Almost partition identities

Significance

In the past, what we call “almost partition identities” were immediate corollaries of classical theta series identities, such as Jacobi’s triple product. Consequently, they were always of this form: “The number of partitions of n from a certain class with a given partition statistic even” equals “the number of partitions of n from the same class with the given partitions statistic odd.” This paper opens possibilities for research in this area, relying on subtle results in basic hypergeometric series and mock modular forms.

Abstract

An almost partition identity is an identity for partition numbers that is true asymptotically $100\%$ of the time and fails infinitely often. We prove a kind of almost partition identity, namely that the number of parts in all self-conjugate partitions of $n$ is almost always equal to the number of partitions of $n$ in which no odd part is repeated and there is exactly one even part (possibly repeated). Not only does the identity fail infinitely often, but also, the error grows without bound. In addition, we prove several identities involving the number of parts in restricted partitions. We show that the difference in the number of parts in all self-conjugate partitions of $n$ and the number of parts in all partitions of $n$ into distinct odd parts equals the number of partitions of $n$ in which no odd part is repeated, the smallest part is odd, and there is exactly one even part (possibly repeated). We provide both analytic and combinatorial proofs of this identity.

1. Introduction

A partition of a positive integer $n$ is a nonincreasing sequence of positive integers that add up to $n$. We write a partition as $\lambda =\left({\lambda }_1,{\lambda }_2,\dots ,{\lambda }_k\right)$ with ${\lambda }_1\ge {\lambda }_2\ge \cdots \ge {\lambda }_k$. The integers ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_k$ are called the parts of $\lambda$. The number of parts of $\lambda$, denoted $\ell \left(\lambda \right)$, is called the length of $\lambda$. We refer to ${\lambda }_1$ as the largest part of $\lambda$.

As usual, $p\left(n\right)$ denotes the number of partitions of $n$, and $p\left(n\mid \text{condition}\right)$ denotes the number of partitions of $n$ satisfying the condition.

The Ferrers diagram of $\lambda$ is an array of left-justified boxes with ${\lambda }_i$ boxes in row $i$. By abuse of notation, when there is no confusion, we also refer to the Ferrers diagram as $\lambda$. The conjugate of a partition $\lambda$, denoted $\lambda \prime$, is the partition with rows that are precisely the columns of $\lambda$. A partition $\lambda$ is called self-conjugate if $\lambda =\lambda \prime$.

Partition identities have a long history starting with Euler’s famous theorem

$$p\left(n\mid \text{parts odd}\right)=p\left(n\mid \text{parts distinct}\right).$$ [1]

Indeed, ref. 1 provides a survey of the subject up through 1972; the subject has flourished since then.

Somewhat in the shadow of partition identities exists related results, which we call almost partition identities. The first of these is Legendre’s interpretation of Euler’s Pentagonal Number Theorem (2):

$$p\left(n\mid \text{even \# of parts, all distinct}\right)=p\left(n\mid \text{odd \# of parts, all distinct}\right),$$ [2]

except when $n=j\left(3j\pm 1\right)/2$, in which case the two sides differ by $\pm 1$. Thus, [2] is an almost partition identity (i.e., [2] is asymptotically valid $100\%$ of the time), failing for $n\in \left[1,N\right]$ on the order of $\sqrt{N}$ times.

Part of the reason that almost partition identities attract little interest is that, until now, they have been immediate corollaries of a $q$ series or theta series identity and are of the form

$$p\left(n\mid \text{even number of parts subject to}\hspace{0.17em}X\right)=p\left(n\mid \text{odd number of parts subject to}\hspace{0.17em}X\right).$$ [3]

For example, Legendre’s theorem [2] can be generalized by similarly interpreting Jacobi’s triple-product identity (ref. 3, p. 21):

$$\sum _{n=-\infty }^{\infty }{\left(-1\right)}^n{q}^{k\left(\genfrac{}{}{0.0pt}{}{n}{2}\right)+jn}=\prod _{n=0}^{\infty }\left(1-q^{kn+j}\right)\left(1-q^{kn+k-j}\right)\left(1-q^{k\left(n+1\right)}\right).$$ [4]

This work was inspired by the recent work on the difference in the total number of parts in all partitions of $n$ into odd parts and the total number of parts in all partitions of $n$ into distinct parts (4–7). It is natural to consider differences in the number of parts in partitions involved in other identities.

We consider the well-known identity (refs. 3, p. 14, example 8 and 8):

$$p\left(n\mid \text{parts odd distinct}\right)=p\left(n\mid \text{self}-\text{conjugate}\right).$$

We follow the notation of ref. 9. Let $s_o\left(n\right)$ denote the total number of parts in all partitions of $n$ into distinct odd parts, and let $s_c\left(n\right)$ denote the total number of parts in all self-conjugate partitions of $n$. We denote by $\mathcal{O}\mathcal{E}\left(n\right)$ the set of partitions of $n$ in which no odd part is repeated and there is exactly one even part (possibly repeated). Let $OE\left(n\right)=\left|\mathcal{O}\mathcal{E}\left(n\right)\right|$. We denote by $\mathcal{O}\mathcal{E}\prime \left(n\right)$ the subset of partitions in $\mathcal{O}\mathcal{E}\left(n\right)$ with smallest part odd and let $OE\prime \left(n\right)=|\mathcal{O}\mathcal{E}\prime \left(n\right)|$. Then, the following identity holds.

Theorem 1.

For $n\ge 0$, we have

$$s_c\left(n\right)-s_o\left(n\right)=OE\prime \left(n\right).$$

In section 2, we give both analytic and combinatorial proofs of this theorem.

Example 1:

The self-conjugate partitions of 10 are $\left(\mathrm{5,2,1,1,1}\right)$ and $\left(\mathrm{4,3,2,1}\right)$; so, $s_c\left(10\right)=9$. The partitions of 10 into odd distinct parts are $\left(\mathrm{9,1}\right)$ and $\left(\mathrm{7,3}\right)$; so, $s_o\left(10\right)=4$. We have $\mathcal{O}\mathcal{E}\prime \left(10\right)=\left\{\left(\mathrm{7,2,1}\right),\left(\mathrm{6,3,1}\right),\left(\mathrm{5,4,1}\right),\left(\mathrm{5,2,2,1}\right),\left(\mathrm{3,2,2,2,1}\right)\right\}$. Thus, $OE\prime \left(10\right)=5=s_c\left(10\right)-s_o\left(10\right)$.

A modification of the combinatorial proof of Theorem 1 leads to the following identity.

Theorem 2.

Let $\mathcal{A}\left(n\right)=\left\{\lambda \in \mathcal{O}\mathcal{E}\left(n\right)\mid \lambda \hspace{0.17em}\text{has at least one odd part}\right\}$. Let $\mathcal{O}{\mathcal{E}}_2\left(n\right)$ be the subset of $\mathcal{A}\left(n\right)$ of partitions for which half the size of the even part is greater than the number of odd parts. Let $\mathcal{O}\mathcal{E}″\left(n\right)$ be subset of $\mathcal{A}\left(n\right)$ of partitions with smallest part even. Then, for $n\ge 0$,

$$\left|\mathcal{O}{\mathcal{E}}_2\left(n\right)\right|=|\mathcal{O}\mathcal{E}″\left(n\right)|.$$

In section 3, we prove the following asymptotic result about $s_c\left(n\right)$. Note that this result is an almost partition identity that is not of the form [3].

Theorem 3.

The identity

$$s_c\left(n\right)=OE\left(n\right)$$ [5]

is true for almost all $n$. Explicitly, if $N\left(x\right)$ is the number of times that [5] is true for $n<x$, then

$$\underset{x\to \infty }{lim}\frac{N\left(x\right)}{x}=1.$$

Surprisingly, identity [5] fails infinitely often, and the error grows without bound. We explain this in section 3.

Example 2:

The only self-conjugate partition of 7 is $\left(\mathrm{4,2,1,1}\right)$, and therefore, $s_c\left(7\right)=4$. We have $\mathcal{O}\mathcal{E}\left(7\right)=\left\{\left(\mathrm{6,1}\right),\left(\mathrm{5,2}\right),\left(\mathrm{4,3}\right),\left(\mathrm{3,2,2}\right),\left(\mathrm{2,2,2,1}\right)\right\}$, and therefore, $OE\left(7\right)=5$. However, the self-conjugate partitions of 12 are $\left(\mathrm{6,2,1,1,1,1}\right),\left(\mathrm{5,3,2,1,1}\right),\left(\mathrm{4,4,2,2}\right)$; so, $s_c\left(12\right)=15$. We have

$\mathcal{O}\mathcal{E}\left(12\right)=\left\{\left(12\right),\left(\mathrm{9,2,1}\right),\left(\mathrm{8,3,1}\right),\left(\mathrm{7,4,1}\right),\left(\mathrm{7,3,2}\right),\left(\mathrm{7,2,2,1}\right),\left(\mathrm{6,6}\right),\left(\mathrm{6,5,1}\right),\left(\mathrm{5,4,3}\right),\left(\mathrm{5,3,2,2}\right),\left(\mathrm{5,2,2,2,1}\right),\left(\mathrm{4,4,4}\right),\left(\mathrm{4,4,3,1}\right),\left(\mathrm{3,2,2,2,2,1}\right),\left(\mathrm{2,2,2,2,2,2}\right)\right\}$; hence, $OE\left(12\right)=15$.

As a consequence of Theorems 1 and 3, we have the following asymptotic result.

Corollary.

Let $\widetilde{OE}\left(n\right)=|\mathcal{O}\mathcal{E}\left(n\right)\backslash \mathcal{O}\mathcal{E}\prime \left(n\right)|$ be the number of partitions in $\mathcal{O}\mathcal{E}\left(n\right)$ with smallest part even. Then, the identity

$$s_o\left(n\right)=\widetilde{OE}\left(n\right)$$ [6]

is true for almost all $n$ (as in Theorem 3).

We denote by $\mathcal{O}\left(n\right)$ the set of partitions of $n$ into odd parts and by $\mathcal{D}\left(n\right)$ the set of partitions of $n$ into distinct parts. Let $SD\left(n\right)$ denote the sum of the largest parts in all partitions of $n$ into distinct parts: that is,

$$SD\left(n\right)=\sum _{\pi \in \mathcal{D}\left(n\right)}{\pi }_1.$$

Given a partition $\pi$, let

$$h\left(\pi \right)=\frac{{\pi }_1-1}{2},$$

and let

$$SO\left(n\right)=\sum _{\pi \in \mathcal{O}\left(n\right)}h\left(\pi \right).$$

Let $O\#\left(n\right)$ denote the total number of parts in the partitions of $n$ into odd parts: that is,

$$O\#\left(n\right)=\sum _{\pi \in \mathcal{O}\left(n\right)}\ell \left(\pi \right).$$

In section 4, we give analytic and combinatorial proofs of the following theorem.

Theorem 4.

For $n\ge 0$, we have

$$SD\left(n\right)-SO\left(n\right)=O\#\left(n\right).$$

Example 3:

Since $\mathcal{D}\left(6\right)=\left\{\left(6\right),\left(\mathrm{5,1}\right),\left(\mathrm{4,2}\right),\left(\mathrm{3,2,1}\right)\right\}$, we have $SD\left(6\right)=6+5+4+3=18$. However, $\mathcal{O}\left(6\right)=\left\{\left(\mathrm{5,1}\right),\left(\mathrm{3,3}\right),\left(\mathrm{3,1,1,1}\right),\left(\mathrm{1,1,1,1,1,1}\right)\right\}$. Hence, $SO\left(6\right)=\left(5-1\right)/2+\left(3-1\right)/2+\left(3-1\right)/2+\left(1-1\right)/2=2+1+1+0=4$. We also have $O\#\left(6\right)=2+2+4+6=14$. Thus, $SD\left(6\right)-SO\left(6\right)=O\#\left(6\right).$

Finally, in section 4, we also show that $O\#\left(n\right)$ is almost always even. Specifically, we have the following theorem.

Theorem 5.

$O\#\left(n\right)$ is odd if and only if $n$ is an odd generalized pentagonal number.

2. Proofs of Theorems 1 and 2

2.1. Analytic Proof of Theorem 1

We use the notation $Df\left(z,q\right)$ to mean the derivative of $f\left(z,q\right)$ with respect to $z$ evaluated at $z=1$: that is,

$$Df\left(z,q\right)≔{\left.\left(\frac{\partial }{\partial z}f\left(z,q\right)\right)\right|}_{z=1}.$$

Note that, if $f\left(z,q\right)$ is a partition generating function wherein the exponent of $q$ keeps track of the number being partitioned and the exponent of $z$ counts the number of parts, then $Df\left(z,q\right)$ is the generating function for the number of parts in the partitions considered.

We use the fact that

$$D\left(1-z\right)f\left(z,q\right)=-f\left(1,q\right).$$

We denote the generating functions for $s_c\left(n\right)$ and $s_o\left(n\right)$ by $S_c\left(q\right)$ and $S_o\left(q\right)$, respectively. Thus,

$$S_c\left(q\right)=\sum _{n\ge 0}{s}_c\left(n\right)q^n$$

and

$$S_o\left(q\right)=\sum _{n\ge 0}{s}_o\left(n\right)q^n.$$

Recall that the generating function for the number of partitions into distinct odd parts with the exponent of $z$ keeping track of the number of parts is

$$\sum _{m\ge 0}\frac{z^m{q}^{m^2}}{{\left(q^2;q^2\right)}_m},$$

and for the number of self-conjugate partitions, the generating function is

$$\sum _{m\ge 0}\frac{z^m{q}^{m^2}}{{\left(z{q}^2;q^2\right)}_m}.$$

Here, ${\left(a;q\right)}_n=\left(1-a\right)\left(1-aq\right)\left(1-a{q}^2\right)\cdots \left(1-a{q}^{n-1}\right),$ if $n\ge 1$, and ${\left(a;q\right)}_0=1$.

Hence,

$$\begin{array}{ll}\hfill S_c\left(q\right)-S_o\left(q\right)& =D\left(\sum _{m\ge 0}\frac{z^m{q}^{m^2}}{{\left(z{q}^2;q^2\right)}_m}-\sum _{m\ge 0}\frac{z^m{q}^{m^2}}{{\left(q^2;q^2\right)}_m}\right)\hfill \\ \hfill & =\sum _{m\ge 0}\left(\frac{m{q}^{m^2}}{{\left(q^2;q^2\right)}_m}+D\frac{q^{m^2}}{{\left(z{q}^2;q^2\right)}_m}-\frac{m{q}^{m^2}}{{\left(q^2;q^2\right)}_m}\right)\hfill \\ \hfill & =D\sum _{m\ge 0}\frac{q^{m^2}}{{\left(z{q}^2;q^2\right)}_m}.\hfill \end{array}$$ [7]

Now,

$$\begin{array}{ll}\hfill \sum _{m\ge 0}\frac{q^{m^2}}{{\left(z{q}^2;q^2\right)}_m}& \hspace{0.17em}=\hspace{0.17em}\underset{\tau \to 0}{lim}\sum _{m\ge 0}\frac{{\left(-\frac{q}{\tau };q^2\right)}_m{\left(q^2;q^2\right)}_m{\tau }^m}{{\left(q^2;q^2\right)}_m{\left(z{q}^2;q^2\right)}_m}\hfill \\ \hfill & =\underset{\tau \to 0}{lim}\frac{{\left(q^2;q^2\right)}_{\infty }{\left(-q;q^2\right)}_{\infty }}{{\left(z{q}^2;q^2\right)}_{\infty }{\left(\tau ;q^2\right)}_{\infty }}\sum _{m\ge 0}\frac{{\left(z;q^2\right)}_m{\left(\tau ;q^2\right)}_m{q}^{2m}}{{\left(q^2;q^2\right)}_m{\left(-q;q^2\right)}_m}.\hfill \end{array}$$

The last equality was obtained from Heine’s transformation (ref. 3, p. 19, corollary 2.3) by first replacing $q$ by $q^2$ and then substituting $a=-\frac{q}{\tau },b=q^2,c=z{q}^2,t=\tau$.

Therefore,

$$\begin{array}{ll}\hfill \sum _{m\ge 0}\frac{q^{m^2}}{{\left(z{q}^2;q^2\right)}_m}& =\frac{{\left(q^2;q^2\right)}_{\infty }{\left(-q;q^2\right)}_{\infty }}{{\left(z{q}^2;q^2\right)}_{\infty }}\hfill \\ \hfill & +\frac{{\left(q^2;q^2\right)}_{\infty }{\left(-q;q^2\right)}_{\infty }}{{\left(z{q}^2;q^2\right)}_{\infty }}\left(1-z\right)\sum _{m\ge 1}\frac{{\left(z{q}^2;q^2\right)}_{m-1}{q}^{2m}}{{\left(q^2;q^2\right)}_m{\left(-q;q^2\right)}_m}.\hfill \end{array}$$ [8]

We now apply $D$ to [8] to obtain

$$\begin{array}{ll}\hfill S_c\left(q\right)-S_o\left(q\right)& ={\left(-q;q^2\right)}_{\infty }\sum _{m\ge 1}\frac{q^{2m}}{1-q^{2m}}-{\left(-q;q^2\right)}_{\infty }\sum _{m\ge 1}\frac{q^{2m}}{\left(1-q^{2m}\right){\left(-q;q^2\right)}_m}\hfill \\ \hfill & ={\left(-q;q^2\right)}_{\infty }\sum _{m\ge 1}\frac{q^{2m}}{1-q^{2m}}-\sum _{m\ge 1}\frac{q^{2m}{\left(-q^{2m+1};q^2\right)}_{\infty }}{1-q^{2m}}.\hfill \end{array}$$ [9]

Now, the first expression in [9] is the generating function for partitions in which odd parts are distinct and exactly one even part (possibly repeated) appears as a part. The second expression generates the same type of partitions with the added condition that the smallest part is even. Hence, the expression in [9] is the generating function for partitions in which odd parts are distinct, the smallest part is odd, and exactly one even part (possibly repeated) appears as a part.

2.2. Combinatorial Proof of Theorem 1

Recall the combinatorial proof of the identity

$$p\left(n\mid \text{parts odd distinct}\right)=p\left(n\mid \text{self}-\text{conjugate}\right).$$

If $\lambda$ is a partition with distinct odd parts, each part of $\lambda$ is bent into a self-conjugate hook. The resulting hooks are nested so that they become the principal hooks of a self-conjugate partition $\mu$. We set $\varphi \left(\lambda \right)=\mu$. For the inverse transformation ${\varphi }^{-1}$, take a self-conjugate partition $\mu$, and straighten all principal hooks to form the parts of ${\varphi }^{-1}\left(\mu \right)$, a partition with distinct odd parts. The number of parts in $\lambda$ is equal to the size of the Durfee square of $\varphi \left(\lambda \right)$. The Durfee square of a partition is the largest square that fits inside the Ferrers diagram of the partition. Thus, $\ell \left(\varphi \left(\lambda \right)\right)-\ell \left(\lambda \right)$ equals the number of parts below the Durfee square of $\varphi \left(\lambda \right)$.

If we denote by $s\left(n\mid \text{condition}\right)$ the number of parts in all partitions of $n$ satisfying the given condition, we have

$$s_c\left(n\right)-s_o\left(n\right)=\sum _{k=1}^{\lfloor \sqrt{n}\rfloor }s\left(\left(n-k^2\right)/2\mid \text{parts no larger than}k\right).$$ [10]

However, by doubling the size of each part, we see that

$$s\left(\left(n-k^2\right)/2\mid \text{parts no larger than}\hspace{0.17em}k\right)=s\left(n-k^2\mid \text{even parts no larger than}\hspace{0.17em}2k\right).$$

Thus,

$$s_c\left(n\right)-s_o\left(n\right)=s\left(n-k^2\mid \text{even parts no larger than}\hspace{0.17em}2k\right),$$

as suggested by [7] in the analytic proof of Theorem 1.

If $\mu$ is a self-conjugate partition, let $\alpha$ be the partition formed by the parts below the Durfee square of $\mu$. Then, the partition to the right of the Durfee square is $\alpha \prime$.

Definition:

A marked partition is a partition with exactly one part below the Durfee square overlined.

Notice that marked partitions are different from overpartitions where only the first occurrence of a part may be overlined.

We denote by $\overline{\mathcal{C}}\left(n\right)$ the set of marked self-conjugate partitions. For example, the self-conjugate partition $\mu =\left(\mathrm{6,6,5,3,3,2}\right)$ contributes three partitions to $\overline{\mathcal{C}}\left(25\right)$, namely $\left(\mathrm{6,6,5},\overline{3},\mathrm{3,2}\right),\left(\mathrm{6,6,5,3},\overline{3},2\right)$, and $\left(\mathrm{6,6,5,3,3},\overline{2}\right)$, and three is also the number of parts below the Durfee square in $\mu$. It follows from [10] that

$$s_c\left(n\right)-s_o\left(n\right)=\left|\overline{\mathcal{C}}\left(n\right)\right|.$$

To prove the theorem combinatorially, we create a bijection between $\overline{\mathcal{C}}\left(n\right)$ and $\mathcal{O}\mathcal{E}\prime \left(n\right)$. Recall that $\mathcal{O}\mathcal{E}\prime \left(n\right)$ is the set of partitions of $n$ in which no odd part is repeated, the smallest part is odd, and there is exactly one even part (possibly repeated). We do this in two steps.

(i) Let $\mathcal{O}{\mathcal{E}}_1\left(n\right)$ be the subset of $\mathcal{O}\mathcal{E}\left(n\right)$ consisting of partitions of $n$ in which no odd part is repeated, there is exactly one even part $2k$ (possibly repeated), and $k$ is at most the number of odd parts. Note that each partition in $\mathcal{O}{\mathcal{E}}_1\left(n\right)$ has at least one odd part. We create a bijection $\psi$ from $\overline{\mathcal{C}}\left(n\right)$ to $\mathcal{O}{\mathcal{E}}_1\left(n\right)$.

Start with a marked partition $\mu \in \overline{\mathcal{C}}\left(n\right)$ with marked part $\overline{{\mu }_j}$. Remove the $j$th row (i.e., the row corresponding to $\overline{{\mu }_j}$) and all rows of size ${\mu }_j$ above $\overline{{\mu }_j}$ from the Ferrers diagram of $\mu$. Also, remove the same number of columns of length ${\mu }_j$ from $\mu$ (these columns will be to the right of the Durfee square of $\mu$). Denote the obtained partition by $\pi$. Then, $\pi$ is a self-conjugate partition, and ${\varphi }^{-1}\left(\pi \right)$ is a partition with distinct odd parts. Merge each removed row with the corresponding removed column to form even parts, obtaining a partition $\nu$ with equal even parts. Let $\psi \left(\mu \right)={\varphi }^{-1}\left(\pi \right)\cup \nu$ be the partition formed from the parts of ${\varphi }^{-1}\left(\pi \right)$ and $\nu$. Then, $\psi \left(\mu \right)\in \mathcal{O}{\mathcal{E}}_1\left(n\right)$.

For example, $\psi \left(\left(\mathrm{6,6,5},\overline{3},\mathrm{3,2}\right)\right)=\left(\mathrm{9,7,6,3}\right)$, $\psi \left(\left(\mathrm{6,6,5,3},\overline{3},2\right)\right)=\left(\mathrm{7,6,6,5,1}\right)$, and $\psi \left(\left(\mathrm{6,6,5,3,3},\overline{2}\right)\right)=\left(\mathrm{9,7,5,4}\right)$.

The transformation $\psi$ is invertible. For a partition $\xi \in \mathcal{O}{\mathcal{E}}_1\left(n\right)$, first remove the even parts to obtain a partition $\delta$ with distinct odd parts. Split each even part of $\xi$ into two equal parts. Insert half of the resulting parts in the appropriate place under the Durfee square of $\varphi \left(\delta \right)$ and the other half as columns to the right of the Durfee square of $\varphi \left(\delta \right)$. Overline the last inserted row. The resulting partition, ${\psi }^{-1}\left(\xi \right)$, is in $\overline{\mathcal{C}}\left(n\right)$.

(ii) Next, we create a bijection $\chi$ from $\mathcal{O}{\mathcal{E}}_1\left(n\right)$ to $\mathcal{O}\mathcal{E}\prime \left(n\right)$. Recall that the 2-modular MacMahon diagram ${\left[\lambda \right]}_2$ of a partition $\lambda$ is obtained from collapsing two adjacent squares in a row of the Ferrers diagram of $\lambda$ whenever possible and labeling the new square with 2. If ${\lambda }_j$ is odd, the last square in ${\left[\lambda \right]}_2$ is labeled with 1.

For example, if $\lambda =\left(\mathrm{7,4,4,3,1}\right)$, the 2-modular MacMahon diagram ${\left[\lambda \right]}_2$ is given below:

For a partition $\lambda$ in which no odd part is repeated, exactly one even part $2k$ (possibly repeated) occurs, and $k$ is larger than the number of odd parts, we create a modified 2-modular MacMahon diagram ${\left[\lambda \right]}_{2\prime }$ as follows. Denote by $e\left(\lambda \right)$ the partition consisting of the even parts of $\lambda$ and by $o\left(\lambda \right)$ the partition consisting of the odd parts of $\lambda$. Then, ${\left[\lambda \right]}_{2\prime }$ is the diagram obtained by placing the conjugate of ${\left[o\left(\lambda \right)\right]}_2$ below ${\left[e\left(\lambda \right)\right]}_2$. Since $k>\ell \left(o\left(\lambda \right)\right)$, this is the 2-modular MacMahon diagram of a partition.

For example, if $\lambda =\left(\mathrm{8,8,8,5,1}\right)$, the modified 2-modular MacMahon diagram ${\left[\lambda \right]}_{2\prime }$ of $\lambda$ is given below:

Start with $\lambda \in \mathcal{O}{\mathcal{E}}_1\left(n\right)$. Let $2k$ be the size of the even part, and let $m=\ell \left(o\left(\lambda \right)\right)$, the number of odd parts of $\lambda$. Then, $k\le m$. If the smallest part of $\lambda$ is odd, let $\chi \left(\lambda \right)=\lambda \in \mathcal{O}\mathcal{E}\prime \left(n\right)$. If the smallest part of $\lambda$ is even, we will modify ${\left[\lambda \right]}_2$ as follows. Denote by $a$ the smallest odd part of $\lambda$. It corresponds to a part of length $\frac{a+1}{2}$ in ${\left[\lambda \right]}_2$. Since the smallest part in $\lambda$ is even, we have $\frac{a+1}{2}>k$. We can write uniquely $\frac{a+1}{2}-k=q\ell \left(\lambda \right)+r$ with $0<r\le \ell \left(\lambda \right)$. Now, in ${\left[\lambda \right]}_2$, we remove $q\ell \left(\lambda \right)$ boxes labeled 2 from each of the first $m$ rows, and we add $qm$ boxes labeled 2 to each of the $\ell \left(\lambda \right)$ rows of the resulting diagram. The conjugate of the final diagram is the modified 2-modular MacMahon diagram ${\left[\mu \right]}_{2\prime }$ of a partition $\mu$ with $k+qm$ even parts of size $2\ell \left(\lambda \right)$, exactly $m$ odd parts (and thus, $m<\ell \left(\lambda \right)$), and smallest part equal to $2r-1\le 2\ell \left(\lambda \right)-1$. We let $\chi \left(\lambda \right)=\mu \in \mathcal{O}\mathcal{E}\prime \left(n\right)$.

Example 4:

If $\lambda =\left(\mathrm{37,27,25,4,4}\right)$, we have $\ell \left(\lambda \right)=5,k=2,m=3,a=25$. Then, $\frac{a+1}{2}-k=13-2=11$. The 2-modular MacMahon diagram ${\left[\lambda \right]}_2$ of $\lambda$ is

We write $11=2\cdot 5+1$. We remove 10 boxes labeled 2 from each of the parts ending in 1 to obtain

and add $2\cdot 3$ boxes labeled 2 to each part in the diagram above to obtain the following diagram:

The conjugate of this diagram is

which is the modified 2-modular MacMahon diagram for

$\chi \left(\lambda \right)=\left(\mathrm{13,12,12,12,12,12,12,12,12,3,1}\right)$.

The transformation $\chi$ is invertible. To see this, start with $\mu \in \mathcal{O}\mathcal{E}\prime \left(n\right)$, a partition in which no odd part is repeated, the smallest part is odd, and there is exactly one even part $2k$ (possibly repeated). Let $m$ be the number of odd parts in $\mu$. If $k\le m$, then ${\chi }^{-1}\left(\mu \right)=\mu$. If $k>m$, consider the conjugate ${\left[\nu \right]}_2$ of the modified 2-modular MacMahon diagram ${\left[\mu \right]}_{2\prime }$. Denote by $d$ the number of even parts in $\mu$, and write $d$ uniquely as $d=qm+r$ with $0<r\le m$. Remove $qm$ boxes labeled 2 from each row of ${\left[\nu \right]}_2$. We removed a total of $qmk$ boxes. Next, add $qk$ boxes to each of the first $m$ rows of the obtained diagram. Note that these are precisely the rows ending in a box labeled 1. We obtain a 2-modular MacMahon diagram ${\left[\lambda \right]}_2$ for a partition $\lambda$. The partition $\lambda$ has no repeated odd parts, one even part equal to $2r$ and repeated $k-m$ times. Since $r\le m$, $\lambda \in \mathcal{O}{\mathcal{E}}_1\left(n\right)$, and we take ${\chi }^{-1}\left(\mu \right)=\lambda$.

Then, the transformation $\psi ○\chi$ gives a bijection between $\overline{\mathcal{C}}\left(n\right)$ and $\mathcal{O}\mathcal{E}\prime \left(n\right)$.

Remark 1:

As noted at the beginning of this section, the excess in the number of parts between $\varphi \left(\lambda \right)$ and $\lambda$ equals $t-k$, where ${\lambda }_1=2t-1$ and $\ell \left(\lambda \right)=k$. This is also the $M_2-\text{rank}$ of $\lambda$ as defined in section 5 of ref. 10. If we denote the $M_2-\text{rank}$ of $\lambda$ by $m_2\left(\lambda \right)$, we have

$$s_c\left(n\right)-s_o\left(n\right)=\sum m_2\left(\lambda \right),$$

where the sum runs over all partitions of $n$ with distinct odd parts.

2.3. Combinatorial Proof of Theorem 2

The set $\mathcal{A}\left(n\right)$ is the disjoint union of $\mathcal{O}{\mathcal{E}}_1\left(n\right)$ and $\mathcal{O}{\mathcal{E}}_2\left(n\right)$. It is also the disjoint union of $\mathcal{O}\mathcal{E}\prime \left(n\right)$ and $\mathcal{O}\mathcal{E}″\left(n\right)$. In step (ii) of the combinatorial proof of Theorem 1, we showed that $|\mathcal{O}\mathcal{E}\prime \left(n\right)|=\left|\mathcal{O}{\mathcal{E}}_1\left(n\right)\right|$. Then, $\left|\mathcal{O}{\mathcal{E}}_2\left(n\right)\right|=|\mathcal{O}\mathcal{E}″\left(n\right)|$.

We also give a bijection $\zeta$ between $\mathcal{O}{\mathcal{E}}_2\left(n\right)$ and $\mathcal{O}\mathcal{E}″\left(n\right)$.

Let $\lambda \in \mathcal{O}{\mathcal{E}}_2\left(n\right)$, and consider ${\left[\lambda \right]}_{2\prime }$ the modified 2-modular MacMahon diagram of $\lambda$. The conjugate of ${\left[\lambda \right]}_{2\prime }$ is the 2-modular MacMahon diagram ${\left[\mu \right]}_2$ of a partition $\mu \in \mathcal{O}\mathcal{E}″\left(n\right)$. We set $\zeta \left(\lambda \right)=\mu$.

Example 5:

If $\lambda =\left(\mathrm{8,8,8,5,1}\right)$, the modified 2-modular MacMahon diagram is

and its conjugate is

giving $\zeta \left(\lambda \right)=\left(\mathrm{11,7,6,6}\right)$.

The transformation $\zeta$ is clearly reversible, and therefore, $\left|\mathcal{O}{\mathcal{E}}_2\left(n\right)\right|=|\mathcal{O}\mathcal{E}″\left(n\right)|$.

3. Proof of Theorem 3

As in the proof of Theorem 1, we write the generating function for self-conjugate partitions as a limit. We have

$$\sum _{m\ge 0}\frac{z^m{q}^{m^2}}{{\left(z{q}^2;q^2\right)}_m}=\underset{\tau \to 0}{lim}\sum _{m\ge 0}\frac{{\left(-\frac{q}{\tau };q^2\right)}_m{\left(q^2;q^2\right)}_m{z}^m{\tau }^m}{{\left(q^2;q^2\right)}_m{\left(z{q}^2;q^2\right)}_m}.$$

Next, we apply the transformation on the last line of p. 38 of ref. 3, in which we first replace $q$ by $q^2$ and then, substitute $a=-\frac{q}{\tau },b=q^2,c=z{q}^2,t=z\tau$. Finally, we take the limit as $\tau \to 0$. We obtain

$$\sum _{m\ge 0}\frac{z^m{q}^{m^2}}{{\left(z{q}^2;q^2\right)}_m}=\left(1-z\right)\sum _{m\ge 0}{\left(-q;q^2\right)}_m{z}^m.$$ [11]

To find $D\left(1-z\right){\sum }_{m\ge 0}{\left(-q;q^2\right)}_m{z}^m$, we use ref. 11, proposition 2.1 and theorem 1. When using theorem 1 in ref. 11, we first replace $q$ by $q^2$ and then, set $a=0$ and $t=-q$. Therefore,

$$S_c\left(q\right)=D\sum _{m\ge 0}\frac{z^m{q}^{m^2}}{{\left(z{q}^2;q^2\right)}_m}=\sum _{n=1}^{\infty }\frac{q^{n^2}}{{\left(-q;q^2\right)}_n}+{\left(-q;q^2\right)}_{\infty }\sum _{m\ge 1}\frac{q^{2m}}{1-q^{2m}}.$$ [12]

The fact that asymptotically $100\%$ of the coefficients in

$$\sum _{n=1}^{\infty }\frac{q^{n^2}}{{\left(-q;q^2\right)}_n}$$

are zero follows from Theorem 5 and section 3 of ref. 12.

To explain the failure of [5], let $\mathcal{B}\left(n\right)$ be the set of partitions of $n$ into odd parts with no gap (i.e., if $2k-1$ appears as a part, so does every odd part less than $2k-1$). Let $r_e\left(\lambda \right)$ denote the number of parts in $\lambda$ that are repeated an even number of times, and set

$$b\left(n\right)=\sum _{\lambda \in \mathcal{B}}{\left(-1\right)}^{r_e\left(\lambda \right)}.$$

Then, $\sum _{n=1}^{\infty }\frac{q^{n^2}}{{\left(-q;q^2\right)}_n}$ is the generating function for $b\left(n\right)$, and [5] fails whenever $b\left(n\right)\ne \hspace{0.17em}0$. Moreover, from ref. 12, we have $lim\hspace{0.17em}sup\hspace{0.17em}b\left(n\right)=\infty$.

4. Proofs of Theorems 4 and 5

4.1. Analytic Proof of Theorem 4

We first introduce some notation and recall two identities from Ramanujan’s “lost” notebook, which are proved in ref. 13.

Denote by

$$S\left(q\right)={\left(-q;q\right)}_{\infty }=\frac{1}{{\left(q;q^2\right)}_{\infty }}$$

the generating function for partitions into odd parts (and also, for partitions into distinct parts), by

$$D\left(q\right)=-\frac{1}{2}+\sum _{n=1}^{\infty }\frac{q^n}{1-q^n}$$

the generating sum for partitions into indistinct parts (adjusted by the constant $-1/2$), and by

$$R\left(q\right)=\sum _{n=0}^{\infty }\frac{q^{n\left(n+1\right)/2}}{{\left(-q;q\right)}_n}$$

the generating function for the difference between the number of partitions of $n$ into distinct parts with even rank and the number of partitions of $n$ into distinct parts with odd rank. The rank of a partition $\lambda$ equals ${\lambda }_1-\ell \left(\lambda \right)$, the difference between the largest part and the number of parts. From ref. 13, equations 1.4 and 1.5, we have

$$\sum _{n=0}^{\infty }\left(S\left(q\right)-{\left(-q;q\right)}_n\right)\hspace{0.17em}=\hspace{0.17em}S\left(q\right)D\left(q\right)+\frac{1}{2}R\left(q\right)$$ [13]

and

$$\sum _{n=0}^{\infty }\left(S\left(q\right)-\frac{1}{{\left(q;q^2\right)}_n}\right)\hspace{0.17em}=\hspace{0.17em}S\left(q\right)D\left(q^2\right)+\frac{1}{2}R\left(q\right).$$ [14]

Now, we subtract [14] from [13]. Notice that ${\left(-q;q\right)}_{\infty }-{\left(-q;q\right)}_n$ is the generating function for partitions in $\mathcal{D}\left(n\right)$ with parts larger than $n$. When summing, each partition of $\mathcal{D}\left(n\right)$ is counted as many times as its largest part. Hence, the left-hand side of [13] is the generating function for $SD\left(n\right)$. Similarly, the left-hand side of [14] is the generating function for $SO\left(n\right)$.

The right-hand side of [13] minus the right-hand side of [14] equals

$$S\left(q\right)\left(D\left(q\right)-D\left(q^2\right)\right)=\frac{1}{{\left(q;q^2\right)}_{\infty }}\sum _{n=1}^{\infty }\frac{q^{2n-1}}{1-q^{2n-1}}.$$

This is the generating function for $O\#\left(n\right)$ (as shown in the proof of Theorem 5).

4.2. Combinatorial Proof of Theorem 4

Recall Sylvester’s bijection (14). Start with $\pi \in \mathcal{O}\left(n\right)$, and consider the 2-modular MacMahon diagram ${\left[\pi \right]}_2$. For each successive principal hook, create two parts: one equal to the number of boxes in the hook and one equal to the number of boxes labeled 2 in the hook. The partition $\mu$ formed by all of the new parts is in $\mathcal{D}\left(n\right)$.

For example, if $\pi =\left(\mathrm{9,7,7,3}\right)$, the 2-modular MacMahon diagram ${\left[\pi \right]}_2$ is

and the corresponding partition with distinct parts is $\mu =\left(\mathrm{8,7,5,3,2,1}\right)$.

The largest part in $\mu$ is

$${\mu }_1=\frac{{\pi }_1-1}{2}+\ell \left(\pi \right).$$

Summing after all partitions in $\mathcal{O}\left(n\right)$ finishes the proof.

4.3. Analytic Proof of Theorem 5

We have

$$\begin{array}{lll}\hfill \sum _{n\ge 0}O\#\left(n\right)q^n& =D\frac{1}{{\left(zq;q^2\right)}_{\infty }}\hfill & \hfill \\ \hfill & =\frac{1}{{\left(q;q^2\right)}_{\infty }}\sum _{n\ge 1}\frac{q^{2n-1}}{1-q^{2n-1}}\hfill & \hfill \\ \hfill & \equiv \frac{1}{{\left(q;q^2\right)}_{\infty }}\sum _{n\ge 1}\frac{\left(2n-1\right)q^{2n-1}}{1-q^{2n-1}}\hfill & \hfill \left(mod2\right)\\ \hfill & \equiv q\left(\frac{d}{dq}\right)\frac{1}{{\left(q;q^2\right)}_{\infty }}\hfill & \hfill \left(mod2\right)\\ \hfill & \equiv q\left(\frac{d}{dq}\right){\left(-q;q\right)}_{\infty }\hfill & \hfill \left(mod2\right)\\ \hfill & \equiv q\left(\frac{d}{dq}\right){\left(q;q\right)}_{\infty }\hfill & \hfill \left(mod2\right)\\ \hfill & \equiv q\left(\frac{d}{dq}\right)\sum _{n=-\infty }^{\infty }{\left(-1\right)}^n{q}^{n\left(3n-1\right)/2}\hfill & \hfill \left(mod2\right)\\ \hfill & \equiv q\sum _{n=-\infty }^{\infty }{\left(-1\right)}^n\frac{n\left(3n-1\right)}{2}{q}^{n\left(3n-1\right)/2-1}\hfill & \hfill \left(mod2\right)\\ \hfill & \equiv \sum _{\begin{array}{c}n=-\infty \\ n\left(3n-1\right)/2\text{odd}\end{array}}^{\infty }{q}^{n\left(3n-1\right)/2}\hfill & \hfill \left(mod2\right).\end{array}$$

Remark 2:

If $\lambda \in \mathcal{O}\left(n\right)$, then $\ell \left(\lambda \right)\hspace{0.17em}\equiv \hspace{0.17em}n\left(mod2\right)$. Therefore, if $n$ is even, then $O\#\left(n\right)$ is even. If $n$ is odd, then $O\#\left(n\right)\hspace{0.17em}\equiv \hspace{0.17em}Q\left(n\right)\left(mod2\right)$, where $Q\left(n\right)$ is the number of partitions of $n$ into odd parts. By Euler’s Pentagonal Number Theorem, $Q\left(n\right)$ is odd if and only if $n$ is an odd generalized pentagonal number.

5. Concluding Remarks

The inspiration for this article came from ref. 4, where the difference in the number of parts in all partitions counted by the respective sides of Euler’s identity is shown to be equal to the number of partitions similar to those counted by one side of the identity, the condition on the parts being just slightly modified. In this article, we considered the analogous problem for the identity

$$p\left(n\mid \text{parts odd distinct}\right)=p\left(n\mid \text{self}-\text{conjugate}\right).$$

In Theorem 1, we showed that the difference $s_c\left(n\right)-s_o\left(n\right)$ in the number of parts in all partitions counted by the respective sides of the identity is equal to the number of partitions similar to those counted on the left [i.e., all odd parts must be distinct, but there is one even part (possibly repeated), and the smallest part must be odd]. It is natural to ask if $s_c\left(n\right)-s_o\left(n\right)$ is also equal to the number of partitions similar to those counted on the right of the identity with a relaxation of the condition. Indeed, in the combinatorial proof of Theorem 1, we showed that $s_c\left(n\right)-s_o\left(n\right)=\left|\mathcal{O}{\mathcal{E}}_1\left(n\right)\right|$, where $\mathcal{O}{\mathcal{E}}_1\left(n\right)$ is the set of partitions of $n$ in which no odd part is repeated, there is exactly one even part $2k$ (possibly repeated), and $k$ is at most the number of odd parts. Given a partition $\lambda \in \mathcal{O}{\mathcal{E}}_1\left(n\right)$, we take the partition formed by the odd parts, $o\left(\lambda \right)$, and transform it into a self-conjugate partition $\nu$. If $2k$ is the even part of $\lambda$ and it is repeated $d$ times, we add $2d$ to each of the first $k$ parts of $\nu$. The obtained partition, $\mu$, is almost self-conjugate (i.e., the Frobenius symbol of $\mu$ is such that the difference between the first row and the second row is a sequence consisting of $2d$ repeated $k$ times and zero repeated $m-k$ times, where $m$ the number of odd parts in $\lambda$). Note that, in the Frobenius symbol of a self-conjugate partition, the two rows are equal. The described transformation is reversible. Therefore, $s_c\left(n\right)-s_o\left(n\right)$ is also the number of almost self-conjugate partitions of $n$ in which the difference in the rows of the Frobenius symbol is of the form $\left(a,a,\dots ,a,\mathrm{0,0},\dots ,0\right)$, with $a$ even, positive, and appearing at least once.