A student asked me the other day the following rather homework-looking question: Given a natural number , how many solutions does the equation

have for and natural numbers?

The question has a very easy answer: Simply notice that and that any like this determines a unique such that is a solution. So, there are solutions if is even (as can be any of ), and there are solutions if is odd.

I didn’t tell the student what the answer is, but I asked what he had tried so far. Among what he showed me there was a piece of paper in which somebody else had scribbled

which caught my interest, and is the reason for this posting.

I don’t think the student had seen the connection between this product of two series, let’s call it , and his question. If we denote by the number of solutions to the equation, the series is the generating function of the sequence , i.e.,

To see this, notice that the coefficient of in is precisely the number of ways we can write as a product of a term from the first series and a term from the second one, i.e., it is the number of solutions with and natural numbers to the equation or, equivalently, . That is to say, the coefficient of in is exactly .

This gives us a purely algebraic (analytic?) way of solving the question, even if there is no understanding of how to approach it from a combinatorial point of view.

Both series on the product that makes up are geometric series, so we have

that of course coincides with the formula we obtained earlier by combinatorial considerations.

The question the student had is a very simple example of a problem about integer partitions, a beautiful area of mathematics that I hope I am not misconstruing by thinking of it as a branch of combinatorial number theory. The technique of generating functions is a very useful and powerful combinatorial tool that I have always found quite nice although, granted, its use is a bit of an overkill for the question at hand. At the same time, this technique provides us with a (standard) method for solving any problem of the same kind: For fixed natural numbers , find for each the number of tuples of natural numbers such that

One can then go further to study the much subtler partition function and its relatives.

(And I still don’t know the name of the student, who didn’t bother to introduce himself, and I have no idea who suggested to him to look at to begin with.)

Well, I got the partial solution finally, here is the question that I tried to solve by the equation x + 2y = n, x,y,n > 0

Let n be a positive integer. Harry’s school year has n school days. Harry has budget of exactly $n for buying exactly one snack per day at school. There are only two types of snacks available: M&M for $1. 00 per packet, or a pair of bananas at $2.00 per pair. The following restrictions must apply to Harry.

(1) Harry must spend all $n on snacks during the school year.

(2) Harry does not have to buy a snack each school day.

At the end of the school year, Harry must report how many times he bought M&M, and how many times he bought bananas. How many different reports are possible?

Anyway, thank you for your posting 🙂 It helps me to understand the solution better.

[…] while ago I posted a short note on integer partitions and generating functions. I am adding a link here, as it is obviously connected to the combinatorial applications of power series we have been […]

Let $C$ be the standard Cantor middle-third set. As a consequence of the Baire category theorem, there are numbers $r$ such that $C+r$ consists solely of irrational numbers, see here. What would be an explicit example of a number $r$ with this property? Short of an explicit example, are there any references addressing this question? A natural approach would […]

Suppose $M$ is an inner model (of $\mathsf{ZF}$) with the same reals as $V$, and let $A\subseteq \mathbb R$ be a set of reals in $M$. Suppose further that $A$ is determined in $M$. Under these assumptions, $A$ is also determined in $V$. The point is that since winning strategies are coded by reals, and any possible run of the game for $A$ is coded by a real, […]

Yes. This is obvious if there are no such cardinals. (I assume that the natural numbers of the universe of sets are the true natural numbers. Otherwise, the answer is no, and there is not much else to do.) Assume now that there are such cardinals, and that "large cardinal axiom" is something reasonable (so, provably in $\mathsf{ZFC}$, the relevant […]

Please send an email to mathrev@ams.org, explaining the issue. (This is our all-purpose email address; any mistakes you discover, not just regarding references, you can let us know there.) Give us some time, I promise we'll get to it. However, if it seems as if the request somehow fell through the cracks, you can always contact one of your friendly edit […]

The characterization mentioned by Mohammad in his answer really dates back to Lev Bukovský in the early 70s, and, as Ralf and Fabiana recognize in their note, has nothing to do with $L$ or with reals (in their note, they indicate that after proving their result, they realized they had essentially rediscovered Bukovský's theorem). See MR0332477 (48 #1080 […]

The paper MR1029909 (91b:03090). Mekler, Alan H.; Shelah, Saharon. The consistency strength of "every stationary set reflects". Israel J. Math. 67 (1989), no. 3, 353–366, that you mention in the question actually provides the relevant references and explains the key idea of the argument. Note first that $\kappa$ is assumed regular. They refer to MR […]

Start with Conway's base 13 function $c $ (whose range on any interval is all of $\mathbb R $), which is everywhere discontinuous, and note that if $f $ only takes values $0$ and $1$, then $c+f $ is again everywhere discontinuous (since its range on any interval is unbounded). Now note that there are $2^\mathfrak c $ such functions $f $: the characteris […]

Yes, there are such sets. To describe an example, let's start with simpler tasks. If we just want $P\ne\emptyset$ with $P^1=\emptyset$, take $P$ to be a singleton. If we want $P^1\ne\emptyset$ and $P^2=\emptyset$, take $P$ to be a strictly increasing sequence together with its limit $a$. Then $P^1=\{a\}$. If we want $P^2\ne\emptyset$ and $P^3=\emptyset$ […]

The result was proved by Kenneth J. Falconer. The reference is MR0629593 (82m:05031). Falconer, K. J. The realization of distances in measurable subsets covering $R^n$. J. Combin. Theory Ser. A 31 (1981), no. 2, 184–189. The argument is relatively simple, you need a decent understanding of the Lebesgue density theorem, and some basic properties of Lebesgue m […]

No, not even $\mathsf{DC}$ suffices for this. Here, $\mathsf{DC}$ is the axiom of dependent choice, which is strictly stronger than countable choice. For instance, it is a theorem of $\mathsf{ZF}$ that for any set $X$, the set $\mathcal{WO}(X)$ of subsets of $X$ that are well-orderable has size strictly larger than the size of $X$. This is a result of Tarski […]

I’m the student asked the question 😉

Well, I got the partial solution finally, here is the question that I tried to solve by the equation x + 2y = n, x,y,n > 0

Let n be a positive integer. Harry’s school year has n school days. Harry has budget of exactly $n for buying exactly one snack per day at school. There are only two types of snacks available: M&M for $1. 00 per packet, or a pair of bananas at $2.00 per pair. The following restrictions must apply to Harry.

(1) Harry must spend all $n on snacks during the school year.

(2) Harry does not have to buy a snack each school day.

At the end of the school year, Harry must report how many times he bought M&M, and how many times he bought bananas. How many different reports are possible?

Anyway, thank you for your posting 🙂 It helps me to understand the solution better.

Hi Eliot,

I’m glad this helped. Marion mentioned to me the `M&Ms problem’ the other day, I figured this was the same question without distractions.

[…] while ago I posted a short note on integer partitions and generating functions. I am adding a link here, as it is obviously connected to the combinatorial applications of power series we have been […]