 Yeah well it's nice to be back after two years and it's amazing what has happened with this seminar. It's extraordinary. I think we all greatly appreciate how it's kept us together over this difficult period and some people are going back into the world. I'm not personally but yeah so I'm looking forward to more number 30 web seminar. So today's talk, here we go for technology, there we go, is dedicated to the memory of three three gentlemen who this is the first Canadian number theory association meeting in 1989, a picture courtesy of Hugh Williams and the gentleman left is Richard Guy, John Salfridge in the middle and Alf van de Porten. Guy in van de Porten will feature in this talk. Maybe I should mention you can see that it was a dancing time back in 1989, the first C&T. You can also see the propensity in Canada for facial hair growth. I think the cold weather brings a lot of facial hair and you see Richard Guy has tamed his to go up his face which is rather interesting. Anyway besides that so Richard Guy for those who didn't know him you may know his book Unsolved Problems in Number Theory. He was the great problem master of the second half of 20th century in number theory and the dates there, 1916 to 2020 are not wrong. He died at the age of 104, fairly recently and Carl Pomerance and I had the opportunity to write a obituary for him in the notes says and I took that as an opportunity to try and read through a lot of his papers. I mean I sort of knew him best obviously as the problem man and somebody who liked to party at conferences but yeah it was it was really interesting reading through a lot of his papers and seeing his fascination and he was somebody who really loved certain problems. He was just fascinated by it wasn't necessarily improving all his conjectures. In fact he really focused on conjectures he didn't understand what to do or didn't even necessarily know what he should conjecture which is a wonderful thing to do to really just want to understand the difficult problem and I got kind of hooked on one of these problems which I'm calling linear division sequences which he has about four or five papers on. He also has beautiful papers with Andrew Brenner on Tying Squares of Triangles and Aliquot sequences that's better known in the most effective number theory that work. Anyway but we're going to talk today about recurrent sequences so we'll start with the grandfather of them all the Fibonacci numbers which I hope you know well of course they're linear recurrent sequences because they satisfy linear recurrence equation and the property that's important for us today is they also satisfy that the nth element of the sequence divides the nth element of the sequence whenever m divides in. If you ever played Fibonacci numbers you'd probably come across this perhaps notice it yourself and a sequence that does this is called the division sequence I mean you put them together that's what I'm calling a linear division sequence or linear divisibility sequence and these have been explored somewhat in the literature but let me before getting into the history talk a little bit more about some examples. So our big question will be can we determine all linear division sequences can we classify them all? So this was one of Richard's favorite questions and I also want to note the other property that's great about them is that you have this GCD of f m and f n it's not individual by the f of the GCD of m and n but it's actually equal which is kind of cool and what I'm not going to talk about today are elliptic divisibility sequences so if you look at a point p on a elliptic curve defined over q and you look at 2p, 3p, 4p and and your recurrent sequence if you like is going to be the set of denominators the z n's then they actually have this property that z m divides z n whenever m divides n and you if you know your elliptic curve theory you can sort of see that because if after m iterations you're at the point of infinity over fp then you will be at 2m iterations 3m iterations etc. So it's not hard to see that the z coordinates of the of the multiples of ration points and elliptic curve are a division sequence but the the process is not linear so it's not going to fall into the rubric of what I'm doing but this great work this is something that was spotted by Morgan Ward back in the 1940s and there's you know this beautiful work by Silverman, Poon and Einstein, Everest, Ward and Kate Stange recently that's not what I'm talking about there. So let me give some more examples of linear division sequences so the mesenn numbers it's pretty obvious that 2 to the n minus 1 divides 2 to the n minus 1 whenever m divides n maybe if you're not a big person on recurrent sequences it's not so obvious it is a recurrent sequence but there you can see the recurrence relation and what you notice there's an underlying polynomial x squared minus 3x plus 2 and the roots of that are the 2 and the 1. So in 1878 as I mentioned in the abstract the very first journal in the United States of America in mathematics was the American Journal of Mathematics and the first edition was 1875 to 1878 and Lucca in French no less wrote 80 odd pages on recurrent sequences and did a lot of things actually really interesting things and yeah so one of the things he noted was the sort of general formula alpha n minus beta n over alpha minus beta so you see the 10 numbers are alpha equals 2 beta equals 1 and you get the mesenn numbers and you I mean what Lucca notes is you can always take conjugate quadratics and so for instance you get the Fibonacci numbers or I've got this wrong that should be plus root 5 it's obviously it's not imaginary but anyway you get the Fibonacci numbers by plugging in there. So why do Lucca sequences always give a division sequence well one way to view it is you're taking the polynomial x dn minus y to the n over x minus 1 I'm writing it like this and then just using that just like we did for mesenn numbers that x dm minus y to dm always divides x dn minus y to dm. So that's the simple proof and I'll notice it's also actually always gives a recurrent sequence because if you take this x and y you can again you can see these the kind of roots of a quadratic equation that forms a quadratic equation x squared minus x plus y times capital x plus x y and so you know if any given alpha and beta that are that are conjugate you're going to get a nice thing over the integers here but in general you can think of as also as a polynomial linear division sequence where you're substituting in values of alpha and beta. So some less interesting examples of linear recurrence sequences would just be a constant times a to the n and that there's the linear recurrence of just one term. One that that is not if until you see it obvious is that n squared for instance satisfies this recurrence. If you go to like Tim Gower's lectures on additive combinatorics he's often using identities like this we're looking at arithmetic progressions and squares and for higher degree and in fact you know these these coefficients one three three one you should recognize obviously is binomial coefficients and in fact any degree d polynomial satisfies the binomial coefficient recurrence of degree d plus one. So what's kind of a bit surprising first site is that you can take any two linear division sequences and you can multiply them together and get a new one. So for instance if I take two to the n minus one f and well let me just point out multiplying two division sequences is kind of obvious right if a of m divides a of n and b of m divides b of n then a m b n divides a n b n so it's obvious it's a division sequence it's less obvious it's a linear recurrence but here's the linear recurrence the the smallest one satisfies at least the order one and you can see this comes out from multiplying the polynomial corresponding to let's talk about these polynomials in a second most of you probably know this theory and the polynomial corresponding to the Fibonacci numbers and getting the corresponding polynomial for the product. So let me just give a general proof of linear product linear recurrence is always linear recurrence. So linear recurrence you start with what u zero to uk minus one and then you you just iteratively define things so the Fibonacci's numbers are fn is fn minus one plus fn minus two fk is two and from this we there's a corresponding polynomial so you just take the coefficient so you've got this one and then you bring these c1 to ck on the other side we'll assume ck is non zero otherwise this term wouldn't be there anyway and then you factor this as a polynomial and then there's a remarkable formula it's easy to prove by induction it's not it's a triviality to prove but it's it's when you first see it it's pretty cool that the nth term in sequence is given by you take these alpha i's to the ends and then you have polynomials in n and perhaps the one thing worth saying is once you have these alpha i's knowing these k coefficients of these polynomials is equivalent to knowing the first k terms here you can deduce them from each other so there's a very simple very general theory of how to build linear recurrence sequences working of characteristics polynomials so if i say the word order it means as small as such k um you can define this in the ring of in z but i could also be in the ring of integers of a number field i could be a polynomials i could be a polynomials in many variables um i'm not i'm going to try and avoid as much as i can the gawa theory that really it's living behind what i'm going to talk about today but just to be clear um although i'm looking at sixes over the integers i'm going to want to factor this polynomial and use the root so i'm in the number field um there it's obviously gawa because it's a splitting field extension and um i'll assume it has ring of integers a okay so let's have a look at multiplying linear recurrence sequences so what we said is if we have a linear recurrence sequence you get some representation like that and that gives you all the terms now if i multiply these two together it's pretty obvious i'm going to get another thing like that right the two polynomials are multiplied together and i'll get alpha i times beta j to the n so it's obvious the product is also linear recurrence so what we did with two to the n minus one times f of n um but um we also as i noted a minute ago um that if you have uh two numbers two sequences that are division sequences and their product is also division sequence so it's clear that if you take two linear division sequences so linear recurrence plus division sequence multiplying together you get another one um but here's the first remarkable theorem i wanted to talk about which is um the hadamard quotient theorem and that says something rather surprising that rather multiply together suppose you have a of it so always i'm going to assume a of n and b of n are each linear recurrence sequences and i've got a of n and b of n and let's suppose the quotient is always an integer okay then the corollary then it's extraordinary that you can show that this actually is a linear recurrence um this this quotient i'm not sure why it's called the hadamard quotient theorem maybe it was asked by a hadamard i'm not sure it was first proved um by van de porten in 1982 and and you know i'm a big fan of alphand de porten and he brought a lot of new ideas into the subject but there's some very tricky little details in the subject this proof was had some issues and he sorted out roomily a few years later um and it's a hard paper i mean it's hard to do so um we'll talk a little bit about some of the issues that that van de porten had trouble with later um i also had trouble with them so quite familiar with them um and uh this was vastly improved this theorem by kovay and zanie in 2002 who showed that actually you don't need this always to be true you just need it for infinitely many n and then at the very least you've got that this is a linear recurrence sequence on some arithmetic progression some sub progression and that's best possible so this is a remarkable i mean they actually van de porten had conjectured that sometimes this should be true but it's a long way from conjecture to proof and this is an amazing step like kovay and zanie oops i just mentioned that it's irrelevant to me today another great theorem from zanie on linear recurrence these are probably the best you know the best theorems on linear recurrence sequences in the last my career i think um so zanie proved this wonderful conjecture that if you've got a linear recurrence sequence in every term in it is a d-th power then actually you've got some sort of d-th root so there's some cn so that an cn to the d and that cn is a linear recurrence sequence um yeah so that's kind of remarkable result so um let me tell you about kovay and zanie's main new idea and actually all of their work and zanie's work and others come into this idea and i perhaps should mention that um well let me let me say it in a minute so the let me explain this so we're going to take a finely generated multiplicative subgroup of the algebraic closure of the rational so maybe i'll take all multiples of two three five and seven okay so in other words i can have two to the hundred three to the minus 27 five to the 19 seven to the 11 and a similar number for you and for v and what i'm interested in so i put n for norm here let's just think about in the integers suppose i've got some big product of primes minus one here and where the primes is from some fixed sets that's what i mean by finitely generated similarly with v then the question is can the gcd of u minus one and v minus one be big and um the answer is no it can't be very big so here i just mean the maximum of u and v to the epsilon so um these really can't get big except there is obviously a counter example to that suppose that v was u squared then we see u minus one and u squared minus one had a big common factor so the only counter example is if there's some sort of multiplicative relation between u and v so yeah so that's the theorem um it was inspired by there earlier work work of a lot of people kind of got involved in this area at around that time bujo i think was the first paper of kvayan zanier um the joint paper silverman had ideas in this direction uh hanandes and florian luca and there's i'm probably missing some it was quite the rage but this is kind of the some ways the ultimate form you can generalize this in some ways but it's a corollary of this um so let me leave that up and oh so again i've got n so i actually haven't been very careful because i said the gcd but i had denominators in you right i said you could have three to the minus 100 so what i mean is you take the gcd of the numerators of these two and that's the same in an algebraic number field but that's you've got a little bit careful in algebraic number field because you know a rational in an algebraic number field doesn't have a well-defined numerator right i mean you can't you can't if you've got if i give you a fraction algebraic number field that's not reduced that there's a non-trivial gcd of numeration denominator the gcd might be a non-principal ideal so i can't divide out by it so you gotta be it's it's a tricky business the idea of reduced fractions in in algebraic number fields so here we've got to just look at the prime ideal divisants of the numerator and then you can have a well-defined notion of norm okay so let me give you the corollary that's relevant to us that if you have two multiplicatively independent algebraic numbers u and v then okay so my gamma now is just going to be well generated by u and v and i look at utn minus one and v to the n minus one and then this gcd of the numerators thinking of it as the gcd involving the prime ideals that divide both that is less than e to the epsilon n so it can't grow exponentially fast with n so that's the wonderful theorem um and what we saw a minute ago when i described lucar sequences is is that um i referred back to uh doing them in terms of polynomials so um that's going to be a theme here so in fact there was a an analogous result in with polynomials a little bit stronger because oh so you can work with polynomials a little bit more easily so there's some uniformity in here in elon rudnick and then our host um alina astaffi she proved a similar result um with general polynomial ring which i also need in this um with more uniformity in other words uh i think elon rudnick they the polynomials involved they this constant ball features of the polynomial beyond just the degree and stuff like very simple things and alina's result just involves the degree essentially it's a tiny bit more complicated but it's very uniform okay so um what i want to look at is um this observation that if i have two division sequences then the gcd is also a division sequence um i mean if a m divides a n and b m divides b n then if you think about the gcd then it obviously divides the gcd of a and b and similarly the lcm so the question then is are they also linear recurrent sequences and what's kind of cool is if one is then the other is because remember we proved that a and if a and b and about division sequences then the product is now if this guy's are a current sequence uh then it divides that and the hadamon quotient third and says if i divide a and b and buy that gcd i get a new um linear recurrent sequence so that's kind of cool and you have to work feature one of them to work it out um so i'll just work for gcd and i'm going to use an example and this will give me an opportunity to apply covaya danier so with two to the n well we know about two to the n minus one in the utn minus three to the n minus one and fn involves now i've got them right the two roots one plus root five or two one minus root five over two and you know the numerator of this is like the fibonacci number um and here it is the messen number and covaya danier says since these things are are independent multiplicatively which not hard to prove then this cannot grow exponentially this gcd now let's just remind ourselves any linear recurrent sequence has this sort of formulation and then just by some sort of mean square argument it's easy to show that the size of un must grow at least as fast as the maximum of the alpha rise to the n not always but and that you mean you could have cancellation occasionally but for infinitely many n it has to grow that fast so in particular if some alpha i was greater than one then the un's would have exponential growth but that contradicts what we get from covaya danier so we know none of the alpha rise can be greater than one so each of them is less than or equal to one and then um because we're working on the integers we know we have for each alpha all its conjugates have absolute value less than or equal to one and there's a famous term of chronicle that says that means that they all have absolute value one so um well we're calling un slow growing and what you see from this finger if all the alpha rise of absolute value one then the main growth comes out of the polynomials so we certainly have that un doesn't grow faster than n to some power d but you can just play with this so i mean if p is a quadratic residue mod five then we know that p divides f p minus one obviously so that's sort of thermo-as-little theorem for fibonacci numbers and we know that p divides two to the p minus one minus one so p divides this gcd whenever p is plus or minus one mod five and from that i can construct and here's a construction and an where un actually grows faster than than polynomially um so this is where i take those p minus ones where all the p's are plus or minus one mod five i take their lcm and um all their prime factors are less than root x so if you know the proof we did 30 years ago of pomerance and all sort of Carmichael numbers this sort of thing came up that was exactly this construction okay so we've proved that this can't be a linear recurrent sequence because it can grow too fast and Kovaya Zanier says that this linear recurrence sequence can't grow fast. Contradiction okay i want to put on the on the line at bottom line in red okay so um yeah so the question is classify all linear division sequences in the integers and Marshall Hall um was really the first to write a paper on this specifically this problem 1936 where he tried to just do third order so you have three terms just second orders are easy enough for for luca sequences um so he brings up the general problem and you know there were older works that kind of worked in this problem even back to luca but without specifying the problem but it's sort of obviously there as Paul mentions so it's a bit hard to date this problem um in uh the book which is that's the key book on the subject by Everest and the point by Linsky and Ward they just give up trying to give a date to it just call it very old um so there's not many theorems actually um so there's one that's kind of interesting will come up in a bit later that any linear visibility sequence divides the product of certain luca sequences so that gives some hint that that is that tied in with luca sequences somehow um so these are linear division sequences of order two and then maybe times some constant times entity we'll see a proof of this later there've been a lot of other papers on the subject including Richard Guy um but the main output that's been relevant to the theory are are surprising and inspiring examples so let me very quickly do a little classification work if you ends the linear division sequence then obviously if I divide by you one these are all integers I get another in division sequence so I might sort of assume you one is one I can do the same trick for any d it's kind of interesting turns out to be a useful tool if um u zero isn't zero we have un divides divides u zero always so there's only finite many possibilities for un and that means there's finitely many vectors like that but remember un is defined in terms of these vectors so that'll actually make it periodic okay so if it's periodic then you can do something like if uh b and a have a common gcd with the period I'll always use m for the period in this talk then u of a divides u of r a right a divides r a which is u of b u of b divides u of s b which is u of a so u and a and u and b are plus or minus each other so with a periodic thing you get something um very very simple really everything's defined in terms of what happens between one and m and those guys just depend on the gcd of a and m so I just want to make one example which is if I pick one everywhere except integers divisible by three that's obviously a recurrent sequence linear recurrence sequence it's obviously a division sequence but it's not the product of loop class sequences so the uh periodic linear division sequences make it a little difficult to classify things but one might guess maybe every linear division sequence is a product of c n to the d's and periodic linear division sequences and lucar sequences I think that was kind of what people thought especially after this paper of uh Bezzavine uh and uh Vendor Porton I mentioned him in the beginning but Vendor Porton um I mentioned this stuff about slow growth and I'm just going to talk about some trivial other trivial trivialities so this first part u n is sigma r a u d with sigma a is plus or minus one is the periodics but I could multiply through by um by some sort of polynomial and still get a linear division sequence and this classifies all of them like that it's not a very hard proof and if other boring ones is I could also multiply through by a constant to the n minus one and then over the periodic ones constant to the n over d minus one so there are constructions from periodicity out of polynomial out of c to the n minus one have respect for the periodicity and this is what I'm going to call a boring linear division sequence so um we might guess that all linear division sequences are the product of a boring linear division sequence and a lucar sequence well we might have guessed that and I should say I think literature pretty much did guess that um but then um there's a very surprising paper in 2014 so in 2014 Yuta Bala proved looked at this sequence so this is known as lucas sequence or the companion sequence the Fibonacci numbers so it starts two one three four but satisfies the same linear recurrence as the Fibonacci's and here's the explicit formulas for both of them and uh Bala proved that it's a linear division sequences sequence and I'll note that it's not a product of lucar sequences one can prove that um and boring linear division division sequences so that that conjecture that had sort of been around and half enunciated in papers is wrong from this example um let me just show you my proof of why this is a linear division sequence well I'm going to write this as a polynomial now from beta and here what I'm going to do well the x plus y the alpha n to the plus beta at the end and the x cube minus y cube gives you this thing um because it was f3n here and um yeah so we get this linear recurrence sequence out of this ratio and um yeah so now before remember I just used x minus y and I said x dm minus m y to the m divides x m minus y t and it's obvious that's true is it is it obvious that when I plug in m and n in here where m divides m that f with x dm y to the m divides f with x n y t and I think the answer is not at first site um but here's the trick I can actually view this as the least common multiple of these two polynomial so when I'm working over z of x1 so I said I we proved before that with 2 to the n minus 1 and f of n but it's not necessarily the case that when you take the gcd or lcm you get a new linear division sequence or new linear recurrence was the problem but it turns out in polynomials you do so in polynomial rings you can take gcds and you can take the lcm's of linear division sequences and get new ones and so that's what you get here I mean it's pretty obvious this is going to give you as described a a linear recurrence and then from what we said earlier about lcm's it's obvious it will retain the fact it's a division sequence inside the ring of polynomials and so this is a proof of Barlow's result and so what we have is there's a nice structure of polynomials in the reserve integers that you can take gcds and lcm's of linear division sequences of homogenous polynomials and get linear division sequences and then you can substitute in and get something for the integer sequences that's pretty cool that gives us kind of a new tool that wasn't there so what we've seen is lucar sequences of specializations of this thing and Barlow's example is the specialization using f is this lcm and then once you see this you think well if lcm's work and that should do more so um yeah and I can de-homogenize just think of it as stuff happening in in polynomial rings so um yeah so I can just generalize this as much as I like I mean finally many of these none, in case none zero and this is going to be a linear division sequence when I you know do this sort of action with it um and in fact Koschkin in 2019 showed these are the only linear division sequences in c of t of this precise one we actually wrote it in a different form so it takes some work to show my version his are the same but they are um in fact there are more um linear division sequences in c of t they're not all of this form but this they're kind of all a similar idea I'll describe a little bit more about them in a minute so maybe all linear division sequences are part of a boring linear vision sequence and not lucar sequences but which we know now is wrong but specializations of polynomial linear division sequences so let's just think about how do we go from a linear division sequence over the polynomials to a linear division sequence in z well we start with these kinds of things so I'm actually going to take a bunch of these polynomials and here's here I'm showing you how you can get more polynomials that are linear division sequences um in a polynomial ring because I could bung in a algebraic number in there like a root of unity so I can throw in some roots of unity take the LCM this is also a linear division sequence um okay so I have some polynomials that you know when I put in these n powers I get a linear division sequence I kind of homogenize back to have two variables then I can specialize with putting an alpha and beta and then I have linear division sequence in some number field in fact the ring of integers of some number field and then I take the norm to get introduced so fairly straightforward program so let's just I want to look at this last step about taking norms so if I've got a linear division sequence in um some number a ring of integers of some number field and I can just take the norm and get a linear division sequence in z so let's try it for a favorite example I'll take LCM by space n over alpha minus beta where alpha and beta are you know from Fibonacci numbers and when I take the norm I'm projecting this together twice so I get f of n squared so this isn't actually going to give me everything um so I mean it's obvious what the problem here is that the gawa element swaps out from beta so I need to take a set of representatives of the gawa group modular stabilizer of these two elements so it's not a big deal but it's a minor technicality one's got it sort out let me show you another one that is really quite fun so when I'm working over the integers and I've got six to the n minus three to the n of six minus three so six and three of my my roots characteristic polynomial then I can sort of see that I can I can just get rid of this big GCD that's lurking there so let me show you that an example in a number field so I'm going to look at hello there's a question is it it is not hard to bound the degree of a product to linear current sequences but if a of n and b of n of linear recurrence isn't it um yes yes you can but let me not get into that there but there's an algorithm by which you can proceed to find the ratio yeah um that was in response to the chat anyway let me continue with this sorry about that so yeah so well I'm doing the same sort of thing here but I have this problem now that if I wanted to divide out by the GCD um it's non-trivial there is there is a GCD of three and one plus root minus five over two but it's actually a non-principle ideal and it's prime so I can't like divide take out some principle factor um so that okay who cares but now I'm going to take the norm and um let's take the norm so I do that times its conjugate and then when I look at the terms here I see a three in every term a three to the n right so now I can divide out by three to the n but I couldn't divide out back here so when I do that I get you know something straightforward and it turns out it's linear division sequence it's a bit weird at first sight see these four terms the other one that gives you one but you can see from the construction it is likely to be um so it turns out you can't divide out GCDs from individual terms when you you do this kind of construction but you come from the product um and of course I took this example from everybody's favorite textbook example of uh non-unique factorization which is where three times two is one plus root five one minus root five you can see all the terms there and that that gives you this nice example so you could object that maybe I could just say well I know that uh the class number here is two so I could square the ideal and get a principle ideal and so every second term I could divide out a GCD in general I would have to work with the class number and so you could do something you could you could jury rig something up like that but I think we rather complicated to keep track of it all okay so let me give you let me just talk a little bit more about this example I got it from a paper of Guy in Williams Richard Guy was 95 years old at the time um and still doing fantastic research so he was interested in following problem which is um how do you place dominoes on a four by n minus one rectangle it turns out n minus one is the right normalization um so yeah so I mean four by one you can just put the two n by n but once soon as you've got four by two you've got all sorts of ways to place them and it turns out that um four by two there's 11 ways and um yeah so there's this this linear this uh recurrence relation for the number of ways and um the characteristic polynomial factors in an interesting way and what Guy in Williams noted is that um this is an example where um the roots alpha beta gamma delta satisfy this right it's kind of obvious alpha delta is beta gamma this is the conjugate of that and so they noted this is linear division sequence and I think they thought and and maybe and I thought someone looked at the paper that this showed that Lucas sequences are wrong but it is actually an example of what we had in the last slide rather strangely so as I said in the last slide you can't see the product until you multiply things scale things up and then you divide out this divisor so there's tricky business around here so we have to our general algorithm we take something like this we take the norm well we don't quite take the norm we have to take a set of representatives of the gawa group modular stabilizer we divide through by an appropriate gcd and we get what we might call polynomial generated linear division sequence in the integers and so maybe the conjecture is all linear division sequences is the product of boring linear division sequence and a polynomial generated linear division sequence um well I haven't talked about the generate sequences where some uh some uh ratio of roots in the characteristic polynomial is a root of unity so here's a annoying sequence so it's one on the odd numbers t to the n minus one on even numbers and this has general formula here you can see the t to the n minus t to the n their their difference is a uh root of unity multiplicatively and so you tend to get structures like this is a nice one if I don't even go the other way because we can combine them but all of these are degenerate and we don't like degenerate things because it can cause an amazing calculation cancellation in the formula so here when n is odd these two terms cancel and so you get something like that and so it's hard to work with so typically in literature people work non-degenerate case which leads to no cancellation but I'm not going to do that for technical reasons but often people will go to the non-degenerate so um you can't I'm running out of time so I'm not going to go into this length but if you've got something degenerate you can understand what happens on each earth under progression and rebuild things just by the normal kind of way you work with such things um so you might ask are all linear division sequences the product of boring linear division sequence polynomial generated linear division sequences but constructed like this on each arithmetic progression mod m which we saw might be necessary and the answer is yes um so you understand now why the literature has had problems with this question in that it's very very hard to decide what the eventual answer is and you come up with a perfectly plausible idea and then you find an example that takes you to another direction so it took me a long time to do this um and the proof is terrible so um you can simplify this a bit and you can come up with a following classification theorem that every linear division sequence in the integers is the product of a boring one and specializations of linear division sequences in the homogenous polynomials divided by appropriate gcds okay so let me give you an idea of how we attack this problem so I'm just going to work in the simplest possible case now where the we're all integers the things we're taking powers to the constants are probably integers so I'm not dealing with complications I want to show you Barbaro's 2017 proof of Bayes of Ampeto and Vandeporten so here's the idea is um if you end divides if you end divides you two and you three and you four and it's a division sequence then um what we have is that each of these when I put an n two and three and they're all congruent to zero mod u n which I can write as a vandemonde system right they're just a the two n eight to the three n eight to the four n times these coefficients and if I assume these coefficients have gcd one which I might as well just pull out a factor um then we see that the determinant's got to be zero mod u n and so that gives us this theorem of Bayes of Ampeto and Vandeporten that u n divides the uh the determinant of the Vandemonde so that's pretty cool but kind of the subject was kind of stuck with that theorem and which makes you think oh well you know it's all about um it's all about lucas sequences I mean that's what they look at like things like this um so that's what we need to understand but it was very hard to just see what we had to remove from here to get the u n so my first observation was to approach this a little bit differently the same idea was um suppose that well what we see here sorry let me go back if I've got a prime p dividing u n it divides this product so it divides one of these terms so a i to the n a j to the n are congruent mod p for some i and j so what I'm going to do is I'm going to group the a i to the n by the residue class mod p so when I do that um I have a partition of one k I've reorganized them so the the distinct residue classes come the first few and what's going to be useful is to for each of these sets is to take the sum of the coefficients up there now look at u of r n so I take this is just partitioning the original sum up here but now into these classes now mod p these a i to the n's are a j to the n's and then I've just got the sum over the coefficients so I've got another van der monde system it's a little different because now I've just got these h things but in this case all of these a j to the n's were chosen to be different so this determinants no longer zero that means each of these g i j is a zero mod p so now we get new information now if g i j is not equal to zero then there's only finitely many primes dividing it so I'm just going to throw away I'm not going to work with the primes that divide some non-zero sum of coefficients I'll only work with the other primes and in that case when a prime divides u n we've got this deduction but it's not a prime that divides a non-zero g i j so the only possibility is that g of i j is zero so what have I proved I prove it for um that there are subsets of these coefficients whose sum is zero so very quickly when I just plug that in I can because the subset sum is zero I just plug subtract zero times a j n and what I see is that u n is the visible by this big gcd and that's the visible by p and so what if p is a large prime so Govindy and Zanie said if you have a gcd of things like this that's big then the only way it can happen is if they're multiplicatively dependent and actually it's easy to use that theorem and say if there's a bunch of different things like this you get the same result so by Govindy Zanie all of these are multiplicatively dependent so that's a bit of a shocker you kind of have that so what does that tell you that if they're multiplicatively dependent then I can just plug in and then I'll just think of this as a variable the r to the n it's a polynomial I get this and what I see is that this u n is the visible by r to the n minus one so again what have I shown what have we shown we've looked at these sets where these things are congruent by p we're looking at this big gcd we know it must look like r to the n minus one the gcd so we actually have r to the n minus one divides u of n it's sort of like we're factoring out a polynomial and if you like we're factoring out something that looks a bit like a lucar sequence so that's sort of the remarkable thing that happens now I did this with one prime but you could have many primes at whose product is bigger than e to the epsilon n so um questions can we account for the whole size of all of u of n like this and well to make this work in general we need something to do with algebraic numbers we're going to use prime ideals then there's a notion of size of prime ideals which is a tiny bit tricky you've got these complicated height functions you've got to work with um you need a version of van de monday that works for when you've got polynomials in here but actually that turns out not to be a big deal that's done in the literature and it can be done nicely the tough part is what if u of n is actually the product of many small primes to very high powers and that that is about 80 percent of the proof is dealing with one case okay so let me leave that attack and show you a second attack so what we just saw is that um we're going to if we got division sequence we're going to find for some partitions of the terms the ai over aj are multiplicatively dependent so for instance if I look at this sequence which is um a linear division sequence then the we can use the 6 is 2 times 3 and then we see the factor of 1 minus 2 to the n times 1 minus 3 to the n so in general I can write any sum of g i n alpha i to the n exactly as g i n and then the alpha i I can write in terms of well here it was easy enough to see the uh the basis of the alpha i's multiplicatively it's just 2 and 3 in general there's some multiplicative basis for the alpha i's and we may have some roots of unity involved but they will be associated for period and these are all very independent so actually I can work with this like I'm working with a polynomial so I can just replace the better to the ends by independent variables so what I'm going to do is try and understand the divisibility of the u n's by understanding the visibility of a corresponding polynomial so here's here's where I've kind of set it up I'm going quite fast I'm sorry I've got about five minutes I need to get this done so um let's have a look so we've taken the u n and we've written these g i n's in terms of this where we've got these we alphas are um in terms of uh a basis a multiplicative basis oops so uh let's have a look if I've got u n is v n times w n it's in your current sequences then it means these polynomials look like u a equals v a times w a and um well um oh so in in response to the question that was in the chat you could use this to uh to bound the degree of the quotient but um yeah so by the hadamard quotient theorem if u n divides u k n okay then I can write u k n as u n times v n for some linear occurrence and then I can just use these polynomials and I get something like this so I've got this u of x divides u of x okay well forget the subscript depending on the roots of unity maybe I should but exposition purposes I should just assume there are no roots of unity so we're going to take these polynomials u of x we know they must divide u of x the k for every k and that is going to be a very weird class of polynomials so that's what we need to get into and um interestingly that was a subject of interest to people in the 1920s and 30s Rick Goran uh McCall and then there's a more recent work of Everest and Vanderpotent but Goran's really I mean Rick gets all the credit but actually Goran's paper is very beautiful um if you like this subject look at his paper um so this is what we want to understand is if we we put in nth powers in place of these um how does this factor in fact in in written Goran they they actually have different powers and different variables but we don't need that so one thing to notice if I've got a irreducible thing like this and and in factors then I can just replace x by x to the k and get boring factorizations what I might call old factorizations so we're only interested in new factorizations the first time a factorization appears and the theorem of McCall the thing that's missing is that if f has um at least three monomials then there are only finitely many new factorizations so um we find ourselves in the situation that um there's not much new can happen here and in fact all the new factors have at least three monomials so in the case we have two monomials so y and z are products of these exercise then when I take the nth power I simply get this and we can factor that into binomials so binomials factor into binomials when you take the nth powers for the variables and greater than or equal to three monomials only factor into greater than or equal to three so in some sense there's a complete understanding and what this means in particular this finitely many new factorizations is if you plug in a power in here at some point the irreducible factors can not reduce any further when you plug in values I'm going so fast but a few people are understanding so let me just see how this works um if I have this polynomial and I have a u of x n divides u of x to n for all n then I'm going to claim that u of x actually factors is a product of monomials to binomials and so here's the idea I'm going to write it as product of monomials binomials times the irreducible factors will bigot with with three or more monomials then just plugging in that mu so from that point onwards you you get no new factorizations I get binomials times multi-nomials if you like divides binomials times multi-nomials so comparing irreducible factors this divides that now they don't reduce any further they got the same number of irreducible factors so you can just pair them um right I mean if if f of x divides g of x and they have the same number of irreducible factors they're the same but this clearly has bigger degree unless they're constant so we've just proved that f must be a constant so u is the product only of monomials and binomials so finally what do these binomials look like well something like this so we end up with u of n looks like c n to the d times some gamma zero at the end where gamma zero is some multiplication of the betas which would multiplication of the alphas and then some polynomials because we're taking factors like this these binomial factors in some gamma j to the m when I plug back in the beta and so this actually we all we need is f of u of n dividing u of two and to get a structural like this using rit sterum so finally the final theorem is if u of n is a linear division sequence of integers then there's a boring linear division sequence there's polynomial linear division sequences and multiplicatively independent algebraic integers so that we actually have a formula for it um and maybe I wouldn't go on so let me stop there