Difference between revisions of "1986 USAMO Problems/Problem 5"

(Solution)
(Solution)
Line 35: Line 35:
 
So,  
 
So,  
  
<cmath>\begin{align*}
+
<cmath>T(n) = \sum_{k=1}^n k\cdot (\text{\# of partitions of n that contain a k}).</cmath>
T(n) &= 1\cdot (\text{\# of partitions of n that contain a 1})\\ &+ 2\cdot (\text{\# of partitions of n that contain a 2})\\ &\cdots \\ &+ n\cdot (\text{\# of partitions of n that contain a n})
 
\end{align*}</cmath>
 
  
 
However, the number of partitions of n that contain a <math>i</math> is the same as the total number of partitions of <math>n-i,</math> so  
 
However, the number of partitions of n that contain a <math>i</math> is the same as the total number of partitions of <math>n-i,</math> so  

Revision as of 14:11, 30 August 2018

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

\[S(n) = \sum_{k=1}^n k\cdot (\text{\# of partitions of n-k with no 1's}).\]

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

Note that there is no $(1+x+x^2+x^3+\ldots)$ 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 $F(x)=c_0+c_1x+c_2x^2+c_3x^3+\ldots.$ We wish to compute $S(n)=1\cdot c_{n-1}+2\cdot c_{n-2}+\ldots+n\cdot c_0.$

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

\begin{align*} G(x) &= (x+2x^2+3x^3+4x^4+\ldots)(c_0+c_1x+c_2x^2+c_3x^3+\ldots)\\ &= x(1+2x+3x^2+4x^3+\ldots) \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 $1\cdot c_{n-1}+2\cdot c_{n-2}+\ldots+n\cdot c_0,$ 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,

\[T(n) = \sum_{k=1}^n k\cdot (\text{\# of partitions of n that contain a k}).\]

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\cdot (\text{\# of partitions of n-1})\\ &+ 2\cdot (\text{\# of partitions of n-2})\\ &\ldots \\ &+ n\cdot (\text{\# of partitions of 0}) \end{align*}

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

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

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

\begin{align*} H(x) &= (x+x^2+x^3+x^4+\ldots)(d_0+d_1x+d_2x^2+d_3x^3+\ldots) \\ &= x(1+x+x^2+x^3+\ldots) \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 $d_{n-1}+d_{n-2}+\ldots+d_0,$ 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

See Also

1986 USAMO (ProblemsResources)
Preceded by
Problem 4
Followed by
Last Question
1 2 3 4 5
All USAMO Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS