Pentagonal Number Theorem

Euler's pentagonal number theorem is one of the most remarkable identities in partition theory and q-series. It provides an explicit formula for the infinite product $\prod_{k=1}^\infty (1-q^k)$ and leads to efficient algorithms for computing partition numbers.

TheoremEuler's Pentagonal Number Theorem

$\prod_{k=1}^\infty (1 - q^k) = \sum_{n=-\infty}^\infty (-1)^n q^{n(3n-1)/2} = 1 + \sum_{n=1}^\infty (-1)^n (q^{n(3n-1)/2} + q^{n(3n+1)/2})$

The exponents $\frac{n(3n\pm 1)}{2}$ are called pentagonal numbers. For $n = 1, 2, 3, \ldots$ these give: $1, 2, 5, 7, 12, 15, 22, 26, 35, 40, \ldots$

The name comes from the geometric representation: the $n$ -th pentagonal number counts dots in a regular pentagon with $n$ dots on each side.

ExampleFirst Few Terms

Expanding the product to verify: $\prod_{k=1}^5 (1-q^k) = (1-q)(1-q^2)(1-q^3)(1-q^4)(1-q^5)$ $= 1 - q - q^2 + q^5 + q^7 - q^{12} - q^{15} + \cdots$

The exponents $1, 2, 5, 7, 12, 15$ match the pentagonal numbers exactly, with alternating signs in pairs: $+, -, -, +, +, -, -, +, +, \ldots$

ProofCombinatorial Proof (Sketch)

Consider partitions into distinct parts. We establish a sign-reversing involution on "almost all" such partitions, leaving only those corresponding to pentagonal numbers.

For a partition $\lambda$ into distinct parts, draw its Ferrers diagram. Identify:

The base: the smallest part (bottom row)
The slope: the longest sequence of consecutive integers starting from the smallest

Define an involution that either:

Moves the base to become a new slope column on the right, or
Removes the rightmost slope column to become a new base

This pairs up most partitions with opposite signs. The partitions that cannot be paired are exactly those where the transformation is undefined—these occur at pentagonal numbers with the prescribed signs.

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TheoremPartition Recurrence

The pentagonal number theorem gives a recurrence for $p(n)$ , the number of partitions of $n$ : $p(n) = p(n-1) + p(n-2) - p(n-5) - p(n-7) + p(n-12) + p(n-15) - \cdots$

More precisely: $p(n) = \sum_{k=1}^\infty (-1)^{k+1} [p(n - k(3k-1)/2) + p(n - k(3k+1)/2)]$

where $p(m) = 0$ for $m < 0$ and $p(0) = 1$ .

This recurrence follows from: $\left(\sum_{n=0}^\infty p(n)q^n\right) \cdot \left(\prod_{k=1}^\infty (1-q^k)\right) = 1$

ExampleComputing Partitions

Calculate $p(7)$ using the recurrence: $p(7) = p(6) + p(5) - p(2) - p(0)$ $= 11 + 7 - 2 - 1 = 15$

The pentagonal numbers used are $1, 2, 5, 7$ . Since $12 > 7$ , we stop. We need:

$p(6) = 11$ (can be computed similarly)
$p(5) = 7$
$p(2) = 2$
$p(0) = 1$

This gives $p(7) = 15$ efficiently without listing all partitions.

Remark

The pentagonal number theorem connects to:

Dedekind eta function: $\eta(\tau) = q^{1/24}\prod_{n=1}^\infty (1-q^n)$ where $q = e^{2\pi i\tau}$ is a modular form
Rogers-Ramanujan identities: Similar q-series identities with deep connections to Lie algebras
Partition congruences: Ramanujan's congruences like $p(5n+4) \equiv 0 \pmod{5}$
Statistical mechanics: Partition functions in physics

TheoremJacobi Triple Product

A generalization of the pentagonal number theorem: $\prod_{n=1}^\infty (1-q^{2n})(1+zq^{2n-1})(1+z^{-1}q^{2n-1}) = \sum_{n=-\infty}^\infty z^n q^{n^2}$

Setting $z = -q$ and simplifying recovers the pentagonal number theorem. The triple product is fundamental in the theory of theta functions and elliptic functions.

ExampleApplication to Partition Identities

Using the pentagonal theorem, we can prove Euler's partition theorem (distinct parts = odd parts) by manipulating q-series: $\prod_{k=1}^\infty (1+q^k) = \frac{\prod_{k=1}^\infty (1-q^{2k})}{\prod_{k=1}^\infty (1-q^k)} = \prod_{k=1}^\infty \frac{1}{1-q^{2k-1}}$

The middle equality uses the pentagonal theorem implicitly through product manipulations.

The pentagonal number theorem exemplifies how combinatorial identities, analytic functions, and algebraic structures intertwine in the study of partitions and q-series.