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 […]

(As I pointed out in a comment) yes, partial Woodinness is common in arguments in inner model theory. Accordingly, you obtain determinacy results addressing specific pointclasses (typically, well beyond projective). To illustrate this, let me "randomly" highlight two examples: See here for $\Sigma^1_2$-Woodin cardinals and, more generally, the noti […]

I am not sure which statement you heard as the "Ultimate $L$ axiom," but I will assume it is the following version: There is a proper class of Woodin cardinals, and for all sentences $\varphi$ that hold in $V$, there is a universally Baire set $A\subseteq{\mathbb R}$ such that, letting $\theta=\Theta^{L(A,{\mathbb R})}$, we have that $HOD^{L(A,{\ma […]

A Wadge initial segment (of $\mathcal P(\mathbb R)$) is a subset $\Gamma$ of $\mathcal P(\mathbb R)$ such that whenever $A\in\Gamma$ and $B\le_W A$, where $\le_W$ denotes Wadge reducibility, then $B\in\Gamma$. Note that if $\Gamma\subseteq\mathcal P(\mathbb R)$ and $L(\Gamma,\mathbb R)\models \Gamma=\mathcal P(\mathbb R)$, then $\Gamma$ is a Wadge initial se […]

Craig: For a while, there was some research on improving bounds on the number of variables or degree of unsolvable Diophantine equations. Unfortunately, I never got around to cataloging the known results in any systematic way, so all I can offer is some pointers to relevant references, but I am not sure of what the current records are. Perhaps the first pape […]

Yes. Consider, for instance, Conway's base 13 function $c$, or any function that is everywhere discontinuous and has range $\mathbb R$ in every interval. Pick continuous bijections $f_n:\mathbb R\to(-1/n,1/n)$ for $n\in\mathbb N^+$. Pick a strictly decreasing sequence $(x_n)_{n\ge1}$ converging to $0$. Define $f$ by setting $f(x)=0$ if $x=0$ or $\pm x_n […]

All proofs of the Bernstein-Cantor-Schroeder theorem that I know either directly or with very little work produce an explicit bijection from any given pair of injections. There is an obvious injection from $[0,1]$ to $C[0,1]$ mapping each $t$ to the function constantly equal to $t$, so the question reduces to finding an explicit injection from $C[0,1]$ to $[ […]

One way we formalize this "limitation" idea is via interpretative power. John Steel describes this approach carefully in several places, so you may want to read what he says, in particular at Solomon Feferman, Harvey M. Friedman, Penelope Maddy, and John R. Steel. Does mathematics need new axioms?, The Bulletin of Symbolic Logic, 6 (4), (2000), 401 […]

"There are" examples of discontinuous homomorphisms between Banach algebras. However, the quotes are there because the question is independent of the usual axioms of set theory. I quote from the introduction to W. Hugh Woodin, "A discontinuous homomorphism from $C(X)$ without CH", J. London Math. Soc. (2) 48 (1993), no. 2, 299-315, MR1231 […]

This is Hausdorff's formula. Recall that $\tau^\lambda$ is the cardinality of the set ${}^\lambda\tau$ of functions $f\!:\lambda\to\tau$, and that $\kappa^+$ is regular for all $\kappa$. Now, there are two possibilities: If $\alpha\ge\tau$, then $2^\alpha\le\tau^\alpha\le(2^\alpha)^\alpha=2^\alpha$, so $\tau^\alpha=2^\alpha$. In particular, if $\alpha\g […]

Fix a model $M$ of a theory for which it makes sense to talk about $\omega$ ($M$ does not need to be a model of set theory, it could even be simply an ordered set with a minimum in which every element has an immediate successor and every element other than the minimum has an immediate predecessor; in this case we could identify $\omega^M$ with $M$ itself). W […]

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 […]