Tanya Khovanova's Math Blog

Robert Calderbank and Ingrid Daubechies jointly taught a course called “The Theory of Games” at Princeton University in the spring. When I heard about it I envied the students of Princeton — what a team to learn from!

Here is a glimpse of this course — a problem on Evolutionarily Stable Strategy from their midterm exam with a poem written by Ingrid:

On an island far far away, with wonderful beaches
Lived a star-bellied people of Seuss-imagin’d Sneetches.

Others liked it there too — they loved the beachy smell,
From their boats they would yell “Can we live here as well?”
But it wasn’t to be — steadfast was the “No” to the Snootches:
For their name could and would rhyme only with booches …

Until with some Lorxes they came!
These now also enter’d the game;
A momentous change this wrought
As they found, after deep thought.

Can YOU tell me now
How oh yes, how?
In what groupings or factions
Or gaggles and fractions
They all settled down?

Sneetches and Snootches only:

Sneetches and Snootches and Lorxes:

Find all the ESSes, in both cases.

To get to the national swimming championship, you need to win the state running championships.

What? Is that a joke? Perhaps you’re having the same reaction. Because this is exactly what is happening with math competitions. The official USA math competition has three rounds: AMC, AIME and USAMO.

AMC is a multiple-choice competition with 25 problems in 75 minutes. To be good at it, you need speed, accuracy and the ability to guess well.

AIME is 3 hours long and has 15 problems. The problems are a different level of difficulty and guessing will not help you. Though AIME is also multiple-choice, unlike AMC where you choose out of 5, in AIME you choose out of 1,000. But you still need speed and accuracy. A small arithmetic mistake will cost you the whole problem.

USAMO is a competition that runs for 9 hours and has 6 problems. The problems are much harder and you have to write proofs. Proofs? What proofs? Where are the proofs coming from? It is like you got to the national swimming championship because you are a great runner, but you do not know how to swim.

As the result of this system of selection, the USA team at the International Math Olympiad has diverse skills: these kids are good at all three types of the math competitions. It is like taking an Olympic triathlon team to the Olympic swimming event.

However, the US loses by selecting in this way. There are many kids who are great mathematicians: they may be good at difficult problems but not that good at speed-racing problems. An arithmetic mistake costs you only one point at IMO, but a whole problem at AIME. There are kids who are deep mathematicians prone to small arithmetic mistakes. They could get a gold medal at IMO, but they can’t pass AMC or AIME.

Not only that. As many AMC problems are boring and do not require ideas, AMC might discourage kids from all math competitions at an early stage.

I will write later with my ideas about how to change AMC. Right now I would like to offer a solution to a smaller problem. I am sure that the US math team organizers know many cases in which a non-senior kid does great at USAMO and is potentially a team member for the next year’s US IMO team, but, oops, next year he can’t pass AMC.

I suggest the following: USAMO participants are allowed to go to next year’s AIME no matter what their AMC scores are. USAMO winners are allowed to go to the next year’s USAMO no matter what their AIME results are. This way kids who have proved that they are great swimmers do not need to compete in running again.

I was interested for some time in the divisibility of odd Fibonacci numbers. Fibonacci numbers are closely related to Lucas numbers. Lucas numbers L_n follow the same recurrence as Fibonacci numbers: L_n = L_n-1 + L_n-2, but with a different start: L₀ = 2, L₁ = 1. Obviously, every third Fibonacci and every third Lucas number is even.

Primes that divide odd Lucas numbers divide odd Fibonacci numbers. Let us prove this. Suppose a prime p divides an odd Lucas number L_k, then we can use the famous identity F_2k = F_kL_k. We see that p divides F_2k. The fact that L_k is odd means that k is not divisible by 3 and so is 2k. Thus F_2k is an odd Fibonacci that is divisible by p.

We see that primes that divide odd Lucas numbers are a subset of primes dividing odd Fibonacci numbers. It is easy to see that it is a proper subset. The smallest prime that divides an odd Fibonacci and doesn’t divide any Lucas number is 5.

The next natural question to ask: is there a prime number that divides an odd Fibonacci, doesn’t divide an odd Lucas number, but divides an even Lucas number? Below I show that such a prime doesn’t exist. In other words, that the set of prime factors of odd Lucas numbers is the intersection of the set of prime factors of odd Fibonacci numbers with the set of prime factors of all Lucas numbers.

Let us consider a Fibonacci-like sequence a_n in a sense that a_n = a_n-1 + a_n-2 for n > 1. Let me denote by b_n, the sequence that is a_n modulo a prime number p. The sequence b_n has to be periodic. It could happen that b_n is never 0. For example, Lucas numbers are never divisible by 5. Suppose the sequence b_n contains a zero. Let us denote r(p) the difference between the indices of two neighboring zeroes in b_n. We can prove that r(p) is well-defined, meaning that is doesn’t depend on where you choose your neighboring zeroes. Moreover r(p) doesn’t depend on the sequence you start with (see 9 Divides no Odd Fibonacci for the proof). The number r(p) is called the rank of apparition of the prime p.

As the Fibonacci sequence starts with a zero, then the terms divisible by a prime p are exactly the terms with indices that are multiples of the corresponding rank of apparition. The Lucas sequence doesn’t start with a zero, but we know that L_-n = (-1)ⁿL_n. That means, if a Lucas number is ever divisible by a prime p, then the smallest positive index for such a number has to be r(p)/2. Also, the indices of the Lucas numbers divisible by p will be odd multiples of r(p)/2.

We know that the indices of even Fibonacci or Lucas numbers are multiples of 3. Hence, a prime number p that divides an odd Fibonacci number must have a rank of apparition that is not divisible by 3, That means that if p ever divides a Lucas number, then it divides a Lucas number with an index of r(p)/2, which is not divisible by 3, meaning that p divides an odd Lucas number. QED.

I ran an experiment. I copied multiple choices from the 2007 AMC 8 into a file and asked my son Sergei to try to guess the answers, looking only at the choices. I allowed him to keep several choices. The score I assigned depended on the number of leftover choices. If the leftover choices didn’t contain the right answer, the score for the problem was zero. Otherwise, it was scaled according to the number of choices he left. For example, if he had four choices left and the right answer was among them he got 1/4 of a point. Here are the choices:

1. 9, 10, 11, 12, 13.
2. 2/5, 1/2, 5/4, 5/3, 5/2.
3. 2, 5, 7, 10, 12.
4. 12, 15, 18, 30, 36.
5. 24, 25, 26, 27, 28.
6. 7, 17, 34, 41, 80.
7. 25, 26, 29, 33, 36.
8. 3, 4.5, 6, 9, 18.
9. 1, 2, 3, 4, cannot be determined.
10. 13, 20, 24, 28, 30.
11. Choose picture: I, II, III, IV, cannot be determined.
12. 1:1, 6:5, 3:2, 2:1, 3:1.
13. 503, 1006, 1504, 1507, 1510.
14. 5, 8, 13, 14, 18.
15. a+c < b, ab < c, a+b < c, ac < b, b/c = a.
16. Choose picture: 1, 2, 3, 4, 5.
17. 25, 35, 40, 45, 50.
18. 2, 5, 6, 8, 10.
19. 2, 64, 79, 96, 131.
20. 48, 50, 53, 54, 60.
21. 2/7, 3/8, 1/2, 4/7, 5/8.
22. 2, 4.5, 5, 6.2, 7.
23. 4, 6, 8, 10, 12.
24. 1/4, 1/3, 1/2, 2/3, 3/4.
25. 17/36, 35/72, 1/2, 37/72, 19/36.

It is clear that if you keep all choices, your score will be 5, which is the expected score for AMC if you are randomly guessing the answers. Sergei’s total score was 7.77, which is noticeably better than the expected 5.

There were two questions where Sergei felt that he knew the answer exactly. First was question number two with choices: 2/5, 1/2, 5/4, 5/3, 5/2. All but one of the choices has a 5 in it, so 1/2 must be wrong. Numbers 2/5 and 5/2 are inverses of each other, so if organizers expect you to make a mistake by inverting the right answer, then one of these choices must be the right answer. But 5/4 and 5/3 are better suited as a miscalculation of 5/2, than of 2/5. His choice was 5/2, and it was correct. The second question for which he was sure of the answer was question 19, with his answer 79. I still do not have a clue why.

Sergei’s result wasn’t too much better than just guessing. That means that AMC 8 organizers do a reasonably good job of hiding the real answer. You can try it at home and see if you can improve on Sergei’s result. I will publish the right answers as a comment to this essay in a week or so.

My knowledge about vampires comes mostly from the two TV series Buffy The Vampire Slayer and Angel. If you saw these series you would know that vampires can’t stand the sun. Therefore, they can’t get any tan at all and should be very pale. Angel doesn’t look pale but I never saw him going to a tanning spa. Nor did I ever see him taking vitamin D, as he should if he’s avoiding the sun.

But this is not why I’m confused about vampires. My biggest concerns are about vampires that are numbers.

Vampire numbers were invented by Clifford A. Pickover, who said:

If we are to believe best-selling novelist Anne Rice, vampires resemble humans in many respects, but live secret lives hidden among the rest of us mortals. Consider a numerical metaphor for vampires. I call numbers like 2187 vampire numbers because they’re formed when two progenitor numbers 27 and 81 are multiplied together (27 * 81 = 2187). Note that the vampire, 2187, contains the same digits as both parents, except that these digits are subtly hidden, scrambled in some fashion.

Some people call the parents of a vampire number fangs. Why would anyone call their parents fangs? I guess some parents are good at blood sucking and because they have all the power, they make the lives of their children a misery. So which name shall we use: parents or fangs?

Why should parents have the same number of digits? Maybe it’s a gesture of gender equality. But there is no mathematical reason to be politically correct, that is, for parents to have the same number of digits. For example, 126 is 61 times 2 and thus is the product of two numbers made from its digits. Pickover calls 126 a pseudovampire. So a pseudovampire with asymmetrical fangs, is a disfigured vampire, one whose fangs have a different number of digits. Have you ever seen fangs with digits?

In the first book where vampires appeared Keys to Infinity the vampire numbers are called true vampire numbers as opposed to pseudovampire numbers.

We can add a zero at the end of a pseudovampire to get another pseudovampire, a trivial if obvious observation. To keep the parents equal, we can add two zeroes at the end of a vampire to get another vampire. Adding zeroes is not a very intellectual operation, but a vampire that can’t be created by adding zeroes to another vampire is more basic and, thus, more interesting. In the book Wonders of Numbers: Adventures in Mathematics, Mind, and Meaning a vampire where one of the multiplicands doesn’t have trailing zeroes is called a true vampire, as opposed to just a vampire. Thus, the trueness of vampires changes from book to book, adding some more confusion. It looks like the second definition of a true vampire is more widely adopted, so I will stick to it.

By analogy, we should call pseudovampires that do not end in zeroes, true pseudovampires. It’s interesting to note that by adding zeroes we can get a true vampire from any pseudovampire that is not a vampire. You see how easy it is to build equality? Just add zeroes.

A true vampire might not be true as a pseudovampire. For example, a vampire number 1260 = 20 * 61 is generated by adding a zero to a pseudovampire 126 = 2 * 61. In this case, the pseudovampire is truer than the vampire. Why does something more basic get a prefix “pseudo”?

Here’s another question. Why do vampires have to have two fangs? Can a vampire have three fangs? For example, 11439 = 9 * 31 * 41. This generalization of vampires should be called mutant vampires. Or multi-gender vampires.

To create more confusion, a mutant vampire can, at the same time, be a simple vampire: 1395 = 31 * 9 * 5 = 15 * 93.

Of course, nothing prevents a mutant vampire from being politically correct, that is, to have multiple and equal parents with the same number of digits, as in 197925 = 29 * 75 * 91.

People continue creating a mess with vampires. For example, a definition of a prime vampire number is floating around the Internet. When you look at this name, your first reaction is that a prime vampire is a prime number. But a vampire is never prime as it is always a product of numbers. By definition a prime vampire is a vampire with prime multiplicands, for example 124483 = 281 * 443. So “prime vampire number” is a very bad name. We should call these vampires prime-fanged vampires — this would be much more straightforward.

To eliminate some of this confusion, we mathematicians should go back and rename vampires consistently. But in the meantime, check out the illustration of vampire numbers shown above that I found at flickr.com with this description:

Like the count von Count in Sesame Street, there is a tradition that vampires suffer terribly from arithromania: the compulsion to count things. To keep vampires from wreaking murderous havoc at night, poppy seeds were strewn about their resting places. On waking, the vampire would be compelled to count the seeds. It would take him all night, and keep him from mischief.

My knowledge about vampires comes mostly from the two TV series Buffy The Vampire Slayer and Angel . If you saw these series you would know that vampires can’t stand the sun. Therefore, they can’t get any tan at all and should be very pale. Angel doesn’t look pale but I never saw him going to a tanning spa. Nor did I ever see him taking vitamin D, as he should if he’s avoiding the sun.

But this is not why I’m confused about vampires. My biggest concerns are about vampires that are numbers.

Vampire numbers were invented by Clifford A. Pickover, who said:

If we are to believe best-selling novelist Anne Rice , vampires resemble humans in many respects, but live secret lives hidden among the rest of us mortals. Consider a numerical metaphor for vampires. I call numbers like 2187 vampire numbers because they’re formed when two progenitor numbers 27 and 81 are multiplied together (27 * 81 = 2187). Note that the vampire, 2187, contains the same digits as both parents, except that these digits are subtly hidden, scrambled in some fashion.

Some people call the parents of a vampire number fangs. Why would anyone call their parents fangs? I guess some parents are good at blood sucking and because they have all the power, they make the lives of their children a misery. So which name shall we use: parents or fangs?

Why should parents have the same number of digits? Maybe it’s a gesture of gender equality. But there is no mathematical reason to be politically correct, that is, for parents to have the same number of digits. For example, 126 is 61 times 2 and thus is the product of two numbers made from its digits. Pickover calls 126 a pseudovampire. So a pseudovampire with asymmetrical fangs, is a disfigured vampire, one whose fangs have a different number of digits. Have you ever seen fangs with digits?

In the first book where vampires appeared Keys to Infinity the vampire numbers are called true vampire numbers as opposed to pseudovampire numbers.

We can add a zero at the end of a pseudovampire to get another pseudovampire, a trivial if obvious observation. To keep the parents equal, we can add two zeroes at the end of a vampire to get another vampire. Adding zeroes is not a very intellectual operation, but a vampire that can’t be created by adding zeroes to another vampire is more basic and, thus, more interesting. In the book Wonders of Numbers: Adventures in Mathematics, Mind, and Meaning a vampire where one of the multiplicands doesn’t have trailing zeroes is called a true vampire, as opposed to just a vampire. Thus, the trueness of vampires changes from book to book, adding some more confusion. It looks like the second definition of a true vampire is more widely adopted, so I will stick to it.

By analogy, we should call pseudovampires that do not end in zeroes, true pseudovampires. It’s interesting to note that by adding zeroes we can get a true vampire from any pseudovampire that is not a vampire. You see how easy it is to build equality? Just add zeroes.

A true vampire might not be true as a pseudovampire. For example, a vampire number 1260 = 20 * 61 is generated by adding a zero to a pseudovampire 126 = 2 * 61. In this case, the pseudovampire is truer than the vampire. Why does something more basic get a prefix “pseudo”?

Here’s another question. Why do vampires have to have two fangs? Can a vampire have three fangs? For example, 11439 = 9 * 31 * 41. This generalization of vampires should be called mutant vampires. Or multi-gender vampires.

To create more confusion, a mutant vampire can, at the same time, be a simple vampire: 1395 = 31 * 9 * 5 = 15 * 93.

Of course, nothing prevents a mutant vampire from being politically correct, that is, to have multiple and equal parents with the same number of digits, as in 197925 = 29 * 75 * 91.

People continue creating a mess with vampires. For example, a definition of a prime vampire number is floating around the Internet. When you look at this name, your first reaction is that a prime vampire is a prime number. But a vampire is never prime as it is always a product of numbers. By definition a prime vampire is a vampire with prime multiplicands, for example 124483 = 281 * 443. So “prime vampire number” is a very bad name. We should call these vampires prime-fanged vampires — this would be much more straightforward.

To eliminate some of this confusion, we mathematicians should go back and rename vampires consistently. But in the meantime, check out the illustration of vampire numbers shown above that I found at flickr.com with this description:

Like the count von Count in Sesame Street, there is a tradition that vampires suffer terribly from arithromania: the compulsion to count things. To keep vampires from wreaking murderous havoc at night, poppy seeds were strewn about their resting places. On waking, the vampire would be compelled to count the seeds. It would take him all night, and keep him from mischief.

This puzzle is a generalization of a problem from the 1977 USSR math Olympiad:

At the beginning of the game you are given a polynomial, which has 1 as its leading coefficient and 1 as its constant term. Two people play. On your turn you assign a real value to one of the unknown coefficients. The person that goes last wins if the polynomial has no real roots at the end. Who wins?

It is clear that if the last person’s goal is for the polynomial to have a root, then the game is trivial: in this case, he can always make 1 a root with the last move. Also, an odd degree polynomial always has a real root. Therefore, to make the game interesting we should assume that the degree of the polynomial is even.

Though I can’t imaging myself ever being interested in playing this game, figuring out the strategy is a lot of fun.

Once I witnessed John H. Conway factoring large numbers in his head. Impressed, I stared at him. Encouraged by my interest, he told me that if I ever want to be able to factor large numbers, I should know all the primes below one thousand.

The secret to knowing all such primes is to remember the composites, he continued. Obviously, we don’t need to remember trivial composites — the ones divisible by 2, 3, 5, or 11. Also, everyone knows all the squares below one thousand, so we can count squares as trivial composites. We only need to remember the non-trivial composites. There are not that many of them below one thousand — only 70. I mean, 70 is nothing compared to the number of primes: 168.

So, I need to remember the following seventy numbers:

91, 119, 133, 161, 203, 217, 221, 247, 259, 287, 299, 301, 323, 329, 343, 371, 377, 391, 403, 413, 427, 437, 469, 481, 493, 497, 511, 527, 533, 551, 553, 559, 581, 589, 611, 623, 629, 637, 667, 679, 689, 697, 703, 707, 713, 721, 731, 749, 763, 767, 779, 791, 793, 799, 817, 833, 851, 871, 889, 893, 899, 901, 917, 923, 931, 943, 949, 959, 973, 989.

If you are very ambitious and plan to learn the primes up to 50,000, then the trick of learning non-trivial composites instead of primes is of no use to you. Indeed, for larger numbers the density of primes goes down, while the density of non-trivial composites stays about the same or increases very slightly due to a smaller number of squares.

The turning point is around 11,625: the number of primes below 11,625 equals the number of non-trivial composites below it. So, compare your ambition to 11,625 and tailor your path of learning accordingly.

If you are lazy, you can learn primes only up to 100. In this case your path is clear; you should stick with remembering non-trivial composites, for you need to remember only one number: 91.

For his Master’s thesis, my son Alexey Radul wrote the Symmetriad — a computer program to compute and display highly symmetric objects.

The objects the Symmetriad is after are the 4D generalizations of Platonic and Archimedean solids. His thesis contains a picture gallery made with the Symmetriad, and these are my three favorite pictures.

The pictures have cute titles. The problem is that I still haven’t installed the WordPress plugin that allows me to put captions under the pictures. That is why you have to guess which title matches which picture. The titles are: “The Planets are Aligned,” “Great Jaws,” and “Diamonds are Forever.”

If you are wondering why one of the pictures shows a non-connected object, in fact the object is connected, but some of the edges are white, so that you can better see 3D cells of the object.

Subsequent to the publication of this thesis, Alexey enhanced his software to make even more dramatic pictures. The following pictures have no titles, so feel free to suggest some:

I think it would be nice to publish a book with all these pictures. As a book on symmetries should be published on a symmetric date, the next opportunity would be 01.02.2010.

I love Grand Tour puzzles more than I love Sudoku. You are given a graph, for example, a square grid like the one on the left. Some edges are highlighted. You need to find a closed path that visits each vertex exactly once and includes the highlighted edges as part of the path. Mathematically speaking you need to reconstruct a Hamiltonian circuit on a graph, once you are given a part of it. The highlighted edges are chosen to guarantee a unique solution to the puzzle.

On the left you can see a sample grand tour problem with its solution. This puzzle was designed by Glenn Iba. On Glenn’s Grand Tour Puzzle Page you can find many grand tour puzzles of varying levels of difficulty. The puzzles are playable. That is, you can click or unclick an edge. You can also branch out in a different color, which is especially useful for difficult puzzles when you want to test a hypothesis. I just want to warn you: these puzzles are addictive — I couldn’t stop playing until I solved all of them.

Below there is a simple grand tour puzzle from Glenn’s collection, but this time on a triangular grid:

You do not need a grid to construct a puzzle. But these puzzles look very natural on grids. I tried to analyze square grid puzzles a little bit. The first important point is that for square grids with odd number of vertices on each side of the square, Hamiltonian cycles do not exist. This point is easier to prove for directed Hamiltonian cycles. You can make a directed cycle from an undirected one by choosing a direction. If you have a directed cycle on a square grid, then the number of edges pointing up should be the same as the number of edges pointing down. We can say the same thing about edges on the cycle pointing left or right. Hence, the number of edges of a Hamiltonian cycle on a square grid should be even. At the same time, the number of edges of any Hamiltonian cycle equals the number of vertices.

I just proved that you need only consider square grids with an even number of vertices on each side. For square grids with two vertices on each side, there is only one Hamiltonian cycle, namely the border of the square. The only grand tour puzzle for this grid won’t have highlighted edges at all. For a square grid with four vertices on each side there are only two different Hamiltonian cycles up to isomorpshisms:

If we count all the reflections and rotations, we will get six Hamiltonian cycles. The next picture shows all 11 grand tour puzzles on this grid. If we count rotations and reflections, we will get 66 different grand tour puzzles.

Below are the sequences associated with this puzzle. Except for one case, I do not know if these sequences are in the Online Encyclopedia for Integer Sequences. I don’t know because I only counted two terms of each sequence, and this information is not sufficient to identify the sequence.

• A003763 Number of Hamiltonian cycles on 2n by 2n square grid of points. The sequence starts 1, 6, 1072, ….
• Number of Hamiltonian cycles up to isomorphism on 2n by 2n square grid of points. The sequence starts 1, 2, ….
• Number of Grand Tour puzzles on 2n by 2n square grid of points. The sequence starts 1, 11, ….
• Number of Grand Tour puzzles up to isomorphism on 2n by 2n square grid of points. The sequence starts 1, 66, ….
• The smallest number of edges in a Grand Tour puzzle on 2n by 2n square grid of points. The sequence starts 0, 2, ….
• The largest number of edges in a Grand Tour puzzle on 2n by 2n square grid of points. The sequence starts 0, 4, ….

If you look at Glenn Iba’s 6 by 6 square grid puzzles, you can see that the smallest number of edges is not more than 6. And the largest number of edges is no less than 12.

You can also make similar sequences for a triangular grid.

I invite you to calculate these sequences and submit them to the OEIS, if they are not already there.

I’m not kidding; there is such a paper. It is titled, “A Headache-causing Problem” and its authors are Conway (J.H.), Paterson (M.S.), and Moscow (U.S.S.R.). The acknowledgements in the paper shed some light on how Moscow became a mathematician:

The work described here was carried out when the first and second named authors enjoyed the hospitality of the third. The second and third authors are indebted to the first for expository details. The first and third authors gratefully remark that without the constant stimulation and witty encouragement of the second author this paper

[The next part was meant to be on the following page, Conway told me, but the editor missed the humor and just continued the sentence…]

was completed.

As a consequence of this joke, Moscow is envied by many mathematicians as it has an Erdős number of 2. Now wait for a couple of hundred years, and Moscow will be the only living mathematician with an Erdős number of 2. I can just imagine future mathematicians trying to persuade Moscow to coauthor papers with them, because this will be the only way for them to score an Erdős number of 3.

Even though I lived there for 30 years, I had no idea that Moscow had a talent for math. Of course, this talent only emerged when Moscow was more than 800 years old.

This wonderful paper by Moscow was very difficult to find. It was presented to Hendrik W. Lenstra on the occasion of his doctoral examination. It was published in 1977 in a book titled “Een pak met een korte broek,” which in Dutch means, “A Book in Short Trousers.”

I tried to find it on the Internet — it wasn’t there. I asked John Conway — it took him quite some time to find it. Here is the picture of John Conway searching for a headache-causing problem. Luckily for you and me, he found it. To save you from another headache, I am uploading the scan of it in pdf format here: A Headache-causing Problem by J.H. Conway, M.S. Paterson, and U.S.S.R. Moscow.

I hope that Moscow will not start complaining that I never asked its permission to post the paper. Some might argue that Moscow, U.S.S.R., doesn’t exist anymore, but I would counter that it exists, but with a changed name. If Moscow tries to sue me, I hope it’s not because it is still bitter that I left it behind in 1990.

Hey Moscow, it’s time we were friends again. Would you like to co-author a paper with me?