 Thank you very much. It's really a pleasure to be back. And I just wanted to thank Alina, Michael, and Philip for keeping this amazing seminar series running. I know it's a lot of work. So thank you very much for doing this for the community. If you have questions, just interrupt me or put them in the chat. And Alina, could I ask you if I might not see the chat? So if it comes in, if you could just let me know. So this is a little fun project that I'd like to share with you on a very basic question on smallest denominators. So here's the problem. So we look at an interval, just a second. We look at an interval of length delta centered at a point x. And we look at all rational points that are in there. Of course, we have infinitely many. And we ask, what is the smallest denominator of all rational points that fall into this interval? And now let's shift this interval across, say, x from 0 to 1 and see what the distribution of those smallest denominators is. So what I've plotted here is just a few examples of interval sizes where we have on the x-axis here just the center point x. And you see how it moves through. And as the size, the length of the interval becomes smaller and smaller, you see a more and more complicated function emerge. And of course, as maybe it's not a surprise to you, it very much depends on where the interval is. And circle method people will recognize pictures like that. So that's the question. What is this function, the smallest denominator function? So recently, Shannon Haynes looked at this problem and said, well, let's work out the expected value for this function. So in other words, just integrate over x. And what they proved is that it's 16 over pi squared times 1 over the square root of the length of the interval plus an error term that you see here. And the constant 16 over pi squared is kind of an interesting one, right? It has pi squared in it and so on. So the question is really where this is coming from. And I'd like in this talk to shed a little more light on that question. But also, of course- So actually, can I ask, isn't it in the case that log of delta as delta becomes small, becomes very big? So how is that an error term? Ah, so the log of delta, yeah, you can put it as log of 1 over delta if you want, right? Delta to the minus 1 half even grows faster than log delta. Are you happy with that answer? So log delta grows slower than delta to the minus 1 half. OK, thanks, yeah. Yeah? OK, so and here is just a little illustration. That's sort of 16 over pi squared here. And you see it sort of makes sense, yeah? So the little numerical illustration that I've given you is in the right ballpark of the Chen Haines result. Now, there has been a conjecture from the 70s that I'm now going to talk about, where instead of looking at x as a continuous variable, we just sample on discrete points of the form j over n. And in order to explain that to you, I'm just going to redefine slightly my smallest denominator. So I'm not going to take it in an interval centered at x, but just as an interval where the left point is x. And I'll do this because what I want to do now is I want to simply divide my unit interval up into these small boxes. And I'm just asking now what is the, so this is 1 over n, 2 over n, and so on. And I'm just going to ask what is the smallest denominator in each of these boxes. And if you could say, well, isn't that the same thing? Because x and j over n are very close to each other. Well, this is a rapidly varying function, the smallest denominator function. And it's sort of illustrated here. You lose a few values. For instance, this one you don't get. And here you see a whole host of values. So by sampling of the order of x being j over n and the interval of size 1 over n, it's actually a more subtle problem. And indeed, following the Chen-Hains paper, Balazar and Martin proved that if we take the expectation of the smallest denominator function at those discrete points, we still get the same answer. So I've written it in this way. So we first have this discrete average. And then remember the size of the interval delta is now 1 over n. So that is again over delta to the minus 1 half. And they prove this result with an error term. And the proof is more involved than Chen-Hains, in both Clustermann sums and some refined estimates. Precisely for the reason that, as I explained, it's a more subtle evaluation now. Right. So we have the expectation value. And so a natural question is, well, do we understand the whole distribution of our smallest denominators? And here is a theorem that says yes, we do. And the limit distribution is given by this nice formula here. So let's just recap. So we have the smallest denominator. The expectation was going like d to the minus 1 half, if you remember. So I'm normalizing by d to the 1 half to make this a quantity that fluctuates around a constant. And we're asking, what is the probability that for random accents, a sub-interval d of 0, 1, what's the probability and the limit of small delta? And that's the answer. Now, here is a little numerical experiment where I've partitioned the unit interval into the 3,000 points that appeared in my abstract. So we have, again, our unit interval. And we're just sampling now the 3,000 points. So that's 1 over 3,000 here. In each point, we choose the smallest denominator, and that's the histogram. And you see it follows pretty nicely the curve. Unfortunately, my little laptop couldn't do more than 3,000 points without breaking down. But I always do like doing experiments that have a sort of, that actually show also the fluctuations around the limit. So now, how can we prove this theorem? And I should have mentioned there's also a nice paper by Albert Artilis. And Albert is in the audience who proved the existence of the limit and expressed the limit in terms of a certain random variable on the space of lattices. And I'll come back to that in a minute. But the nice question here is, where is this formula coming from? And that's the thing I'm going to explain to you next. But before I do this, let me just say this is not the first time I've seen this distribution. In my work with Andreas Stoenbergson 10 years ago, we found exactly the same distribution in the shortest cycle length of a large random circulant directed graph of in and out degree 2. So here are two examples of such graphs, or actually four examples, this one, this one, this one, and that one. And what we do, so a circulant graph is basically a graph where you connect the end vertices you have in a circulant fashion by telling you which neighbors you connect. So for instance, C8 here connects every other neighbor in this way. And the three refers to connecting the third, which is drawn here. And this one here works similarly. And now this is the distribution corresponding to the directed graph. And you see that these guys here have exactly in each vertex two coming in and two coming out. I won't say more about this, just a little advertisement to have a look at this paper that proves such limit laws for all sorts of geometric quantities, such as a diameter, cycle length, et cetera, four circulant graphs. OK, so let me show you the proof of the theorem for the limit distribution and explain to you also what the limit distribution is. Apart from the explicit formula, what I'll show you is that it's connected to the statistics of ferry fractions. So here is the ferry sequence of level Q. So all denominators reduce fractions with denominator at most capital Q. And we can write it in this way. So Z2 hat here denotes the set of primitive lattice points. And you may recall that the number of points in our ferry sequence of level Q grows like Q squared. And here we have exactly 1 over Z to 2, so 3 over pi squared. Now, the key points in this observation is that you can relate the distribution of the smallest denominator to the so-called void statistics of ferry fractions. And that solves the problem in one step, because we can then refer to the large body of work that we have on the distribution of ferry fractions. So the smallest denominator is bigger than L times delta to the minus 1 half if and only if the set of primitive lattice points so that Q is less than L to the minus 1 half. And we have a rational point in this interval is empty. So I hope that that makes sense to everyone. So if I want my smallest denominator to be bigger than this quantity here, I can only have that if I don't have any denominator that's smaller. That's all this says. So that's the key observation to make everything work. And now that you stare at this and you can convince yourself that that condition here is equivalent to the ferry fractions of level Q not intersecting this particular interval here, if we choose the level to be exactly L times delta to the minus 1 half and the length of this interval, well, that was just exactly given by delta. So you just need to tweak things in such a way that this all works. So if you plug this in here, the Q and the S, you'll see that the second equivalence that we have here is indeed true. So in other words, to recap, the smallest denominator in the interval around x is larger than L times d to the minus 1 half if and only if the ferry fraction doesn't intersect this interval here. And now remember that x is random. And when x is random, the probability that this happens is called the void statistics. And as delta goes to 0, Q goes to infinity. Remember, L is fixed here in all of this. The limit of the void statistics for the ferry fraction was proved in a very beautiful paper by Kargayev and Zyglians Zygliansky in 1997. And they showed that the void statistics converges to the so-called Hall distribution. And so this would be the density. The Hall distribution is the density of the void statistics. So the probability to having a void bigger than S, that's exactly what we need here, is therefore given by the integrated density, the distribution function. And we have an explicit formula for the Hall distribution that was calculated by Hall in 1970 already. And so now we just need to remember what S was. S was this thing here. So we plug it in. So that's our limit distribution. And in order to get the density, we just need to take the derivative of this with respect to L. And you will get the formula for eta of S that I had on the previous slide here, not on the previous on this slide, by plugging in the Hall distribution and working it all out. So that's the end of the mystery. So by using this equivalence in essentially one step, you understand that the distribution of the smallest denominators are simply related to the gaps in the Hall sequence through this relation. Great. Now, of course, it is having the convergence in distribution doesn't imply the convergence of moments. And that was precisely what Chen and Haynes proved and also the proof of the Kruswig-Meier conjecture that I had on the previous slide by Balazar and Martin. So we need to do some extra work to get this. And what I'll show you now is that we can use the techniques that I'm going to show you to even get convergence of moments and therefore get an extension of the theorems that were just restricted to the expectation value. So what this theorem E here tells you is that indeed, we don't just get convergence of the expectation, but for all moments, we have a restriction here on the moment. And that has to do with the fact that the limit distribution that I had given by the density eta has a heavy tail. So actually, moments with real part alpha greater equal to 2 or larger will diverge. So the limit will be infinite. So with this normalization here, we need to restrict alpha to real parts strictly less than 2. And then we get convergence of moments, both for the continuous average and also the discrete sampling here that was studied. And there was sort of the challenge of a crucifix-mayer conjecture. And we also have an explicit expression for the limiting moment. And this function here is a close relative to the function computed by Kargayev and Zeljanski for the moments of the void distribution of faring fractions. You've seen that the two are related. For the convergence in law, the two, as I've shown you, are actually equivalent. For the moments, they are not because now you're, if you like, integrating over L, so L is no longer fixed. So the theorems of Chen and Haynes and Kargayev and Zeljanski are not equivalent, even though they're quite close in spirit. Good. So the nice observation is that this limit function here, m alpha, is elementary apart from the beta function here that we have. And that's Euler's integral of the first kind, which has explicit expressions. And you can work it out. So in particular, if you look at the expectation value, which corresponds to alpha equal to 1, you'll get your 16 over pi squared. You don't need to do a big calculation to find that. So everything's consistent. That's great. And in fact, you can do more generalized moments. So for instance, you could say, well, I'm going to combine my alpha moment with a log moment like that, where n can be any integer. And you can prove a limit theorem for this. And it was sort of fun to work out some of these things. So here's a list of the mixed alpha moment and nth log moment. So here, if alpha is 0, that'll give you the first log moment, the 0th log moment, which is of course 1, because it normalizes the probability density. But then here, you get this formula, for instance. So it's all quite pretty, what you see. You get the Catalan constant for certain negative moments. Anyway, that was all fun. But now let's take a gear up and actually ask the question in higher dimensions and explain to you how and can prove these theorems, not using analytic number theory techniques, but techniques from homogeneous dynamics, which to my mind is certainly an easier first step in higher dimensions, because we do not have good continued fractions available. The analytic theory is more complicated. But having said that, all the results, I'm going to present in principle, could also be proved by analytic techniques. I'm not going to use any egotic theory that cannot be made effective. Now, before I go into this, are there any questions about what I've talked about so far? As I said, I'm going to explain now really what's behind the statistics of smallest denominators and also the corresponding distributions of fairy sequences. Everyone happy? Good. OK. Right. So let's think about higher dimensions and how we can set the problem up there. So again, we look now at rational numbers in Qn. And instead of having an interval, we choose an arbitrary test set A that we're going to fix. So think of A as being just a ball. We scale it down by delta and ask, what is the smallest denominator of all rational vectors in that test set shifted by X? So X, you can think again as the center or the endpoint of the intervals we looked at before. And we have this question. Now, I'm just going to get up and switch the light back on because I've been too quiet. This might happen a few times. OK. So that's what we're interested in here. So that's our main object. And the playground will be the space of lattices, which we can realize as this quotient space here, g mod gamma, where g is the group of n plus 1 times n plus 1 matrices with real coefficients and determinant 1 modulo, the modular group, which stabilizes the cubic lattice. So you can think of g mod gamma as parametrizing the space of lattices. g has a harm measure, which Minkowski showed we can normalize in such a way that it becomes a probability measure on g mod gamma. And now here, this guy is a cone, if you like, with cross-section A. So the picture you want to have in mind here is that this sort of looks like this. And this would be A. I've drawn it as a disk, where here we have X. This is the X direction. And here we have the Y direction. And I hope this makes sense. So this is an Rn plus 1. And now I'm defining a certain probability measure, P0A. The 0 here just stands for the fact that I don't want to have any point of this lattice intersecting my cylinder. Now, this is not quite a lattice. It's the primitive lattice points in Zn plus 1. g here is an element in my group g, modulo gamma. So that distorts my cubic lattice. And by taking the measure of g for which this holds, I'm effectively looking at a random lattice here. And I'm asking for the probability that a randomly linearly distorted lattice does not intersect the cylinder. Now, why is this distribution important? Well, it's important because it gives us the limiting distribution for the minimal denominator. And again, there's been a paper by Albert Artilis who explained exactly this using the space of lattices without reference to the fairy fractions, though. And I'm going to just show you again, as I've done previously, and how the fairy fractions explain this in one step. OK, so what does this theorem say? Well, for any test set, think of it as a ball and any d here. Boundary of measure 0, of course. Otherwise, this might not work. And any L greater than 0, we have a convergence in law. So the probability that q min appropriately normalized. And now we don't have the square root of d here, but we need to adjust our normalization according to the dimension we're in by this factor has a limit distribution. And this limit distribution, EL, is given by this formula. In other words, it's given by the probability that a random primitive lattice does not intersect our cone. We know that the tail of this distribution is power-like as to the minus n. That was proved by Andreas Trömbergson in 2011. There's also a beautiful paper by Afrija and Margules, to prove this as an upper bound for very general sets A. And so this means in particular that if we now just remember the relationship between S and L, which is this, this would be S here, we get this kind of decay for our limit distribution. And you see, we can only have finitely many moments. So the L to the n plus first moment would diverge. Now, the proof of this theorem proceeds exactly as before. So remember, this is the definition of our fairy sequence now in higher dimension. So the fairy sequence level q is exactly given by those fractional points with denominator at most capital Q. Now, the growth of the sequence goes like q to the n plus 1. Remember, in one dimension, it was q squared. Now it's q to the n plus 1. And we use exactly the same argument as in the one-dimensional case, observing that the smallest denominator can only be bigger than L normalized in this way if and only if the fairy sequence doesn't intersect with this set. And here is, again, the relation that you need to work out between qS on one hand and L and delta on the other. And now we are done because we know everything about the fairy statistics in higher dimensions as well. I worked this out in about 10 years ago, drawing heavily on my joint work with Andreas Trömböckson, where we looked at much more general questions on distribution of lattice points in randomly sheared and rotated set. So that's that. And now, well, let's look at the moments. And now I'm going to give you the proof on actually how you what you need to do in addition to understand the moments. So this is now in higher dimensions if you realize, again, with this new normalization factor. And we simply take the smallest denominator integrated over x, or raise it to the power alpha, integrate over x, and we get the limit distribution here. And as I said earlier, this guy here decays like 1 over L plus n plus 2, actually, it should be. Let me just see here. There might be a mistake here. I'll double check that. I'll get back to you tomorrow on this. So in order to have a finite moment, you need to restrict alpha here. And that's the correct restriction. So how do you prove such a statement? Well, if the real part of alpha is 0, then you can, you don't need to do anything in addition because you already have convergence in distribution. And the reason for this is, so if you understand the probability of Q min being larger than L, you can use a usual approximation argument of bounded continuous functions by characteristic function to prove a statement like that. So that's the usual weak convergence statement. So for any bounded continuous function f, we have this convergence. And now you simply realize that if the real part of alpha is 0, this will be a bounded continuous function, and you're done. Now, if the real part of alpha is bigger than 0, then you need to do something more. You need to work harder. And the first step is to realize that you can write the alpha's moment and integration by parts here as an integral over your distribution, which is here. So this gives you the alpha's moment. And so in order to prove convergence of this integral, you need to ensure that you don't get too much weighted infinity of your distribution. And that's not just guaranteed if you have convergence in distribution. So you need an estimate that holds for all l and delta uniformly. And now this is the kind of estimate you need to prove here. So you need to exclude that you have accumulation of mass at infinity. So you integrate from r to infinity, the kind of object you want to work out, and then take the limb soup over delta to 0 and then r to infinity. And if this is 0, it guarantees that you don't have that escape of mass. And then you can commute the limits delta to 0 and r to infinity. So then you can just simply replace in the limit the integral over l. And you get that the alpha's moment is just given by the alpha's moment of the limiting distribution. OK, so the way this works is just to recall, we have this equivalence that I've showed you before. So this is equivalent to the ferry fractions not having a void distribution. And now something really quite pretty happens. So you remember that the ferry fraction not having a certain point in this interval can be lifted to a higher dimensional setting. And that's exactly where the argument of the, not the argument, where exactly this observation comes in that the ferry fraction not intersecting this test set is exactly equivalent of the primitive lattice points not intersecting my cylinder. And that, in turn, is now equivalent of, so what you do here is you multiply on both sides with an appropriate alpha s. And then you see that this condition is equivalent to this condition here, where now your cone is a standard cone that's homothetically dilated by l. So if I had a blackboard, I will do that calculation. But I think you can do that calculation yourself to get from here to there. And then what you need to estimate in order to get this escape of mass bound, well, you need to estimate the measure of x for which this condition is true. And the point is that when l is very large, you look at a very large set that is not allowed to intersect this primitive lattice. So our cone contains a ball because we assumed that A has non-empty interior. And we can understand the shape of lattices that avoid a large ball of radius roughly l. If you, these by Mahler's compactness criterion, these balls, these lattices that avoid a ball of large radius need to be high up in the cusp. And so what we need to now do is we need to look at these shared lattices here that do this. And you can estimate the measure of those x for which you are high up in the cusp by a calculation that Vion Kim, who is also here in the audience and I did recently, and you can show that this is exactly bounded above by l to the minus and l plus 1. So the geometric picture here is really that you are now in the space of lattices and you can geometrically interpret this action, the shearing action, by x as a horosphere. So what we're really estimating is the part of the horosphere that's high up in the cusp in the modular space and getting a uniform bound on that. So that's what you need to do if you want to prove this. And that then completes the proof of the upper bound at infinity. So we have no escape of mass provided alpha is less than n plus 1. And if you have a negative moment, you need to do exactly the same thing. Now you do your integration by parts the other way around and you need to estimate now the set of x for which your minimal denominator is less than l. So what changes from the previous consideration is now that we need to avoid escape of mass near 0, because if the moment is negative, this is bad near 0. It explodes. And so we just need to make sure that the measure of this set is very, very small for small l. Now since we look at the minimal or the smallest denominators less than something, this now translates into the condition that our fairy fraction does have a point in that interval, and therefore that our primitive lattice does intersect with the comb. That in turn for small l again means we have to have a short vector in our lattice and we can again, by a different calculation now, estimate the set of x for which our lattice, this lattice here has a short vector of length less than l. And that gives us the corresponding estimate that we have here. So now the measure decays like l to the n plus 1 for very small l. And again, we get the correct estimate and everything's fine. Good. So we need to get the moments to really prove those additional escape of mass estimates. And that's not a surprise, but it is a significant additional input to go from the convergence in distribution to the convergence of moments. So now I want to come to the discrete sampling aspect, which was the recent proof of this longstanding conjecture from the 1970s. And I want to show you that this also works in higher dimensions. So as you've seen before, we had to look at the measure of x for which a shear had a certain distributional property. And we also had this matrix appear, which was related to our small parameter delta. And what's really underlying all the theorems about the convergence in distribution of this fairy statistics that I showed you on the previous slide are equidistribution theorems on the space of lattices of this form where you integrate, there should be a dx here, where you integrate the horosphere that corresponds to this action. And then you sort of translate it by this diagonal action here, the aq. And as q tends to infinity, one can prove, and these are theorems that go back to Margulis' thesis and a very beautiful paper of Eskin and McMullen who built on that, you can prove that these kind of averages equidistribute with respect to the high probability measure on g mod gamma. And so that's an old story. But if we now move to the discrete average case, we need to replace this continuous average over x by a sum over the j over these small elements. And so instead of having now a continuous average over our whole cycle, I just sum, I still transport everything by this diagonal action. And my claim is now that we still get equidistribution. And once you have this theorem i, everything follows exactly in the same way as before in the continuous average. So the theorems we have known already for 10 years can then just be adapted to also the discrete averages. And it's a non-trivial thing to move from here to there. And I'm going to show you now why it is non-trivial and how, nevertheless, ergodic theory really helps us to understand if you like why the two averages have to be the same in a sort of very nice robust fashion. So what I'd like to show you is that theorem i is true, which means that this kind of average here converges to the same average as in the continuous limit. So I'd like to relate the two in some way. So the first thing I do is I define a probability measure nu j, and this is again a typo that should be nu i. So I define a sequence of probability measures nu i by simply fixing a subsequence of integers n i here. These are my scales over which I sum. And so I have a sequence of probability measures. And what I would like to prove is that any such subsequence converges to my limiting high measure mu. Good. Now the key observation now is that we have a very nice commutation relation here where we can commute through the h and the a. So again, that's very simple linear algebra. But what happens here is that the object that I'm interested in, which is the x0 plus n to the minus 1j, I'll take this part here and I'll convert it through and you get exactly this thing. So why is this important? Well, you've seen in my theorem that I had this assumption here that q to the n plus 1 has to be less or equal to n to the power n. And you see here why this is important, because then this thing here will be of constant or decreasing order. What this also means in geometric terms is that actually as you vary the points J, so as you go from J to J plus 1 and so on, they are a finer distance apart. And so that's already a good thing because that means we can use the same escape of mass estimates as in the continuous setting. But how can we use that connection? Because they're finer distance apart generally to actually reduce the problem to the continuous average. So it's easy if we choose the sequence of q's and i's, the q i's and n i's in such a way that this thing goes to 0 because then actually they're not just a finer distance apart, but they're a very small distance apart. And you can, if you like, apply a sort of Riemann summation argument. So you can approximate the continuous average. You can use the continuous average to approximate the discrete average. So that's when this thing here converges to 0 along a certain subsequence. So let's assume that doesn't happen, but it converges to a positive limit tau 0. If that happens, then my points are a finer distance apart. So how can I then use this type of Riemann sum argument? I can't, right? Because they are just really a finer distance apart. I need something a little more sophisticated. And the sophisticated argument will use the ergodicity of the action. Now, I won't go through the full details here. I'll put those slides online. And you can look at them at your leisure. But this is a really pretty argument that was shown to be by Manfred Einziedler. And the idea here is that you can use the ergodicity of various actions to show that the limits have to coincide. And the argument is the following. So we're interested in the limit of this sequence of measures, nu j. So the first thing I do is I define a slightly fettant nu j. And I'm sorry, this should be nu i in this way. So I'll just, you remember, I have, this is a discrete measure. I'm just summing over those points at j here. And now I just smooth it slightly by epsilon. So I'm bringing it a little closer to the continuous average, but not quite. And epsilon will be very, very small. This here is my continuous average. I want to show that these two have the same limit. And I'm also defining the complementary probability measure, which is defined in such a way that if I add those up, so the one I'm interested in, the nu epsilon, and the complementary measure, I'll get exactly mu. Now, we use an argument that says that, well, let's suppose, first of all, my fettant measure is converging to a limit along a certain subsequence. I know, of course, by the equidistribution that I already have for the continuous average, that mu i will always converge to mu. So along the same subsequence, we have also this convergence. And by this relation here, I also know now that mu i epsilon bar will converge to the right limit. And because my mu i's and mu i's and mu i bars satisfy this relation, also the limit has to satisfy this relationship. Now we know that mu is ergodic with respect to this so-called horospherical action up here by this subgroup. We also know that mu epsilon is ergodic with respect to this action. And if you have an ergodic measure, you know that it's extremal. So it can't be written as a linear combination of other invariant measures. And this means that the three measures have to be the same and have to be harm measure. And then the last point that you don't just need to check is that the fettant measure, mu i epsilon, is actually close to the original measure you're interested in. And so that's it. So this is the ergodic theoretic argument. If you want to do this with analytic number theory methods, you have to do harmonic analysis. And it's much more complicated. But it can also be done. So it'll involve in the one-dimensional setting clusterman sums to resolve this difference between the discrete and the continuous average. And in higher dimensions, it will involve the higher dimensional analogs of clusterman sums. And there are results in the literature which would give you that. OK. So I think I'm coming to the end of the talk. Just two more slides. The first one is just to say, we can, using this equivalence, prove analogous statements also for the higher dimensional distance function to ferry fraction. So you say, OK, what is the distance of a given point x to the ferry fractions? In other words, to the nearest rational point with denominator at most q. So that's what's written here. This is something that Kargayev and Sigliansky mentioned that result earlier looked in 1997. And the techniques that I showed you prove that you also have the convergence of the beta moment of this distance function to ferry points, provided your moment isn't too large. And another statistics that I really like is what we call the pigeonhole statistics. So the pigeonhole statistics, say, for a sequence of numbers in the unit interval, fractional parts of your favorite sequence, works as follows. So you, again, partition your interval into n boxes. So each box has size 1 over n. And you look at the first, say, n elements of your sequence. You can also look at the first sn element of your sequence. That's actually what we're doing here. Or maybe I'm, yeah, well, let me not do this because I think I'm looking at the first n over s integer part elements of my sequence. So you have n boxes, and you ask yourself, where are the elements of my sequence going to be? So there might be two here, there might be none in the second box, then there might be three here. And you're simply asking, well, how many boxes are there that have no element of my sequence in it? How many elements, how many boxes are there that have one, that have two, et cetera? So that's the pigeonhole statistics. And we can, again, use our discrete average to show that the pigeonhole statistics for the fairy sequence converges has a limit. And this limit is, again, given by the probability that a random primitive lattice intersects this cone that I had on the previous slides in exactly k points. So I think I'm done. Here are references. The main references, this paper highlighted in yellow, where you find everything that I've described here. The main upshot of my lecture was, I hope to illustrate to you, that this beautiful problem of finding the smallest denominator is actually related to some really quite sophisticated mathematics, surprisingly, maybe, but that it also relates to a whole body of work on the fairy statistics that was started with Hall in the 1970s and that many people worked on. I only mention a few here, but there's a lot of work also by Boca Zaresco and their coworkers that can connect it to this beautiful problem. And you see some beautiful limit distributions that, certainly in one dimension, you can analyze explicitly. In higher dimension, it's still a big open challenge. We don't really know much about those lattice distributions, except for the tail estimates that I've showed you, the heavy tails. And so it's rather difficult to get explicit formulas for those. Good. So that's all I had to say. I'm going to go back and switch the light back on. And while I do that, you can maybe think about if you have any questions for me on this. Thank you.