joshuazelinsky

New preprint of a paper with Tim McCormack is up https://arxiv.org/abs/2312.11661 ! Let's talk about it!

Everyone knows how to calculate the average, or arithmetic mean of a list of numbers. You add them all up and then divide by the number of things on the list. So, if you want the average of 3, 9, and 27, you add them up to get (3+9+27)/3= 13. But this is not the only sort of average you can take of a list of numbers . Another notion of average in the sense of something which represents a typical value is the geometric mean, where one multiplies all the numbers together and then takes the nth root where n is the number of items on the list. So in our case we would take (3*9*27)^(1/3) = 9. (To make sure this makes sense, we need to insist that our numbers chosen are always positive.)

Essentially, the arithmetic mean is asking to find a number m such that if you add m up n times, one gets the same result as the sum of the numbers on the list. The geometric mean is asking to find a number where when one multiplies that number by itself m times one gets the same number as the product of the numbers on the list. In our example, 13+13+13= 3+9+27, and 9*9*9= 3*9*27.

Now, one thing you may notice is that 9 is less than 13. In fact, this always happens and the arithmetic mean is always greater than the geometric mean, with the exception when all the numbers on your list are equal. This is known as the arithmetic-mean-geometric mean inequality, or the AM-GM inequality.

A brief digression: Oyestein Ore was a mathematician who among other things was the doctoral supervisor for Grace Hopper, and also was the writer of "Number Theory and Its History" which in my opinion remains the best introduction to number theory for people with no background. (My own biases come in part from having read it in 8th and 9th grade and it contributing to my own long-term research interests.) Ore was interested in understanding the positive divisors of numbers. One thing he did is looked at the geometric mean of a set of the positive divisors of a number, which we will call G(n). For example, 10 has divisors 1,2,5 and 10. And so the geometric mean of the divisors is (1*2*5*10)^(1/4)= 100^(1/4) = √10. In fact, Ore was able to show that the geometric mean of the divisors of a number is always the square root of the number. This is a fun exercise if you have not seen it before. (Hint: Try to pair the divisors up that multiply to n. So in our example above 1 pairs with 10 while 2 pairs with 5.)

Note that if write tau(n) as the number of positive divisors of n, and sigma(n) as the sum of the divisors of n. (So for example, tau(10)=4, and sigma(10)=1+2+5+10=18.) Then the average of the divisors is sigma(n)/tau(n). Ore combined this observation with the AM-GM inequality conclude that one must therefore have √n < sigma(n)/tau(n). However, this inequality is essentially trivial, since it is the same as tau(n)√n < sigma(n), and it turns out that for large n, tau(n) is very small compared to n, and obviously sigma(n) is in general bigger than n.

And to some extent this should not be surprising: the AM-GM inequality is a statement about all real numbers. It cannot "know" much of anything about collections of divisors since it is only taking into account them being a list of real numbers. However, there is another version of the AM-GM inequality, the weighted AM-GM which allows one to take averages where some numbers on your list "weigh" more than others. The analogy that may help here is that this is essentially how you take an average when you have multiple grades that count a different amount, and so each is weighed accordingly. So the question arises: can one use the weighted version of the AM-GM inequality with suitably chosen weights to get interesting number theoretic inequalities?

This paper by Tim McCormack and me addresses this question. We show that the answer is yes, but but only weakly. But along the way, we prove a bunch of new inequalities, about Zaremba's function among other things. Zaremba's function, z(n), is the sum of (ln d)/d where d ranges over the divisors of n. So for example, z(4)= (ln 1)/1 + (ln 2)/2 + (ln 4)/4 .

One fun thing that this is connected to is a class of pseudoperfect numbers that we prove new things about. Recall a number is said to be perfect if when we add up all the divisors less than the number we get the number Equivalently when we add up all the divisors we get twice the number. For example, 6 is perfect because 1+2+3=6, or equivalently, 1+2+3+6=12=2(6). A number n is said to be pseudoperfect if when you add some subset of the divisors you get twice the number. For example, 12 is not perfect, because all its divisors add up to too much: 1+2+3+4+6+12=28. But if we are allowed to forget about 4, then it looks perfect: 1+2+3+6+12=24. So pseudoperfect numbers are numbers where we are allowed to cheat. However, pseudoperfect numbers are in some sense way too common. Lots of numbers are pseudoperfect. But we can restrict things a bit, to insist that our divisors look a lot like an actual set of divisors for a number could be using the same pairing rule we used to prove that the geometric mean of the divisors of n is √n . We'll say that a number n is strongly pseudoperfect if there is a subset S of its divisors which add up to 2n and where a given divisor d is in S if and only if n/d is in S. For example, 12 would fail to be pseudoperfect above, because we dropped 4, but 4/12=3 is in our set S. In contrast, 36 is strongly pseudoperfect because we could take the set 1 +2+3 +12+18+36=2(36). We prove that strongly pseudoperfect numbers satisfy some interesting behavior, including that they end up looking a lot closer to perfect numbers than general pseudoperfect numbers.

Strongly pseudoperfect numbers are much rarer than general pseudoperfect numbers. Whenever a number n is pseudoperfect, mn is pseudoperfect for any m. This is another good exercise. However, this is not the case for strongly pseudoperfect numbers. In particular, if n is strongly pseudoperfect and p is a sufficiently large prime, then pn is not strongly pseudoperfect.

We don't know that much about strongly pseudoperfect numbers. There are no strongly pseudoperfect n numbers which are 2 mod 3 (that is where n leaves a remainder of 2 when divided by 3), and there are also no strongly pseuoperfect numbers which are 3 mod 4. Are these the only mod restrictions on strongly pseudoperfect numbers or are there others?

Also, 60, 120, 240, 480, 960 are all strongly pseudoperfect. Does this pattern continue indefinitely?