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Here are a few ideas that could potentially be useful for Open Problem 1b (unimodality of ). However, most of them probably still need lots of extra work and are very incomplete.
Part I, notation
For no particular reason, let us call some singular if it is noninvertible and not an atom. Because is just , it follows that is singular iff we can write it as where the both have at least a nonzero element.
Because , we must have Hence, if , then implying that Similarly It follows that
Part II, classification of singular sets (small cases)
What I did next was to find possible "structures" of singular for cases where was small.
So if , then because and thus contains . But is invertible, so is neither an atom nor is singular.
If , then However, if such that is a nonzero element of , then so this implies that as so that But this is impossible, so can't be singular if
Now consider , so Then we can write where the represents some unknown element. Then all the elements must also be in , as established previously. Hence WLOG we let be in the first component set : Then, let : Then, so we must have in order to satisfy Thus one singular form would be
Next, consider , in which We use the same - setup as before: Here we have two cases.
Case 4a
If , then WLOG let as Then and clearly , , , and must all be distinct. Thus and since Thus another singular form here is
Case 4b
If , then is in both , so clearly Thus, WLOG let's set Now we have Since , we clearly aren't done yet, so let's add some other to any of the sets, WLOG Hence so so inspection reveals that must be in order for to be In that case our singular set form is However, this is just a unique case of the one from Case 4a.
To conclude, if , we have significant singular form:
Finally, we tackle the case where which implies that (This is by far the most complicated case.) Once again, use our - construction is and we also have two big cases.
Case 5a
If , let WLOG. Then, at most four elements of are known, so we must add some element to one of the . WLOG, add to , so that Then Then exactly one of and is in ; otherwise, would be Inspection reveals that the only possibilities for are , , and , so the possible constructions of are and However, if we let and , note that so the latter two cases above are the same. Hence we have two structures here for : and
Case 5b
If , things get a whole lot more complicated. Specifically, we have so must include which WLOG we let be equal to Then we use the same strategy: add some to some , WLOG Then there are two more cases based on whether .
Case 5bi
If , then so Since , so WLOG let If so, so it follows that and But this is exactly what we found in 5a, so this isn't anything new.
Case 5bii
Here Thus so so WLOG let If so, we must add a new element to either set. If we add it to , then so , and it is easy to see by inspection that , so that Similarly, if we add to , then giving Again, inspection gives , so Hence in both cases, which again is just a special case of 5a. Hence, again we have nothing new.
Thus, the only two distinct general forms for are and
I feel like the process for findings these forms are recursive in some way. Maybe the number of unique forms as increases will follow a linear recursions, such as that of the Fibonacci sequence. But I don't know yet, since I'm still thinking about the which is even more complicated than before.
Part III, computations
Here I was able to use the results from Part II to bash out the first values of by hand, for positive integers The method I used was basically just noting that and , and brute-forcing everything else when The results I got were
which seems to agree with the conjecture of unimodality. If I get to I might be able to get a gauge on the bound for Problem 1.1 found by , as
Part IV, finite differences???
This is one final whimsical thought that I had. The idea is that if the are unimodal, then interpreting this as a function of for a fixed will give a concave-down graph. If so, then the second difference of the must be nonpositive, much like for concave-down functions. If so, the second finite difference is I'm not sure how to continue with this though. I'll keep trying this method, though I'm doubtful of how useful this strategy is.
--WJH
Part I, notation
For no particular reason, let us call some singular if it is noninvertible and not an atom. Because is just , it follows that is singular iff we can write it as where the both have at least a nonzero element.
Because , we must have Hence, if , then implying that Similarly It follows that
Part II, classification of singular sets (small cases)
What I did next was to find possible "structures" of singular for cases where was small.
So if , then because and thus contains . But is invertible, so is neither an atom nor is singular.
If , then However, if such that is a nonzero element of , then so this implies that as so that But this is impossible, so can't be singular if
Now consider , so Then we can write where the represents some unknown element. Then all the elements must also be in , as established previously. Hence WLOG we let be in the first component set : Then, let : Then, so we must have in order to satisfy Thus one singular form would be
Next, consider , in which We use the same - setup as before: Here we have two cases.
Case 4a
If , then WLOG let as Then and clearly , , , and must all be distinct. Thus and since Thus another singular form here is
Case 4b
If , then is in both , so clearly Thus, WLOG let's set Now we have Since , we clearly aren't done yet, so let's add some other to any of the sets, WLOG Hence so so inspection reveals that must be in order for to be In that case our singular set form is However, this is just a unique case of the one from Case 4a.
To conclude, if , we have significant singular form:
Finally, we tackle the case where which implies that (This is by far the most complicated case.) Once again, use our - construction is and we also have two big cases.
Case 5a
If , let WLOG. Then, at most four elements of are known, so we must add some element to one of the . WLOG, add to , so that Then Then exactly one of and is in ; otherwise, would be Inspection reveals that the only possibilities for are , , and , so the possible constructions of are and However, if we let and , note that so the latter two cases above are the same. Hence we have two structures here for : and
Case 5b
If , things get a whole lot more complicated. Specifically, we have so must include which WLOG we let be equal to Then we use the same strategy: add some to some , WLOG Then there are two more cases based on whether .
Case 5bi
If , then so Since , so WLOG let If so, so it follows that and But this is exactly what we found in 5a, so this isn't anything new.
Case 5bii
Here Thus so so WLOG let If so, we must add a new element to either set. If we add it to , then so , and it is easy to see by inspection that , so that Similarly, if we add to , then giving Again, inspection gives , so Hence in both cases, which again is just a special case of 5a. Hence, again we have nothing new.
Thus, the only two distinct general forms for are and
I feel like the process for findings these forms are recursive in some way. Maybe the number of unique forms as increases will follow a linear recursions, such as that of the Fibonacci sequence. But I don't know yet, since I'm still thinking about the which is even more complicated than before.
Part III, computations
Here I was able to use the results from Part II to bash out the first values of by hand, for positive integers The method I used was basically just noting that and , and brute-forcing everything else when The results I got were
which seems to agree with the conjecture of unimodality. If I get to I might be able to get a gauge on the bound for Problem 1.1 found by , as
Part IV, finite differences???
This is one final whimsical thought that I had. The idea is that if the are unimodal, then interpreting this as a function of for a fixed will give a concave-down graph. If so, then the second difference of the must be nonpositive, much like for concave-down functions. If so, the second finite difference is I'm not sure how to continue with this though. I'll keep trying this method, though I'm doubtful of how useful this strategy is.
--WJH
This post has been edited 6 times. Last edited by smileapple, Jun 26, 2023, 12:54 AM