 So, thanks everyone for making it enough today. Today we have Hua Huang from Georgia Southern University, and he'll be telling us about coloring, tiling, and counting compositions. So please take us away. All right. Thank you very much. It's always good to visit South Carolina, even if it says virtually. So I chose this topic. Well, one of the reasons is just that it's hard to find new things to talk about, since there are so many people here who has heard my talk many times. So this is relatively new. And another major reason is that my collaborator, Brian Hopkins, made most of the slides already so I figure he did such a good job so I'll just use it. If you see anything you really like, it's probably from him. All right, so let's see. All right, so an integer composition of a number n is an order sum of positive integers. So this n is also multiple, well, also always a positive integer. And the t-word is being ordered. So that's what separates this from partitions. So for example, if you look at all the compositions of the number three, you can have a single three or two plus one or one plus two or one plus one plus one. The order matters, which is why two plus one, one plus two are different. And we use this capital C of three to denote the total number of compositions of n, in this case n equals three. The reason that we use the new instead of n is because for technical reason we have to use n in the terminology of something else. And in the past, some people really don't like us using n for different things. So new here, but I'm still tempted to say n all the time. All right, now, for the majority of this talk, actually for most of almost all of this talk, we will be interested in looking at the tiling representation of a composition. This is to say that if you are looking at the composition of the integer three, we take a one by three board here. And then for each tile, it corresponds to the size of the part, right. So here you can have two plus one here. We have a one by two board, two tile and one by one tile. So this is a very straightforward way of representing integer compositions. Well, there's there's nothing difficult about it. But it's interesting that how many combinatorial insights one can gain using this tiling representation. For instance, among, well many other things that one already know about competition, we know that the total number of competitions is a power of two. And when you look at the tiling representation of a composition, you look at this big board here. And within this big board there are, well within this big board of size one by n, you'll have n minus one invisible vertical lines here separating the square cells, right. And when you look at the tiling representation of this competition, each of these little vertical line could either be a joint connecting two cells into a joint bigger part, or a cut separating part from part, right. And among this m minus one lines, each one could be either join or cut, giving you two to the minus one choices. And of course, if you are looking for the number of compositions with specific number of parts, that is to say that you'll have specific number of cuts here, right. Then all that matters is out of this m minus one invisible lines, which K minus one of them to be cut to be cuts, which give you k parts. So, that gives the number of compositions with specific number of parts. Among many other interesting things about compositions. There are these few that's related to ship not numbers. And I'll take this opportunity to introduce this notation here, see with bracket one to denotes the number of conditions with parts one to. This means a number of conditions with no part of site one. This one means number of conditions with only odd parts. So we're not going to prove all this. Just one of them, I guess, he'll illustrate the basic ideas of this kind of combinatorial arguments. So we'll take the last one here, where we consider compositions with all cards. And then equals one, you'll have just one tape one possibilities one goes to still one, three, four, five, so on and so forth. And we want to show that the number of possibilities is not numbers and as you can see one was two for five so far so good. And once, when you move on to the next level of six. This is one ankles. This is one ankle four and five and this is my ankle six. So all we have to do is to show that it actually satisfies the same recursion as a big numbers. So we take the number of compositions with odd parts of four and five, and we do the following. So if it was a five, we just add a square at the end, very simple. And you get a composition of six with still only odd parts because we added a part of site one, and in the end with a part of site one. And if it was a part of a composition of four, you just expand the last part, which was odd, but expanded by two more for one becomes three, three, five, so forth. And that gives you a combination with only odd parts of six, but this time it ends with a part that is at least three. So that's a simple example of how one can use this piling representation to prove interesting identities of competitions. So what we're interested in is in color compositions. And this is where we use N for. And so the official definition says that each part of site and had in colors. So now if you look at competitions of three now. So, the original three originally we had just one three right but now that this part of size three has three different colors denoted by subgroups. So we have three sub one sub two sub three. And then for the combination two plus one. This one only have one choice because it's all size one, but this tool has two possibilities now for that to the one to the two, so forth. So the number of any color combination of three is eight. Yeah, if I can come right. All right. So, and again, so this has been a topic of some research work in the, in the recent years. But what we're interested in is how piling representation. And so this is a very clever way of representing any color combination for tidings where I just put a dot in each tile. And the position of this dot represents a color right so because this pile of size three as three possibility possible positions for the dot that corresponds to the three different colors right. So sub one sub two sub three corresponds to the location of the dot at the first second or third cell of this tile. Similarly for the other. So this was introduced by Hawkins. And we're happy to report that every single thing presented in this talk can be proved using this representation. So, which make him pretty excited. All right. So first as an example, the number of any color compositions right so is every other feedback number so it's not not a power of two, even though we had an eight before, but it's not the power of two but just every other particular numbers. And there is a very, so this was first proofs through I believe generating functions, when it was introduced. But with spotted piling you have this very new proof where you take all the spotted piling representations of the entire combination. And then you expand it twice as large. Okay, so imagine that you take this and you'll pull it horizontally so that it twice as wide as before. And then for every single solid line that separate separate parts and firing dot we turn that into a solid line here. So in this particular example, this one, which was originally at the second location right but after it was expanded twice as large, it went there. And then this thought which was originally at one half location right because it was in the middle of the first cell. Now after expanding to twice as large it was at the first position. And this one come from that dot right there. So you can see that the resulted composition, regular composition no color involved, must have only all the parts, because the, because there has to be a line between our two dots, right. So the parts in this new composition is defined by these lines, we came from a dot and the line here, which is always a half integer, which gives you an odd number after it was doubled. So you see all the competitions, regular competitions with only odd parts, which we have actually just proved that is business numbers, right. And of course, the, these competitions are only all even numbers, let it go every other is not numbers. So, this is this proof is not by me or anything that I just think it's really nice. And among other things, recursion has been studied for the end color competitions. And this is what this is a rep, one of the representative results where you can represent the number of entire competitions as well, a linear combination of the previous ones. And this was proved in a very much more complicated but very similar way as what we did before. So this is a bijection to show that the destroy union of any kind of conditions of n and in color combination of n minus four has a one to one map to be in color competition of these guys, and it's all being a part expanding a part is sort of very elementary operations. So it's not necessarily short but but but it's but nothing complicated was involved. Now there are other works about any kind of conditions and conditions in general, and most of the work so far seems to be focusing on, you know, the number of conditions number of conditions allowing odd parts, or not allowing certain parts on so forth. So it's all about the different part sizes, but not colors. So what we wanted to do in the, well, wanted to look at in the rest of this talk is to look at any kind of conversations with restricted colors, though not part sizes, but colors. Okay. So, you know, you can list the number of theoretical motivations for studying this. But, of course, also, it's just interesting to look at elementary comment or approves. Okay, so we can formulate the, the, the, I don't really want to call it main results but the, the, the fundamental results as the following three. So, if you allow colors, C1 C2 so on and so forth. So, and a collection a finite collection of colors are allowed only they are allowed. So we have this recursion of the number of color combinations. And if a finite collection of colors are forbidden. Then we have this recursion. And if the number of, I'm sorry, and if we do allow infinity menu and forbidden for the many colors at the same time. But, but, but the colors has to satisfy some modular condition. Then, then we have this recursion. So we're not actually going to prove any of this today, but just know that they are approved through elementary approaches through directions. And some of them involved in many cases. So, some manipulation is needed, but they are all based on the spotted hiding representation of color combination. So, of course, once you get this recursion, you can generate a sequence and can do quite a few different fun things. In particular, you can pretty much just generate sequences for whatever you want. We can allow one particular color, two colors, conceptual colors, so on and so forth, or forbidden, forbid whatever choices of colors or modular cases. And in this line of work, a lot of times, once we find out a way, whether it is through generating function or recursion to generate the sequences, we actually punted into this thing called online encyclopedia of intersequences or yes. And see, see what other people say about this sequence. And a lot of times they count all sorts of weird things and we like finding connections between them. And that's exactly what we did with this, this recursion. So we can generate all this. And in the rest of this talk, I picked a few special cases where we found some other interesting objects that appear to be counted by the same sequence or have direct formulas and so forth. And we tried to explain our connection through this body-piling representation. So these are just some of the many cases that one can look at. I think I picked them mainly because we have nice pictures ready, so I don't have to show them. All right. So the first case is manual only allow colors one and two. And the recursion comes from our theorem. And it turns out that there is a bijection between this and color conditions, allowing only color one and two to currently worse. So, with lengths, one less, so m minus one, currently worse, but avoid zero two and two zero. This bijection is actually very easy to do. You look at the spotted piling representation of an n-color combination here. And you look at all the cuts in this representation and we turn that into a one. And we look at all the joints in this representation. If it is in a tile where the color is two, we label it as two. If it is in a tile where the color is one, we label it as zero. So absolutely nothing fancy is involved here. And the point is that any two and zero must be separated by one, right? So within each tile, you either have all zero or two. To get from zero to two or two to zero, you have to have a one, maybe more than one one, that's okay. So this gives you the primary words of lengths, same number of assets, the regular vertical lines, and avoiding zero to zero. So, quite simple. In color compositions, allowing only colors one and two, there is a direct formula. And we found out that we can generate this, and generalize this direct formula to allowing not just colors one to one and two, but colors one to all the way up to C. I wasn't sure if I should show this, but this is a formula, okay, so there's that. And in there, this thing is defined recursively that way. So I don't think it's super exciting to come up with a formula this complicated. I guess what what was interesting to us is that the way to explain this formula comes directly from how you build up this and color competitions. So I'll just try to explain a little bit of it. Not everything. In the original formula, the number of color combinations, allowing only color one to see, right. And so what do we do is, we look at the number of cells in the title representation that was after the spot. So, let's say that if you have a tile like this, and the spot is here, who are looking at all these cells. And then there are this many of them distributed over this many parts. Okay, so there are this many ways to distribute those cells. And then we just say, okay, so the number of ways to allocate the remaining parts is that and that in turn is recursively defined. Like, like this. So where each of this is just a number of ways to to allocate empty squares in blocks of certain total size that allows colors up to a certain point. So that's how this was built up. Again, the formula itself wasn't that exciting, but we were happy that it can be explained commentary, even though I'm not sure how well I explained that just now. All right. Okay, so another interesting case I promised was when we do not allow color to. And it turns out that the number of color conditions not allowing color to is the same as the number of this sort of combinations with parts congruent to two module three. So this is a very bizarre collection of objects really. And I had no clue how to ever find a bijection between them. So my collaborator he had this brilliant idea that I think some of us might find interesting and you might be able to use it somewhere else is half open while allowing the parts of a composition to be half open half closed or open on both sides. So here's an example. Now this is not an example of the bijection anything, just the example of the kind of half open parts that we're allowing. So you put a dot on the left or the right on both sides to denote that our part is open on the left or right on both sides. And then when you put this kind of parts together to adjacent parts, they are both open at the joint then they are combined. If it's close on one side or both sides, then it's separate. So basically you somehow ended up with this, this, this, this together, then they combine into that. So this concept, I'm not even sure we ever gave the name but this kind of items were used in the bijection between these two collections of objects. And the bijection goes as follows. So we start with a composition with no color two thing. And for any part of size one, we just map it to a three, but open on the right. Anything not all size one, but with color one, we map it to a triple of parts with a left open to a two and something else and then the remaining stuff. And for the third type, which is a part of size at least three, I think, with color history. So the thing is that color one is already handled, which means our color has to get these three, because we don't allow color to do this, this could really just be taken and go straight here. So this is the kind of parts with color history. And we separated into three parts. Well, we map it to three parts. Similarly to be above except that the second part is always going to be larger than two, right, which is important when we map them back. And we add a queue for technical reason. And we, and then we combine them. So here's an example. Without examples, this is impossible to really understand but if you look at this example and look at this one here. So that gives you three. And this one here. And you do that three there. Right. So this one's never stand alone essentially in the, in the final result, because the series they generated are open on the left. And yeah, and they will always be followed by this, which is a left open to right. And then whatever copies of three will be combined together with that to and give you a part of size. To multiple three. And then for this to is color one here is generated to that it generates these three parts. And then you get this thing. So this doesn't help that much. If you want to understand why is this a bijection. So the universe goes as follows. That's quite a lot to read. Well, let me just show the example. Alright, so basically, what happened is that we take the first part, fourth part, seven parts on so forth for all these parts until the last part. So the reason is that with parts that congruent to model three of, of a common of integer three and plus two, you're guaranteed to have three times something plus one parts. So, so this is always the last part. And for each of this, we break it into into a bunch of threes, and the two. Okay, so each part is congruent to more than three. So for example, this eight, right. So if I think of these threes and two, and then when we map it back, these two threes will match once, and that's this two ones here, right. So for each of this, so for the tools here, of course, they just stay as two and there are no three for once generated, right, and that gives you this two here and this two here. And that five gives you a three and a two. The three gives you that one and this two. So basically this takes care of all these ones. And, and leaves this choose with two extra parts afterwards right so this two together with those two choose a group together. This two together with that five and choose a group together, so so forth. And each of these little groups of triples of things that start with you are much back to one part, depending on whether the second part is true or not. So you already saw that from the original map. The main difference is that whether the second part is true or not, that's X here. Then you can map it to a part with color one or color not one. So that was more or less the the bijection. So you can see, it's quite difficult to understand at first, but it was really interesting because of the use of half open stuff. Also about any color composition and color compositions with no color to direct formula. So again, like all this stuff we're just shown there with a cruise. Some people get this before some people can just gas formula, using the computer program and looking at data. It's a pattern that I do not possess. But for us, you know, we want to prove it, and we started writing representation. So what we do is, let's see. We start from. Well, we start from this side of the identity. So basically we look at a one by m plus k rectangle, where white. Oh yeah. So we start with this one by m plus k board, and we label the squares, we label, we label three k of the squares as, you know, a one a two a three k. So we label them so on the next page, there's a picture which will make much more sense. But on this board we label three k of the squares. And then for every three consecutive label squares, we group them into a tile, and then depending on the needle. So we label cell, we, we, we either remove the first cell or the last cell and generate a new part. So this is impossible to learn without pictures. So, let me show the picture. So that's that so there is our one by n plus k, why n plus k in a second. So how do all these little cells, we chose three k of them to be marked by these little circles. And then basically number 123 gives you a part, number 456 gives you another part, so on so forth, right. And in between this part if there is some empty spot between three and four between six and seven between nine and 10. And then for the extra cells, they are all held with this square with a spot inside, so they're all filled with that. And then for each of these we're going to turn it into a part corresponding to a tile in the spot is hiding representation. And of course we do not want color to their thing. So we choose the color, so we chose the middle mark cell to be the color to be the spot. And if it actually happens to be color to we just chop the head off right so we remove that one. So that it becomes one. And if it's not, and if it is not a color to say like this one, then we chop the tail off and this becomes, well, this is the same color, but the part size is reduced by one. So we put them together to get a combination in color combination with no color to here. So the no color to part is quite easy to understand. And you can also see that we do allow all the other possible colors other than color to because everything else is preserved. And when you think about the universe of this map, what you want to do is you want to take all this stuff. One sub lines basically and keep them as blank cells right so eventually they end up there. And then for each part, we generate a tile of size one more than it. And that's what this plus that that's where this class came from, where you either add it to the front, if it was a color one. Or you add it to the end, it's not a color one. And there you have it. And what does this have to do with our direct formula because the direct formula was just a summation of something to 3m right. So this is our 3m right. So if you have k parts in your original and color, just kidding, if you have k parts of size bigger than one in the original in color combination, then each of them corresponds to this one of these triplets of Mark cells. Three times that many, and the size of this board increased by exactly that many because each of these tiles increased by one from here to there, right. So there go and plus k to 3k where k can be from zero to whatever you see the species formula of this. Okay. Right. Another interesting that one can observe is that sometimes. So, completely different type of objects can be related to better so like here. A counts the number of color combinations with odd colors. And becomes number of color conditions where you don't have one. Okay, so no part of one, no one size. And the projection is constructed, as you would expect from the collection of entire conditions with all the parts to the combination of that this joint union of that. So here's how it's described. So if we start. So if we go backwards start from the combination of the where we don't have once right for all the short ones, the ones of n minus one, you just add a one at the end. And after that you combine that new collection which is of length and no, of course, but and with one combined that we see original and color competitions with no one in it. And then all we have to do is to look at every even colored part, right, because we don't want you on color part in, in, in this collection. So if there is an even colored part you just chop the head off as a one part, and the rest is whether the rest remains the same but it becomes a part of size, one less and color one less. So here's a picture. What we have here this in this first row here is a color composition with no part of size one. Right. And this is all n minus one. So this belongs to that that right there. And you add a, add a one at the end, and that gives you something is the prime of them. And then you look in here and you look for everything with even color, and that will be this one and that one. So anything with an even color you chop your head off, make a separate part of size one, and the remaining part is your part of odd color. And that's that. So, yeah. So the next course, if you start with any kind of combinations with odd colors, right, all we have to use also basically looks like something like this. And then every time you hit a one, you just check if the next one has. Well, actually, every time you hit a one, you just combine with the next part there's no need. I'm sorry. So this one combines with the next one, and it gives you a even color to part right and that one combines this one. So this way you are guaranteed to get rid of all the part of size one, unless you have one part of size one in the very end, which you drop to get to a combination in that set. Right. So, yeah, so I guess I want a little faster than I should, but hopefully the idea is up here. And I think that's all I wanted to talk about. Thank you for your attention. Thanks, if we could all thank our speaker in some way. And are there any questions. I have a potentially silly question. So when when you had this summation over binomial coefficients where only three K is down there have you tried summing with third roots of unities does that help at all. Is this what you're talking about. Yeah, yeah, yeah, yeah. So here have you tried to like multiply the sum of unities because then every third is real and the other two like cancel. Does it ever come up when you see module three. Try that. And thinking about it now I have no idea what will come up. Because there is also this song. I see. And we definitely didn't even try. So it's, it's interesting. It's an interesting idea. I just, I don't know what I have seen a formula remotely like this only once in my life and there it helped help to do some sort of a binomial theorem with third roots of unit is which was it would just pick up every third term. Yeah, I don't know because this top number changes. Yeah, well, that's like one over one minus six kind of things like negative. Yeah. So I have no answer. Other questions from other questions. Thank you again for an excellent talk. And thanks everyone for coming out. So I think we'll go ahead and break there. Thank you for having me. Bye. Bye.