五边形数定理

亦即 (1-x)(1-x²)(1-x³)⋯ =1-x-x²+x⁵+x⁷-x¹²-x¹⁵+x²²+x²⁶+⋯.

欧拉函数展开后，有些次方项被消去，只留下次方项为1, 2, 5, 7, 12,...的项次，留下来的次方恰为广义五边形数。

若将上式视为幂级数，其收敛半径为1，不过若只是当作形式幂级数来考虑，就不会考虑其收敛半径。

和分割函数的关系

欧拉函数的倒数是分割函数的母函数，亦即： 1/(φ (x))=∑ ₖ₌₀^(∞)p(k)x^(k)

其中 p(k) 为k的分割函数。

上式配合五边形数定理，可以得到 (1-x-x²+x⁵+x⁷-x¹²-x¹⁵+x²²+x²⁶+⋯)(1+p(1)x+p(2)x²+p(3)x³+⋯)=1

考虑 xⁿ 项的系数，在 n>0 时，等式右侧的系数均为0，比较等式二侧的系数，可得 p(n)-p(n-1)-p(n-2)+p(n-5)+p(n-7)+⋯ =0

因此可得到分割函数p(n)的递归式 p(n)=p(n-1)+p(n-2)-p(n-5)-p(n-7)+⋯

以n=10为例 p(10)=p(9)+p(8)-p(5)-p(3)=30+22-7-3=42

In mathematics, Euler's pentagonal number theorem relates the product and series representations of the Euler function. It states that （公式略）

In other words, (1-x)(1-x²)(1-x³)⋯ =1-x-x²+x⁵+x⁷-x¹²-x¹⁵+x²²+x²⁶-⋯.

The exponents 1, 2, 5, 7, 12,... on the right hand side are given by the formula gk = k(3k − 1)/2 for k = 1, −1, 2, −2, 3,... and are called (generalized) pentagonal numbers (sequence A001318 in the OEIS). (The constant term 1 corresponds to k=0.)

This holds as an identity of convergent power series for |x|<1, and also as an identity of formal power series.

A striking feature of this formula is the amount of cancellation in the expansion of the product.

Relation with partitions

The identity implies a recurrence for calculating p(n), the number of partitions of n: p(n)=p(n-1)+p(n-2)-p(n-5)-p(n-7)+⋯

or more formally, p(n)=∑ _(k≠ 0)(-1)^(k-1)p(n-gₖ)

where the summation is over all nonzero integers k (positive and negative) and gₖ is the kth generalized pentagonal number. Since p(n)=0 for all n<0, the apparently infinite series on the right has only finitely many non-zero terms, enabling an efficient calculation of p(n).

Franklin's bijective proof

The theorem can be interpreted combinatorially in terms of partitions. In particular, the left hand side is a generating function for the number of partitions of n into an even number of distinct parts minus the number of partitions of n into an odd number of distinct parts. Each partition of n into an even number of distinct parts contributes +1 to the coefficient of xn; each partition into an odd number of distinct parts contributes −1. (The article on unrestricted partition functions discusses this type of generating function.)

For example, the coefficient of x5 is +1 because there are two ways to split 5 into an even number of distinct parts (4 + 1 and 3 + 2), but only one way to do so for an odd number of distinct parts (the one-part partition 5). However, the coefficient of x12 is −1 because there are seven ways to partition 12 into an even number of distinct parts, but there are eight ways to partition 12 into an odd number of distinct parts, and 7 − 8 = −1.

This interpretation leads to a proof of the identity by canceling pairs of matched terms (involution method). Consider the Ferrers diagram of any partition of n into distinct parts. For example, the diagram below shows n = 20 and the partition 20 = 7 + 6 + 4 + 3.

Let m be the number of elements in the smallest row of the diagram (m = 3 in the above example). Let s be the number of elements in the rightmost 45 degree line of the diagram (s = 2 dots in red above, since 7 − 1 = 6, but 6 − 1 > 4). If m > s, take the rightmost 45-degree line and move it to form a new row, as in the matching diagram below.

If m ≤ s (as in our newly formed diagram where m = 2, s = 5) we may reverse the process by moving the bottom row to form a new 45 degree line (adding 1 element to each of the first m rows), taking us back to the first diagram.

Partition recurrence

We can rephrase the above proof, using integer partitions, which we denote as: n=λ ₁+λ ₂+⋯ +λ _(ℓ),

where λ ₁≥ λ ₂≥ … ≥ λ _(ℓ)>0.

The number of partitions of n is the partition function p(n) having generating function: ∑ ₙ₌₀^(∞)p(n)xⁿ=∏ ₖ₌₁^(∞)(1-x^(k))⁻¹

Note that is the reciprocal of the product on the left hand side of our identity: (∑ ₙ₌₀^(∞)p(n)xⁿ)· (∏ ₙ₌₁^(∞)(1-xⁿ))=1

Let us denote the expansion of our product by ∏ ₙ₌₁^(∞)(1-xⁿ)=∑ ₙ₌₀^(∞)aₙxⁿ,

so that (∑ ₙ₌₀^(∞)p(n)xⁿ)· (∑ ₙ₌₀^(∞)aₙxⁿ)=1.

Multiplying out the left hand side and equating coefficients on the two sides, we obtain

a0 p(0) = 1 and ∑ ᵢ₌₀ⁿp(n-i)aᵢ=0 for all n≥ 1. This gives a recurrence relation defining p(n) in terms of an, and vice versa a recurrence for an in terms of p(n). Thus, our desired result: （公式略）

for i≥ 1 is equivalent to the identity ∑ ᵢ(-1)ⁱp(n-gᵢ)=0, where gᵢ:= 1/2(3i²-i) and i ranges over all integers such that gᵢ≤ n (this range includes both positive and negative i, so as to use both kinds of generalized pentagonal numbers). This in turn means: ∑ _(i even)p(n-gᵢ)=∑ _(i odd)p(n-gᵢ).

In terms of sets of partitions, this is equivalent to saying that the following sets are of equal cardinality: （公式略） and （公式略）

where P(n) denotes the set of all partitions of n.

All that remains is to give a bijection from one set to the other, which is accomplished by the function φ from X to Y which maps the partition （公式略） to the partition λ '=φ (λ) defined by: （公式略）

This is an involution (a self-inverse mapping), and thus in particular a bijection, which proves our claim and the identity.