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Euler's pentagonal recurrence for partition numbers

Statement

Let $p (n) = 0$ for $n < 0$ and $p (0) = 1$ . Then for every integer $n > 0$ ,

$p (n) = p (n - 1) + p (n - 2) - p (n - 5) - p (n - 7) + p (n - 12) + p (n - 15) - \dots,$

where the offsets are the generalized pentagonal numbers $r (3 r - 1) / 2$ and the sum stops once the offset exceeds $n$ .

Equivalently, for every integer $n$ ,

$\sum_{r \in Z} (- 1)^{r} p ⁣ (n - \frac{r (3 r - 1)}{2}) = 0$

when $n > 0$ , while the same sum is $1$ at $n = 0$ .

Facts & Assumptions

Given: the partition series $P (x) = \sum_{n \geq 0} p (n) x^{n}$ .

[L1]

The partition generating function is $P (x) = \prod_{m \geq 1} (1 - x^{m})^{- 1}$ (Integer partitions have generating function $\prod_{n \geq 1} (1 - x^{n})^{- 1}$ ).

[L2]

Euler's pentagonal theorem gives $\prod_{m \geq 1} (1 - x^{m}) = \sum_{r \in Z} (- 1)^{r} x^{r (3 r - 1) / 2}$ (Euler's pentagonal number theorem by Franklin's involution).

[L3]

Coefficients of products are Cauchy sums, and equality of formal series is coefficientwise (Coefficient extraction is $R$ -linear, separates formal series, shifts under multiplication by $x^{k}$ , and converts products to finite convolution).

Proof

technique · coefficient comparison

1.1

Multiply the identities of [L1] and [L2]. The two products are reciprocals, so $P (x) \sum_{r \in Z} (- 1)^{r} x^{r (3 r - 1) / 2} = 1$ .

L1L2algebra

2.1

By [L3], the coefficient of $x^{n}$ on the left side of step 1.1 is $\sum_{r \in Z} (- 1)^{r} p ⁣ (n - \frac{r (3 r - 1)}{2})$ , with the convention $p (m) = 0$ for $m < 0$ . The coefficient on the right side is $0$ when $n > 0$ and $1$ when $n = 0$ . This proves the equivalent formulation.

step 1.1L3

3.1

For $n > 0$ , isolate the $r = 0$ term in step 2.1. The remaining nonzero terms occur in the pairs $r = \pm 1, \pm 2, \dots$ , whose offsets are $1, 2, 5, 7, 12, 15, \dots$ . Moving them to the other side yields the displayed recurrence for $p (n)$ .

step 2.1algebra∎

Depends on

Used by

Euler's pentagonal recurrence computes p(10)=42 Example

Dependency tree · two levels

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Sources

Darij Grinberg, Enumerative Combinatorics: class notes (standard reference, not scraped)