Tanya Khovanova's Math Blog

There is a big difference between evaluating exercise DVDs and reviewing movies. You are supposed to use exercise DVDs many times. So the value of the DVD changes over time. An exercise DVD that is too difficult at the first try could become a lot of fun later. Alternatively, one that explains everything in detail can be great at the beginning, but it will become boring after several viewings.

Smart DVD producers probably know some common rules. The number of people who use a DVD for the first time is much bigger than the number of people who use it for the hundredth time. Users often post reviews and ratings of products they have bought. Therefore, the proportion of reviews by the first-time watchers is much higher than by the hundredth-time watchers. This means that to get better ratings the DVD producers should target the first-time watchers. Is this why we have so many boring exercise DVDs?

In my opinion, exercise DVDs should have two parts. One part explains everything by breaking the routine down into elements and the other part allows people who have learned the routine to do it without interruption.

Keep your eyes open for my upcoming web page with reviews of dance exercise DVDs that I own. These reviews will address both first-time users and every-day-for-a-year users.

Everyone knows that math education in public schools in this country is pathetic. If you looked at this problem from an economics point of view, the first question would be, “Qui prodest?.”

Who profits from bad math education? I know one place — the lottery. People who understand how the lottery works rarely buy tickets. They might buy an occasional ticket as entertainment, but never as an investment. No wonder they say that the lottery is a tax on people bad at math.

Huge money from lotteries goes to states and towns, and a big portion of that goes to education. That means towns, schools and math teachers have direct financial incentive not to provide good math education. This conflict of interest creates a situation in which, in the long run, it is profitable for schools to hire very poor math teachers or cut their math programs.

The situation is unethical. I think that lottery organizers should at least pretend that they are resolving this conflict and spend part of the lottery money to educate people not to play the lottery.

I did it. I handed in my resignation letter to my boss. I’m resigning effective Jan 3, 2008. If you want to know why I’m waiting until next year, I can give you several reasons.

• First, Christmas-time is usually the most enjoyable work time because no one is there. It is quiet.
• Second, I’m superstitious: I believe the way I greet the New Year determines how the New Year is going to go, so I want to be employed at the strike of the midnight clock.
• Last but not least, it appears that to get my company’s annual profit-sharing bonus, I have to be employed on December 31^st.

I am happy and sad at the same time. In four and a half years I’ve made a lot of friends and accomplished a lot professionally. Now it is my time to move forward. Where is forward? It is in the direction of a cemetery, but I would rather be doing something more meaningful to me than battle management while I am slowly crawling there.

Do you know that 1210 is the smallest autobiographical number? You probably do not know what an autobiographical number is. You are right if you think that such a number should be a pompous self-centered number whose only purpose in life is to describe itself.

Here is the formal definition. An autobiographical number is a number N such that the first digit of N counts how many zeroes are in N, the second digit counts how many ones are in N and so on. In our example, 1210 has 1 zero, 2 ones, 1 two and 0 threes.

Let us find all autobiographical numbers using the “zoom-in” method.

1. By definition, the autobiographies can’t have more than 10 digits. It is nice to know that these egotistical numbers can’t be too grand.
2. The sum of the digits in an autobiography equals the number of the digits. Consequently, the sum of the digits will not be more than 10.
3. The first digit is the number of zeroes. As you know, self-respecting integers do not start with a zero. Hence, the number of zeroes is not a zero.
4. Subtracting statement “c” from statement “b” above, we get a resulting statement that the sum of all the digits, except for the first one, is equal to the number of non-zero digits plus 1.
5. That means, other than the first digit, the set of all other non-zero digits consists of several ones and 1 two.
6. Furthermore, the number of ones is either 0, 1 or 2.

Now we continue zooming in in three different directions depending on the number of ones. In this blog entry, I will consider only the case in which there are no ones; I leave the other two cases to the reader.

• If the number of ones is zero, then the only non-zero non-first digit of such a number is 2.
• This 2 should be included in the autobiography; since the third digit of the number is not zero, it must be 2.
• The number has 2 twos.
• It must be 2020.

Here is the full set of autobiographical numbers: 1210, 2020, 21200, 3211000, 42101000, 521001000, 6210001000.

This is the sequence A104786 in the Online Encyclopedia of Integer Sequences (OEIS), where I first encountered the autobiographical numbers.

Autobiographical numbers are very cute numbers. But there is a problem with their name. If there is a notion of an autobiography of a number, then it would be logical to expect that there is a notion of a biography of a number. What would be the logical candidate for a biography of a number? Let us say that given a number N, its biography is another number M such that the first digit of M is the number of zeroes in N, the second digit of M is the number of ones in N and so on.

Of course, for a number to have a biography, we need to assume that none of its digit is present more than nine times. Still there are several problems with the definition of a biography.

The first problem is that if N doesn’t have zeroes, its biography starts with a zero. As numbers don’t start with 0, that biography is not a number! Furthermore, if N starts with 0, it can have a biography but N is not a number. Luckily for this article, a digit string starting with zeroes can’t be an autobiographical string, because the number of zeroes is not a zero. It is a relief that those illegitimate strings that are trying to pretend to be numbers can’t actually be autobiographical.

The second problem with biographies is that a number can have many biographies. Indeed, if a number doesn’t have nines, you can remove or add zeroes at the end of a biography to get another biography of the same number. Since mathematicians like to define things uniquely, we might consider it a problem if a number has several biographies. In real life it is possible to have many biographies of a person. So the second problem is not a big problem. I will call the shortest possible biography of a number the curriculum vitae and the longest possible biography the complete life story.

The third problem is that numbers with the same digits in different permutations have the same biographies. So in a sense a biography follows the life not of a number, but rather the set of its digits.

Suppose for now we allow a biography to start with 0. Also, let us choose the curriculum vitae — the shortest biography in case there could be several. Let us build a sequence of CVs. As an example, we start with 0. Zero’s CV is 1, one’s CV is 01, continuing that we get the following sequence: 0, 1, 01, 11, 02, 101, 12, 011, 12, 011, 12, …. You can see that the CVs’ sequence fell into a cycle in this case. I tried sequences of CVs starting with many numbers. I found that they fall into two cycles. One cycle is described above and another one is: 22, 002, 201, 111, 03, 1001, 22. Can you find another cycle or, alternatively, can you prove that all the numbers that allow the sequence of CVs converge to only these two cycles?

Let us build the sequence of complete biographies, that is, life stories, starting with 0: 0, 1000000000, 9100000000, 8100000001, 7200000010, 7110000100, 6300000100, 7101001000, 6300000100, …. We see that this sequence falls into a cycle of length two. The members of this cycle are legitimate numbers. These numbers are too shy to advertise themselves. But Alice praises Bob, because Bob praises Alice. It’s a very advantageous flattery pattern! I will call such a pair a mutually-praising pair. We’ve already seen mutually-praising strings: 12 and 001. Two other examples of number pairs thriving on each others’ compliments are, first, 130 and 1101, and second, 2210 and 11200.

I would like to become a professor of mathematics. How can I get to academia? I was told that applicants are measured by the number of papers they write. They expect about 3 papers per year starting after the Ph.D. I got my Ph.D. 20 years ago. I published 6 papers after my Ph.D. papers. That means I urgently need to come up with 54 papers.

There are several problems with working in industry and trying to publish at the same time:

• Some results derived at work you can’t publish because they are proprietary.
• Some results derived at work are classified.
• Journals want simulations; that means you may have to depend on colleagues.
• If your project is closed before your group finishes your simulations, you have to put your paper on hold.
• Also, if your paper is not directly a part of your project, you can’t use your official work time to write your paper.

Because of these obstacles, the papers I have started at my job are on hold. It’s unlikely that I’d be allowed to finish them during work hours.

So I started writing non-job-related papers on my weekends. I started doing this seriously a year ago. It goes very slowly and I hope to publish three papers soon, but my speed needs to be much higher than that to catch up with the 54 papers I didn’t have time to write while being a single mom and providing for my family.

So, I came up with this idea: to quit my job and write papers. I do not have enough money to support this idea for very long. Certainly, not enough time for 54 papers. We probably can survive on my savings for half a year. My goal is to write as many papers as I can in half a year and see what my real speed is. This way I can at least prove to myself that I am a mathematician for real.

The only problem is that my savings were meant for a down payment on my first house. I’ve been asking myself for awhile: What is more important− a dream job or a dream house? I just realized today that I will never be happy if I am not happy at my job and I am quite happy with the apartment I am renting now. I guess this is it − I just have to take the plunge.

Are you afraid of Friday the 13th? Here is my only Friday the 13th story.

It was Friday the 13^th, and I was listening to the psychologist Joy Browne’s show. Joy asked her listeners to call in with stories of interesting things that happened to them on Friday the 13th. I wondered why I didn’t remember any stories about Friday the 13^th.

At the end of her show, I went to pick up my mail, where I found a book kindly sent to me by Princeton University Press named Nonplussed!: Mathematical Proof of Implausible Ideas by Julian Havil. I opened this book to a random page, and it was Chapter 13. That was not such a big deal by itself, but in addition, Chapter 13 was titled “Friday the 13th”.

One of the things Julian Havil discussed in the chapter is how often the 13th of the month falls on different days of the week. You might remember from your elementary school education that the Gregorian calendar repeats an entire identical day-of-the-week cycle every 400 years. Hence, it is just a matter of calculation to check on which day of the week the 13th falls the most often.

Can you guess the answer? I am sure you can apply some meta-thinking and derive that there is one special day of the week on which the 13th most frequently falls. You might even guess by now that that day is Friday. Otherwise, why would I write this blog entry? Or what would Mr. Havil have to say in the whole chapter of the aforementioned book?

As we can see, this calculation increases the worry for people who suffer from paraskevidekatriaphobia — the fear of Friday the 13th. The 13th falls on Friday more often than on any other day.

Should we be worried?

Mathematically, the difference between the number of Fridays the 13th and, say, Thursdays the 13th is so small that it can only be observed when we look at the 400 year lifetime of the Gregorian calendar. Many countries have yet to experience the full cycle of the Gregorian calendar. For example, Russia adopted the calendar only in the 20th century. Is this why I am not so very afraid?

On second thought, for people of my generation, who are unlikely to live until the year 2100, the situation is slightly different. In the years between 1901 and 2099 our calendar has a days-of-the-week cycle of 28 years. You can calculate and check that in the period of 28 years, the 13th falls on any day of the week with the same probability. Hence, in events happening around my life time, there is not much to worry about, because Friday is no more special than any other day.

On third thought, a particular individual might see more Fridays on the 13th in his lifetime depending on the exact date of his birth. In my own life up to today, Monday is the most frequently occurring 13th. Maybe that’s why I do not like Mondays.

For 25 years my children were my priority. I made several decisions in my life that benefited my family, but “harmed” my mathematical career. I do not regret any of my choices. After all, being a single mom made me a more confident, stronger person. Maybe this will help my career in a long run.

Now that my youngest son is 16 years old, my life can’t revolve around him anymore. Now I must think about the meaning of my life, beyond bringing up children. The only thing I want to do is mathematics. I am actually doing some math on weekends, but I really want to do it full-time. My tasks at my job are getting further and further from mathematics and research. In short, I feel that my job doesn’t fit me at this stage of my life.

I really should find another job. I am somewhat scared of change though. I think that the first thing to do is to try to turn around the situation at my current job. There is a lot of interesting mathematics in battle management. The problem is to match a math problem to a charge number. That is, I would need to convince my management that the algorithms we design need a sound mathematical basis.

Here is my decision: I will try to find some tasks at my work that include mathematics and see how I can change my situation there by the end of the year. If I don’t succeed, I will have to think of something else. Let the Web be my witness. I will report the results to you soon.

I stumbled upon the following sentence in the MathWorld article on the Fibonacci numbers: “No odd Fibonacci number is divisible by 17.” I started wondering if there are other numbers like that. Of course there are — no odd Fibonacci number is divisible by 2. But then, an odd number need not be a Fibonacci number in order not to be divisible by 2.

So, let’s forget about 2 and think about odd numbers. How do we know that the infinite Fibonacci sequence never produces an odd number that is divisible by 17? Is 17 the only such odd number? Is 17 the smallest such odd number? If there are many such odd numbers, how do we calculate the corresponding sequence?

We’ll start with a general question: How can we approach puzzles about the divisibility of Fibonacci numbers? Suppose K is an integer. Consider the sequence a_K(n) = F_n(mod K), of Fibonacci numbers modulo K. The cool thing about this sequence is that it is periodic. If it is not immediately obvious to you, think of what happens when a pair of consecutive numbers in the sequence a_K(n) gets repeated. As a bonus for thinking you will get an upper bound estimate for this period.

Let us denote the period of a_K(n) by P_K. By the way, this period is called a Pisano period. From the periodicity and the fact that a_K(0) = 0, we see right away that there are infinitely many Fibonacci numbers divisible by K. Are there odd numbers among them? If we trust MathWorld, then all of the infinitely many Fibonacci numbers divisible by 17 will be even.

How do we examine the divisibility by K for odd Fibonacci numbers? Let us look at the Fibonacci sequence modulo 2. As we just proved, this sequence is periodic. Indeed, every third Fibonacci number is even. And the evenness of a Fibonacci number is equivalent to this number having an index divisible by 3.

Now that we know the indices of even Fibonacci numbers we can come back to the sequence a_K(n). In order to prove that no odd Fibonacci number is divisible by K, it is enough to check that all the zeroes in the sequence a_K(n) have indices divisible by 3. We already have one zero in this sequence at index 0, which is by divisible by 3. Because the sequence a_K(n) is periodic, it will start repeating itself at a_K(P_K) . Hence, we need to check that P_K is divisible by 3 and all the zeroes up to a_K(P_K) have indices divisible by 3. When K = 17 it is not hard to do the calculations manually. If you’d like, try this exercise. To encourage (or perhaps to discourage) you, here’s an estimate of the scope of the work for this exercise: the Pisano period for K = 17 is 36.

After I checked that no odd Fibonacci number is ever divisible by 17, I wanted to find the standard solution for this statement and followed the trail in MathWorld. MathWorld sent me on a library trip where I found the proof of the statement in the book Mathematical Gems III by Ross Honsberger. There was a proof there alright, but it was tailored to 17 and didn’t help me with my questions about other such odd numbers.

The method we developed for 17 can be used to check any other number. I trusted this task to my computer. To speed up my program, I used the fact that the Pisano period for K is never more than 6K. Here is the sequence calculated by my trustworthy computer, which I programmed with, I hope, equal trustworthiness:

• A133246 Odd numbers n with the property that no odd Fibonacci number is divisible by n.
  9, 17, 19, 23, 27, 31, 45, 51, 53, …

The sequence shows that 9 is the smallest odd number that no odd Fibonacci is ever divisible by, and 17 is the smallest odd prime with this same property. Here is a trick question for you: Why is this property of 17 more famous than the same property of 9?

Let us look at the sequence again. Is this sequence infinite? Obviously, it should include all multiples of 9 − hence, it is infinite. What about prime numbers in this sequence? Is there an infinite number of primes such that no odd Fibonacci number is divisible by them? While I do not know the answer, it’s worth investigating this question a little bit further.

From now on, let K be an odd prime. Let us look at the zeroes of the sequence a_K(n) more closely. Suppose a zero first appears at the m-th place of a_K(n). Then a_K(m+1) = a_K(m+2) = a. In this case the sequence starting from the m-th place is proportional modulo K to the sequence a_K(n) starting from the 0-th index. Namely, a_K(n+m) = a*a_K(n) (mod K). As a is mutually prime with K, then a_K(n+m) = 0 iff a_K(n) = 0. From here, for any index g that is a multiple of m, a_K(g) = 0. Furthermore, there are no other zeroes in the sequence a_K(n). Hence, the appearances of 0 in the sequence a_K(n) are periodic with period m.

By the way, m is called a fundamental period; and we just proved that the Pisano period is a multiple of the fundamental period for prime K. Hence, the fact that no odd Fibonacci number is divisible by K is equivalent to the fact that the fundamental period is not divisible by 3. It is like saying that if the smallest positive Fibonacci number divisible by an odd prime K is even, then no odd Fibonacci number is divisible by K.

If the remainder of the fundamental period modulo 3 were random, we would expect that about every third prime number would not divide any odd Fibonacci numbers. In reality there are 561 such primes among the first 1,500 primes (including 2). This is somewhat more than one third. This gives me hope that there is a non-random reason for such primes to exist. Consequently, it may be possible to prove that the sequence of prime numbers that do not divide odd Fibonacci numbers is infinite.

Can you prove that?

Why didn’t I think of this before?

I want to share some ideas about mathematics and about my life as a mathematician. In this blog, you’ll read about such things as the properties of numbers and sequences and how mathematicians approach practical things.

I started my life as a genius girl mathematician, winning silver and gold medals at the International Math Olympiad (IMO) as a teenager. My PhD is from Moscow State University. When I got married, I wanted to have a family and mathematics at the same time, but being a woman, this affected my mathematics career. Now my kids are growing up and mathematics is becoming more and more important in my life. This is why I decided to start this blog.