 apparently climbing now in the Canadian Rockies and is not listening to this talk. I'm at a hotel and using the internet connection there and hopefully that'll last for my whole talk. So practical numbers. One story begins with Fibonacci in the year 1202. In his famous book, Libre Abachi, the same book where he introduced the Hindu Arabic numerals to the West. He was interested in unit fractions, sometimes called Egyptian fractions, one over N. And he noticed that some numbers like 12, every fraction, up 112, 12th, up to 11th, 12th, or even 12th, 12th can be written as a sum of distinct unit fractions where the denominators are divisors of 12. So it's distinct unit fractions, denominators, divisors of 12. For example, 5th, 12th, you can see is 1, 6th plus 1, 4th. It's also 1, 3rd plus 1. I can't get the sound. Why? Did someone say something? Anyway, if you take these equations and you multiply them through by 12, you can see that the numbers from one to 12 can be written as a sum of distinct divisors of 12. And that's the definition of a practical number. Or you could take Fibonacci's definition if you'd like, but he didn't call them practical. They were called that by Srinivasan in a short article that he wrote in 1948. And he had a dig at the metric system, I guess what he didn't like. He said that the subdivisions of money, weight, and measures were often done with such numbers as 4, 12, 16, 20, 28. They're usually thought to be so inconvenient as to deserve replacement with powers of 10. But he liked these numbers, such as 12 and so on, because of this property of, I suppose, I'm not sure why it's so practical to have every number from one up to N being represented as a sum of distinct divisors, but that's what he called them. And maybe it was the Fibonacci thought as well. Anyway, he began a study of the multiplicative nature of practical numbers. And he didn't come up with a criterion. And this was done independently in 1954 by Bonnie Stewart and Vatslav Serpinski in 1955. And essentially, this is what their definition is, their criterion is. It's a recursive definition of a set of numbers. And the recursive definition says that the set contains the number one. And also, if it contains a number M, then it contains all of the numbers M times P for all primes P up to sigma M plus one. Sigma of M is the sum of divisors, of all of the divisors of M. So, and this recursive definition gives a set of numbers, and that's the set of practical numbers. Here's Fibonacci, Srinivasan, and Serpinski. So from the criterion, we can see that every power of two is practical, because I guess because one is practical and the criterion says you can go up to sigma one plus one, which is two for the next prime. And once you have a prime, you can put in any power of it you want. So the powers of two are practical. The fact that you can represent numbers smaller than powers of two by divisors of powers of two, is just the binary representation of a number. Also, the criterion shows that every practical number after one is an even number. So let's see if we can prove this elementary criterion of Stuart and Serpinski. So first, we should note that the criterion is obviously necessary. If the prime P is bigger than sigma M plus one, then you cannot write P minus one, which is bigger than sigma of M with divisors of MP. You can't use a P itself as that's too big and all of the divisors of M together add up only to sigma of M. And that's smaller than P minus one. So there's a necessary condition that P is less than or equal to sigma of M plus one. And then to prove it, we're going to look actually at a stronger statement, improve a stronger statement that not only can numbers up to M be represented as sums of divisors, but every number up to sigma M P to the alpha. So we're going to assume, we're gonna prove in fact that for a practical number M, you can get every number up to sigma of M and then we're gonna prove that by induction. So let's start out by assuming that every number up to sigma of M is a subset sum of divisors of M by way of induction. And let's let P be a prime not dividing M with P of the right size, less than or equal to sigma M plus one. And we're gonna prove it for all of the numbers, MP to the alpha again by induction on alpha. So this is a double induction. Okay, so if alpha equals zero, our induction hypothesis is that everything is cool at M, so M is good. Now we'll assume it holds at some alpha and we'll look at this interval, I sub A. Here A is just an integer less than or equal to sigma of M. The left-hand endpoint of the interval is A times P to the alpha plus one. And the right-hand end of the interval is that number plus sigma MP to the alpha. Now you'll notice that A times P to the alpha plus one, A being less than or equal to sigma of M by our induction hypothesis means that A is the sum of distinct divisors of M. And here P to the alpha plus one is a divisor of MP to the alpha plus one. So this is a divisor, this can be represented as a sum of divisors of MP to the alpha plus one. Plus, if we throw in divisors of MP to the alpha, by induction hypothesis, we can get every number up to this point, we can hit every number in this interval with divisors of MP to the alpha plus one. But does that get every number up to sigma of MP to the alpha plus one? Well, first of all, note that since we're getting all the numbers up to MP to the alpha plus one, it means that this interval wraps around to the next interval at A plus one. And so let's look at sigma of MP to the alpha. It's greater than or equal to P to the alpha plus one minus one because of our assumption that sigma of M is greater than or equal to P minus one and this formula for sigma of P to the alpha. And so these intervals, I sub A can be glued together and they keep on going until you reach the end at this point and this point is equal to sigma of MP to the alpha plus one. Maybe a little garbled here, but this is how this criterion is proved. So one can wonder if it's a practical number theorem giving an asymptotic for N of X. So that's the notation I'll use for the number of practical numbers up to X. So Srinivasan began this also. He computed that N of 200 is 50 and he wondered if maybe this fraction, 50 over 200, whether that is gonna be going to zero when you count up to higher levels or not. And 1950, Erdisch claimed that in fact it does that N of X is little low of X. Erdisch didn't supply a proof but I believe this is what he was thinking using some ideas that he used frequently. So he was quite familiar with the theorem of Hardy and Ramanujan that number N is usually close to log log N prime factors. And this is true whether you count distinct prime factors or not, you can count with or without multiplicity. So let's look at a little bit higher than log log N, 1.1 times log log N. So the Hardy-Ramanujan theorem says that asymptotically all integers have fewer than 1.1 log log N prime factors. And there's only density zero of them with more than 1.1 log log N prime factors. So let's assume now our N is in the majority set with at most 1.1 log log N prime factors. That means that the number of divisors event is at most the power set on these primes. I'm assuming these primes occur with repetitions. Nevertheless, I can just take the power set on a set of this size, two to that power and I get the number of divisors, an upper bound for the number of divisors. And so now let's look at the power set of the set of divisors. Look at all possible subset sums of divisors. So that's two to the two to the 1.1 log log N. Certainly a big number is of the double exponential, but log log N is pretty small. So how does this compare with N? Well, if you bring the two up to the top as log two, you have a 1.1 log two. 1.1 log two is less than one. And if you bring log N down to the middle here, so you have two to the log N to an exponent smaller than one. So that's even smaller than entity epsilon. So they're just not enough divisors to have enough subset sums to be equal to all of the numbers up to N. And so if it has at most 1.1 log log N prime factors, it's not gonna be practical. So that's a proof of the Erdisch claim. Now, this just says little O of X. However, here are our heroes, Ramon and Sean Hardy and Erdisch. If you look at a more quantitative version of the Hardy-Raman and Sean theorem, you actually get X divided by a power of log X as an upper bound for the number of practical numbers. The exponent here, eta is this expression, which is actually sort of famous in the anatomy of integers about 0.086. So upper bound is X divided by log X to the 0.086. But point, that's not the right exponent to have here. And Gerard Tenenbaum and a couple of papers in 1986 and 1995 show that it's basically one. The exponent here should be one. It's X over log X, you had a positive power of log log X, upper bound and a negative power of log log X in the lower bound. But the main shape of this is X over log X. A couple of years later, Eric Seah got rid of the log-log factors. This notation here looks like a smile and a frown is some notation in analytic number theory. That means that n of X is less than a constant times X over log X. And it's also greater than another positive constant times X over log X. So its order of magnitude of n of X is X over log X. That is the correct order of magnitude. Now this is all going towards a conjecture that Maurice Margenstern had made a few years earlier in 1991. And he did some computing. He computed up to 10 to the 13th and found that the number of practical numbers up to that level was looking like a constant times X over log X, where the constant is about 1.341. And so he made this conjecture that there should be some constant and n of X should be asymptotically a constant times X over log X. So this was proved in 2015 by Andreas Weingartner. And he didn't compute the constant in his 2015 paper, but just earlier this year he published a proof that the constant is actually very close to what Margenstern thought it might be. He said 1.341, Andreas shows 1.33607. Here is sort of a closed form for the constant. It's not closed because it involves an infinite sum. And the sum is over practical numbers. And at first glance you might even think that this sum is a divergent series because if here you have a one over n ranging over practical numbers, the practical numbers are distributed sort of like primes, constant times X over log X, the sum of the reciprocals of the primes is infinite. However, this one over n is multiplied, well, see what is it multiplied by? It's multiplied by this thing. Well, what is this sum? Well, by essentially the prime number theorem where Margenstern's this sum over the primes up to sigma of n plus one is the log of sigma of n. So here's the log of sigma of n minus log n. Well, that's the log of sigma n over n. So that actually can be pretty big, making this divergence look even more troubling. However, here we cut it down a bit. Now we have a one over log sigma of n and that's enough cutting it down to get it to converge slowly. It was somewhat of a triumph for Andreas to get so many decimal places for this series to compute that series, some because it converges so slowly. So you gotta use some extra cleverness. So Lola Thompson in 2012 and her dissertation gave a generalization of practical numbers. So we don't consider numbers, we consider polynomials. So here we look at the polynomial t to the n minus one. I'm calling the variable t rather than x because I'm using x as a bound to count up to. So you have t to the n minus one and we can look at its divisors over the integers. For example, it has the divisor t minus one. And we'll notice that t to the 12 minus one has divisors of every degree up to 12. How can we see that? Well, let's look at the irreducible factors of t to the 12 minus one. These are the cyclotomic polynomials, phi sub d of t where d ranges over the divisors of 12. So one, two, three, four, six and 12. Those six cyclotomic polynomials have degrees one, one, two, two and four. And the question is, can you use these weights to make up every number up to 12? Well, it's obviously, obviously you can. That's easy. And so this polynomial has divisors of all degrees up to 12. So she called such a number n to be phi practical. This is the Euler phi function. The reason she called it this is because phi of d is the degree of the d cyclotomic polynomial. So the criterion, the arithmetic criterion is that for every integer m up to n there's a subset of the divisors of n such that if you add the phi of d for d ranging over that subset, you will get n. So you can ask if there's an analog of the Sierpinski-Stewart criterion for the phi practical numbers. And it turns out that it's quite tricky which means that it's not just a simple analog. So for example, with practical numbers, if you have one and you peel off the top prime, it's still practical, but it's not true for the phi practicals. So for the number 315, three squared times five times seven that is a phi practical number. But if you peel off the seven and are left with 45 that is not phi practical. In fact, around 22, 23, those degrees cannot be represented as divisors of t to the 45 minus one. Another example is nine. Nine is not a phi practical number, three is. So this is an example where if you increase a prime from a phi practical, it doesn't stay as phi practical. So that's not true for the practical numbers. So nevertheless, Thompson was able to use say a machinery to show an analog of its theorem that the number of phi practical numbers up to x is a border of magnitude x over log x. And she conjectured that there is also an asymptotic and she did some computing and it seemed like here the constant was near to one. So she even wondered if, in fact, it should be asymptotic to the number of primes up to x. So a few years later with a Weingartner and myself, the three of us proved Thompson's conjecture and we didn't actually compute C but we gave the heuristic that it is close to one but not one, it's slightly smaller. So another application of the ideas behind practical numbers is numbers with dense divisors. These were defined by Tenenbaum and written about by Seah and Weingartner. Number with dense divisors. Here is the sequence of all divisors of n. Start with one, go to the next one and so on up to n. And look at the ratios of consecutive divisors. d sub i divided by d sub i minus one. If all these ratios are less than or equal to two, then that's the definition of a number having dense divisors. These are, in fact, we're used in the studies of practical numbers and it's not very hard to see that any number which has dense divisors will be practical. Now this condition on dense divisors is similar to the Erdisch-Propinquity problem. So Erdisch wondered about the density of integers which just has two consecutive divisors, d d prime with the ratio between one and two. Is that true? Well, it's obviously true for a positive proportion of numbers. For example, it's true for all multiples of six because every multiple of six has divisors two and three and the ratio of three over two is bigger than one and less than or equal to two. So there is a positive proportion of numbers that have such a pair of divisors and Erdisch conjectured that in fact it's density one. This was finally proved in a joint paper of Meyer and Tenenbaum in 1984. But then again, dense divisors is not that there should exist one pair but that every pair of consecutive divisors should have that relationship. So there's a multiplicative criterion for the integers with dense divisors and this criterion is useful in the study of practical numbers because it's very similar to the criterion of Sierpinski and Stewart. So for Sierpinski and Stewart, if M is practical, you can take every prime P up to sigma M plus one and throw it in. With numbers with dense divisors, you can take every prime P up to two M and throw it in. In the study of practical numbers, one can generalize this where you replace two with a larger number and that also plays a role. We now know after their work that there's a distribution for numbers with dense divisors, it's again of the shape constant times X over log X. So because of this expression X over log X, it seems natural, X over log X is so much associated with prime numbers, so it seems natural to ask similar questions for practical numbers as one asks for prime numbers. So for example, the famous conjecture is are there infinitely many twin primes? So let's ask for twin practicals. Well, the practical numbers after one are even numbers so we can't expect N and N plus one both to be practical infinitely often but we can expect perhaps N and N plus two to be practical simultaneously infinitely often. Morgenstern proved that this is in fact the case that there were infinitely many. And then in a new paper that I have with Feingartner, we got estimates for how many there are up to X. The conjecture is that the upper bound is tight except for the constant. By the way, this notation less than less than an analytic number theory is the same as big O where this says that N sub two of X is less than or equal to a constant times X over log squared X. And it means it's greater than or equal to a constant times the slower bound. So you can ask how we prove these theorems. So for example, on the upper bound, I can't see my screen here. Are we looking at, oh yeah, good. So we're looking at the upper bound. So we write a pair of twin practicals. We pull off initial divisors of N and N plus two by an initial divisor, I mean all the primes in M are less than or equal to all of the primes in Q and the same thing for M prime and Q prime. And I'll choose M and M prime to be in this interval since these numbers are practical, they have divisors in this interval, X to the one seven to X to the one third. And then we fix a pair M prime of practical numbers in this interval. And we then count primes, sorry, numbers Q. Q has all large primes, because they're all larger than the primes in M such, so Q satisfies a few properties. That satisfies the property that QM plus two is a multiple of M prime. And it satisfies the property that if I divide that QM plus two by M prime, I'll get a number that purports to be a Q prime and that only has large prime factors. So this argument would get X over log squared X times the power of log log X. And we have some technical hurdles to overcome the power of log log X and get the X over log X squared. So how do we show, how do we get the lower bit to show that there are X divided by that power of log X number of twin practicals up to X? So we do it as follows pretty much. We choose two practical numbers in this interval between square root of two, sorry, one half square root of X and square root of X. And we choose a pair of practical numbers in that interval that are almost co-prime. Their GCD is two. And which is essentially as low as a GCD can be between two practical numbers larger than one. And so given such a pair, there are going to be a one and a two such that a one M one minus a two M two is two. And a one is small, a one is smaller than M two and a two is smaller than M one. And so we have multiples of practical numbers. And the multiple is small enough to guarantee that the product is again practical. So these are twin practicals. Well, that looks like we've proved an even stronger theorem. We have X over log squared X pairs of practical numbers in this interval. With this GCD criterion and for each pair, we choose a one and a two, and we get twin practicals up to X. So why doesn't this give X over log squared X? Well, the trouble is, is that we could be over counting that the same twin pair could arise in multiple ways from this expression. Well, we, we know that there are a lot of things that handle this by first not choosing any practical numbers in this interval. We only choose those which don't have too many prime factors. And we can use sort of analogs of the Hardy-Romanijan theorem to show that we can keep the total number of prime factors of M one and M two. That's big omega counts the total number of prime factors with multiplicity. We can keep them under control. And this function that takes M one and M two to A one is going to be more or less one to one. In fact, it's at most two to one using the inequality that we have and the fact that we have the practical numbers in this interval. So it means that we can again discard we're not gonna hit numbers too often with large values, large numbers of prime factors. So we can assume that the number of prime factors of A one and A two are under control. So with those two under control, it means the total number of divisors of A one M one and A two M two is power of log. And that tracing effect gets that power of log we had in that previous theorem. So another connection between primes and practical numbers brings in the primes themselves. So a couple of years ago, Gwo and Weingartner looked at shifted primes that are practical. So P is a prime number. P minus one is the shifted prime, one less than the prime P. And we wonder if this even number is practical. And they proved the count for that again with the lower bound and upper bound. You might recognize this exponent which we saw earlier. So in the same paper that I have with Weingartner recently, we showed that this exponent here on log X is not 0.086, but in fact one. So that's the upper bound. Presumably again, this is the truth and the lower bound we improved the exponent there. For our upper bound proof, we write a practical number as M times Q. We're now Q is the largest prime factor event. And if N is one less than a prime, then plus one is prime. For a given number M, here we have M. We ask for primes Q going up to say X over M with MQ plus one also prime. So here we have a sieve with two primality conditions. And this essentially gives us a log squared X. And you can see that this expression is just X over log squared X. So again, there are some technicalities because Q might be small. And then the sieving interval was not so long but we handle those cases where M is smooth and therefore Q is small. Or when this is a factor that comes up in the sieve, you don't have to wonder when that's large. So we have to handle those technicalities to prove our upper bound. So for the lower bound, here's how we did that. So we take a number N that's practical. We take, we look at primes up to N squared where P is one mod N. Suppose there are plenty of such primes. And if we write P minus one is A times N, that'll be practical automatically because A is going to be smaller than N. N is practical, so this number is automatically practical. However, we don't know that if we take primes up to N squared, we don't know even if any of them will be one mod N. However, if we're just a little bit below for our modulus, it's just a little bit below square root of X like square root of X divided by high power of log X. Then we can use the Bombiery-Vinogrado theorem to show that there are many pairs P and N where P is one mod N. But the trouble now is that A is not going to be smaller than N and it could be a bit bigger than N. Well, that's still okay so long as A doesn't have any big prime factors. So we use an upper bound in the sieve to rule out the case when A has big prime factors and we are home. Now, we could have used instead of the Bombiery-Vinogrado theorem, a new theorem of James Maynard which is an improvement of the Bombiery-Vinogrado theorem in the case when you have a fixed residue class, which is the case that we'd be looking at. We're looking at the residue class one. Actually, it was sort of hard to find this photo of Vinogrado on the web. There's another Soviet number theorist named Vinogrado, I am Vinogrado, he's more famous, has more photos on the web, but apparently this is our AI Vinogrado. The notation less than less than that we've been using is often called the Vinogrado notation. I'm not sure to whom it's due, it might be to him or to I am. For prime numbers, we have the Goldbach conjecture. Even numbers should be a sum of two primes. So, a Morgenstern conjecture, well, he tried to conjecture that for practicals. Practicals are even, so if you take two practicals, the sum will be even. So he conjectured that every even number is the sum of two practicals. And he wanted to say, well, what about odds? Don't leave them out. He said that every odd number should be the sum of a prime and a practical. So Giuseppe Melfi proved the Morgenstern conjecture for the even numbers. He proved that every even number is the sum of two practical numbers. And now in our paper with Von Gardner, we essentially did the odd part. We did it for sufficiently large odd numbers. So we're trying to prove that a large odd number is the sum of a prime and a practical. So let's let A be an odd number in the interval between X and two X. And let's let N run over the practical numbers that are a bit smaller than square root of X. And use the Bambieri-Vinogrado theorem to find primes P up to X in the residue class A mod N. Okay, so you notice here that A is bigger than X and P is less than or equal to X. So we have these double inequalities. And so therefore A minus P is greater than zero. And A minus P is gonna be divisible by N. So it means that A minus P is a multiple of N. This multiple B is not much bigger than N. And again, we can use SIV methods to rule out the case where B has a large prime factor. So this proves that A is the sum of P and BN. And when B doesn't have a large prime factor, BN is practical. And that proves the conjecture for sufficiently large odd numbers. So you might wonder if you could use these ideas to prove the full conjecture that there aren't any exceptions. Well, that would be tricky to do with the Bambieri-Vinogrado theorem. I'm not sure if it would be tricky with Maynard's theorem, I'm guessing so. But the Bambieri-Vinogrado theorem depends upon Siegel's theorem. Siegel's theorem is the quintessential non-effective, ineffective theorem in analytic number theory. You cannot make the constants explicit unless you assume the extended Riemann hypothesis. And if you assume the extended Riemann hypothesis, we actually have a very strong prime number theorem for residue classes. We have this wonderful inequality for the number of primes up to X in the residue class A mod N. That's what you'd expect when over the Euler-Phi function of N times logarithmic integral of X. And that difference is smaller than square root of X times log of N squared X, starting right from the beginning. So if you assume this prime number theorem, we can prove that the Marginstern conjecture holds perhaps at some starting point like E to the 10,000. So now, for the computationalists in the audience, your job is cut out for you. Just check all odd numbers up to E to the 10,000 and see that they're all the sum of a prime and a practical. Well, Tomas Alevera, a silver, got it started for you. We can show it up to 10 to the nine that it holds. But actually, we were able to get further than that. And let me describe this neat method of going further. Take a power of two. And for each odd residue class A mod that power of two compute the least prime in that residue class. So for example, suppose K is four. So we have two to the four is 16. And we have the odd residue classes up to 16. There are only three residue classes up to 16 that are not prime already. Those are one, nine and 15. And if you look at the first prime that's one mod 16, that's 17. The first prime that's nine mod 16, that's 41. And the first prime that's 15 mod 16, that's 31. So the largest of these least primes is 41. So that's gonna immediately imply our theorem in intervals we'll see in a minute. Linux theorem says that the largest of these least primes is bounded, but it doesn't give us that it's small. Numerically in practice, it will be small. Say smaller than this bound, K squared times two to the K. So that might be tractable to search up to this point for K of moderate size. So once this largest of the least primes is found, we've proved the Marginstern conjecture for every odd number in the interval above P and up to this next power of two to the two K plus one. Because if A is an odd number in this range and look at the residue class A mod two to the K and take a prime in that residue class that's less than or equal to P and look at A plus Q times two to the KB, B is small and this is gonna be practical. So we have the odd number A as a prime plus a practical. So in our toy example, just doing that mental calculation to get 41, we immediately have that every odd number up to 512 is representable as a prime and a practical. We actually implemented this up to two to the 53. If we could prove that every number up to two to the 53 is sum of a prime and a practical. You don't have to even go to this height K squared times two to the K. If you only go to three K, two to the K, then it should be conjecturally only a small fraction of unrepresented residue classes, one in 5,000 or less. And then in those find those classes and then search over them individually. Finally, let's look at this problem. So Landau has a famous conjecture that between consecutive squares is always a prime. Can you prove this for practical numbers? Well, this was done in 1984 by Miriam Hausman and Harold Shapiro and a small improvement was found by Melthy. One can ask, in fact, if there's always a practical number in the short interval x up to x plus x to the epsilon and we don't have that. We don't have that for any epsilon smaller than one half. So that would follow from an argument based upon conjectures about smooth numbers. And so Granville has a famous survey about smooth numbers and in it he has a featured conjecture which says that in the interval x up to x plus x to the beta is going to be an x to the alpha smooth number here alpha and beta are any fixed positive numbers and x is sufficiently large in terms of alpha and beta. If we have that, then we, we're looking at x up to x plus x to the epsilon. We can take the first power of two above x to the epsilon over two and consider the multiples of that power of two that in this interval and we want this d to be x to the epsilon over two smooth. So by this featured Granville conjecture though there will be b's in this interval that are so smooth and then that b times two to the k will be practical but this may not be the only way to prove the conjecture about short intervals. Anyway, going to these number theory web seminars week after week, I see that people are doing really wonderful things. I'm not sure what I'm talking about is so wonderful but at least it's practical. Thank you for your attention.