# 1986 USAMO Problems/Problem 5

## Problem

By a partition $\pi$ of an integer $n\ge 1$, we mean here a representation of $n$ as a sum of one or more positive integers where the summands must be put in nondecreasing order. (E.g., if $n=4$, then the partitions $\pi$ are $1+1+1+1$, $1+1+2$, $1+3, 2+2$, and $4$).

For any partition $\pi$, define $A(\pi)$ to be the number of $1$'s which appear in $\pi$, and define $B(\pi)$ to be the number of distinct integers which appear in $\pi$. (E.g., if $n=13$ and $\pi$ is the partition $1+1+2+2+2+5$, then $A(\pi)=2$ and $B(\pi) = 3$).

Prove that, for any fixed $n$, the sum of $A(\pi)$ over all partitions of $\pi$ of $n$ is equal to the sum of $B(\pi)$ over all partitions of $\pi$ of $n$.

## Solution

Let $S(n) = \sum\limits_{\pi} A(\pi)$ and let $T(n) = \sum\limits_{\pi} B(\pi)$. We will use generating functions to approach this problem -- specifically, we will show that the generating functions of $S(n)$ and $T(n)$ are equal.

Let us start by finding the generating function of $S(n)$. This function counts the total number of 1's in all the partitions of $n$. Another way to count this is by counting the number of partitions of $n$ that contain $x$ 1's and multiplying this by $x$, then summing for $1\leq x \leq n$. However, the number of partitions of $n$ that contain $x$ 1's is the same as the number of partitions of $n-x$ that contain no 1's, so

\begin{align*}
S(n) &= 1*(\text{\# of partitions of n-1 with no 1's})\\ &+ 2*(\text{\# of partitions of n-2 with no 1's})\\ &\begin{center}\cdots\end{center} \\ &+ n*(\text{\# of partitions of 0 with no 1's})
\end{align*} (Error compiling LaTeX. ! Missing \$ inserted.)

The number of partitions of $m$ with no 1's is the coefficient of $x^m$ in

\begin{align*} F(x) &= (1+x^2+x^4+\cdots)(1+x^3+x^6+\cdots)(1+x^4+x^8+\cdots)\cdots \\ &= \prod\limits_{i=2}^{\infty}\frac{1}{1-x^i} \end{align*}

Note that there is no $(1+x+x^2+x^3+\cdots)$ term in $F(x)$ because we cannot have any 1's in the partition.

Let $c_m$ be the coefficient of $x^m$ in the expansion of $F(x)$, so we can rewrite it as $F(x)=c_0+c_1x+c_2x^2+c_3x^3+\cdots$. We wish to compute $S(n)=1*c_{n-1}+2*c_{n-2}+\cdots+n*c_0$.

Consider the power series $G(x)=(x+2x^2+3x^3+4x^4+\cdots)F(x)$.

\begin{align*} G(x) &= (x+2x^2+3x^3+4x^4+\cdots)(c_0+c_1x+c_2x^2+c_3x^3+\cdots)\\ &= x(1+2x+3x^2+4x^3+\cdots) \prod\limits_{i=2}^{\infty}\frac{1}{1-x^i}\\ &= x\cdot\frac{1}{(1-x)^2} \prod\limits_{i=2}^{\infty}\frac{1}{1-x^i}\\ &= \frac{x}{1-x} \prod\limits_{i=1}^{\infty}\frac{1}{1-x^i} \end{align*}

If we expand the first line, we see that the coefficient of $x^n$ in $G(x)$ for any $n$ is $1*c_{n-1}+2*c_{n-2}+\cdots+n*c_0$, which is exactly $S(n)$! So by definition, $G(x)=\frac{x}{1-x} \prod\limits_{i=1}^{\infty}\frac{1}{1-x^i}$ is the generating function of $S(n)$.

Now let's find the generating function of $T(n)$. Notice that counting the number of distinct elements in each partition and summing over all partitions is equivalent to counting how many partitions of $n$ contain i and then summing for all $1\leq i \leq n$.

So,

\begin{align*} T(n) &= 1*(\text{\# of partitions of n that contain a 1})\\ &+ 2*(\text{\# of partitions of n that contain a 2})\\ &\cdots \\ &+ n*(\text{\# of partitions of n that contain a n}) \end{align*}

However, the number of partitions of n that contain a $i$ is the same as the total number of partitions of $n-i$, so \begin{align*} T(n) &= 1*(\text{\# of partitions of n-1})\\ &+ 2*(\text{\# of partitions of n-2})\\ &\cdots \\ &+ n*(\text{\# of partitions of 0}) \end{align*}

The generating function for the number of partitions is \begin{align*} P(x) &= (1+x+x^2+\cdots)(1+x^2+x^4+\cdots)(1+x^3+x^6+\cdots)\cdots \\ &= \prod\limits_{i=1}^{\infty} \frac{1}{1-x^i}. \end{align*}

Let's write the expansion of $P(x)$ as $P(x)=d_0+d_1x+d_2x^2+d_3x^3+\cdots$, so we wish to find $T(n)=d_0+d_1+d_2+\cdots+d_{n-1}$.

Consider the power series $H(x)=(x+x^2+x^3+x^4+\cdots)P(x)$.

\begin{align*} H(x) &= (x+x^2+x^3+x^4+\cdots)(d_0+d_1x+d_2x^2+d_3x^3+\cdots) \\ &= x(1+x+x^2+x^3+\cdots) \prod\limits_{i=1}^{\infty} \frac{1}{1-x^i}\\ &= x\cdot \frac{1}{1-x}\prod\limits_{i=1}^{\infty} \frac{1}{1-x^i}\\ &= \frac{x}{1-x} \prod\limits_{i=1}^{\infty} \frac{1}{1-x^i} \end{align*}

If we expand the first line, we see that the coefficient of $x^n$ in $H(x)$ for any $n$ is $d_{n-1}+d_{n-2}+\cdots+d_0$, which is precisely $T(n)$. This means $H(x)$ is the generating function of $T(n)$.

Thus, the generating functions of $S(n)$ and $T(n)$ are the same, so $S(n)=T(n)$ for all $n$ and we are done.

~Peggy