Partition

A partition of a positive integer is a way of expressing it as the sum of other positive integers. Typically, one disregards the order of the summands. For example, there are three partitions of 3: $3 = 2+1 =1+1+1$ .

There is no known, simple formula that gives the number of partitions of a number. There is, however, a rather ugly formula discovered by G. H. Hardy, J. E. Littlewood, and Srinivasa Ramanujan. However, this formula is rather unwieldy: it is not even known for which values of $\displaystyle{n}$ the number of partitions of $\displaystyle{n}$ is even, despite the presence of a formula!

A more fruitful way of studying partition numbers is through generating functions. The generating function for the partitions is given by $P(x)=\prod_{n=1}^\infty \frac{1}{1-x^n}$ . Partitions can also be studied by using the Jacobi theta function, in particular the triple product. The generating function approach and the theta function approach can be used to study many variants of the partition function, such as the number of ways to write a number n as the sum of odd parts, or of distinct parts, or of parts congruent to $1\pmod 3$ , etc.

Ferrers Diagrams

A Ferrers Diagram is a way to represent a certain partition of a number. The diagram consists of rows of dots. Each row represents a different addend in the partition. The rows are ordered in non-increasing order so that that the row with the most dots is on the top and the row with the least dots is on the bottom.

For example, 9 can be partitioned into 4 + 3 + 1 + 1 which would be represented by the following Ferrers Diagram:

4	$\bullet$	$\bullet$	$\bullet$	$\bullet$
3	$\bullet$	$\bullet$	$\bullet$
1	$\bullet$
1	$\bullet$

The Conjugate

The conjugate of a Ferrers Diagram is formed by reflecting the diagram across its diagonal (the one starting in the top left of the diagram). This can also be interpreted as exchanging the rows for the columns. For example, consider our example from before but this time let's count the number of dots in each column:

	4	2	2	1
4	$\bullet$	$\bullet$	$\bullet$	$\bullet$
3	$\bullet$	$\bullet$	$\bullet$
1	$\bullet$
1	$\bullet$

The original partition is 4 + 3 + 1 + 1 and the conjugate is 4 + 2 + 2 + 1.

Generating Functions

Generating Functions can be used to deal with problems involving partitions. First we will find the generating function for the number of ways to partition an integer $n$ .

Consider partitioning $n$ into addends that are equal to just 1. The generating function for this is $1 + x + x^2 + \cdots$ since there is only one way to represent $n$ as the sum of 1's.

Consider partitioning numbers using just 2's as addends. There's 1 way to partition 0 into 2's, 0 ways to partition 1 into 2's, 1 way to partition 2 into 2's and so forth. Therefore, the generating function is $1 + x^2 + x^4 +\cdots$ .

We can proceed in this manner to find that the generating function for the number of ways to partition $n$ into addends equal to $k$ is $1 + x^k + x^{2k} + x^{3k} + \cdots$ .

Now, multiplying each of these generating functions will give us the generating function for partitioning $n$ in general which is

$(1 + x + x^2 + \cdots + )(1 + x^2 + x^4 + \cdots )(1 + x^3 + x^6 + \cdots)\cdots.$

Using the formula for the sum of an infinite geometric sequence we can express this in the more compact form:

$\frac 1{1-x}\cdot \frac 1{1-x^2}\cdot \frac 1{1-x^3}\cdots$

Formulas

An interesting theorem is that the numbers of partitions consisitng of only consecutive positive integers of n is the number of odd divisors of n.

Proof: Let $a$ be the smallest integer in such a partition, and $k$ be the number of integer in this partition, then we have:

$n = a + (a+1) + (a+2) + \dots + (a+k-1)$

$n = ka + (1 + 2 + \dots + k-1)$

$n = ka + \frac{k(k-1)}{2}$

$2n = 2ka + k(k-1)$

$2ka = 2n-k(k-1)$

$2ka = 2n - k^2 + k$

$a = \frac{2n-k^2 + k}{2k}$

and finally:

$a = \frac{n}{k} + \frac{1-k}{2}$

Let's allow negative integers in our partition for a moment, and let $s$ denote the number of odd divisors of $n$ . Now we consider the two possible cases:

Case 1: $\frac{1-k}{2}$ is an integer.

Since the expression on the right hand side must be an integer, this implies that $k | n$ . Also, for $\frac{1-k}{2}$ to be an integer, $1-k$ must be divisible by 2, which implies that $k$ is odd. Therefore, as long as $k$ is an odd divisor of $n$ , we can find $k$ consecutive integers starting with $a$ that adds up to $n$ . Therefore, in this case the number of solutions is equal to $s$ .

Case 2: $\frac{1-k}{2}$ is not an integer.

For the right hand side to still be an integer, we must have $\frac{n}{k} = \frac{t}{2}$ for some odd integer $t$ . Since $\frac{1-k}{2}$ is not an integer, $1-k$ must be odd, and therefore $k$ must be even. expressing $n$ in the form $2^a\cdot b$ with $b$ being an odd integer, we see that $\frac{n}{k} = \frac{t}{2}$ if and only if $k = 2^{a+1} \cdot d$ , where $d$ is any odd divisor of $n$ . As the result, the number of solutions in this case is also equal to $s$ .

The two cases combined gives us $2s$ solutions. However, this includes parititions with negative integers. However, since $n$ is positive, the partition with negative integers must be of the form:

$-a - (a-1) - (a-2) - \dots - 1 + 1 + 2 +\dots + a + a+1 + \dots + b$

Where $b > a$ . As the result, we can cancel out the negative integers with the corresponding additive inverses to produce a valid partition. Similarity, we can go from a valid partition to one that contains negative integers by adding all the positive integers before the smallest member, along with the corresponding additive inverses. As the result, we established a bijection between valid partitions with the partitions containing negative integers.

Since our count includes all the possible partitions, and the number of valid partitions is equal to the number of partitions that contain negative integers, the numer of valid partitions must be exactly half of our original count. Therfore, the total number of such partitions is $s$ , which is the number of odd divisors of $n$ . Q.E.D.

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Partition

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Ferrers Diagrams

The Conjugate

Generating Functions

Formulas

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