 Hey, thank you very much for this invitation to speak in this wonderful seminar. I'm going to discuss a recent joint work with Collin Faverjion, who is working with me in Lyon. And the talk will be divided into two parts. So in the first part, I'm going to discuss Mahler's method, which is a part of transcendental number theory. And I want to present some recent results in that theory. And then in the second part of the talk, I will explain how to use these new results to solve problems that are related to the first and back conjecture from the tater and that involve a finite automata. So let's start with transcendental number theory and Mahler's method. So much of transcendental number theory is concerned with the values of a complex analytic function with algebraic coefficients at algebraic points. So the situation, for instance, that you have a several function f1 of z, fn of z, which are complex analytic function with algebraic coefficients. And let alpha be an algebraic point. And of course, if there is an algebraic or linear relation of the field of rational function between this function f1 and fn, you can specialize this relation at the point alpha, of course, assuming that the function f1, fn are well defined at alpha, to get a similar relation of the field of algebraic numbers between the values of the complex numbers f1 of alpha, fn of alpha. And the transcendental number theory try to go the other way. So for instance, if you are able to prove that the function exponential of alpha1z up to exponential of alpha and z are algebraically independent, then you can deduce that the numbers, yeah, e of alpha1, e of alphan are algebraically independent with alpha1 and alphan are algebraic numbers. So this is a typical example of a situation where you go the other way. But it's also known for a while, so more than 100 years ago, that this cannot be true in general. So for instance, you can construct a transcendental function that is an entire function with algebraic coefficients, but that takes algebraic values at all algebraic points on the complex plane. But still, there are a few known frameworks where the converse assumption is essentially true. But in that case, you have to put some more structure on the function f1, fn. And usually, they have to be related by some kind of system of functional equations. And typically, they can be related by some linear differential equations or by linear difference equation. Probably the main non-example of such framework is the theory of Z-Gal e function that was developed by Z-Gal and Shilovsky, mainly. And that is related with an entire function like the exponential function that are related by some system of linear differential equation with polynomial coefficients. And Manor's method that was initiated exactly at the same time, Z-Gal introduced the e functions, provide another such framework. And this is the framework I'm going to discuss now. So I will introduce the m functions in analogy to the e function, so m stem for Manor. And the very basic example is given by the sum of the z to the 2 to the m. So it's easy to see that this function, f of z, satisfies the inhomogeneous relation. f of z square is just f of z minus z. So when you look at f of z square, it's just a shift of the series. And you get this relation. And Manor used in a clever but elementary way as this equation to prove that the values of this function are transcendental for all algebraic, non-zero algebraic number in the open unit piece. This equation, so it's inhomogeneous because we have this minus z here. But of course, you can turn it into a homogeneous relation of order 2 instead of an inhomogeneous relation of order 1. And you get this following equation that involve f of z, f of z square, and f of z4. And this motivates the following definition. We fix q, a natural number, at this equal to 2. And we say that a Poirre series with algebraic coefficients is a q-matter function or an m-tube function. If you can find polynomials with algebraic coefficient p0, pm of z, which are not all 0 and f of z satisfy the following linear difference equation. So in the setting of e function, for instance, we study similar equations with the derivative of the function f, the successive derivative of the function f. And here, the derivatives are replaced by the action of the morphism, the operator z give to the q. So we use the power of q, iterated m times. And we will just say that f is an m-function. If it is an m-tube function for some q, and we do not want to say more about this particular q. In this definition, it's very important that we are considering solutions to this equation that are Poirre series. For instance, the logarithm satisfy, of course, log of z, q is q log z. So it satisfies such equation. But it's not an m-function. And the theory, unfortunately, does not apply to this logarithm, because it's not analytic around 0. So what happens with m-function? So they are defined as Poirre series. But in fact, such Poirre series are always convergence. So they are analytic in some neighborhood of 0. And we can prove using the equation that an m-function is always can be continued meromorphically. In the open unit disk. And then you have the following nice dichotomy. So either the function is a rational function, in which case it can be continued meromorphically in the wall-open complex plane. Or if not, the Poirre series, an m-function, is transcendental. It has the unit circle as a natural boundary. And it is meromorphic in the open unit disk, but cannot be continued outside this open unit disk. So in particular, it doesn't satisfy any linear differential equation, because it has infinitely many singularities. And recently, in a joint work with Charlotte Ardouin and Thomas Dreyfus, we have been proved that an m-function, when it's not rational, does not satisfy any algebraic differential equation. So it's called hyper-transcendental. And this is much harder to prove. And for this, we use some Galois series attached to this equation. But let's come back to transcendental number series. So the very first question you can imagine about this m-function is the following fundamental question. So if you choose one m-function, transcendental, an alpha and algebraic point where f is well-defined, can we decide whether f of alpha is transcendental or not? So a simple question, but not that easy. Essentially, Mahler's already answer to the question is the case of order 1 equation. He normally knows how to order 1 equation. But we would like to have an answer for general equations. So as usual, and this is a case also with linear difference equation, so you can move from equations to linear system. So if you have an equation of order 1, you can move to a system of size m. You can move to a system of size m using the so-called companion matrix. And for instance, if you have this equation of order 2, we already saw, you can turn into a 2 by 2 linear system where now we had some coordinate at the f of this square. And the matrix of this system is a 2 by 2 matrix whose coefficients are a rational function with algebraic coefficients. And this is a general fact. So what we want to study are such Mahler's system. So we consider a q-Mahler system is a system of the form yz is az y to zq, where az is just a matrix, m by m matrix that is invertible. And the coefficient of rational function is algebraic coefficients. And once we have a system like this, what we want to consider is a vector solution, f1 of z, fm of z, whose coordinates are q-Mahler functions. So I'm q function. So they are power series solution to this system. OK. So what can Mahler method can do for us? The first thing is that we have to define some bad points. And by this, I mean the singular point with respect to the systems. And there are exactly the points that can be mapped under the operator z give to zq to some pole of the coefficient of the matrix az or some pole of the coefficient of the matrix az minus 1. So that has a point that can be sent by the operator to some place where the matrix is not defined or the matrix is not invertible. But once you remove these bad points, you have the following very nice results that was obtained by Nishoka in 1990. And that is arguably a major theorem in this theory. It says the following thing. You pick an algebraic number alpha in the open unit disk that is non-zero. And you assume that it's a regular point. And so under this general assumption, you get that the transcendence degree of the function f1 of z, fm of z, the vector of solution to your systems over the field of rational function is the same as the transcendence degree of the values f1 of alpha, fm of alpha over the field of algebraic numbers. So in particular, if you are able to prove that the function f1, fm are algebraically independent, then you get that the values are algebraically independent. In that case, each value is transcendental in particular. So this would solve the first problem in that case. But unfortunately, this is not always the case that the vector solution of algebraically independent coordinate. So in general, this is not the case. And you have to compute exactly this transcendence degree. An important thing to note here is that it's very important to focus on a regular point. So I'm going to show you the following examples. So if you consider the symphony product, the function g of z, that is defined as a product over n of the 1 minus 2z3 to the n. So it's a transcendental function. That is an n3 function. It satisfies a very simple equation of order 1. So you can see this as a 1 by 1 system. So the matrix has just one coefficient that is 1 minus 2z3. And so the transcendence degree for this system, the transcendence degree of this function, you only have one function. So the transcendence degree is 1, because the function is transcendental. But now you see that if you evaluate the function at point alpha such that alpha 3 to the n is 1 half for some n. So there is one term in the infinite product that vanishes. So the whole product will vanish. And in that case, the transcendence degree of the value is 0. So you see that the theorem is for the conclusion of the theorem does not hold, because the transcendence degree on the right is 1, and the transcendence degree on the left is 0. And the point is that these particular points are exactly the singular points for the system. So the singular points are already the point where this theorem is not true. OK. So Nishoka's theorem is a really nice theorem. It's the exact analog of the Ziegler-Chilotski's theorem for E function. It is exactly the same rule. But this result is a quantitative statement. So it can be rephrased as an equality of dimension, of cruel dimension between some rings. So it just says that there are no more algebraic relations between the functions and between the values. And this is a kind of deficiency, because the typical situation is that we have one M function we are interested in. And so we put it into a system. And then we have other function. We don't know too much about them. So in this situation, the only thing we can know if the function is transcendental is that the transcendence degree on the right is at least 1, because we start with one transcendental function. So the conclusion is that the transcendence degree on the left is at least 1, saying that one of the number here is transcendental, at least one of the number is transcendental, but we do not know which one. So you cannot say that you want to use the values f1 of alpha is transcendental. And recently, this quantitative statement has been refined by Philippon as a qualitative statement. So the qualitative statement is the following one. So imagine that under the same assumptions that we start with a system, a vector of solution, and a regular point. So alpha is assumed to be a regular point in this theorem. So imagine you have an homogeneous relation between the values f1 of alpha, fm of alpha. Then this relation can be lifted as a relation between the function f1 of z, fm of z, so that if you specialize at alpha, you get the original relation. So this theorem really take care of any relation between the values. And this is clearly the situation we, the best situation we can hope, as I explained at the very beginning of the talk. Here I put the homogeneous part in red because it's a stronger result than the result without this homogeneous part. Because for instance, it says that if the function are linearly independent, then the values are linearly independent. Because an homogeneous relation has to be lifted in an homogeneous relation of same degree. And the Philippon proved these results without this homogeneous part. And this complement has been given in my joint work with Coulomb-Faverson. So I would like to mention here that again, Nishoka's theorem was the analog of the Ziegler-Schildowski theorem. And here, similar results were first proved in the realm of E functions. So by Schildowski and then Nestorenko and Schildowski. And the precise is exactly the same results in the context of E function was first pushed by Fitzbakers using some results of André. And then it has been re-proved by André using some new kind of Galois theory. And in all these cases, the lifting theorem are deduced from Nishoka's theorem. So you start with Nishoka's theorem and prove that if Nishoka's theorem is true, then you can prove this lifting theorem. And more recently, Galoisian proof in the same spirit as André was provided by Nagy and Zamuely in the context of M function. So going from Nishoka to Philippon's theorem. Once you have this lifting theorem, you can, in fact, solve the very first question. So this is what we did with the Kolempha version. So we prove the following results. So you start with the F of Z and M function and alpha and algebraic point. And you just assume that F is well-defined at alpha. So here's the main feature is that your M function is arbitrary. And the point is also arbitrary. So we do not say that alpha has to be regular, a regular point with respect to some hidden system. So just the function is well-defined at alpha. And we consider the number of it generated by the coefficient of F and alpha. So this is, I have to say that when you have an M function, so because of the equation, so the question always generate a finite extension of Q. So the field generated by the question and alpha is always the number of it. And then the conclusion is that we have the following dichotomy. So either the value of the F of alpha belong to this number of it K or F of alpha is transcendental, okay? So the function is arbitrary, the point is arbitrary, but the price to pay is that we have a dichotomy. We cannot prove that F of alpha is transcendental and this is not true in general. I already show you one example where the function vanishes, but we have this dichotomy. And already in the case, K equals Q. So when the question of the previous series are rational numbers and alpha is a rational number, so this result was conjectured by Cobham in 1968. And in that case, the conclusion is that F of alpha is either rational or transcendental. And in many cases, you are able to prove that it's not rational, so you did use that it is transcendental. And more than this, the proof is effective. So it provides an algorithm to decide this alternative. So by this, I mean, if you give me an M function, so the way you give me an M function is you give to me an equation, a value equation, with sufficiently many initial coefficient of the power series so that it is uniquely defined, and a point alpha. And if you give me this finite amount of data, the algorithm will say whether F of alpha is transcendental, and if not, it will compute the value of alpha alpha. Okay, so with this, we can say that the first fundamental question is completely solved. So we know really well how to manage the transcendence of the value of one matter function at some algebraic point. But the reason that I would like to focus on today is of a different nature. So the framework is as follow. So what we do is we choose several matter functions, so say F1, F2, and they can be associated with some operator QI that may be different. And then we choose a bunch of points, alpha one, alpha R, which are algebraic points, which are non-zero in the opinion in this. And we just ask that the function are well defined at this point. And again, we consider the field that is generated by all the question of all this function and all the points alpha. So again, I say just K is just a number field. So this is our new framework. So we have several function and several points. I recall that some complex numbers, alpha one, alpha R are say to be multiplicatively independent if there is no non-zero tuple of integers and one and R, so that the product alpha one and one and two alpha R and R is equal to one. So if you choose two, three and five, they are multiplicatively independent, but the number two, five and 10 are not, okay? And with this definition, I can state our main result. So in fact, there are two serums in ones. So we assume we are in this framework with the function fI and the point alpha I and we assume over one of these two situations. So the first situation is when the point alpha I are multiplicatively independent. And the second situation is when the parameters qI associated with the mallor operators are pairwise multiplicatively independent. And in each situation, the conclusion is that the number f1 of alpha one, f2 of alpha two up to fR of alpha R are algebraically independent over the field of algebraic numbers, unless one of them belong, at least one of them belong to the number field k, okay? So the first comment here is that in the case a equal one, so if you only have one function, so the condition I and two are both trivial. So what you obtain is exactly the previous serum. So the dichotomy either f of alpha is belong to k or f of alpha is transcendental. So the case a equal one is a previous result. Again, the main feature here and the main difficulty to prove these results come from the fact that the fIs they satisfy arbitrary mallor equations. You don't know they are not order one equation also. Solution of order one equation, but they satisfy arbitrary order equations. The points are arbitrary. The qI here are essentially arbitrary. So you can have several kinds of transformation. And a remarkable feature here also is that if you think a bit about initial k serum of Philippon, the lifting serum by Philippon. So those results say that if the function are algebraic independent, for instance, then the values are. So you have to check that some function are independent in some sense. And here there is nothing to do like that. So you get for free directly to algebraic independence of the values. The point is that the theory I described before, the initial k serum or Philippon lifting serum do not apply at all when you want to prove this serum. So and instead, so you have to develop similar results, but in a more general framework. I just would like to give few words on this framework. So the point is that instead of staying with function of one variables, what we have to do to prove this serum so is to develop a general theory. So results like initial k serum of Philippon lifting serum for formal series of several variables that satisfy some kind of matter equation but in this more general setting. And the results here is just a by-product of this more general theory. So just to give you a very rough idea of what it is, some methods in several variables, what you do is instead of looking the operator, the give to the Q, you replace this by a monomial operator. So it means that you will replace the one, the N, and the one is replaced by some monomial into the one, the N. So there is a matrix that define N by N matrix that define such transformation. So for instance, if you choose this matrix here, T is one, two, three, four. So it will map Z1 to Z1, Z2 square, and Z2 to Z1, Z3 to Z2, four. And then again, you can consider some linear systems associated related to this particular transformation and whose coefficients are given by a matrix AZ and whose coefficients are rational function in several variables. But when you try to prove say, Nishokas theorem in this framework, so there are a number of new difficulties you have to overcome. During the late, maybe 1970s and 1980s, several mathematicians have tried to prove this Nishokas theorem in such context and including Kubota, Nishoka, Lockston, and Vanderpotten. There is also a very nice contribution by David Masser. And but people were a bit stuck. And in the end, they were only able to deal with system like this, but essentially when the matrix AZ is a diagonal matrix. And then Nishokas introduced new tools that was introduced by Nester and Co and Philippon in the setting of a transcendental number theory at that time to prove in the case of one variable is theorem, Earth's theorem, so the Nishokas theorem. But this new tools do not apply well in the multivariate setting. And what we did with Collin-Farragand is that we introduced a number of new ingredients to be able to deal with general matrices, in particular a new application of the Hilbert Nusch theorem sets. So it would be interesting to say more on the way we can prove such results, but instead I would like to move to application and try to see, to give some motivation to prove our main theorem. So I recall you the main theorem and for me there are two main interests for that. So the first motivation really comes from transcendental numbers theory. So usually it's so hard to prove that some numbers are transcendental or algebraically independent that it's always good to have a framework where we are able to prove such general results. So that's the first point. And the second point is that when we start this with Collin-Farragand, we already had some application in mind and I'm going to describe you this application and I would like to criticize a bit on this matter equation because people are usually much more interested by e-functions because they are a solution of linear differential equations. They appear quite naturally in physics and many branch of mathematics, but these matter equations have also their own interest even if it's a bit different and they are more related to expansion of number in some integer basis. And this is what I'm going to explain now. We can go to this second part of the talk and that is related to this first and back conjectures and finite automata. So the starting point here is very vague heuristic that says that when you expand natural or real numbers in a multiplicatively independent basis such as two and three or two and 10 that should have no common structure. And the other comment is that it seems very, very difficult to confirm this heuristic principle in some way or another, okay? So this is a place where we have a lot of conjecture. We can formulate many problems, but they are usually really hard to solve. And of course, this heuristic now is very vague and we have to formalize it a little bit. And the very first step is to try to find out some good problems related to this heuristic to give you just an idea of what I mean by this. You can consider the following binary number. This is a two-month number and it's a number that is defined by its binary expansion. It's a real number and you just define the nth digit of this number in a following way. So you write n in base two and you count the number of ones in this binary expansion of n. And if the number of one is even, then the digits, the nth digit is zero. And if the number of one is odd, then the nth digit will be one. So like this, you define this two-month number and the point is that it has a very predictable expansion. So I guess that if I give you say 100 of the first 100 digits and if you start thinking a bit, you will find out a way to produce the other digits. So you will be able to complete the expansion. But now if you consider the same numbers and move it to base 10 and look at its base 10 expansion. So I guess that even if I give you a billion digits to look at, you will be in trouble to find out some rules to describe these digits. So everything seems to be just has to be break. And we would like to have explanation for this phenomenon. So at the end of the 1960s, Fierstein-Beg provide a series of conjectures that became quite famous. And I'm going now to, and they take place in the dynamical setting. So I'm going now to give one of them which is my favorite one. So the dynamical setting is the following one. So you consider a very simple map, but that leads to some tricky dynamics. So you consider the multiplication by q on the circle. And we are going to consider the orbit of a point under this map. Here there is a well-known dictionary between the dynamical properties, so of the orbit of x and the combinatorial properties of the base q expansion of x. So x, q here is a natural number, that is equal to two. And so the combinatorial properties of the expansion of x in base q are encoded into this orbit. So for instance, the orbit is finite if and only if the expansion of x in base q is eventually periodic. And in which case x is a rational number. We have also that the orbit is dense if and only if every block of digits occur infinitely often in the expansion of x. And the orbit is uniformly distributed with respect to the Lebesgue measure if and only if x is a normal number in base q. So every block of digits occur with uniform frequency. So we have this very nice dictionary. And the idea is that if the orbit is small in some sense, so the expansion of x in base q is quite simple. And if the orbit is large, the expansion of x in base q is quite complicated. Then first term that's suggested the following conjecture. So you choose two multiplicatively independent basis p and q and a real number that is assumed to be irrational. The conjecture says that in that case, if you consider the closure of the two orbits, so under tp and tq of x and take the house of dimension, then the sum of the two house of dimension is actually equal to one. It's worth mentioning that the two assumptions, so the fact that the bases are multiplicatively independent and the fact that x is irrational are really needed. So you really need this assumption of the way the conjecture is forced. And for instance, if x is rational number, then the orbit are finite, so the sum of the two house of dimension is just zero. And so you see if the first orbit, so imagine the first orbit is small, so the house of dimension is small, then the conjecture says that the house of dimension in the other base has to be large, so the x has to have a quite complex expansion in base q. So this conjecture beautifully expressed and in a very compact way is the fact that if the number x has a low complexity in one base and if it is irrational, it should have a high complexity in every other independent basis. Now let's see what can be said on this conjecture. So the first thing is that it is true for almost every x with respect to Lebesgue-Majure. And this is a consequence of the vehicle of ergodic serum because if you choose the map tq and that with the Lebesgue-Majure, so you get an ergodic dynamical system, so almost every orbit are uniformly distributed. And in that case, the orbit is dense and so the house of dimension is just one and it's the same almost surely in the other base. So the sum of the two house of dimension is equal to two almost surely, okay? And in fact, the most interesting part of this conjecture is when x has a simple expansion and especially when x has zero entropy, so when the orbit of x has a zero topological entropy and there is a very nice way to see this, zero entropy in base q, it just means that if you count the number of block of digits of size n you see in the expansion of x in base q, you know that this number is at most q to the n, but if it's less than an exponential, if it's sub exponential with respect to n, then you obtain zero entropy and zero entropy is equivalent to house of dimension zero for the closure of the orbit. So you see now that it's in the first base say p, so the house of dimension is zero, then the conjecture predicts that in the other base, the house of dimension is exactly one and in that case, we know that the orbit has to be dense, but if the orbit is dense, this means that in the other base, all block of digit has to occur. So in the other base, you have full entropy, okay? And this is really where the conjecture is stronger. So if you come back to the binary two most number, what you get is that it's not hard to prove and it's when now that it has zero entropy in base two, as I told you, it has a very simple expansion base two. So when you move to base 10, so the conjecture predicts that it should have full entropy. So every block of digit has to occur infinitely often and this should expand is, this would provide a very nice explanation for the phenomenon I mentioned before, okay? Boris, we have a question in the chat room by Russell Delacroix, maybe just unmute and ask away. Sure, if the microphone works, do you hear me correctly? Yes. Ah, yeah, very good. Hi, hi, Boris. So the question is whether in this conjecture, there are examples of X, where both quantities are neither zero, no, one, OPX, the house of dimension is intermediate for both bases. Oh, for both bases, I think that there are some words of Yang usual, I think, but I don't have this in my mind right now. I would say yes, not probably not explicit example, but existence of numbers with intermediate house of dimension, but I would have to check to be sure. Okay, thanks. But no explicit example, for sure. Yeah, should I go back? Come on, okay, go on. Recently, so we have this very wonderful conjecture for Stenberg and recently and independently, Pablo Schmeckin and Meng Wu, they prove quite remarkable results in this direction. So as it proves that the set of exception to first in the conjecture has house of dimension zero. So it's a little very nice result because as I told you, we already knew that the set of exception has measure zero. So it's a small set. And now we know that it's even smaller than this. It has house of dimension zero. So it's a little bit situation similar to the little wood conjecture in that approximation. And but nevertheless, so this theorem is quite remarkable, but Stenberg conjecture remains far out of reach of the current method. So you can think at first glance that we are almost proving with this Schmeckin Wu theorem the conjecture, but it's not at all the case. And even worse than this, so the most interesting part of Schmeckin Wu conjecture escaped to this theorem. So there is a result by Maudui and Morera proving that if you consider the collection of all numbers with zero entropy in some bases, in some base, so all these numbers, they form a set of house of dimension zero. So all the numbers like the two most numbers, they could all be in the set of exception. So the theorem say nothing about this particular numbers. And this is a reason why we are interested in finding problems that share the same heuristic, but maybe are more tractable than this first and very conjecture. And in our work with Collin Faber-Jean, what we did is we moved to from the dynamical settings to a computational point of view. So in that case, you are interested in the so-called computable numbers. And there are the numbers that whose expansion in some base can be computed by a general Turing machine. Most of the number you are interested in and arithmetic are computable, but among computable numbers, some are easier to compute than others. And this allowed to find a kind of hierarchy between computable numbers and to define some that are especially simple. And the notion of simple number, we are going to consider as a notion of automatic read numbers. So while computable numbers can be generated by general Turing machines, so automatic numbers have expansion that can be generated by just a finite automaton that can be seen as a very rudimentary Turing machine. And that's why they are considering as simple numbers or numbers of low complexity. More precisely, we say that a real number X is automatic in base B. If there exists a finite automaton that takes as input the expansion of n and produces as output the nth digit of X in the base B. I don't want to give a formal definition, but if we come back to the binary Turing most number, I already mentioned several times. So it's an example of an automatic number in base two. So which means that this sequence of digits here can be produced by a finite automaton. In this particular case, we only have an automaton with two states, A and B. And one of these states, state A is the initial states. That's why we have this arrow here. And such machine has the ability to read a finite string of zero and ones. And so the machine will work as follows. You start with n, the natural number n, you write n in base two. And like this, you get a finite string of zero and ones. And the machine is going to read this string starting from left to right, digit by digit. So the reading of the number n would correspond to a finite pass in the graph. Now, if you end the reading in state B, the machine will output a one. And if you end the reading of the natural number n in state A, the machine will output a zero. And we can prove that this machine exactly computes the binary two most number. In the general case, of course, an automatic numbers can be produced by such a graph, but with an arbitrary number of states. So it can be much more complicated. And but in this particular examples, we only have these two states. So once we have this notion of simple numbers, there is a very natural conjecture. It's exactly as with first and very conjecture. We start with two multiplicatively independent bases, P and Q, and one irrational number X. And the natural conjecture is that if X is simpler in base P, it cannot be simpler in base Q. So in our case, it means that if X is automatic in base P, it cannot be produced by an automaton in a base Q. Okay? And in fact, this conjecture is a very special instance of first and very conjecture because it's known that when X is automatic in base P, the entropy of X in base P is zero. So first and very conjecture predicts that the entropy of X in base Q is full, is one. And in that case, it cannot be automatic anymore. Okay? So this is a very special instance of first and very conjecture, but on the other way, this was completely open. So for instance, a special case of this conjecture was conjectured almost 20 years ago by Aloush and Shalit when on book on automatic sequences. And in fact, not a single number, ring number was known to be automatic in some base and prove not to be not automatic in another base. So even if you choose the specific example of the two most number, which is one of the most basic example, it's automatic in base two and it was not known whether it is not automatic in base 10 or not automatic in base three. Okay? Now, because I'm interested in transcendental number and algebraic independence, we can provide a stronger conjecture saying that the situation is we have two multiplicatively independent bases, but now we've picked two number X1 and X2. Each is automatic with effect to a given base and they are both irrational. So conjecture one predicts that they cannot be equal. X1 is not equal to X2. And what we expect is that in fact, X1 and X2 are algebraically independent. So of course, conjecture two implies conjecture one and conjecture two say for instance that if you start with a two most number and you can take any polynomial with algebraic coefficients in the two most numbers and then taking n's roots, you will never end with something that is automatic in base 10. Okay? And you can even go one step further by saying you can choose X1, XR, some irrational automatic numbers with respect to multiplicatively independent bases B1 and R. And then again, your conclusion, the expected conclusion is that X1, XR are algebraically independent. And of course in the case A equal two, you get, you obtain conjecture two. So we have three conjecture and conjecture three imply conjecture two that implies conjecture one. Okay? So now what is the connection with the Mahler's method and the first part of the talk? So it comes here. It was noticed already in 1968 by Coban, but there is a fundamental connection between this finite automata and so with automatic numbers and M function. So the point is that if X is an automatic number in base B, so you have this sequence of digits A0, A1, A2 that is produced by a finite automaton, then the generating series here F of Z satisfy a Mahler equation. So it's an M function. So in particular, so not that the AI here are just between zero in the set zero, one up to B minus one. So there are just the digits. So there exists an M function, F of Z, whose questions are rational. And that is when evaluated as a point one over B which is the rational point, give my automatic number X, okay? So if you remember already the dichotomy we had, so we already know that such number because the question are rational and the point is rational. So these automatic numbers are either rational or transcendental. So, and in particular, this means that if you consider a irrational algebraic number like square root of two, it's expansion in base 10 cannot be produced by a finite automaton, okay? So this question, this was a conjecture for a while and the fact that the square root of two is not generated by a finite automaton was first proved in a joint work with Jan Bugeot using the Schmid subspace theorem. And here we obtain a new proof using this matters method. And each approach has its own advantage. And so the matters, the subspace theorem approach is much more flexible somehow but it was hardly with several numbers while matters method is well suitable to prove algebraic independence of numbers. But once you have this connection, what you can say is that all problems concerning algebraic relation between automatic numbers, they can be restated and even extended as problems concerning algebraic relation between values of M function at algebraic points. And I have to say here that when X is automatic but generated by an automaton with a large number of states, then the M function you get satisfy an equation with large order. So when you want to deal with general automata, you have to understand general M function. Once we have this connection, so we are almost ready to prove the main problems I mentioned. So I would just want to finish the talk by showing you that. So if we come back to our main results of the first part, so the assumption was that we have some function F1FR which are pro series and which are M functions. And then we have points alpha one, alpha R which are algebraic in the opening disk and non zero. So that the FI are well defined at alpha. And then we consider the number of generated by all the coefficient and all the points. And the point I of the first part of the main theorem was that if we assume that the alpha one, alpha R of the point are multiplicatively independent, then we can deduce that the values are algebraically independent unless one of them belong to the number of K. So with this, you can easily deduce the three conjecture. So solve the three conjecture I mentioned. So, and we are just going to see how to put the conjecture three, which I remember you implies conjecture two and conjecture one. So you start with X1, XR, some irrational automatic numbers with respect to some multiplicatively independent basis V1, VR, that's the setting of conjecture three. So we have several automatic numbers in different basis, but the basis are independent. By the connection provided by Cobam, we know that XI is automatic in base BI, so which means that there exists some M function FI with rational coefficient so that XI is just FI at the point one over BI. So you set alpha I to be one over BI and you see that the function, the equation are rational and the point is rational so we can have K is just Q. So now because the BI are assumed to be multiplicatively independent, their inverse are also multiplicatively independent. So the number alpha I, the numbers alpha one, alpha R are multiplicatively independent and the number are assumed to be irrational. So none of them belong to the number of K because K is just Q. So the conclusion is that our number X1, XR are algebraically independent and this was precisely what conjecture three was saying. Okay, so this proof conjecture three and just conjecture two and just conjecture one. So that's the right place to stop. I just would like to add one small thing is that in the main CRM there are two points so there are also point two and point two has specific application in particular to expansion not of real numbers independent basis but to natural numbers in independent basis. For instance, if you look at the power of two the sequence of power of two in base two they have a very nice expansion, a one followed by a bunch of zero. But if you move to base 10 it's much harder to recognize if a number is a power of two in base 10 and you can use the point two of our main CRM to have results in this direction. Okay, so thank you very much for your attention.