 Thanks again everybody for making it out. Today we have Rigoberto Florez from the Citadel who will be talking to us about some enumerations on non-decreasing dyke paths. Go ahead, Rigoberto. Okay, by the way, I'm not used to listening to Rigoberto. I used to say Rigo. Oh, okay, sorry about that. No, no, it's okay. It's okay, but I'm not used to. And then like people say Thomas, people prefer Tom, right? I prefer my short name. And okay, again, thank you for the invitation here with you. And I enjoy a lot your seminars. And now it's online, which is much better for me. Okay, this is a work with several co-workers one is Eva. And actually this is a picture that Eva and Lardo were here in how many years? Eight? God, it's a lot. Okay, and this is José Luis and the Andrew. I don't know what is located, maybe in China somewhere. And then, okay, those are my co-authors. And I want to start this talk with the FioNachie numbers. But it's supposed that we'll be about a path, but I want to start FioNachie numbers because FioNachie numbers is present in many, many results that we had here. And FioNachie numbers actually are very beautiful and nice that they are present here. Okay, let's start with the idea. What is a big word? A big word is a word in the XY alphabet and the length, same number of Ys, same number of Xs. And the idea is that every initial supath should have no more Ys than X. So if you see here, satisfactory condition. And this part also, see that. But in this part, no, here, so far it's good. But in this part, we have more Ys than X, so this is not a big word. If you choose a length, eight, and you do all possible combinations to satisfy the conditions that I gave you, we have 14 big words. And that 14 is actually a special number. If you recognize the number Yd, I don't know Yd. And then there it is. And also, you can represent big words into the balanced parenthesis, where the Xs are open parenthesis and Ys are the closed parenthesis. There is another representation for big words, and we call big path. And big path is just the geometric representation of big words, where the old X is actually representing the north is a steps, and Y is representing the south is a steps, and we have here the representation of the path. And the 14 big words, there is a variation, of course, in between the path, both here and here we have also 14 paths. And then there is another representation, if you want, also variation, is big words and binary trees. And there is also more representation using trees, but here I brought just the binary trees. And I'm going to show you how is the binary trees. The other is more popular than this. For the reason I want to show you how is the representation, support that we want this represented as a tree. So it's the number and every north is a step is an edge that is going, this is ordered. So it's going to the left. So if you go up, up, up, this is going down, down, down. And when you go here, south is a step, this is going to the right and here. And this is another down. And because it's binary, this is busy. So we have to go to the previous one and go here down. And this is up. So from here, you go to the right. And this is down, go here. And this is another down. And this is busy, you go here. And that way you can represent using binary trees, the big path. And so to count the big path, we use the Catalan numbers. And I told you before 14 was a special number because it's the fourth Catalan number. Now, because we want to study several aspects, I'm going to talk to you about some of the aspects that we are interested here. One is, for example, the peaks and peaks is the X and Y is here. All valleys are Y and X, you see. And a pyramid is a short or a short path of this form. And we have, this is a pyramid, but this is maximal. But here we have another pyramid that is maximal, but that is within a maximal pyramid. Now we are going to define what we want to study here today is the non-decreasing big path. So what we are going to do is we are going to, from left to right, we are going to collect all vertices of all valleys. And you select the Y coordinate of that vertex here, Y coordinate. And if the Y coordinates form a non-decreasing sequence, we are going to say that this is a non-decreasing big path. And that's what we are going to study today, only non-decreasing big path. So we are going to use this notation whenever we see D, calligraphic D, N, means the set of all non-decreasing big paths are length to N. And in this case, if you count, we have 13. 13, and in the previous case, we have 14. So some of the decreases are not, some of the big paths are not decreasing. And in 1987, Barcuci, at all, they proved that the total number of non-decreasing big path is given by this pionacci number, as I told you at the beginning, many times the pionacci numbers are here. And this is 13, of course, is a pionacci number. And now, this is Barcuci, but now the idea is if we have the set of non-decreasing big path, what we want to ask or we want to work. So there are many aspects that we can ask. One is, for example, count the number of peaks, right? Or maybe we can count the number of valleys, or we can count the height of peaks or height of valleys, or the area under the path, or we can just, the area of every pyramid, for area of every pyramid, we say that is the weight of the pyramid. So there are many questions that we can ask. But here, I'm just speaking a few, and I'm going to show you what is the technique that we used to prove that. And the same technique, most of the time, works for all cases, right? Yes, depends on the case, but it's the same technique. So here is, we are going to, I pick to work the total number of peaks in non-decreasing big path. And we say, okay, the total number of peaks in non-decreasing big path is given by this number. Again, you see if you are matching, and it's a combination. And the number of peaks is basically the number of pyramids, but let's count peaks, one, two, three, four, five, six, and so on, and we obtain what? 32. And the question is how to prove that that is true. So this is the proof that I'm going to show you here, there's only one, and the other, we'll use the same technique. So we are going to take a dn, and we are going to separate that into two sets. One is b, and the other is a. The four, the four means is the length, right? Because this is length eight. So this is four and four. So, and in this part, we are going to have all paths that they have at least one valley at the ground level. So at least one valley at the ground level. And in this part, we are going to have all paths that they don't have valleys at ground level. Later, we are going to refer this with another name. Here we call bn, but the other name later will be primitive path, but not yet. So let's count first this part, how many peaks here. And the idea is, because we don't have valleys at ground level, we can take out, delete this step up, and this step down, and we send that one into this. We do the same here, and all of this. And when we do that, we observe that we keep the same number of peaks. So these variations doesn't affect the number of peaks. And this variation just sent bn into the previous set of non-vexing depart. So instead to count the valleys at the peaks here, we are going to count the peaks here. And we are going to say, okay, let pn, the total number of peaks in d minus one. So because this variation, now we know how many peaks are there here, is pn minus one. Now let's work with the set A. In the same set A, we subdivide again the set into more sets. And first, we are going to say, okay, these are all paths with the first pyramid has height one, height one. And this is, you see, length four and height one. This is the first pyramid, height two, and this is height three. So there is a natural variation from these sets into this set is just you delete the first pyramid. And so you send all of this after the deletion into this. And you send all of this after the deletion into this and same thing with this. So this gives us an idea how to count the total number of peaks. So to count the total number of peaks here is basically the total number of this part. But we know the answer for this because is one of the peaks, right? But after that we had to add what? One for this, one for this, one for this, one for this, one for this. But that addition is actually equal to the cardinality of this. So now we have the total numbers here, plus the number of peaks after the deletion and plus the number of the first pyramid. So basically pn is equal to this. And when we simplify, we obtain the number that we had before. Okay, now the second proof is using a generative function. And the same idea is also works for many, for many other topics that we study here. So for the generative function, we need some parameters. So first of all, let's say that the cardinality of dn is this. And we are going to say lp, like the semilength of the path. And now we have the generative function. So the generative function is, this is depending on the length. So and we say, okay, the path here in dn, but when we have this simplified is basically this cardinality. So this is the generative function for the total number of non-decreasing the path of the n. And okay, let me go a little bit here. So if we have length one, we have one pyramid. When the length is four, but semilength is two, we have two. And when the semilength is three, we have five. And it looks that it's few on each number. Of course, we knew that answer before, but that is using a generative functions. But that isn't what we want to prove. We want to prove is just the total number of peaks, not how many non-decreasing paths. So let's construct the generative function for the total number of peaks. So this is, we need an extra parameter. And we say this is the number of peaks. And we need a variable to track the number of the peaks in the path. And we put this. And now this is what we are going to use to count the total number of peaks. And this is the generative function. And for the generative function, when we set this equal to one, we obtain the total number of non-decreasing the path here. This is the generative function. Okay, let's see how is the proof of this generating function. So the proof of the generative function is you remember how we decompose the path before? So we decompose the path in two ways. One is we have all paths that they don't have any valley at the ground level. And the other path that they have at least one valley at ground level is here. So it means this pyramid cannot be empty. So because we want one valley at least. So this is, we need at least one length equal to x. And we don't want this path be no empty. So we are going to consider this here in cases this empty. So we have this. Now with this, we can create the generative function. And the generative function is for here the length is one. And the tracking of peaks is one. So we say this is x, y. For this part, here we don't have peaks, but we have length. This is the length. And times the generative function for all paths in this way we put in this form. And in this part, so we are tracking the number of peaks here, but the number of peaks is only one. And the number of pyramids, this is fixed. So we need one x, x. And the other is maybe all one or two or three or infinite. So this is one, one minus x to count all pyramids here times the generative function for this. And when we simplify this, we obtain the generative function that we were looking for. So that is basically what we do for all proof that we have here. We try to use the same technique in both cases. One we call generative function, the other is combinatorics proof. Okay. But as for all these papers that we have been working, we want always connection with other objects of combinatorics. And the first connection that we have here, very natural, you are going to see why it's natural, is using a polyominoids. And there are two, there are many definitions of polyominoids. One is given by math words, right? That is a generalization of the domino camset. And the other is just a set of squares here. One together, you put together. But what happened? We need that when we go around, right, and they pass around of this or cycle, they don't overlap. That is the first definition of polyomino and the other is from math words. Our interest is in the general polyomino. Our interest is a special polyomino. And one is the column convex. And the column convex means when we have a vertical line within the polyomino, that vertical lines to be totally included in the polyomino. So this is completely this column convex. But this is not because if we have a vertical line here, that is not totally included in the polyomino. So this is not column convex. And we need directed. Directed means from the lowest point on the left, you can go wherever you want with a path. But this path is using only north or east a step. So it satisfies this condition, which is directed. And this is good because wherever you want to go, you go up, up, up, this, up, this. But here is not because if you want to go to this part, you can go up, up, but you can reach this point. So what we are interested here is just this is the P or this is the polyomino. And let's see what is the relationship in between this. This is the polyomino and non decreasing the path. First of all, Barcucci, he proved that the total number of this is a polyomino is a pionacci number. But this is not a pionacci number. It's this pionacci number, which is actually the same pionacci number that counts non decreasing the path. So when we have one square is one, all polyminots, when we have two squares is two. And we have three squares is five. And we have one, two, three, four is 13. So that is what he proved. Actually, they prove. Okay, later, Doge and Prudinger, they found a connection in between this and non decreasing the path. And the variation was given by this, they create levels and they said, okay, this is the level zero, level one, level two, level three, level four. And that is what we have here, level zero and upper level four. Same thing for this, level one, two, three, and so on. And with this, they actually he proved this by example, but this is very simple, right? Very simple. And maybe I just start from zero to upper level four. And then you go from four to one and from one to three, and from one to three, and from three to two, from two to five, from five to two, from two to five, from five to three, from three to six, this is six, from six to three. And then, from six to three, and three to five. And then this is the last one. And that's to complete the path, you just send that one to zero. That is a deposition that the Proudinger and Roach prove it. And the idea is we have the total number of peaks. And the first question is, is there an easy, easy, very easy interpretation of the total number of peaks here and the total number of something here? And we say the something is what? The total number of columns and that is the deposition. So the total number of peaks here is actually for the total number of columns in polyominoids. So we see here one, two, three, four, five, six, seven, nine, eight, nine, and ten. And that is what we have here. One, two, three, four, five, six, seven, and ten. So and this is interesting that connection is because at this point there are a lot of open questions in polyominoids and we believe, we don't know yet, but we believe that maybe instead of proof that question in polyominoid, maybe it's possible to prove the questions here in a non-deficient device. That's the future project to work. Okay. Remember that I told you a little bit about primitive path? Now let's see another idea here to count. And it's the primitive path. And primitive path, we are going to say that a non-deficient path is primitive if the path doesn't have values at the ground level. So all of these are primitive path. And the total number of primitive path is a few natural numbers. And it makes sense because in the previous results, we proved that when we delete one here, it's actually in vegetation with a dn minus one. So that is basically the proof. Now, the idea is to count the total number of non-deficient path that are concatenation with k primitive path. In this case, this is the big path that are concatenation of just one primitive path. But here this is length is four, but two primitive path. You say one, two, one, two, one, two, one, two. And this is three, one, two, three, one, two, three, and this is four. So how we count that, again, using, we use both the generative functions and recursive relations. And using the generative functions, we have this generating function and the closed formula for that generative function is given by this. And then same idea for the proof. Okay, I'll be back on this, but let me introduce first another idea. I'll be back on this in maybe five minutes. So we are going to find the connections into non-deficient path or a primitive path and reordering arrays. I'm going to introduce here the idea of reordering array. A reordering array is basically a generalization of the Pascal triangle. So people who discovered this were working with Pascal triangle and they found that it was possible to generalize this. And this is what is in a lower triangular matrix and that satisfies some conditions. And the first condition is that the first entry here never, never is zero. And the rows here, or let's say columns, we, the people who discovered this, they call, they want an external column and they call this internal columns. So we are going to generate the external columns in one way and the internal columns in another way. So we need a vector with the first entry not equal to zero. And here, whatever, right? Or you can use generating functions and the entries here of these vectors are the coefficients of the variable x. And here we are going to use another infinite vector or a generating function. So how is there any entry here? Let's suppose this is a square that I brought here. This is a square. We are going to do a combination. So we multiply, if we want this square, we have to go one before on the left. So this is the square we go here. And we multiply this green times a zero, this plus this green times a one, this green times a two. And because all of this is zero, we don't care about that part. Basically it's what we are doing here. You see this part. And we combine all of them and we obtain the square. So we have here the square. Now in this part, we do the same. But in this case, we start from one left. Now here we start from one up, one up. And we want this square. So we multiply all of this by, so this times z, z zero, z one, z two. And we have this. Now let me give you an example of that. So the sample is this. So using this generating function and using this generating function, we obtain this vector. And if you want 11, if we want 11, we just multiply seven times one and four times one and we obtain 11. And if we want 13, you multiply five times one, four times one, three times one, you combine and we obtain 13. So now there is one construction, but there is another construction. And the other construction is if we have two generating functions, g and f, but this we need one condition. X is always a common factor of this generating function. And this is just independent. It's no zero, right? Independent is no zero. So we need a condition and we do this. The first column is g times the generating function raised to zero. The second column, actually column zero, right? The first column is g. The generating function raised to one and so on. And that way we obtain the columns of the reordering array. So the reordering array is basically defined using the generating function and this generating function where the columns are here with the conditions that I told you before. Okay, what happened with the back to the total number of paths that can be written as a concatenation of primitive path is given by this. So if you observe here, we have one generating function and super that we don't have this power, right? Just this and we have another generating function. So this is giving us an idea that we, but I, k, we obtain f and this is my g. So with this we can create the reordering array. So now let's go back. So we raise this to zero. We obtain this generating function, this other generating function and so on. So when we take the coefficient of this and we put all of them here, all coefficient of this and we put here and so on, we have now the reordering array for the total number of paths that can be written as k concatenation of k primitive path. So if you observe here, so this is a concatenation of one primitive path is five is here. This is concatenation of two primitive path is here, concatenation of three primitive path is here and four is here. So this is the reordering array for the concatenation of k primitive path. And the question is how come, why is this important? It's important not only because we can collect all that information in images and it's because we can find some properties. So we are going to find some properties or identities for this using reordering array. So remember that 13 is the addition of all of this, right? But five is t1, 42 and three is t3 and the other is one. So how is that? So now the first identity, so for this is because the addition of all of this is equal to this part. We say, okay, the first identity for this is that tm plus one is actually the addition of all this, I mean this. So the other identity is because this is obtained, adding these two, so this is actually this, equivalent to this part, is this number plus this number. So using this big path and reordering array, we can obtain identities for other combinatorial objects. Okay, that is one. Now let's go to another topic that we also come to here is the weight of pyramids. This is very brief, very short. And just I want to give you more ideas of how we did in this paper. So the idea is we are going to evaluate the area of this pyramid, or we can say the weight and the weight we are going to define or is the height or semi-length, right? Semi-length of the pyramid. So this is weight one, the semi-length is here is two, the weight is two, and the weight is one. So here it is. Now we defined g and k is, this is the length of the path, and we say it is the number of pyramids at the floor level that can be, oh, the weight k, right, in a dn. So a floor level and we want a length four and weight one. How many are there? There are 13 and you can see here the, when you count, you obtain 13. But 13 is a few numbers, so don't. A length, weight two, and weight two is here, weight two, and the total is what? Five, again, few numbers. And weight three, and this is the weight, is three, and is two. And the last one is weight four, which is one. So using the same idea, we find the reordinary for that counting. And now from here you can figure out what is the generating function for this part, for g and k. Basically, this times x to the n or to the k. So we have the reordinary. And we can find similar properties using the same technique that we did before. But this is very brief. I'm not going to spend too much time on this. So we were looking for more ideas to work in non-decreasingly path, and then searching for more ideas to work in non-decreasingly path, we read a paper of a solid. It's actually not a paper, it's a thesis. And in 2018, she defined a word in this alphabet. And the word in this alphabet, she defines peaks, and we say, wait a moment. She's defining peaks, and we have peaks in our path. Let's see what happened. And then she defines peaks in words in this way. This is a peak, and this peak is a peak if the number that is before is less than this, and the number that is after that is less than this. So that is a peak. This is not a peak. Five is a peak, because two is less than five, and this is less than five. This is also a peak. So NG also defines what is a symmetric peak. NG says that symmetric peak is if the two here, the two numbers, the two values here are the same, that is symmetric. So these two are the same. This is symmetric, but this is not symmetric. NG defines also a symmetric peak. This is asymmetric, because these two values are not the same. So the idea is how to extend this comp set to non-decreasing the path. And we did this. We defined a symmetric peak, because we have peaks, but we need to define symmetric. And to define symmetric peaks, we say, okay, a peak is symmetric if the maximal pyramid containing the peak, the two values that are forming the maximal pyramid, they are at the same level. So this is symmetric. This is also symmetric, symmetric, another symmetric, and another symmetric. And asymmetric peaks, if the two values are not... We consider this a valid, right? And we consider also this a valid. So if the two values are not at the same level, so this is asymmetric. But when we were searching, if this concept already exists in a big path, we realized that nobody watched this concept before in big path, then we made the decision to work first this idea on big path, because it's more general. And then later, after that, worked this on non-decreasing big path. And that's what we did. So the results is for big path. And later, actually we are working, we just submitted the paper. Yes, I remember that. So we did that for a non-decreasing. So using the technique that we did before, but you see, before we say, okay, all paths, they have a pyramid. In this case, we don't say all paths, they have a first pyramid. We say all paths that they have, what? One primitive path, right? That's how we divide. Using that idea, we found this recursive relation. And then we simplify and give us this crazy recursive relation. And this is our catalan numbers, catalan numbers. And then using this recursive relation, we found this total number of symmetric peaks. And if you count here, all of these are symmetric, and we found 23. And how to find asymmetric? That is very easy, because since we know from literature, the total number of peaks, and we subtract the symmetric peaks, we obtain the asymmetric peaks. So that is not complicated. And using that generating the recursive relation, we found the generating function, and we found a closed formula. What we did after this, we extend this to non-decreasing path, same ideas. But the generating function in non-decreasing path, we're a little bit more different. And we also introduced other concepts, and now the we submitted that paper, I think last week or two weeks ago for publication. And I think that's it. Thank you, Rigo. If we could all thank our speaker in some way, and then we'll open it up for some questions. Okay, do we have any questions for our speaker? Okay, if there are no questions, then before we go, I want to say that next week, the seminar due to some scheduling issues needs to get pushed back an hour. So we will be starting at this time at 3.30 instead of 2.30 next week. I'll say that in the email next week, but just letting everybody know ahead of time. So thanks again, Rigo. And thanks, everybody, and have a good weekend.