 Hi everyone. Hi everyone and welcome to my talk. Good morning or good afternoon or good evening wherever you are. I want to start by thanking Alina, Mike and Philip for giving me the opportunity at speaking at this prestigious seminar. This talk I have already given parts of it in other places, for example, the BAMF meeting in September that Alina was one of the co-organizers and the conference in Debrates in last summer and at the couple of places in South Africa. But hopefully for some of you, some of these things will be new. So let's move on. So my talk is about recent progress on the school and problem. My co-authors are depicted here, Joel Waknin. He's a director of the Max Park Institute for Software Systems in Sarbrookin. And the colleague depicted on the right, James Worrell, he is a professor at Computer Science at Oxford. And there is them, they visited me in South Africa. In January of last year, here is them a picture with Poar Tambo in the airport. Okay, so let's go and let's say what is the school and problem. So the basic question is what is the simplest class of programs for which deciding termination or halting is not obvious. And the answer is simple linear loops. So here is what you do when you're a student and you maybe take the first time a course in computers in computer science and you learn some loops. So you have three initial values for XYZ and while X is not equal to zero, you have to reassign these values of the variables so XYZ goes into 2X plus Y, Y goes into Y plus three minus Z, Z goes into minus four Z plus six. So the general problem is while X start with some vector a and while the first component of the vector X is not equal to zero. So let's move X to M times X where M is sub matrix. Well, this example is not exactly like that. There is also some shift, but let's take this as the simplest example. So the first one is called the school and problem because it's, you need to decide when is X someone different than zero because if X sub one is equal to zero then you stop. The problem is called the positivity problem because while the first coordinate is at least zero then you can continue and if the first coordinate ends up being negative then you have to stop. And what exactly are the school and positivity problems here are the school and positivity problem. So the school and problem says start with the square matrix K by K M. And the question is, is there a positive integer N such that the top right entry of M to the power N is equal to zero and the top right entry is just something we chose we could ask for any particular IJ entry. The question is the same question is that that you are not asking if it is exactly zero but you're asking if for every end into the power and is greater equal to zero. And this boil this can be recast in terms of linear recurrent sequences. The sequence is a sequence for us the members are always going to be in Z or in Q so they're going to be integers or rational numbers, such that they are constant say 182 AK with the property that UN plus case a linear combination of the previous K terms with the coefficient say 182 respectively AK. The example is the Fibonacci numbers which starts with zero one and then every term is the sum of the previous two terms, the length of the minimal recurrence it's called the order of the sequence. And the school and problem now, and be reformulated in terms of zeros of the linear recurrent sequences you are given a linear recurrent sequence. The question is does there exist an n greater equal to zero, such that you sub n is equal to zero. And the positivity problem is, is it the case that for every n greater equal to zero you sub n is greatly equal to zero. And believe it or not, these things are still open in general. Here are some quotes from some famous colleagues. The question is that is it fate outrageous that this problem is still open. And Richard Lipton it calls it mathematical embarrassment the most prominent problem whose decidability status is currently unknown. Now a fact every nearly recurrent sequence can be finitely decomposed into finitely many where they are called non degenerate linearly current sequence. For example, you take the sequence you sub n such that you sub n is zero for all and odd, and you sub n is two for all and even. Then if you think about it, this sequence is constant every two other terms. So then you can say that the minimal recurrence is UN plus two is equal to UN. That's a linearly recurrent sequence of order two. Now what does it mean not generate these linearly recurrent sequence you can associate something which is called the characteristic polynomial, which is going to appear later on slides but what it basically is. You take this relation from here, and you write down the polynomial x to the power k minus a one x to the power k minus one and so on down minus to a sub k. You compute the roots of this polynomial. And then the general theorem says that you sub n is a linear combination with some coefficients, which might be polynomials in N of those distinct roots to the power n. So in the example that I gave you with you sub n plus two is equal to you sub and the polynomial is x square minus one. It has roots plus minus one so the answer is one to the n plus minus one to the end. Now here's where degenerate the alias comes into the picture a sequence is called degenerate if there is this the ratio of these two roots which is a root of unity. If there is a ratio of these two roots which is a root of unity like in the case of one to the n plus minus one to the end then that sequence can collapse on an entire arithmetic progression of indices. All that can happen in some sense, so you take that linear recurrence sequence, you look you say aha there are some roots of unity among these ratios you get rid of them by choosing a sufficiently large number m, such that if all these roots of unity to power m become one, and now you split your sequence along residue classes module M. So you get m of them depending on whether n is zero one or two up to m minus one module of that modules and each one of these itself is a linear recurrence sequence now. And these ones are all non degenerate except that some of them might be identically zero like in the example that I gave you with UN is odd if n is equal to it's any odd UN is zero if n is odd and UN is two if n is even. This so that the theorem is that after you get rid of the degeneracy is the scholar mother leg theorem says that the set of zeros of a non degenerate linear recurrence sequence is fine. And the decidability is somehow equivalent to being able to compute the finite set of zeros of any given non degenerate linear recurrence sequence. Fortunately all known proofs of the scholar mother leg theorem make use of the non constructive techniques so there is the classical proofs, which use pediatric analysis, which somehow pick a prime P, which is a convenient prime. In one section your sequence along some number M, such that let's say all these roots are M is a common order of all these roots, modulo p so alpha the power M is congruent one module of P, which means that alpha to the M power x is congruent one module of P and this one you can express is a periodic function by writing alpha to the MX is being the exponential of the logarithm of itself. And you're applying the logarithm to some number which periodically is equal to one so you can write this number is one plus itself minus one which periodically is equal to zero you expand it. And then you do what you do in calculus. But and then you get some analytic function and then you say okay this analytic function the pedics are compact therefore it has only finite many zeros, but you cannot extract information about how big those zeros are, and they are more modern than using the subspace theorem and so on which gives you information about how many of such zeros. Can there be work of Schmitt and Schlich via so on but again, these are known effectively tells you up a balance which we believe they are very far from the truth because they are double triple exponentials in some parameters of the sequence, which is the sequence cannot have more than this many zeros. Now the school and problem and the positivity problem arise in several areas, for example in theoretical biology population dynamics software verification dynamical systems, weighted automata in games control theory, formal power series and combinatorics and so on. So for example does our example finish. Let's go back to the example that we started with. It turns out that x is never equal to zero module of three and to see why that is like this look at this module of three. So it starts with one zero zero module of three these two numbers are zero so they disappear. The beginning goes from one to two, and then the other two numbers stay zero. So after the first step you start with one zero zero get to zero zero you do it again. You get four zero zero but four is equal to one module of three so you get back to where you were. So in some sense this is a sequence which is periodic of periods three if periods to, and we can see that the only values are one and two so it will never stop it will never be zero. So this is the Fibonacci variant of course you should forget about the fact that the Fibonacci numbers are positive. We will try to use sort of periodic information about it to decide what's what. So if you started the Fibonacci variant that starts with two and one which is the so called Lucas sequence. And this one you can detect it for example module of five it has no zero, because it's a of length four, and it's a period of length four which is 2134 so it takes all the non zero classes module of five but it doesn't take the zero class. How about the shifted Fibonacci sequence I started with zero and one so I already gave it away that there's a zero there but then I may say let me delete that zero and be a bit tricky and I start with one and one. But now if I take it math to I reduce it math to I notice that every third number is even so if any zero math three, then, then f of n is even. If I took it math three, then every fourth multiple every multiple of four, it becomes zero math three module of four every multiple of six. So many families a multiple of six is zero math for what five the period is 20. This is that this one one here and it continues all over there but inside the period of length 20, you have four zeros. So inside this period in fact you have every residue class appears exactly for time. So this means that for the Fibonacci sequence a modular argument can never work in some sense if you have some zero then that zero modularly will be there. It doesn't matter if you decide to delete some terms of your sequence and started very far away once you reduce it module or something. It is always periodic. And the theorem says that it's purely periodic, especially if the modulus is co-prime to the last coefficients of the recurrence. If the period is not co-prime module of the last coefficient of the recurrence it might not be purely periodic think about x sub n plus one equal to x sub n and you start with one. You should get you get powers of two if I reduce this module over 32. Of course it's going to be periodic is going to be equal to zero after a certain number of steps but at the beginning it isn't so it has some pre period. But if the module of m is co-prime to the last coefficient this phenomenon doesn't happen and it's periodic from the very beginning. Fibonacci sequence if you shift it doesn't contain a zero but modularly you can see that it's somehow haunted by the ghost of a zero in its past. Okay, one of the things we like to do it's you like to go back with them. So what does it mean to go back well I think of this recurrence if I want to extend it to the left of zero what do I do well I know how much is you sub zero and I know how much is you are zero and one. What do I have to put to the left of zero such that the sum of that number with zero is one well I have to put the number one and what do I have to put to the left of that one. Such that the sum of that number with the number one is zero is the number minus one and you get this. A minute of reflection will convince you that the numbers is the same numbers that you get as the Fibonacci numbers except that there's a change of sign when n is even. So minus one to the n minus one f sub minus n is equal to minus one to the exponent n minus one times f sub n. Here is the classical arithmetic progression. And this one you can always extend it in the past. Now this sequence, you can go with it in the past, but when you go with it in the past then you're going to get rational numbers you're no longer going to get into chairs. There's two to be zero reversible and this one to not to not be zero reversible meaning you can always go back into the past except that you're not going to end up with integers. You're going to end up with rational numbers, but these rational numbers are not that complicated if you think about it if you have some coefficient a sub k at the very end here. There's no time when you want to extract the next term you're basically dividing into that coefficient so those rational numbers are not that complicated they are just divisible by the primes that divide the last coefficients of your recurrence. So let's formulate the bicycle and problem I take the bilinear recurrence sequence it's over Q. Does there exist an integer n such that you sub n is equal to zero. Let's go and call the character the linear recurrence sequence we will call it simple if it's characteristic roots are simple. That is that minimal polynomial that you have has only simple roots in here the minimal polynomials are x square minus x minus one it has simple roots this is x minus one square because it's an arithmetic regression. It's a linear combination of one and end so this one is not simple and that one is simple. It's a monarchy sequence it's simple. The vast majority of them are simple but of course not all of them are, if you think of it in terms of matrices, then the simple ones correspond to diagonalizable matrices. What can we say about those zeros can we actually compute them, or let's say can we decide if it has a zero what there's various ways in which you can make that decision. The simplest decision would be to have some formula saying that all the largest zero of these linear recurrence sequences smaller than some function, which depends on the length of the sequence the magnitude of the coefficients and the magnitude of the initial terms. So there's a result from 84 and also independent from 85 by me not sure it tie the money and very shagging where they prove that linearly recurrence sequences of order at most for this column problem is decidable. And what sits inside that it's basically a count on the number of roots of maximum absolute value. So if you have a route which is a maximum absolute value and it's by itself and the contribution of that route to power and it's going to beat everything. From some point on which you can write down what it is so that is easy if you have one route which is of absolute maximum absolute value. If you have two roots which are of maximum absolute value this means that they are a root that it's complex conjugate and the ratio of them is not a unity because they are not degenerate. So if you have two competing forces, the root alpha to the power n and the root alpha bar to the power and maybe with some coefficients which might even be polynomial in it. Well it turns out that this can be a little bit smaller than the absolute value of alpha to the end but thanks to the Baker method this can be much smaller you're only losing let's say a polynomial factor in N. So this contribution from these two maximum roots, it's at least the absolute value of this term maybe divided by some power of n, but you're keeping the exponential character growth character of it and because of that you can also solve it. Now you can even do it for three roots but that is characteristic to the fact that the sequence has integer terms, because if you have exactly three roots of maximum absolute value then they are a number it's conjugate and the absolute numbers. So if you write some linear expression in those ones to the power n and you factor out the absolute value of that number to the power n, then you're basically left with some trinomial in the number let's call it T, which is the trinomial it's the number T is going to be e to the i n theta where theta is let's say the argument of alpha so we have some coefficient times e to the i n theta, plus some other coefficient times the conjugate of that which is its reciprocal, plus the third coefficient and that's a quadratic polynomial into which you can factor and when you factor it becomes a product of two binary things T minus something where T is e to the i n theta. So there you can do Baker's method again so because of that they win and actually they win even if you have the same argument shows that if you have at most three roots, which somehow are dominant in some pediatric value valuation then you win, you either use, if they are dominant as complex then you use some infinite valuation and if not you use some some finite valuation and this is the critical ingredient as I explained is Baker's theory. So for linearly recurrent sequence of order at most for the bias column is decided. And we have some contribution last year with his colleagues with Richard Lipton juries new felon is a PhD student of Israel, Israel David person was opposed over that time and start broken. We prove that if the sequence is reversible of order at most seven this column problem is decided, but that's a little gallery primary problem because basically we we used this what we know that it can have let's say, if you have at most three dominant roots and you're fine. But having the notion of reversibility this means that these roots are actually units because the polynomial that you're looking at has the last coefficients plus minus one. And then you can ask how ugly can that polynomial be so that all its roots are units and at least three of them are at least four of them have the same absolute value and you prove that polynomials of degree at most seven cannot have that proper. But there is examples of polynomial as a degree eight that have that problem. This is why our argument breaks down there. And there is an older result of about 10 years ago about positivity, which is decidable for generally current sequences up to five of Israel and Ben. And then if you further impose simplicity, then it's decidable up to nine. Okay, so now since we can't do better. It's time to throw in some conjectures into the pot. So many problems in mathematics and computer science are so subject to various standard conjectures. For example, Riemann's hypothesis, or the belief that factoring is not in polynomial time that depends the security of RSA. The decidability of the first order theory of real arithmetic with the explanation is subject to channels conjecture. So since I mentioned channel conjecture. Let's recall what the channel conjecture is. So channel conjecture the following take n complex numbers which are linearly independent of the queue. And look at the field obtained by adjoining these two n numbers to queue, which are the numbers you start with, and their exponentials. So the theorem that conjecture is that the degree of transcendence of this field is at least and equivalently if you have n complex numbers which are linearly independent over q. And then within this set of two n numbers one can find at least n of them, which are algebraically independent over q. And you can see why the condition of linearly independent over q is because if, for example, one number is the sum of the other two that the exponential of that number is the product of the expression of the other two. So you need some linearly independent condition. Some examples we know that is transcendental we know that pie is transcendental what about E plus pi and E pi. Of course they can't be both algebraic because if they are both algebraic then E plus pi and E pi algebraic so this polynomial, which has E and pi as roots is a polynomial with algebraic coefficients. So this makes one of them transcendental, but which one, probably both of them are transcendental, but we don't know that what about other more complicated examples like this one. So let's apply the channel conjecture with the two numbers one and I pi. The channel conjecture says that within this set one I pi e to the one e to the I pi, there should be two numbers which are algebraically independent, but e to the I pi is negative one. So the only two numbers that can be algebraically independent are those two I pi and E and E is algebraic and I is algebraic so pi and E should be algebraically independent. For any non zero polynomial with rational algebraic coefficients we should have that P of the pair E pi cannot be zero. In particular those examples must all be irrational assuming this conjecture. So in particular channels conjecture implies that there are no algebraic relationships, you can write something like that but that's not an algebraic relation. Okay, what does this have to do with our linear recurrence sequence. So let's start with the linear recurrence sequence let's reverse it. Now we are assuming in general we will assume shortly that it's simple but for now we no longer assume that a sub case plus minus one so that it is reversible but it can be any number at once. But now I take a modulus m which is co prime to a sub k. Since it's co prime to a sub k and thinks that this is a sequence module and it behaves like the interest. So for example this sequence if I reduce it mod three becomes that sequence and if I reduce it mod five it becomes that sequence so the prime three doesn't care that I'm dividing by things. Because for the prime three a sub k, I can divide by it because it's inverse it's also well defined module. There is a fairly wide ranging conjectures formulated in 1937, which is known as the exponential local global principle. And like channels conjecture is wildly believed by number theories but only proven in special case. So what this column conjecture is considered this recurrence relation, and suppose that the sequence is simple there's some reason for that and I'll give you a counter example as to why this conjecture is false for non simple sequences. The conjecture says that the sequence has no zeros. If and only for some integer m which is co prime to K, we have that you sub n is not zero model. So in some sense whether this conjecture says the conjecture says that if the sequence has no zero, then there must be a modules that notices that that witnesses the fact that there is no zero. So, why is it false for non simple sequences well I can give you the one of the easiest example is a sequence of the form two to the power n times to n plus one. This sequence doesn't have any zero even if you go to negative numbers it's still not going to have any zero where it has a rational zero but let's not count that one. Which is at minus one half so I have two to the n plus one, two to the power n multiplied with two and plus one does this have a zero module every m which is my love every m. And I say of course it does because when you look at two to the n times two and plus one. You can put the power of two you're already winning it for free for large and so you can look only at the odd part of M, but then you solve to n plus one is equal to zero module of the odd part of M and you get a solution. So you get infinity main solution so that will have zeros much no any m. Now we proved that the school in problem for linear recurrence sequences of order five is decidable assuming just the school and trajectory. So that is one small step as we said before we could do it up to four now we can do it with five. But which whether they're simple or not. There is a very important theorem in the section. There is an algorithm which takes as inputs, a simple non degenerate linear recurrence sequence and produce it is finite set of zeros. And the termination is guaranteed assuming the school and conjecture and the periodic channel conjecture. And the way this argument works is the following. So there's two problems there problem number one decide if the sequence has a sick as a zero. And number two find all the zeros of that sequence. Well if you only assume the school and conjecture then you can already solve problem number one. And here is how you do you start to algorithms. Two programs one of them is looking for zeros and one of them is looking for witnesses that there are no zeros. So one of them goes through all the modules and go crime to a K and calculates the values of you sub and module and which you can do in finitely many steps because the sequence is periodic. And if the school and conjecture is true then one of the two algorithms will stop either there is a zero and the first algorithm will find it either there's no zero and the second will find a witness that there's no zero. So so far, we can construct just out of the school and conjecture we can find whether there is a zero or not using these two things. Now, what about finding all the zeros. So as I said, well, the important thing to mention here is that we actually implemented this algorithm and this algorithm is only needed to prove termination not correctness. So the algorithm, if it works is going to produce a correctness certificate which is independent of all the conjectures the only thing that is needed for the only way that the only part of the proof that needs those conjectures is to show that this algorithm actually terminates. And it's implemented in this online tool which is called school so if you want to get school and you can you can go here. Or it's accessible from the web page of my colleague Joel working in you go to his web page there is some software that he has he and his team have developed and this is one of them. So what does the school and problem do you're going to be faced with this. You're going to have to promise that the sequence that you're given is simple that you're not trying to trick it by giving it polynomials with multiple roots and so on. You give those values and those the initial values and the coefficients, and it will produce some lists, and it will say for example here are the zeros. This particular example that at zero one and four, and then there's a list, and the list says, these are the only zeros as it can be proved by this sequence of congruences. So, all the other ends which are not one of those they are going to be trapped in some congruence. There is some congruence class module some modules, such which is congruent to that number in which is not a zero for which when is not zero module that module. The key technical tool. The key technical tool is the following started the non degenerate linearly current sequence. Do the algorithm that I said that you do you have these two algorithms one of them which looks for zero one of which looks for witnesses. If you find a witness that it's not zero then you're done. And it's over there's no zeros. So let's suppose the first algorithm finds a zero then what do you do next. Well what you do next you can shift that zero into the origin. Now you have to go further. So how can you go further here is the important lemma assuming the periodic channel conjecture one can compute an integer m, such that if you look at this sequence, only over the multiples of that integer m. Then that sequence has no other zeros. This is the key technical lemma, meaning it has a zero where little n is equal to zero but for little n greater equal to one it has no zero. If you have that you have that then you're basically done. Because why is it then you're done well you compute this number capital M this is what actually the program does it compute this number capital M. Now once you says find this number. It splits and into residue classes module capital M, and says let's go through the residue class, let's say our module capital M well if R is equal to zero, this is exactly the sequence. And the lemma says there is no further zeros there don't bother going that way you should just look at the other M minus one sequences. So now you re-intitiate your problem with every one of these M minus one sequences and you start either looking for zero witnesses and so on and the fact that there's only 590 many zeros will guarantee that this procedure terminates. So where do we assume the periodic channel conjecture we have to assume something about so. As I said you take, as I mentioned in the older proofs of the Schole-Maler-Lech you take some suitable prime P, you associate some analytic function, and you have to look at the zeros of this analytic function. And so you have to look at let's say some of the periodic zero you have a this is you're in zero you have a periodic zero there. And then you have to look at the expansion around the zero, but when you look at the exponential of the logarithm of something times x and you expand that in series. Then of course the first term is zero but what's the next term the next term is basically the first derivative. And the first derivative of the exponential of the log is the log so that the first derivative is a linear combination those books and the second derivative is a quadratic polynomial which is homogeneous. Which is homogeneous in those logs. And the third derivative is a cubic polynomial which is homogeneous in those logs and so on. So, maybe if you're in the first case you can get away by periodic Baker's method and so on. Oh, maybe if it's not zero then I can even bounded and but as you go to quadratic relations among the logs cubic relations among the logs then you are stuck. And so on conditional results and that's where we are assuming the periodic channel conjecture, meaning that if there is no obvious relation between those logs. Then that coefficient is non zero and then you can apply stress months there were more something like that. The resulting sequence is guaranteed not to contain any zeros and then independent correctness certificate can be produced. And this periodic channel conjecture is only needed to ensure termination namely being able to calculate this number capital F. And here is some little example of something that has zeros and then you keep on doing this periodically pro game so you draw circles and circles larger and larger. That's something very complicated at the end. All right. Okay, so let me go over this one. Okay, so now I have another maybe. What is it 20 minutes 15 minutes. Let me talk about the different topics so so far. So my presentation is called progress on the school and problems we made some progress on the school and problem assuming two conjectures. And we basically solved it for simple linear recurrence sequences assuming these two conjectures. Now we said let's initiate a different program. So the program that gave us some conditional results which was based on the Baker method and so on was based on the enumeration of those roots of dominant absolute value so if you don't have too many of those roots then somehow you can have a lot of cancellation. Of course thanks to the subspace there and you know that you don't have enough, you don't have too much calculation for large and so this independently of how many dominant roots you have. You still don't have a lot of cancellation for large and but you don't know what large and means. So this. Tell what we initiated we initiated a different thing instead of saying that the sequence is bad and the integer is good we said that this all the sequences are good and the integer some of them are bad and some of them are good. So the good integers are the integers for which you can test any linearly recurrent sequence whether it has a zero in that integer. And the bad integers are the integers for which we don't know how to do that. So let's say that s is a universal square M set if there is an effective procedure such that given a linear recurrence sequence you outputs whether or not there is an integer any of this set such that you sub n is equal to zero. And we stole this definition from the notion of a universal Hilbert said. What is the universal Hilbert sets so Hilbert's theorem says the following let's start with the polynomial which is a polynomial with rational co efficiency, which is irreducible as a polynomial in two variable for which is not linear or constant in X so excess degree at least two. And then you can ask okay fine let's start evaluating this polynomial in some integers make y equal to some integer then I get some polynomial of one variable. One variable a degree at least two sometimes is irreducible sometimes is not and what can you say about the irreducibility of it. So Hilbert's irreducibility theorem asserts that the set of integers for which P of capital X and n is reducible is of density zero. So most integers that you're going to stick into here, the polynomial as one variable of X is going to keep the property of being irreducible. So that's the difference of measure or density so the number of those ends of absolute value at most T. If I divide that number by T and I make T go to infinity, then this number this limit is zero. And SD Cohen proved that, in fact, this set is. T to the one half multiplied by log T so it's really thin. This, this little off team applied by this limit it's not like the primes but it's more like the square roots. But they are polynomials for which you do get the square roots. For example, x square minus y, if you evaluate why to be a perfect square then this polynomial of course factors and up to T they are about square root of T perfect if why isn't an integer which is not a perfect square then it does affect. So motivated by such a result the universal Hilbert set is an infinite set of engines with the property that it intersects this bad set transversely in only finitely many points for every irreducible polynomials p of x y with values with rational coefficients to prove gave with explicit example, it showed that the set of integers of the form m cube plus this little thing the integer part of double log of the absolute value of m this is a universal Hilbert set. So whenever you take any polynomial of the great list to in X, which is irreducible and you make why to be equal to a member of this guy. Then you get the diophantine equation. And this diophantine equation has only finite many solutions as a diophantine equation in the variable X and integer m. And philosophy and Wilcox constructed a dense universal Hilbert set. So the first one that we constructed was in 2021. So here is this range construction take the following function which is defined on the set of natural numbers except for zero, which takes n and it sends it to something a little bit smaller than log n which should be an integer. So the little bit smaller we chose it to be square root of log n but you can choose log n to two thirds or the one minus f silent and define the following sequence. The zero term is one and the nth term is the factorial event, plus the value in the seat of the sequence in the index f of n, where f of n is given by this. So this definition is somehow recursively calling itself, but since f of n is strictly smaller than and by the time you get to this step, everything else before that has already been defined so. So you can I even list this but look how slowly this grow so basically what S of n is this some huge factorial plus some tiny factorial about like the factorial of the square root of log of n integer part plus something which is the factorial of the square root of log of the previous one so it's basically like a double log and so on so the nth term is basically a sum of a few factorials but each one of them goes down very quickly to zero. And I claim that this one is the universal column set well you see when I say that something is universal column said it's not enough for me to show you the set. I also have to take to explain to you what am I doing how can I make sure how can I compute the, the set of zeros of any linearly current sequence in this sequence because otherwise why would you believe me that this is a universal column set. So here is the important lemma suppose that UN is given by this minimal recurrence, I take delta to be the discriminant of this characteristic polynomial. Actually this delta might be zero it's better to take delta to be the maybe the discriminant of the decomposition field of this polynomial or something like that. So the proposition is the following take this linearly recurrent sequence, take a prime, which is small enough, but it can be any prime other than finitely many bad prime so the bad primes are the primes which divide this coefficient, meaning somewhere in some extension there might be some prime idea that divides the roots, and then divide the discriminant of the field or at the polynomial and so on. So suppose that the prime is not so large it's up to m to the one over D, then you sub n plus m factorial looks like you sub n marginal p. And the proof of this it's very easy it's basically Fermat's theorem in algebraic number field. So what you do you look at let's say alpha is a characteristic root of that sequence. In that sequence it's a linear combination of alpha to this inputs and so I look at alpha to the power n plus m factorial. So this is very small alpha let's say it's invertible much low P this means that it's going to have some order. And the order of it divides something of the form P to the F minus one where F is it say the inertia of some prime ideal sitting over P. But the point is that P to the F minus one is at most P to the D minus one which is a part of m factorial. And then the factorial disappears much low P and it becomes so alpha to the n plus m factorial becomes just alpha to the end. So in particular this congruences through so this is you just write down the characteristic equation and you prove it. So in particular here is why this universal school M said suppose that USF of S of n equal to zero and write down the formula of S of n it's n factorial plus S which is much smaller. So you multiply this previous lemma in here. This previous lemma in here it tells you basically that you can get rid of this factorial and that USS of F of n is going to be zero much low P for a bunch of primes what does it mean a bunch of primes, all but finitely many of them, and all of them to the one over D and you multiply these primes together up to n to the one over D and by the prime number theorem, you get the exponential of log of n, the exponential of n to the one over D. And the exponential of n to the one over D, it's the double exponential of log of n divided by D. But how big is this guy, this guy is basically, let's say the dominant root in S of F of n and S of F of n is like the factorial of F of n. And the factorial of F of n, so it's the exponential, we have one exponential because of the root you get another exponential because of the factorial, but it's basically the double exponential of F of n. And this is why we chose F of n to be here, because this function is much smaller than the exponential of double logarithm of n to the power D. And now you use this argument that says that if an integer divides another integer which is of a smaller size, then the other integer has to be the integer zero. So this means that this being zero forces this being zero. And now it's perfect. Now you have a little machine. If it's zero, then the next one is zero and the next one is there and you keep on going down. But you see now you get too many zeros of this linearly recurrent sequence we have very concrete bounds on how many of them they can be. For example, there's a result of Schlegel by that of Schmitt that says that the number of zeros of a linearly recurrent sequence of order case that most triple expansion in case of most X of X of K of something X of X of X of three K low K, and he doesn't even assume that this sequence has integer coefficients it could be any complex numbers for initial values and recurrence and so on. So, actually this was a cute six page paper it even got some distinguished paper where that the conference in logic in computer science. So this is our set well the set is not very thick. Since the S of n is up to X and it's a bit like factorial that is very, very low. So can we do better. And then we say that yes we can do better. And this is one of my last slides. This is a very complicated slide but I will try to explain the content of it. So basically looks up K of X is going to be the iterated logarithm of X, so this is three fold iterated logarithm of X. Now for X efficiently large. Choose two sets of numbers, a of X and B of X, a of X is up to about square root of log X. And just to make things easy consider primes in here, and you chop off the beginning of a few of them because of some technical reasons. So basically you can you consider a dyadic interval, which starts around something that's little o of log of X. That's like, let's say log of X over the square root of triple log X now with this two statistics take a number between X and two X, and try to represent it in the following way. And then look for representations of the form Q capital P plus a, where Q is a prime in here is an integer in there, and the result capital P is whatever comes out of this equation. Now you might say how do you even do that. Well, it's easy. Let me explain to you how to do that you fix the number and, and you fix a prime Q in here. And then Q, what's left are a and P. Now for this equation to make sense in integers, a has to be known modulo q because q is already fixed and then is fixed so a has to be known modulo q. So you go into this interval and you say, Okay, let me pick up numbers in here, I can pick up any number I want. It just has to be correct module A, but see a goes up the square root of log X. So this interval has size around log X, you lose a little bit so this interval has size roughly log X divided by a time this fudgy. So you certainly have many candidates, all of those all those candidates when you compute the expression and minus a over Q you're going to get integers. And every now and then these integers are primes. And enough of them are primes that said this quadruple log of X or something like that and you have some other technical conditions. Then you say the integer n is good, and you put it in your set, the set of good integers. I claim that the set of good integers is a universal square m set. And why is that. Well, let's go back to that expression with, let's say linear combination of powers of alpha to the n where alpha is some number. So we try to use the same argument as we used before the visibility with some prime p. So the divisibility some prime p is going to be this prime. So I look at alpha to the power n, and I write n as being little q times p plus a well alpha to the power p let's assume that we are in the fortunate case like Fermat's little theory. So alpha to the power p becomes one much low piece so it goes away. So alpha to the power and becomes alpha to the q plus a. So asking a view sub n is equal to zero basically is asking for you in the index q plus a to be equal to zero. So look at q plus a q plus a is little all of log x, and the prime p is around x, so this is the exponential of the log of x. This makes the prime p bigger than you sub q plus a, and you sub q plus a still has to be a multiple of p therefore is equal to zero. It's equal to zero many times because this has many representation, but we have a formula we say that the number of zeros cannot be too big. This means that the number of zero which is at least quadruple log x is bounded by let's say the triple exponential in k. So we get that x or n is bounded by a seven fold exponential in k. And this is true inside the set, of course there is more things to do because maybe alpha to the p is not necessarily equal to alpha might be some conjugate of alpha. So, so there's more things going into here. And why is this a positive density. This is a positive density because you can use Koshy Schwartz. You can use Koshy Schwartz and you compute the sum of the number of representations of any in this way, and you get some obvious up about. And then you compute the second moment, which is the sum of RM square and then it's like doing twin primes you're counting q p plus a equal q prime p prime plus a prime. And this is an upper bound sieve which is not very hard, and you do Koshy Schwartz and you get a positive proportion of them. And this positive proportion is the proportion one, and because basically the bag of numbers that we are associated to it grows a little bit larger than the logarithm of x to every end, which means that we should get enough primes to do that. Okay, if you go through some, for example we contacted young Hendrik ever say and he pointed out some interesting references, and using that we said well in our set which is a positive lower density, all the solutions of all the zeros of this sequence satisfy this inequality, except three is the triple fold explanation in a except five is the quintuple explanation in those parameters where this is if you want the complexity of the sequence how big it is the start how long it is, and how large the coefficients and initial values are. And, and this is what I wanted to show you so I wanted to show something about the school and problem, the positivity problem is even less is known. And that's it I will stop here.