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sniffnoyOK. In this post I'm going to write Arias's conjectures in English, and state what we know about them. For most of them the status will of course be, "We haven't even considered this question." Note that I don't actually know Spanish, but these all seem pretty clear anyway, as they're almost entirely in math notation; Conjecture 3, the one that wasn't, I asked Hunter, he knows some Spanish, he confirmed it said what it looked like. (Well, his commentary on the conjectures is of course in Spanish, but I'm going to just ignore that.)
Unrelatedly, I'm starting to like Arias's notation of ||n|| better than Guy's notation f(n), because I often want to call other functions f. That shouldn't matter here, but I think I'll switch to that for this.
Conjecture 1. For every n, there exists a such that ||3jn||=3(j-a)+||3an|| for every j≥a.
Status: This is true, we proved it quite a while ago.
Conjecture 2. For all natural numbers p and q, there exists a such that ||p(q3j+1)||=3j+1+||p||+||q|| for every j≥a.
Status: As stated, this is easily false, for two different reasons. The first is that it runs into the problem that ||3n|| is not always ||n||+3. For instance, ||107||=16, but ||321||=18. So if we pick q=107, we'll find that for j≥1, ||p(107*3j+1)||≤3(j-1)+18+1+||p||=3j+15+1+||p||<3j+16+1+||p||. The second reason is that it has a problem if p=1. In that case, including ||p|| in the sum is clearly a mistake. Fortunately, these are both easily salvaged. (Including ||q|| if q=1 could also be considered a mistake, I suppose, but it's a submistake of the ||3n|| vs ||n||+3 mistake.)
Salvage of conjecture 2: For all natural numbers p and q, p>1, there exists an a such that ||p(q3j+1)||=3(j-a)+1+||p||+||3aq|| for all a≥j.
Obvious generalization of salvage: For all natural numbers p and q, there exist a,b such that ||p(q3j+1)3k||=3(j-a)+3(k-b)+1+||3bp||+||3aq|| for all a≥j.
(Note that this generalization does *not* follow immediately from conjecture 1 (theorem 1?) + salvage, because the former does not give any bounds on how far out you have to go! And neither does our proof of it. So there's an additional uniformity condition here.)
Status: We know many special cases of this to be true from my computations, but as for the general case, we haven't really worked on it. I have an idea or two but I don't really expect them to work. I also have an idea or two for how to generalize this further - however, I also have computations showing the most obvious generalization is false! Probably salvageable, of course, but I'm guessing we should work on this before worrying about that...
...and, taking a look at these again, I'm realizing, whoops, these are a *lot* less clear than I'd realized. 3 I understand because I asked Hunter about it; 4 is all in Spanish and not so clear; 5-6-7 (really all one thing) aren't so clear either. 8 is entirely clear, but 9-10-11 again I'm not so sure of either.
OK, that was silly! I'm going to stop here for now, then, because I don't want to do 3 without doing the subsequent ones.
-Harry
Unrelatedly, I'm starting to like Arias's notation of ||n|| better than Guy's notation f(n), because I often want to call other functions f. That shouldn't matter here, but I think I'll switch to that for this.
Conjecture 1. For every n, there exists a such that ||3jn||=3(j-a)+||3an|| for every j≥a.
Status: This is true, we proved it quite a while ago.
Conjecture 2. For all natural numbers p and q, there exists a such that ||p(q3j+1)||=3j+1+||p||+||q|| for every j≥a.
Status: As stated, this is easily false, for two different reasons. The first is that it runs into the problem that ||3n|| is not always ||n||+3. For instance, ||107||=16, but ||321||=18. So if we pick q=107, we'll find that for j≥1, ||p(107*3j+1)||≤3(j-1)+18+1+||p||=3j+15+1+||p||<3j+16+1+||p||. The second reason is that it has a problem if p=1. In that case, including ||p|| in the sum is clearly a mistake. Fortunately, these are both easily salvaged. (Including ||q|| if q=1 could also be considered a mistake, I suppose, but it's a submistake of the ||3n|| vs ||n||+3 mistake.)
Salvage of conjecture 2: For all natural numbers p and q, p>1, there exists an a such that ||p(q3j+1)||=3(j-a)+1+||p||+||3aq|| for all a≥j.
Obvious generalization of salvage: For all natural numbers p and q, there exist a,b such that ||p(q3j+1)3k||=3(j-a)+3(k-b)+1+||3bp||+||3aq|| for all a≥j.
(Note that this generalization does *not* follow immediately from conjecture 1 (theorem 1?) + salvage, because the former does not give any bounds on how far out you have to go! And neither does our proof of it. So there's an additional uniformity condition here.)
Status: We know many special cases of this to be true from my computations, but as for the general case, we haven't really worked on it. I have an idea or two but I don't really expect them to work. I also have an idea or two for how to generalize this further - however, I also have computations showing the most obvious generalization is false! Probably salvageable, of course, but I'm guessing we should work on this before worrying about that...
...and, taking a look at these again, I'm realizing, whoops, these are a *lot* less clear than I'd realized. 3 I understand because I asked Hunter about it; 4 is all in Spanish and not so clear; 5-6-7 (really all one thing) aren't so clear either. 8 is entirely clear, but 9-10-11 again I'm not so sure of either.
OK, that was silly! I'm going to stop here for now, then, because I don't want to do 3 without doing the subsequent ones.
-Harry
