Tanya Khovanova's Math Blog

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 have forgotten all the problems from the math Olympiads I participated in, except for one. That problem cost me one point, and as result I didn’t get a perfect score. Here is Problem Number 4 from the 1976 IMO:

Determine the greatest number that is the product of some positive integers, and the sum of these integers is 1976.

The solution goes like this. Consider a partition of 1976. If there is an integer x in the partition greater than 5, then replacing it with two integers x-2 and 2 gives a bigger product. Replacing 4 in the partition by 2 and 2 doesn’t change the product. On the other hand, if there is 1 in the partition, it is profitable to attach it to any other number: to replace 1 and x by x+1.

That means the maximum-product partition of a number greater than one has to consist only of twos and threes. If we have three twos in the partition, we can replace them by two threes, thus increasing the product. That means that our partition should consist of twos and threes, and the number of twos shouldn’t be greater than three.

I lost my point when I made an elementary arithmetic mistake while calculating the remainder of 1976 modulo 3.

Let’s generalize this problem for any number n. If we define the maximum product of partitions a(n) as a function of the number n, we get the sequence A000792 in the Encyclopedia of Integer Sequences. This sequence has other interesting definitions, which appear if we add some structure to partitions of n.

For example, we can suppose that our n is not just a number, but it also represents n objects that are being permuted by the symmetric group S_n. In this case, we can associate some Abelian subgroups of the symmetric group with every partition. Namely, we can divide the set of n elements into disjoint subsets of sizes corresponding to our partition and cycle the elements in every subset. The product of these cycles is an Abelian subgroup, the order of which doesn’t depend on how exactly we partition or how we cycle. This order is the product of the partition numbers. It appears that we can get all maximal Abelian subgroups of the symmetric group in this manner. Thus the sequence a(n) is the maximum size of the maximal Abelian subgroup in a symmetric group S_n.

We can do something else with our n objects. Suppose they are represented by vertices of a graph. We take any partition of n and divide our graph into groups of vertices according to this partition. Next we connect vertices of the graph that belong to different groups. If we take a vertex from every group, then the corresponding subgraph is a clique. We can see that it is a maximal clique as two vertices from the same group are never connected and we can’t add any more vertices. The number of different cliques of this size is the product of the number of elements in every group. We can prove that a(n) is the maximum possible number of maximal cliques in a graph with n vertices.

Even though I can’t return to 1976 to correct my stupid mistake, I love this problem and the corresponding sequence.

How many disjoint letters “O” can you draw on a plane? Clearly, the answer is infinitely many. Now the real question is whether this number can be uncountable. We assume that the letter “O” is just a circle. In this case, if we draw all possible concentric circles, we get an uncountable number of circles. I call letters for which you can build a disjoint uncountable set on the plane stackable letters.

If we do not allow one letter “O” to be drawn inside another, then we can prove that it is not possible to draw more than a countable number of letters “O”. Indeed, we can always find a special point inside each letter “O” with both coordinates being rational numbers. As circles do not overlap, no two circles can share the same special point. The set of ordered pairs of rational numbers is countable, thus this set of letters “O” has to be countable.

Now let us move to a more interesting question: how many disjoint letters “T” can you draw on a plane? Let us define the letter “T” as two perpendicular line segments, where the end point of one of the segments is the middle point of the other. Prove that the letter “T” is unstackable.

Can you analyze the whole alphabet for stackability? It doesn’t matter which alphabet — you just need to define the drawing of every letter in some reasonable manner.

My real question is, can you define the drawings of English letters in such a way that they are recognizable and the number of stackable letters is maximized? What is the biggest number of letters that you can make stackable?

I brought up this subject to my son Sergei one fine morning. As a result we came up with the following breakfast theorem: Stackable letters are topologically equivalent to a line segment or a circle.

And here is our proof. If the letter in question has a part which is topologically equivalent to “T”, then the letter is not stackable, because “T” is not stackable.

Fortunately, or unfortunately, this is not the whole story. The stackability of each letter depends on how you define its shape. There are two groups of letters in the English alphabet that you can make stackable. The first group consists of letters like “V” that can be described as a function on a segment and stacked as a parallel translation of the same letter. The second group, like letter “O”, can be described as a function of an angle in polar coordinates, and stacked as a dilation of the same letter.

It is clear that this doesn’t cover all cases. For example, a piece of a spiral can be stackable — you stack it by rotating it.

I invite my readers to define two different homeomorphic shapes for the same letter — stackable and unstackable.

Romanian Masters in Mathematics might become the most prestigious math competition for high school students. Romania invites teams from the 20 best countries from the previous year’s International Math Olympiad. This way they guarantee that all the competitors are extremely strong and they can give more difficult problems than the usual IMO problems.

This year they held their second competition and my son, Sergei Bernstein, was invited to join the USA team. Here is one of the problems:

For a finite set X of positive integers, define ∑(X) as the sum of arctan(1/x) for all elements x in X. Given a finite set S, such that ∑(S) < π/2, prove that there exists a finite set T such that S is a subset of T and ∑(T) = π/2.

The official solution involved a tedious greedy algorithm. My son, Sergei, got an extra point for his unique solution, which was very different from the official one. Here’s his solution:

To an integer x we associate a complex number x + i, which in polar coordinates is r(cosθ + i sinθ). Note that θ equals arctan(1/x). That means we can interpret ∑(X) as the angle in the polar form of ∏(x+i) — the product of (x+i) for all x in X.

We are given a set S with elements s_j such that ∏(s_j+i) has a positive real part, and we need to find other elements t_k, such that ∏(s_j+i) multiplied by ∏(t_k+i) has real part 0.

But we know that the ∏(s_j+i) = a + bi, for some integers a and b. I claim that we can always find a positive integer y, so that (a + bi)(y + i) has a smaller real part than a.

Note that ∑(S) < π/2, hence a and b are positive. By an ever so slightly modified version of the division algorithm we can find integers q and r such that b = aq + r, 0 < r ≤ a. Now simply set y equal to q + 1 (which is a positive integer). Then the real part of (a + bi)(y + i) equals ay – b = a – r, and is a non-negative number less than a.

Now we just repeat the process, which obviously has a finite number of steps.

My son’s solution wasn’t complete. The problem talked about sets of numbers and that implies that the numbers should be distinct. I leave it to my readers to finish this solution for distinct numbers.

You are visiting your girlfriend and she orders pizza. Your evil girlfriend has perfect eyesight and notices fly droppings in three places on the pizza. She is seeking revenge on you for refusing to babysit her poodle and proceeds to cut the pizza. Assuming that the droppings occurred independently and in random places, what is the probability that she will be able to cut a half-circle pizza slice with all three droppings in one half?

Now that I’ve ruined your appetite, here’s a simpler puzzle you can solve as an appetizer to the one I just gave you above. In this puzzle, you have a bread stick that is, of course, in the shape of a line segment. Your girlfriend notices that the bread stick also has three fly droppings on it and she offers to cut an exact half length out of the middle for you. What is the probability that she can cut it so that you end up with all the droppings?

Can you bear to continue this tasty discussion? If so, I’ll give you an even simpler problem. Try to solve both of the above puzzles — but with only two droppings instead of three.

If I’ve spoiled your appetite so completely that you’ll skip dinner, why not use the extra time to solve the most challenging of these problems with a meatball that has four fly droppings on it. These droppings, too, are in random places and you must calculate the probability that you are able to cut a semi-sphere with all four droppings on one side.

This final puzzle — in a less distasteful setting — was sent to me by Nick Petry.

I have an old book which I value a great deal. The book is called 200 Problems in Linguistics and Mathematics and only 1,550 copies were printed in 1972. Luckily, a new extended edition just appeared on the web. Both editions are in Russian, so I decided to translate some of the problems into English. Here is a sample:

Problem 1. Here are phrases in Swahili with their English translations:

• atakupenda — He will love you.
• nitawapiga — I will beat them.
• atatupenda — He will love us.
• anakupiga — He beats you.
• nitampenda — I will love him.
• unawasumbua — You annoy them.

Translate the following into Swahili:

• You will love them.
• I annoy him.

Problem 2. You are given words in Swahili: mtu, mbuzi, jito, mgeni, jitu and kibuzi. Their translations in a different order are: giant, little goat, guest, goat, person and large river. Make the correspondence.

Problem 3. In Russian the middle name is the patronymic. Thus, the middle initial is the first letter of the father’s first name. And, as in many languages, the first initial is the first letter of the first name. Here are names of males in a family:

• A.N. Petrov
• B.M. Petrov
• G.K. Petrov
• K.M. Petrov
• K.T. Petrov
• M.M. Petrov
• M.N. Petrov
• N.M. Petrov
• N.K. Petrov
• N.T. Petrov
• T.M. Petrov

Draw the family tree of the Petrovs, given that every father has two sons, the patriarch of the family has four grandsons, and his sons have two grandsons each. Prove that the solution is unique.

Problem 4. In Latvian a noun can be one of two genders; furthermore, adjectives agree with nouns in gender, number and case. You are given phrases in either the nominative or the genitive case with their translations:

• silts ezers — warm lake
• melns lauva — black lion
• liela krāsns — big oven
• lielas jūras — big sea’s
• sarkana ezera — red lake’s
• melna kafija — black coffee
• sarkans putns — red bird
• liela kalna — big mountain’s
• sarkanas lapas — red leaf’s
• sarkana pils — red castle
• liels ezers — big lake
• melna putna — black bird’s
• liela lauvas — big lion’s
• silta jūra — warm sea
• melnas kafijas — black coffee’s
• liels kalns — big mountain

Indicate which words are nouns and which are adjectives. Divide Latvian nouns into two groups, so that each group contains words of the same gender.

Problem 5. The Portuguese language takes its roots from Latin. In this problem modern Portuguese words are written on the left and their roots (in Latin and other languages on the right). All the words on the left belong to one of three classes: ancient borrowing, early borrowing and late borrowing.

• chegar — plicare
• praino — plaine
• plátano — platanum
• chão — planum
• plebe — plebem
• cheio — plenum
• prancha — planche

For every Portuguese word, indicate which class it belongs to. (Note that in Portuguese “ch” is pronounced as “sh”.)

I love “Knights and Knaves” logic puzzles. These are puzzles where knights always tell the truth and knaves always lie. A beautiful variation of such a puzzle with Gnyttes and Mnaivvs was given at the 2009 MIT mystery hunt. In this puzzle people’s ability to tell the truth changes during the night depending on the sex partner. You will enjoy figuring out who is a Gnytte or a Mnaivv for each day, who is infected and who slept with whom on each night. Just remember that the ultimate answer to the puzzle is a word or a phrase. So there is one more step after you solve the entire logic part. You do not really need to do this last step, but you might as well. Here we go:

The Sexaholics of Truthteller Planet

Each inhabitant of Veritas 7, better known as the Truthteller Planet, manifests one of two mutations: Gnytte or Mnaivv. Gnyttes always tell the truth, and Mnaivvs always lie. Once born a Gnytte or Mnaivv, the inhabitant can never change…until now.

Veritas 7 is in the midst of an outbreak of a nasty virus dubbed Nallyums Complex II. If an infected inhabitant has sex with another inhabitant of that planet, each one can be converted from Gnytte to Mnaivv, or Mnaivv to Gnytte, as shown below:

• If the sex is heteroverific (1 Gnytte and 1 Mnaivv), both become Mnaivvs.
• If the sex is homoverific (2 Gnyttes or 2 Mnaivvs), both become Gnyttes.

This occurs if either party or both parties have the disease. The disease itself is not transmitted via sex, which is some relief.

Several members of a Veritas 7 village have just contracted Nallyums Complex II. Below are statements from the 15 village residents, taken over the 5-day period since the outbreak. Interviews were taken each morning, and sex occured only at night. Each night, residents paired off to form seven separate copulating couples, with one individual left out. No individual was left out for more than one night.

Can you identify the infected individuals and track their pattern of sexual activity?

A note on wording: If someone refers to sex with someone who was a Gnytte or Mnaivv, they are referring to the individual’s truth-telling status just before sex. If a clue says that two individuals had sex, it means they had sex with each other. “Mutation” refers to the individual’s current status as Gnytte or Mnaivv.

Interviews Day 1

• Artoo: Etrusco is not infected.
• Bendox: Cravulon and Flav are not the same mutation today.
• Cravulon: Either Artoo or Flav is a Gnytte today.
• Dent: Jax-7 and I are both Mnaivvs today.
• Etrusco: Greasemaster is a Mnaivv today.
• Flav: There are at least five Mnaivvs today.
• Greasemaster: Murgatroid is a Gnytte today.
• Holyoid: Etrusco is either an infected Mnaivv or an uninfected Gnytte today.
• Irono: Etrusco is a Mnaivv today.
• Jax-7: Among Artoo, Greasemaster, and Nebulose, exactly one is a Gnytte today.
• Killbot: Bendox is a Gnytte today.
• Lexx: Holyoid and Irono are not the same mutation today.
• Murgatroid: There are at least eight Gnyttes today.
• Nebulose: Murgatroid is a Mnaivv today.
• Oliver: Lexx is a Gnytte today.

Surveillance Night 1

Security cameras revealed that Jax-7 did not have sex with anyone last night, and that Nebulose and Murgatroid had sex.

Interviews Day 2

• Artoo: Last night, I did not have sex with an infected individual.
• Bendox: Last night, Oliver had sex with someone who was a Mnaivv.
• Cravulon: There are at least six Gnyttes today.
• Dent: If there are only five Gnyttes today, then Cravulon and Oliver had sex last night.
• Etrusco: Last night, Bendox had sex with someone who was a Mnaivv.
• Flav: Neither Killbot nor her partner last night is infected.
• Greasemaster: Last night, either Cravulon and Bendox had sex, or Oliver and Etrusco had sex, but not both.
• Holyoid: Last night, I had sex with someone who was a Gnytte.
• Irono: Last night, I did not have sex with Flav.
• Jax-7: Last night, Artoo had sex with Dent.
• Killbot: Last night, I had sex with someone who was a Mnaivv.
• Lexx: Cravulon is infected.
• Murgatroid: Nebulose is not infected.
• Nebulose: Last night, either Cravulon or Flav had sex with Etrusco.
• Oliver: Last night, Irono had sex with someone who was a Mnaivv.

Surveillance Night 2

Security cameras revealed that Holyoid did not have sex with anyone last night, and that Irono and Oliver had sex.

Interviews Day 3

• Artoo: Last night, I had sex with the individual who had sex with Murgatroid on Night 1.
• Bendox: Last night, I had sex with an uninfected individual.
• Cravulon: Last night, Jax-7 had sex with the individual who had sex with Lexx on Night 1.
• Dent: Last night, I had sex with someone who was a Mnaivv.
• Flav: Last night, I had sex with an uninfected individual.
• Holyoid: Neither Oliver nor Irono is infected.
• Irono: Last night, the individual who had sex with Oliver on Night 1 had sex with someone who was a Mnaivv.
• Jax-7: There are more than seven Gnyttes today.
• Killbot: Bendox and Murgatroid are the same mutation today.
• Lexx: Last night, the individual who had sex with Flav on Night 1 had sex with an infected individual.
• Nebulose: The individual who had sex with Etrusco on Night 1 is a Mnaivv today.
• Oliver: Last night, Lexx had sex with the individual who had sex with Bendox on Night 1.

Surveillance Night 3

Security cameras revealed that Dent and Jax-7 had sex.

Interviews Day 4

• Artoo: There are at most seven Gnyttes today.
• Cravulon: Last night, I had sex with someone who has never been a Gnytte.
• Dent: The two sex partners of an individual who was a Mnaivv on the first three days had sex last night.
• Etrusco: Last night, Oliver did not have sex.
• Flav: Last night, I had sex with an infected individual.
• Greasemaster: There are exactly eight infected individuals.
• Killbot: None of the individuals who have been left out of the sexual activity is infected.
• Murgatroid: Last night, I had sex with someone who was a Gnytte on Day 1.
• Nebulose: Last night, I had sex with someone who has had sex with Artoo.
• Oliver: Last night, an individual who had sex with Holyoid during one of the first two nights had sex with an individual who had sex with Jax-7 during one of the first two nights.

Surveillance Night 4

Security cameras have been vandalized.

Interviews Day 5

• Artoo: Last night, I had sex with a Mnaivv, but not the individual who, on Night 3, had sex with the individual who, on Night 2, had sex with the individual who, on Night 1, had sex with Dent.
• Bendox: Last night, Dent had sex with the individual who, on Night 1, had sex with the individual who, on Night 3, had sex with the individual who, on Night 2, had sex with Artoo.
• Cravulon: Last night, Holyoid had sex with the individual who, on Night 3, had sex with the individual who, on Night 2, had sex with the individual who, on Night 1, had sex with Murgatroid.
• Dent: Jax-7 is a Gnytte today.
• Flav: Last night, Cravulon did not have sex with the individual who, on Night 1, had sex with the individual who, on Night 2, had sex with the individual who, on Night 3, had sex with Cravulon.
• Jax-7: Last night, Greasemaster did not have sex.
• Nebulose: Last night, Killbot either had sex with an uninfected individual or did not have sex.

I do not remember where I dug this logic puzzle out from:

Folks living in Trueton always tell the truth. Those who live in Lieberg, always lie. People living in Alterborough alternate strictly between truth and lie. One night 911 got a call: “Fire, help!” The operator couldn’t identify the phone number, so he asked, “Where are you calling from?” The reply was Lieberg.
Assuming no one had overnight guests from another town, where should the firemen go?

After you have solved this problem, you will see that the sentence “Fire, help!” is true. I wonder, if this statement were a lie, how would we interpret it? It could be that it is just a joke and there is no fire and therefore no need for the police. Or it could be that help is needed — perhaps for a robbery, but not for a fire.

However, when you solve this puzzle, you’ll find out that the “Fire, help!” statement is true, so you do not need to wonder what it would mean if the statement were a lie. But I wondered, and as a result I invented several new puzzles.

Here is the first one:

Let’s say that people will call the police only if something is happening and there are only two things that could be happening: fire or robbery. Suppose that when people calling the police need to lie, they replace fire with robbery, and vice versa. Suppose also that when asked about location, people will not say, &quotI am not from Lieberg,” as they could have, but will always reply with one word, which is a name of one of three towns. So, there are two possibilities for the first statement — Fire, Help! or Robbery, Help! — and three possible answers to the question about location. — Trueton, Lieberg and Alterborough. My puzzle in this case is: What answer to the location question will give the biggest headache to the police?

We can branch the original puzzle out in a different way. Here is my second puzzle:

We can assume that only fire, not robbery, could happen in this remote place and when people call the police they either say “Fire, help!” or “We do not have a fire, thank you for your time.” Let us again assume that people will call the police only if something is happening. In this case, what combinations of first and second sentences of the callers will never happen?

My third puzzle is a more complicated version of the second puzzle:

Let’s assume that fires happen with the same probability in every town and that the cost of sending a team of firefighters is identical for every town. Furthermore, the ferocity of all the fires is minuscule and the cost of sending a team is the same, whether or not it turns out that there is a fire. If the police think that the call could have come from several places, they send several teams and the cost multiplies accordingly. The police are obsessed with making charts and the most important number they analyze is the cost they incur, depending on the content of the first sentence of each call.
Given that there are frequent fires, what could the ratio of the cost to police for the calls “Fire, help!” be compared to those that begin with “We do not have a fire, thank you for your time”?

Computational Linguistics Olympiads started in Moscow in 1962. Finally in 2007 the US caught up and now we have the NACLO — North American Computational Linguistics Olympiad.

The problems from past Soviet Olympiads are hard to find, so here I present a translation from Russian of a sample problem from the Moscow Linguistics Olympiad website:

You are given sentences in Niuean language with their translations into English:

1. To lele e manu. — The bird will fly.
2. Kua fano e tama. — The boy is walking.
3. Kua koukou a koe. — You are swimming.
4. Kua fano a ia. — He is walking.
5. Ne kitia he tama a Sione. — The boy saw John.
6. Kua kitia e koe a Pule. — You are seeing Pule.
7. To kitia e Sione a ia. — John will see him.
8. Ne liti e ia e kulï. — He left the dog.
9. Kua kai he kulï e manu. — The dog is eating the bird.

Translate the following sentences into Niuean:

1. John swam.
2. You will eat the dog.
3. Pule is leaving you.
4. The bird will see the boy.
5. The dog is flying.

Here is a baby puzzle. On Monday the baby said A, on Tuesday AU, on Wednesday AUUA, on Thursday AUUAUAAU. What will she say on Saturday?

You can see that this very gifted baby increases her talking capacity twice each day. The first half of what she says repeats her speech of the day before; and the second half is like the first half, but switches every A and U. If the baby continues indefinitely, her text converges to an infinite sequence that mathematicians call the Thue-Morse sequence (A010060). Of course, mathematicians use zeroes and ones instead of A and U, so the sequence looks like 0110100110010110100….

This sequence has many interesting properties. For example, if you replace every zero by 01 and every one by 10 in the Thue-Morse sequence, you will get the Thue-Morse sequence back. You can see that this is so if you code A in the baby’s speech by 01 and U by 10. Thus the Thue-Morse sequence is a fixed point under this substitution. Moreover, the only two fixed points under this substitution are the Thue-Morse sequence and its negation (A010059).

The Thue-Morse sequence possesses many other cool properties. For example, the sequence doesn’t contain substrings 000 and 111. Actually any sequence built from the doubles 01 and 10 can’t contain the triples 000 and 111 because we switch the digit after every odd-indexed term of such a sequence. A more general and less trivial statement is also true for the Thue-Morse sequence: it doesn’t contain any cubes. That is, it doesn’t contain XXX, where X is any binary string.

I stumbled upon this sequence when I was playing with evil and odious numbers invented by John H. Conway. A number is evil if the number of ones in its binary expansion is even, and odious if it’s odd. We can define a function, called the odiousness of a number, in the following way: odiousness(n) is one, if n is odious and 0 otherwise. We can apply the odiousness function to a sequence of non-negative integers term-wise. Now I can describe the Thue-Morse sequence as the odiousness of the sequence of non-negative numbers. Indeed, the odiousness of the number 2ⁿ + k is opposite of the odiousness of k, if k is less than 2ⁿ. That means if we already know the odiousness of the numbers below 2ⁿ, the next 2ⁿ terms of the odiousness sequence is the bitwise negation of the first 2ⁿ terms. So odiousness is built the same way as the Thue-Morse sequence, and you can easily check that the initial terms are the same too.

Let me consider as an example the sequence which is the odiousness of triangular numbers A153638: 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0…. What can we say about this sequence? We can say that the number of zeroes is infinite, as all the terms with indices of the form 2^n-1(2ⁿ+1) are zeroes. Also, the number of ones is infinite because all the terms with indices of the form 2^2n-1(2^2n-1-1) are ones.

Obviously, we can define the evilness of a number or of a sequence with non-negative terms. Namely, the evilness of a number is 1 if the number is evil, and 0 if it is not. The evilness is the bitwise negation of the odiousness. The evilness of the sequence of non-negative integers is the negation of the Thue-Morse sequence. The odiousness sequence of any sequence of zeroes and ones is the sequence itself, and the evilness sequence is its negation.

I would like to define an inverse odiousness operation on binary sequences. Many different sequences can have the same odiousness sequence. In such a case mathematicians usually define the inverse operation as a minimal non-negative sequence whose odiousness is the given sequence. Obviously, the minimal inverse of a binary sequence is the sequence itself, and thus not very interesting. I suggest that we define the inverse as a minimal increasing sequence. In this case the odiousness inverse of the Thue-Morse sequence is the sequence of non-negative numbers.

For example, let me describe the inverse odiousness of the sequence of all ones. Naturally, all the numbers in the sequence must be odious, and by minimality property this is the sequence of odious numbers A000069: 1, 2, 4, 7, 8, 11, 13, 14, 16, 19…. Analogously, the odiousness inverse of the sequence of all zeroes is the sequence of evil numbers A001969: 0, 3, 5, 6, 9, 10, 12, 15, 17, 18, 20….

Let us find the odiousness inverse of the alternating sequence A000035: 0, 1, 0, 1, 0, 1…. This is the lexicographically smallest sequence of numbers changing putridity. By the way, “putridity” is the term suggested by John Conway to encompass odiousness and evilness the same way as parity encompasses oddness and evenness.

The odiousness inverse of the alternating sequence is the sequence A003159: 0, 1, 3, 4, 5, 7, 9, 11, 12, 13…. By my definition we can describe this sequence as indices of terms of the Thue-Morse sequence that are different from the previous term. This sequence can be described in many other ways. For example, the official definition in the OEIS is that this sequence consists of numbers whose binary expansion ends with an even number of zeroes. It is fun to prove that this is the case. It is also fun to show that this sequence can be built by adding numbers to it that are not doubles of previous terms.

Let us look at the first differences of the previous sequence. This is the sequence A026465: 1, 2, 1, 1, 2, 2, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2… — the length of n-th run of identical symbols in the Thue-Morse sequence. As we know that the Thue-Morse sequence doesn’t contain three ones or three zeroes in a row, we can state that the terms of this sequence will continue to be ones or twos.

You can define putridity sequences for any non-negative sequence. Which of them are interesting? I do not know, but I know which of them are not very interesting. For example, the putridity of pronic (oblong) numbers sequence is the same as the putridity of the triangular numbers sequence. This is because pronic numbers are twice triangular numbers and putridity is independent of factors of two. Another uninformative putridity sequence is the odiousness of the powers of two. This sequence consists only of ones.

I bet that my readers can find putridity sequences that are interesting.