Séminaire Lotharingien de Combinatoire 92 (2025), Article B92c

A PROBABILISTIC PROOF OF EULER’S PENTAGONAL NUMBER THEOREM

SHANE CHERN

Abstract. We present a probabilistic proof of Euler’s pentagonal number theorem based on a shuffling model.

To George Andrews, a great advisor, on his ⌊(tanh 1; tanh 1)_{∞}^{−1}⌋-th birthday.

1. Prologue

Euler’s pentagonal number theorem is recognized as one of the pioneering discoveries in the analysis of the infinite. Formally, it is the following series expansion.

Theorem 1.1. We have

∏_{j≥1}(1 − q^{j}) = ∑_{n=−∞}^{∞}(−1)^{n}q^{\frac{n(3n+1)}{2}} .  (1.1)

This result was first communicated from Euler to Daniel Bernoulli in a letter dated November 30, 1740^{1}; however, no proof was included therein. The seemingly earliest demonstration offered by Euler [3] was published in 1760, which is a retelling of the idea in his letter to Goldbach [4, Brief 144] on June 9, 1750. Euler’s proof, as briefly presented at the end of this paper, is inductive, relying on an ingenious arithmetic manipulation of the series.

Ever since its discovery, the pentagonal number theorem has shown broad and profound connections to other fields. As observed by Legendre [8, pp. 132–133, §458], the series expansion (1.1) may be combinatorially interpreted in terms of integer partitions; this interpretation motivated Franklin’s bijective proof [6] half a century later. Also, Goncharova [7] noticed that the pentagonal number theorem arises as a natural implication from the calculation of the cohomology of the negative part of the Witt algebra, and it is just a tip of a big picture that bonds Lie algebras, partition theory, and q-series. We refer the interested reader to a terse exposition by Andrews [1], and a more detailed summary on Euler’s discovery by Bell [2].

2020 Mathematics Subject Classification. 60C05, 05A19, 11P84.

Key words and phrases. Euler’s pentagonal number theorem, probabilistic proof, shuffling model.

1Euler’s original letter, unfortunately, has not been preserved but the lines regarding the pentagonal number theorem were alluded to briefly in Bernoulli’s reply dated January 28, 1741.

See [5, Brief 52].

The objective of the present paper is to interpret the pentagonal number theorem in a different way. To be specific, we propose a shuffling model in the next section and utilize it to provide a new probabilistic proof.

2. A shuffling model

Let ℕ_{+} be the set of positive integers, and ℕ := ℕ_{+} ∪ {0} the set of natural numbers. Given a, b ∈ ℕ, we denote by [a, b] the set of natural numbers n such that a ≤ n ≤ b, and by [a, ∞) the set of natural numbers n such that n ≥ a.

Throughout, we always assume that q is a real number such that 0 < q < 1. In a classical probability model (Ω, P), the sample space Ω describes the set

of all possible outcomes, and the probability function P : Ω → [0, 1] assigns a probability with value taken from [0, 1] to each element in Ω such that

∑_{ω∈Ω} P_{ω} = 1.

We may also attach an event space E, which is a subset of Ω, to the model in order to describe the outcomes of particular interest. By abuse of notation, we denote by P_{E} the total probability of all elements in the event space E. This probability space is written as (Ω, E, P).

Now fix a sequence 𝔰_{k} := (s_{0}, …, s_{k−1}, s_{k}) of k + 1 different natural numbers for a certain k ≥ 0. We define a probability model �_{𝔰_{k}} = (Ω_{𝔰_{k}}, P^{(𝔰_{k})}) where the sample space is

Ω_{𝔰_{k}} = {s_{0}, s_{1}, …, s_{k−1}, s_{k}},

with^{2}

s_{i} = s_{i}(𝔰_{k}) := \begin{cases}(s_{0}, …, s_{k−1}, s_{k}), & if i = 0, \\ (s_{0}, … s_{k−i−1}, s_{k}, s_{k−i}, …, s_{k−1}), & if i ∈ [1, k − 1], \\ (s_{k}, s_{0}, …, s_{k−1}), & if i = k. \end{cases}  (2.1)

2It is notable that throughout this paper, three different fonts for “s” are used, including s, s and 𝔰, and they should be carefully distinguished, especially for the latter two (in bold and Fraktur styles).

The probability function is then assigned by

P_{s_{i}}^{(𝔰_{k})} = \begin{cases}1 − q, & if i = 0, \\ q^{i}(1 − q), & if i ∈ [1, k − 1], \\ q^{k}, & if i = k. \end{cases}  (2.2)

Note that the outcomes in the sample space are sequences permuted by the original one. Intuitively, we swap s_{k} with its preceding entry with a probability q through a step-by-step procedure, and the program is terminated when a certain step of swapping is not executed or s_{k} already reaches the leftmost position.

This procedure more or less aligns with card shuffling. So although less accurate^{3}, in the present paper, for S a subset of ℕ, we say a sequence 𝔰 := (s_{0}, s_{1}, s_{2}, …) is a shuffle (instead of using the usual term of permutation) of S if every number in 𝔰 belongs to S and every number in S appears exactly once in 𝔰.

We then describe our shuffling model � according to a Markov process:

Definition 2.1 (Markov process for the shuffling model �).

• Time 0: We begin with the sequence (0);

• Time k (k ≥ 1): For the sequence 𝔰_{k−1} = (s_{0}, …, s_{k−1}) derived at Time k −1, we append the number k to the right end to get 𝔰̂_{k} = (s_{0}, …, s_{k−1}, k) (so that the new sequence 𝔰̂_{k} is of length k + 1) and then apply the probability model �𝔰̂_{k}.

If we only look at the outcomes of this stochastic process at Time k and denote the associated probability model as �_{k} := (Ω_{k}, P^{(k)}), it is clear that the sample space Ω_{k} covers all shuffles of [0, k]. Let 𝔰_{k} = (s_{0}, … s_{k−i−1}, k, s_{k−i}, …, s_{k−1}) be such a shuffle wherein i ∈ [0, k]. Then by removing k, the sequence 𝔰_{k−1} = (s_{0}, …, s_{k−1}) is a shuffle of [0, k − 1] according to the construction of the Markov process, and more importantly, it is exactly the one derived at Time k − 1 in this Markov process, thereby lying in the sample space of �_{k−1}. We also iteratively have the probability evaluation:

P_{𝔰_{k}}^{(k)} = P_{𝔰_{k−1}}^{(k−1)} · P_{i},

3The term shuffle is usually reserved for permutations where one divides a sequence into a front and a tail part and then forms a permutation in which the order of the elements from the front part is preserved, and also the order of the elements from the tail part (thus imitating a riffle shuffle of cards). For example, in the shuffle algebra, the multiplication is exactly defined using this shuffle.

where P_{𝔰_{k}}^{(k)} denotes the probability of the event 𝔰_{k} with respect to the probability model �_{k} and

P_{i} = \begin{cases}1 − q, & if i = 0, \\ q^{i}(1 − q), & if i ∈ [1, k − 1], \\ q^{k}, & if i = k, \end{cases}

according to the probability model �𝔰̂_{k}.

Now we read our shuffling model � = (Ω, P) as �_{∞}. Then the sample space Ω of � covers all shuffles of ℕ.

Example 2.2. Consider the shuffle (2, 0, 3, 1, 4, 5, 6, 7 …) of ℕ. Then in the Markov process, at Time 0 the resulting event is (0); at Time 1 the resulting event is (0, 1); at Time 2 the resulting event is (2, 0, 1); at Time 3 the resulting event is (2, 0, 3, 1); at Time 4 the resulting event is (2, 0, 3, 1, 4); and so on.

3. Event spaces

In this section, we describe the event spaces of our interest attached to the shuffling model � = (Ω, P).

For 𝔰_{k} = (s_{0}, …, s_{k}) a generic sequence of different natural numbers of length k+1, recall that we have defined a probability model �_{𝔰_{k}} = (Ω_{𝔰_{k}}, P^{(𝔰_{k})}) in the previous section, where we have in particular labeled elements in the sample space Ω_{𝔰_{k}} by s_{0}, …, s_{k} as recorded in (2.1). Now in this model, we further attach an event space E_{𝔰_{k}} ⊂ Ω_{𝔰_{k}} and call the probability space �_{k} = �_{k}(𝔰_{k}) := (Ω_{𝔰_{k}}, E_{𝔰_{k}}, P^{(𝔰_{k})}) a shuffling (restricted to E_{𝔰_{k}}). For any element in Ω_{𝔰_{k}}, we say it is produced by this shuffling �_{k} if it belongs to E_{𝔰_{k}}; otherwise, it is not produced by �_{k}.

We make the following categorization:

• Type ▷: The shuffling �_{k} is leftmost if the event space E_{𝔰_{k}} is {s_{k}};

• Type ◁: The shuffling �_{k} is anti-leftmost if the event space E_{𝔰_{k}} is {s_{0}, …, s_{k−1}};

• Type ♢: The shuffling �_{k} is free if the event space E_{𝔰_{k}} is {s_{0}, …, s_{k−1}, s_{k}}.

Next, we move on to the shuffling model �. Recall that � is defined based on a Markov process introduced in Definition 2.1 of the previous section. Given any outcome 𝔰 ∈ Ω of the model �, it is derived by the infinite chain

𝔰_{0} → 𝔰_{1} → ⋯ → 𝔰_{k} → ⋯,

where, for the moment, 𝔰_{k} is the shuffle obtained at Time k of our Markov process. Now it is notable that 𝔰_{k}, called the projection of 𝔰 onto the model �_{k}, belongs to the sample space Ω_{k}. In light of what has been done at Time k of the Markov process, 𝔰_{k} may also be viewed as an outcome of the model �𝔰̂_{k} where 𝔰̂_{k} = (s_{0}, …, s_{k−1}, k) with (s_{0}, …, s_{k−1}) =: 𝔰_{k−1}.

The takeaway in our analysis is that once 𝔰_{k−1} ∈ Ω_{k−1} is determined at Time k− 1 of the Markov process, we will not only focus on all possible outcomes of the above �𝔰̂_{k}. Instead, we look at a shuffling �_{k} = �_{k}(𝔰̂_{k}) = (Ω𝔰̂_{k}, E𝔰̂_{k}, P^{(𝔰̂_{k})}) restricted to a certain event space E𝔰̂_{k} ⊂ Ω𝔰̂_{k}. More precisely, we require that �_{k} belongs to one of the three types introduced earlier: “leftmost”, “anti-leftmost”, or “free”. By doing so, our attention is indeed restricted to an event space E_{k} ⊂ Ω_{k} attached to the model �_{k}.

Example 3.1. Consider the Markov process where the outcomes at Time 0, 1, and 2 are given by

(0) → (0, 1) → (2, 0, 1) → · · · .

Then at Time 3,

• The outcome produced by a leftmost shuffling �_{3} includes (3, 2, 0, 1);

• The outcomes produced by an anti-leftmost shuffling �_{3} include (2, 3, 0, 1), (2, 0, 3, 1), and (2, 0, 1, 3);

• The outcomes produced by a free shuffling �_{3} include (3, 2, 0, 1), (2, 3, 0, 1), (2, 0, 3, 1), and (2, 0, 1, 3).

The same idea may be repeated at each time of the Markov process.

For the moment, let us mark each index in the time set ℕ_{+} with a chosen subscript:

T := {k_{□_{k}} : k ∈ ℕ_{+}},

where □_{k} ∈ {▷, ◁, ♢}; such T is called a marked time set.

Now starting with E_{0}(T) = {(0)} = Ω_{0}, we may iteratively construct an event space E_{k}(T) ⊂ Ω_{k} attached to the model �_{k} by

E_{k}(T) := ⋃_{𝔰_{k−1}∈E_{k−1}(T)} \begin{cases}outcomes of the shuffling \\ �_{k} = (Ω𝔰̂_{k}, E𝔰̂_{k}, P^{(𝔰̂_{k})}) of Type □_{k} \end{cases} \} ,

where as before we write 𝔰̂_{k} = (s_{0}, …, s_{k−1}, k) with (s_{0}, …, s_{k−1}) =: 𝔰_{k−1}. Eventually, we arrive at an event space attached to our shuffling model � = (Ω, P):

E(T) := E_{∞}(T) ⊂ Ω.

Example 3.2. We fix the marked time set as {1_{♢}, 2_{◁}, 3_{▷}, …}. Since we witness a free (♢) shuffling at Time 1, then

E_{1}({1_{♢}, 2_{◁}, 3_{▷}, …}) = {(0, 1), (1, 0)};

next, since we witness an anti-leftmost (◁) shuffling at Time 2, then

E_{2}({1_{♢}, 2_{◁}, 3_{▷}, …}) = {(0, 1, 2), (0, 2, 1), (1, 0, 2), (1, 2, 0)};

next, since we witness a leftmost (▷) shuffling at Time 3, then

E_{3}({1_{♢}, 2_{◁}, 3_{▷}, …}) = {(3, 0, 1, 2), (3, 0, 2, 1), (3, 1, 0, 2), (3, 1, 2, 0)}.

If we define

I := {k ∈ ℕ_{+} : □_{k} = ▷},
J := {k ∈ ℕ_{+} : □_{k} = ◁},

it is clear that I and J, as subsets of ℕ_{+}, are disjoint. Conversely, suppose I and J are two disjoint subsets of ℕ_{+}. Then we may recover a marked time set {1_{□_{1}}, 2_{□_{2}}, 3_{□_{3}}, …} according to the rule:

□_{k} = \begin{cases}▷, & if k ∈ I, \\ ◁, & if k ∈ J, \\ ♢, & if k ∉ I and k ∉ J. \end{cases}

Accordingly, we have an event space E({1_{□_{1}}, 2_{□_{2}}, 3_{□_{3}}, …}) attached to the model �, and we abbreviate it by the notation:

⟨I_{▷}, J_{◁}⟩ := E({1_{□_{1}}, 2_{□_{2}}, 3_{□_{3}}, …}).

Example 3.3. Consider the shuffle

(6, 1, 0, 4, 2, 3, 5, 7, 8, 9, 10, …) ∈ Ω.

It belongs to the event spaces ⟨∅_{▷}, ∅_{◁}⟩, ⟨{1}_{▷}, ∅_{◁}⟩, ⟨{1, 6}_{▷}, ∅_{◁}⟩, ⟨∅_{▷}, {2}_{◁}⟩, ⟨∅_{▷}, {4, 5}_{◁}⟩, ⟨{1}_{▷}, {2, 4}_{◁}⟩ (and many others) but not the space ⟨{2}_{▷}, ∅_{◁}⟩ (because the projection of shuffles in this space onto �_{2} should be either (2, 0, 1) or (2, 1, 0)) nor ⟨∅_{▷}, {1}_{◁}⟩ (because the projection of shuffles in this space onto �_{1} should be (0, 1)).

We close this section with two observations.

Proposition 3.4. Let I and J be two disjoint subsets of ℕ_{+}, and I′ and J′ another two disjoint subsets of ℕ_{+}. If there exists a positive integer k such that k ∈ I and k ∈ J′, then

⟨I_{▷}, J_{◁}⟩ ∩ ⟨I′_{▷}, J′_{◁}⟩ = ∅.  (3.1)

Proof. Assume that there exists a certain shuffle 𝔰 in both ⟨I_{▷}, J_{◁}⟩ and ⟨I′_{▷}, J′_{◁}⟩. We project 𝔰 onto the model �_{k} to get a shuffle 𝔰_{k} ∈ Ω_{k}. Since 𝔰 ∈ ⟨I_{▷}, J_{◁}⟩ and k ∈ I, the shuffle 𝔰_{k} is produced by a leftmost shuffling at Time k. Meanwhile, we see from the assumptions 𝔰 ∈ ⟨I′_{▷}, J′_{◁}⟩ and k ∈ J′ that 𝔰_{k} is also produced by an anti-leftmost shuffling. However, the two properties cannot occur simultaneously, and hence we are led to a contradiction. □

Next, we recall that the union of two sets S_{1} and S_{2} is written as S_{1} ⊔ S_{2} once S_{1} and S_{2} are disjoint; this union is called a disjoint union.

Proposition 3.5. Let I and J be disjoint subsets of ℕ_{+}. Then for any subset K of ℕ_{+} that is disjoint from both I and J,

⟨I_{▷}, J_{◁}⟩ = ⨆_{K′⊂K} ⟨(I ⊔ K′)_{▷}, (J ⊔ (K\K′))_{◁}⟩.  (3.2)

In particular, if K is a singleton, namely, K = {k} for a certain k ∈ ℕ_{+}, then

⟨I_{▷}, J_{◁}⟩ = ⟨(I ⊔ {k})_{▷}, J_{◁}⟩ ⊔ ⟨I_{▷}, (J ⊔ {k})_{◁}⟩.  (3.3)

Proof. We only need the fact that at an arbitrary Time k ∈ K, the event space of a free shuffling �_{k} is the disjoint union of the event spaces of a leftmost �_{k} and an anti-leftmost �_{k}. □

4. Probabilities

In the previous section, we have determined the desired event spaces ⟨I_{▷}, J_{◁}⟩ attached to the shuffling model � = (Ω, P) where I, J ⊂ ℕ_{+} are disjoint. Now our task is to analyze the probability P_{⟨I_{▷},J_{◁}⟩}.

Recall that the model � is defined according to a Markov process. In particular, at Time k, we focus on a shuffling �_{k} = �_{k}(𝔰̂_{k}), which is a probability space (Ω𝔰̂_{k}, E𝔰̂_{k}, P^{(𝔰̂_{k})}) revolving around a certain event space E𝔰̂_{k}, where 𝔰_{k−1} = (s_{0}, …, s_{k−1}) is the outcome at Time k − 1 and 𝔰̂_{k} = (s_{0}, …, s_{k−1}, k). Note that if �_{k} is of one of the three types categorized in the previous section, the event space E𝔰̂_{k} is also decided accordingly. In light of the probability function (2.2), we compute that

P_{E𝔰̂_{k}}^{(𝔰̂_{k})} = \begin{cases}q^{k}, & if �_{k} is leftmost , \\ 1 − q^{k}, & if �_{k} is anti-leftmost , \\ 1, & if �_{k} is free . \end{cases}

In particular, this probability evaluation is independent of 𝔰̂_{k}, or equivalently, the outcome 𝔰_{k−1} at Time k − 1.

This independence, together with the nature of a Markov process, immediately gives the following probability evaluation for the event space ⟨I_{▷}, J_{◁}⟩ attached to the shuffling model �.

Proposition 4.1. Let I and J be disjoint subsets of ℕ_{+}. Then

P_{⟨I_{▷},J_{◁}⟩} = ∏_{i∈I} q^{i} ∏_{j∈J}(1 − q^{j}).  (4.1)

Proposition 4.1 has two implications.

Corollary 4.2. Let I and J be two disjoint subsets of ℕ_{+} with I being infinite. Then

P_{⟨I_{▷},J_{◁}⟩} = 0.  (4.2)

Corollary 4.3. Let I and I′ be two finite subsets of ℕ_{+}, not necessarily disjoint, and let J be a subset of ℕ_{+} that is disjoint from both I and I′. Suppose that

∑_{i∈I} i = ∑_{i′∈I′} i′.

Then

P_{⟨I_{▷},J_{◁}⟩} = P_{⟨I′_{▷},J_{◁}⟩}.  (4.3)

Finally, we recall that, in a probability model, if two event spaces E_{1} and E_{2} are disjoint, then

P_{E_{1}⊔E_{2}} = P_{E_{1}} + P_{E_{2}}.

Therefore, Proposition 3.5 implies the following relation.

Proposition 4.4. Let I and J be disjoint subsets of ℕ_{+}. Then for any subset K of ℕ_{+} that is disjoint from both I and J,

P_{⟨I_{▷},J_{◁}⟩} = ∑_{K′⊂K} P_{⟨(I⊔K′)_{▷},(J⊔(K\K′))_{◁}⟩}.  (4.4)

In particular, if K is a singleton, namely, K = {k} for a certain k ∈ ℕ_{+}, then

P_{⟨I_{▷},J_{◁}⟩} = P_{⟨(I⊔{k})_{▷},J_{◁}⟩} + P_{⟨I_{▷},(J⊔{k})_{◁}⟩}.  (4.5)

5. The pentagonal number theorem

The key to our probabilistic proof of the pentagonal number theorem is the following relation.

Theorem 5.1. For every nonnegative integer N,

P_{⟨∅_{▷},[1,∞)_{◁}⟩} = ∑_{n=0}^{N}(−1)^{n}(P_{⟨[n+1,2n]_{▷},∅_{◁}⟩} − P_{⟨[n+1,2n+1]_{▷},∅_{◁}⟩})

− (−1)^{N} ∑_{k≥1} P_{⟨[k+1+N,k+1+2N]_{▷},[1+N,k+N]_{◁}⟩}. (5.1)

We show this result by induction on N.

Base case. We need to prove the case where N = 0:

P_{⟨∅_{▷},[1,∞)_{◁}⟩} = P_{⟨∅_{▷},∅_{◁}⟩} − P_{⟨{1}_{▷},∅_{◁}⟩} − ∑_{k≥1} P_{⟨{k+1}_{▷},[1,k]_{◁}⟩}

= P_{⟨∅_{▷},∅_{◁}⟩} − ∑_{k≥1} P_{⟨{k}_{▷},[1,k−1]_{◁}⟩}.

Note that ⟨∅_{▷}, ∅_{◁}⟩ contains all shuffles of ℕ so ⟨∅_{▷}, ∅_{◁}⟩ = Ω. Meanwhile, given an arbitrary shuffle 𝔰 ∈ Ω, we project it onto each model �_{k} to get an outcome shuffle 𝔰_{k} ∈ Ω_{k} at Time k. Then either every 𝔰_{k} is produced by an anti-leftmost shuffling (this case corresponds to ⟨∅_{▷}, [1, ∞)_{◁}⟩) or there exists a unique time k such that 𝔰_{k} is the first shuffle produced by a leftmost shuffling (this case corresponds to ⟨{k}_{▷}, [1, k −1]_{◁}⟩). Furthermore, given two different positive integers k and k′, we assume without loss of generality that k < k′. Then k ∈ {k} and k ∈ [1, k′ − 1]. In light of Proposition 3.4,

⟨{k}_{▷}, [1, k − 1]_{◁}⟩ ∩ ⟨{k′}_{▷}, [1, k′ − 1]_{◁}⟩ = ∅.

Hence,

⟨∅_{▷}, ∅_{◁}⟩ = ⟨∅_{▷}, [1, ∞)_{◁}⟩ ⊔ (⨆_{k≥1} ⟨{k}_{▷}, [1, k − 1]_{◁}⟩) .

Application of Proposition 4.4 leads us to the required relation. □

Inductive step. It suffices to show that, for each nonnegative integer N,

∑_{k≥1} P_{⟨[k+1+N,k+1+2N]_{▷},[1+N,k+N]_{◁}⟩} = P_{⟨[2+N,2+2N]_{▷},∅_{◁}⟩} − P_{⟨[2+N,3+2N]_{▷},∅_{◁}⟩}

− ∑_{k≥1} P_{⟨[k+2+N,k+3+2N]_{▷},[2+N,k+1+N]_{◁}⟩}.

We start by noting that, for each k ≥ 1, the sets [k+1+N, k+1+2N], [2+N, k+N] and {1 + N} are pairwise disjoint. It follows from Proposition 4.4 that

∑_{k≥1} P_{⟨[k+1+N,k+1+2N]_{▷},[1+N,k+N]_{◁}⟩} = ∑_{k≥1} P_{⟨[k+1+N,k+1+2N]_{▷},[2+N,k+N]_{◁}⟩}

− ∑_{k≥1} P_{⟨({1+N}⊔[k+1+N,k+1+2N])_{▷},[2+N,k+N]_{◁}⟩}.

We split the first sum on the right-hand side as

∑_{k≥1} P_{⟨[k+1+N,k+1+2N]_{▷},[2+N,k+N]_{◁}⟩} = P_{⟨[2+N,2+2N]_{▷},∅_{◁}⟩}

+ ∑_{k≥1} P_{⟨[k+2+N,k+2+2N]_{▷},[2+N,k+1+N]_{◁}⟩}.

We further note that the sets [k+2+N, k+2+2N], [2+N, k+N] and {k+1+N} are pairwise disjoint. Another application of Proposition 4.4 yields

∑_{k≥1} P_{⟨[k+2+N,k+2+2N]_{▷},[2+N,k+1+N]_{◁}⟩} = ∑_{k≥1} P_{⟨[k+2+N,k+2+2N]_{▷},[2+N,k+N]_{◁}⟩}

− ∑_{k≥1} P_{⟨[k+1+N,k+2+2N]_{▷},[2+N,k+N]_{◁}⟩}.

Finally, we see from Corollary 4.3 that

∑_{k≥1} P_{⟨({1+N}⊔[k+1+N,k+1+2N])_{▷},[2+N,k+N]_{◁}⟩} = ∑_{k≥1} P_{⟨[k+2+N,k+2+2N]_{▷},[2+N,k+N]_{◁}⟩}.

This implies that

∑_{k≥1} P_{⟨[k+1+N,k+1+2N]_{▷},[1+N,k+N]_{◁}⟩} = P_{⟨[2+N,2+2N]_{▷},∅_{◁}⟩}

− ∑_{k≥1} P_{⟨[k+1+N,k+2+2N]_{▷},[2+N,k+N]_{◁}⟩}.

By singling out the k = 1 term of the sum in the above expression, we obtain the desired part P_{⟨[2+N,3+2N]_{▷},∅_{◁}⟩}. □

Now we are ready to conclude the pentagonal number theorem.

Proof of Theorem 1.1. Letting N → ∞ in (5.1) and recalling Corollary 4.2, we have

P_{⟨∅_{▷},[1,∞)_{◁}⟩} = ∑_{n≥0}(−1)^{n}(P_{⟨[n+1,2n]_{▷},∅_{◁}⟩} − P_{⟨[n+1,2n+1]_{▷},∅_{◁}⟩}).

Application of Proposition 4.1 yields

∏_{j≥1}(1 − q^{j}) = ∑_{n≥0}(−1)^{n}(q^{\frac{n(3n+1)}{2}} − q^{\frac{(n+1)(3n+2)}{2}} ).

This is exactly (1.1). □

6. Epilogue

Before putting down the curtain, let us flash back to Euler’s original proof of the pentagonal number theorem in [3]. Adopting the q-Pochhammer symbol for n ∈ ℕ ∪ {∞},

(A; q)_{n} := ∏_{j=0}^{n−1}(1 − Aq^{j}),

the key observation of Euler, although not explicitly stated, is the following.

Theorem 6.1. For every nonnegative integer N,

(q; q)_{∞} = ∑_{n=0}^{N}(−1)^{n}(q^{\frac{n(3n+1)}{2}} − q^{\frac{(n+1)(3n+2)}{2}} )
− (−1)^{N} ∑_{k≥1} q^{\frac{(1+N)(2k+2+3N)}{2}} (q^{1+N}; q)_{k}.  (6.1)

It is worth noting that our proof can be understood as a translation of Euler’s proof into the language of classical probability theory. To tie the knot, we simply apply Proposition 4.1 and derive that

P_{⟨[k+1+N,k+1+2N]_{▷},[1+N,k+N]_{◁}⟩} = ∏_{i=k+1+N}^{k+1+2N} q^{i} ∏_{j=1+N}^{k+N}(1 − q^{j})
= q^{\frac{(1+N)(2k+2+3N)}{2}} (q^{1+N}; q)_{k}.

In view of (5.1), this is clearly the last piece of expression required for (6.1).

Acknowledgements. This work was supported by the Austrian Science Fund (No. 10.55776/F1002). I would like to thank the anonymous referee for the inspiring suggestions that have improved the quality of this paper to a great extent. I am also grateful to Hugh Thomas for the comments on Proposition 4.4 in an earlier draft.

References

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2. J. Bell, A summary of Euler’s work on the pentagonal number theorem, Arch. Hist. Exact Sci. 64 (2010), no. 3, 301–373.

3. L. Euler, Demonstratio theorematis circa ordinem in summis divisorum observatum, in: Novi Commentarii Academiae Scientiarum Imperialis Petropolitanae, Tom. V, 75–83, Academiae Scientiarvm, Petropoli, 1760.

4. L. Euler, Opera omnia. Series Quarta A: Commercium epistolicum. Vol. IV: Leonhardi Euleri commercium epistolicum cum Christiano Goldbach, Birkhäuser, Basel, 2015.

5. L. Euler, Opera omnia. Series Quarta A: Commercium epistolicum. Vol. III: Briefwechsel mit Daniel Bernoulli. Teil I: Einleitungen, Briefwechsel 1726–1743, Birkhäuser, Cham, 2016.

6. F. Franklin, Sur le développement du produit infini (1 − x)(1 − x^{2})(1 − x^{3})(1 − x^{4}) …, C. R. Acad. Sci. Paris Sér. A 92 (1881), 448–450.

7. L. V. Goncharova, The cohomologies of Lie algebras of formal vector fields on the line, Funct. Anal. Appl. 7 (1973), no. 2, 91–97.

8. A. M. Legendre, Théorie des nombres. Vol. II. Third edition, Chez Firmin Didot Frère, Paris, 1830. (Reprinted: Blanchard, Paris, 1955).

Fakultät für Mathematik, Universität Wien, Oskar-Morgenstern-Platz 1, 1090 Wien, Austria

  Email address: chenxiaohang92@gmail.com, xiaohangc92@univie.ac.at