[0](Actually, thought: My computations show this is true for n with D(n)≤2, but because the boundary between D(n)=2 and D(n)=3 doesn't depend on what ||n|| is mod 3, that means we can rewrite that statement more simply, as follows: If n>g(||n||)/3, then ||3n||=||n||+3. (And thus ||3^k*n||=||n||+3k.)
[3]This *is* the minimal counterexample; the counterexample of minimal defect would be 683*3=2049.

I mean, if cardinals are red, I guess ordinals can be blue, right?

Blue Kirby: Someone thought to buy Smash 64 for the Virtual Console for the house's Wii. I think Dan was a bit surprised I was so good at it. :D I realize I'm not by the standards of old Tufts House - I haven't played in quite a while, I keep failing to smash, I never learned to Z-cancel - but I'm certainly good against the people here. Much better than at Brawl, that's for certain. Need to play more, remember more. Took me a bit to remember how to play Falcon.

Obvious-but-funny-anyway-observation: The controls have been modified to match those of Melee. Hence Z-cancelling would now be done with the L button. I don't expect anyone to start referring to it as "L-cancelling", though.

Blue Water Fangs: I was down at GYGO and saw they had the Xbox version of Metal Slug 3 for cheap, so I picked that up. Hooray, Metal Slug 3 with a non-ridiculous number of continues. Heh - this TVTropes page refers to the Xbox version as a "porting disaster" because of this. Bah! If you can't beat a level in five lives, you aren't beating it properly, I say!

Well-ordering: Let's talk about integer complexity. I finally finished classifying n with d(n)≤21d(2), which means I can classify n with D(n)≤1, and... it's terrible. I decided to not even go through and finish it; it may be easy, but it's still not worth it. {n: D(n)=0} has a nice form, so I guess that's worth noting, but {n: D(n)≤1} really doesn't, so... why bother? (OK, I guess I can't say this for *certain* as I didn't finish, but it seems pretty clear.) Well, at least the statement that D(3n)=1 ⇒ D(n)=1 is kind of nice.

But! Stopping work on that means I can actually get back to work on interesting stuff. A few days ago I had the idea to, y'know, actually formalize the notion of certain types of infinite families of integers we'd been using implicitly the whole damn time, and, well it's been very helpful. For instance, we've been able to finally prove that the set of defects is well-ordered! (It has order type ω^ω.) I was hoping to prove that the set of defects less than n had order type ωⁿ, but the best I have so far is that it's at least ωⁿ, and less then ωⁿ⁺¹. Still - well-ordering! And this is basically the LHF[0]! I definitely think we can milk this concept for a bit more.

Maybe we should contact Arias again and tell him we've proved well-ordering? Then look over his paper more carefully and see if we can prove any more of his conjectures. :D (Well, that we should do anyway...)

-Harry

[0]OK, I suppose nobody who wasn't in Tufts House the past few years is going to recognize that acronym[3]. I don't think I've mentioned Nathan's strange acronym-slang anywhere here previously, have I? Well, this one stands for "low-hanging fruit".
[3]Actually, it does show up on acronymfinder.com, so I suppose there are indeed other people out there using them. But I've never seen it elsewhere personally.

Time required: 45 hours and 15 minutes.
Disk space used: 15944324947 bytes, or about 14.8493 gigabytes.
Conclusions drawn: ...hell if I know, it just finished!

Addendum, 2 hours later: OK, I think we can forget about the database idea... I think I've got most of what I wanted from it.

OOPS! Remember that conjecture from earlier? Well, it's false, but not because something unexpected came up; just because we were not being careful in formulating it in the first place. Instead of number of partitions of n, what we want is number of rooted trees on n+1 nodes. Is this still good enough to prove f not asymptotic to 3log_3, if it's true? Josh says probably.

-Harry