# A Recursive Relation for Bipartition Numbers

Yen-Chi Roger Lin  Department of Mathematics, National Taiwan Normal
University, Taipei 116, Taiwan<EMAIL_ADDRESS>and  Shu-Yen Pan
Department of Mathematics, National Tsing Hua University, Hsinchu 300, Taiwan
<EMAIL_ADDRESS>
###### Abstract.

We establish a recursive relation for the bipartition number $p_{2}(n)$ which
might be regarded as an analogue of Euler’s recursive relation for the
partition number $p(n)$. Two proofs of the main result are proved in this
article. The first one is using the generating function, and the second one is
using combinatoric objects (called “symbols”) created by Lusztig for studying
representation theory of finite classical groups.

###### Key words and phrases:

bipartition number, partition number, generating function, Lusztig’s symbol

###### 2020 Mathematics Subject Classification:

Primary: 05A17, 11P87; Secondary: 20C33

###### Contents

  1. 1 Introduction
  2. 2 Generating Functions
  3. 3 Lusztig’s Symbols
  4. A A Congruence Relation for $p_{2}(n)$

##  1\. Introduction

For a positive integer $n$, let $p(n)$ (resp. $p_{2}(n)$) denote the number of
partitions (resp. bipartitions) of $n$. By convention, we define
$p(0)=p_{2}(0)=1$, and $p(x)=p_{2}(x)=0$ if $x$ is not a non-negative integer.
For $n\geq 1$, Euler obtained the following recursive relation for $p(n)$:

$p(n)=\sum_{k\in{\mathbb{Z}}\smallsetminus\\{0\\}}(-1)^{k-1}p\bigl{(}n-\tfrac{1}{2}k(3k-1)\bigr{)}$

(cf. [And76] corollary 1.8). Our result in this article is the following
recursive relation for $p_{2}(n)$:

######  Theorem 1.1.

For $n\geq 0$, we have

$p_{2}(n)=p\bigl{(}\tfrac{n}{2}\bigr{)}+\sum_{k\in{\mathbb{Z}}\smallsetminus\\{0\\}}(-1)^{k-1}p_{2}\bigl{(}n-k^{2}\bigr{)}.$

We will provide two proofs for the above recursive formula from different
point of views. In Section 2, the theorem is first proved by manipulating the
respective generating functions. In Section 3, we will derive the recursive
relation from Lusztig’s theory of “symbols” and the binomial theorem.

It is well known that the irreducible characters of the Coxeter group of type
$B_{n}$ is parametrized by the set of bipartitions of $n$ (cf. [GP00] §5.5),
just like the set of irreducible characters of the Coxeter group of type
$A_{n}$ (i.e., the symmetric group $S_{n}$) is parametrized by the set of
partitions of $n$. It is not surprising that Theorem 1.1 has applications to
the representation theory of finite classical groups. Two consequences are
given in Remark 3.9 and Remark 3.10.

In the appendix, we provide a new proof of a congruence relation for the
bipartition number $p_{2}(n)$ as a reference.

##  2\. Generating Functions

###  2.1. Partitions and bipartitions

For a nonnegative integer $n$, let ${\mathcal{P}}(n)$ denote the set of
partitions of $n$. By convention, we define ${\mathcal{P}}(0)$ to be the set
of “the empty partition”, and ${\mathcal{P}}(x)$ to be the empty set if $x$ is
not a nonnegative integer. Let $p(n)=|{\mathcal{P}}(n)|$ where $|X|$ denotes
the number of elements of a finite set $X$.

Let ${\mathcal{P}}_{2}(n)$ denote the set of _bi-partitions_ of $n$, i.e.,

$\textstyle{\mathcal{P}}_{2}(n)=\left\\{\,\genfrac{[}{]}{0.0pt}{}{\lambda}{\mu}\mid|\lambda|+|\mu|=n\,\right\\}$

where $|\lambda|$ means the sum of parts of $\lambda$. Let
$p_{2}(n)=|{\mathcal{P}}_{2}(n)|$. We know that $p_{2}(0)=1$, and $p_{2}(x)=0$
if $x$ is not a nonnegative integer. For a bipartition
$\Sigma=\genfrac{[}{]}{0.0pt}{}{\lambda}{\mu}$, its _transpose_ is given by
$\Sigma^{\rm t}=\genfrac{[}{]}{0.0pt}{}{\mu}{\lambda}$. A bipartition $\Sigma$
is called _degenerate_ if $\Sigma^{\rm t}=\Sigma$, it is called _non-
degenerate_ , otherwise. If $\Sigma\in{\mathcal{P}}_{2}(n)$ is degenerate,
then $\Sigma=\genfrac{[}{]}{0.0pt}{}{\lambda}{\lambda}$ for some
$\lambda\in{\mathcal{P}}(\tfrac{n}{2})$. Thus, the number of degenerate
bipartitions in ${\mathcal{P}}_{2}(n)$ is equal to $p(\tfrac{n}{2})$.

###  2.2. First proof of Theorem 1.1

######  Lemma 2.1.

We have

$\sum_{n\in{\mathbb{Z}}}(-1)^{n}z^{n^{2}}=\prod_{k=1}^{\infty}\frac{(1-z^{k})^{2}}{1-z^{2k}}.$

###### Proof.

We start with the _Jacobi’s triple product identity_ (cf. [And76] theorem 2.8)

(2.2)
$\sum_{n\in{\mathbb{Z}}}q^{\frac{n(n+1)}{2}}z^{n}=\prod_{k=1}^{\infty}(1+zq^{k})(1+z^{-1}q^{k-1})(1-q^{k}).$

In both sides of (2.2), we replace $z$ by $z^{-1}$, and replace $q$ by $z^{2}$
to obtain

$\sum_{n\in{\mathbb{Z}}}z^{n^{2}}=\prod_{k=1}^{\infty}(1+z^{2k-1})^{2}(1-z^{2k}).$

Then we replace $z$ by $-z$ to get

$\sum_{n\in{\mathbb{Z}}}(-1)^{n}z^{n^{2}}=\prod_{k=1}^{\infty}(1-z^{2k-1})^{2}(1-z^{2k})=\prod_{k=1}^{\infty}(1-z^{2k-1})(1-z^{k}).$

Now for the right-hand side, we need the _Euler’s identity_ : $p_{\rm
d}(n)=p_{\rm o}(n)$ where $p_{\rm d}(n)$ is the number of partitions into
distinct parts of $n$, and $p_{\rm o}(n)$ is the number of partitions into odd
parts. In terms of generating functions, we have

$\prod_{k=1}^{\infty}(1+z^{k})=\sum_{n=0}^{\infty}p_{\rm
d}(n)z^{n}=\sum_{n=0}^{\infty}p_{\rm
o}(n)z^{n}=\prod_{k=1}^{\infty}\frac{1}{1-z^{2k-1}}.$

Therefore,

$\displaystyle\prod_{k=1}^{\infty}(1-z^{2k-1})(1-z^{k})=\prod_{k=1}^{\infty}\frac{1-z^{k}}{1+z^{k}}$
$\displaystyle=\prod_{k=1}^{\infty}\frac{(1-z^{k})^{2}}{(1+z^{k})(1-z^{k})}$
$\displaystyle=\prod_{k=1}^{\infty}\frac{(1-z^{k})^{2}}{1-z^{2k}},$

and the lemma is proved. ∎

###### First Proof of Theorem 1.1.

Recall that

$\sum_{n=0}^{\infty}p(n)z^{n}=\prod_{k=1}^{\infty}\frac{1}{1-z^{k}}\qquad\text{and}\qquad\sum_{n=0}^{\infty}p_{2}(n)z^{n}=\prod_{k=1}^{\infty}\frac{1}{(1-z^{k})^{2}}.$

By Lemma 2.1, we have

$\left(\sum_{n=0}^{\infty}p_{2}(n)z^{n}\right)\left(\sum_{k\in{\mathbb{Z}}}(-1)^{k}z^{k^{2}}\right)=\sum_{n=0}^{\infty}p(n)z^{2n},$

which implies that

$\sum_{k\in{\mathbb{Z}}}(-1)^{k}p_{2}(n-k^{2})=p\bigl{(}\tfrac{n}{2}\bigr{)},$

i.e., the theorem is proved. ∎

##  3\. Lusztig’s Symbols

###  3.1. Definition of the symbols

In this subsection and the next subsection, we recall the notions of “symbols”
by Lusztig. The materials are modified from [Lus77] §2, §3 (see also [Pan24]
§2).

A _symbol_ is an ordered pair
$\Lambda=\binom{A}{B}=\binom{a_{1},a_{2},\ldots,a_{m_{1}}}{b_{1},b_{2},\ldots,b_{m_{2}}}$
of two finite subsets $A,B$ of nonnegative integers. The elements in each of
$A$ and $B$ are written in decreasing order, i.e.,
$a_{1}>a_{2}>\cdots>a_{m_{1}}$ and $b_{1}>b_{2}>\cdots>b_{m_{2}}$. The sets
$A,B$ in $\Lambda$ are also denoted by $\Lambda^{*},\Lambda_{*}$,
respectively, and called the _first row_ and the _second row_ of $\Lambda$. An
element in $\Lambda^{*}$ or $\Lambda_{*}$ is called an _entry_ of $\Lambda$. A
symbol $\Lambda^{\prime}$ is called a _subsymbol_ of $\Lambda$ and denoted by
$\Lambda^{\prime}\subset\Lambda$ if $\Lambda^{\prime*}\subset\Lambda^{*}$ and
$\Lambda^{\prime}_{*}\subset\Lambda_{*}$. For a subsymbol
$\Lambda^{\prime}\subset\Lambda$, we define the substraction
$\Lambda\smallsetminus\Lambda^{\prime}$ to be
$\binom{\Lambda^{*}\smallsetminus\Lambda^{\prime*}}{\Lambda_{*}\smallsetminus\Lambda^{\prime}_{*}}$.
For two symbols $\Lambda,\Lambda^{\prime}$, their _union_ is defined by
$\Lambda\cup\Lambda^{\prime}=\binom{\Lambda^{*}\cup\Lambda^{\prime*}}{\Lambda_{*}\cup\Lambda^{\prime}_{*}}$.

For a symbol
$\Lambda=\binom{A}{B}=\binom{a_{1},a_{2},\ldots,a_{m_{1}}}{b_{1},b_{2},\ldots,b_{m_{2}}}$,
its _rank_ ${\rm rk}(\Lambda)$, _defect_ ${\rm def}(\Lambda)$ and _transpose_
$\Lambda^{\rm t}$ are defined by

$\displaystyle{\rm rk}(\Lambda)$
$\displaystyle=\sum_{i=1}^{m_{1}}a_{i}+\sum_{j=1}^{m_{2}}b_{j}-\biggl{\lfloor}\biggl{(}\frac{|A|+|B|-1}{2}\biggr{)}^{2}\biggr{\rfloor},$
$\displaystyle{\rm def}(\Lambda)$ $\displaystyle=|A|-|B|,$
$\displaystyle\Lambda^{\rm t}$ $\displaystyle=\binom{B}{A}$

where $\lfloor x\rfloor$ denotes the floor function of a real number $x$. It
is not difficult to check that ${\rm
rk}(\Lambda)\geq\bigl{\lfloor}\bigl{(}\frac{{\rm
def}(\Lambda)}{2}\bigr{)}^{2}\bigr{\rfloor}$ for any symbol $\Lambda$. A
symbol $\Lambda$ is called _degenerate_ if $\Lambda^{\rm t}=\Lambda$; it is
called _non-degenerate_ , otherwise.

On the set of symbols , we define an equivalence relation generated by

$\binom{a_{1},a_{2},\ldots,a_{m_{1}}}{b_{1},b_{2},\ldots,b_{m_{2}}}\sim\binom{a_{1}+1,a_{2}+1,\ldots,a_{m_{1}}+1,0}{b_{1}+1,b_{2}+1,\ldots,b_{m_{2}}+1,0}.$

It is easy to see that the ranks and the defects are invariant in a similarity
class. Let $\Phi_{n,d}$ denote the set of similarity classes of symbols of
rank $n$ and defect $d$. Note that $\Phi_{n,d}=\emptyset$ if
$\bigl{\lfloor}(\frac{d}{2})^{2}\bigr{\rfloor}>n$.

We define a mapping

$\binom{a_{1},a_{2},\ldots,a_{m_{1}}}{b_{1},b_{2},\ldots,b_{m_{2}}}\mapsto\genfrac{[}{]}{0.0pt}{}{a_{1}-(m_{1}-1),a_{2}-(m_{1}-2),\ldots,a_{m_{1}-1}-1,a_{m_{1}}}{b_{1}-(m_{2}-1),b_{2}-(m_{2}-1),\ldots,b_{m_{2}-1}-1,b_{m_{2}}}$

from symbols to bipartitions. It is easy to see that two symbols in the same
similarity class has the same image. So the above mapping will be regarded as
a mapping from the set of similarity classes of symbols to the set of
bipartitions. Moreover, it is not difficult to check that the above mapping
gives a bijection

(3.1)
$\Upsilon\colon\Phi_{n,d}\longrightarrow{\mathcal{P}}_{2}\bigl{(}n-\bigl{\lfloor}\bigl{(}\tfrac{d}{2}\bigr{)}^{2}\bigr{\rfloor}\bigr{)}.$

In particular, $|\Phi_{n,d}|=p_{2}\bigl{(}n-(\tfrac{d}{2})^{2}\bigr{)}$ if $d$
is even. Note that if a symbol $\Lambda$ of rank $n$ is degenerate, then ${\rm
def}(\Lambda)=0$ and $\Upsilon(\Lambda)$ is a degenerate bipartition in
${\mathcal{P}}_{2}(n)$. Finally, we define

(3.2) $\Phi^{+}_{n}=\bigcup_{d\equiv
0\pmod{4}}\Phi_{n,d},\qquad\Phi^{-}_{n}=\bigcup_{d\equiv
2\pmod{4}}\Phi_{n,d},\qquad\Phi_{n}=\Phi^{+}_{n}\cup\Phi^{-}_{n}=\bigcup_{d\text{
even}}\Phi_{n,d}.$

######  Example 3.3.

For $n=4$ and $d$ even, we have the following

$\displaystyle\Phi_{4,0}$
$\displaystyle=\textstyle\bigl{\\{}\,\Lambda,\Lambda^{\rm
t}\mid\Lambda=\binom{4}{0},\binom{3}{1},\binom{4,1}{1,0},\binom{3,2}{1,0},\binom{3,1}{2,0},\binom{3,0}{2,1},\binom{4,2,1}{2,1,0},\binom{3,2,1}{3,1,0},\binom{4,3,2,1}{3,2,1,0}\bigr{\\}}\cup\bigl{\\{}\binom{2}{2},\binom{2,1}{2,1}\,\bigr{\\}},$
$\displaystyle\Phi_{4,2}$
$\displaystyle=\textstyle\bigl{\\{}\binom{4,0}{-},\binom{3,1}{-},\binom{3,2,1}{0},\binom{4,1,0}{1},\binom{3,2,0}{1},\binom{3,1,0}{2},\binom{2,1,0}{3},\binom{4,2,1,0}{2,1},\binom{3,2,1,0}{3,1},\binom{4,3,2,1,0}{3,2,1}\bigr{\\}},$
$\displaystyle\Phi_{4,4}$
$\displaystyle=\textstyle\bigl{\\{}\binom{3,2,1,0}{-}\bigr{\\}},$
$\displaystyle\Phi_{4,-d}$ $\displaystyle=\textstyle\\{\,\Lambda^{\rm
t}\mid\Lambda\in\Phi_{4,d}\,\\},$ $\displaystyle\Phi^{+}_{4}$
$\displaystyle=\Phi_{4,0}\cup\Phi_{4,4}\cup\Phi_{4,-4},$
$\displaystyle\Phi^{-}_{4}$ $\displaystyle=\Phi_{4,2}\cup\Phi_{4,-2}.$

Note that $|\Phi_{4,0}|=20=p_{2}(4)$,
$|\Phi_{4,2}|=|\Phi_{4,-2}|=10=p_{2}(3)$, and
$|\Phi_{4,4}|=|\Phi_{4,-4}|=1=p_{2}(0)$.

###  3.2. Families associated to special symbols

A symbol
$Z=\binom{a_{1},a_{2},\ldots,a_{m}}{b_{1},b_{2},\ldots,b_{m}}\in\Phi_{n,0}$ is
called _special_ if $a_{1}\geq b_{1}\geq a_{2}\geq b_{2}\geq\cdots\geq
a_{m}\geq b_{m}$. For a special symbol $Z$, let $Z_{\rm
I}=Z\smallsetminus\binom{Z^{*}\cap Z_{*}}{Z^{*}\cap Z_{*}}$ be the subsymbol
of “singles” in $Z$. Clearly, ${\rm def}(Z_{\rm I})=0$. The _degree_ $\deg(Z)$
is defined to be $|(Z_{\rm I})^{*}|=|(Z_{\rm I})_{*}|$. Note that a degenerate
symbol is always special. Moreover, a special symbol $Z$ with $\deg(Z)=0$
means that $Z$ is degenerate. For a subsymbol $M\subset Z_{\rm I}$, we define

$\Lambda_{M}=(Z\smallsetminus M)\cup M^{\rm t},$

i.e., $\Lambda_{M}$ is obtained from $Z$ by switching the row position of each
entry of $M$. It is clear that ${\rm def}(\Lambda_{M})=-2\,{\rm def}(M)$. For
a special symbol $Z\in\Phi_{n,0}$, we define

$\Phi_{Z}=\\{\,\Lambda_{M}\mid M\subset Z_{\rm I}\,\\},$

i.e., $\Phi_{Z}$ is the set of symbols of exactly the same entries of $Z$. It
is clear that $|\Phi_{Z}|=2^{2\deg(Z)}$. Moreover, we have

(3.4)
$\displaystyle\begin{split}\Phi_{Z}\cap\Phi_{n,-2d}&=\\{\,\Lambda_{M}\mid
M\subset Z_{\rm I},\ {\rm def}(M)=d\,\\},\\\
\Phi_{n}&=\bigcup_{Z\in\Phi_{n,0},\ \text{ special}}\Phi_{Z}.\end{split}$

######  Example 3.5.

We have the following special symbols $Z$ in $\Phi_{4,0}$:

$Z$ | $\binom{3,1}{2,0}$ | $\binom{4}{0}$ | $\binom{3}{1}$ | $\binom{4,1}{1,0}$ | $\binom{4,2,1}{2,1,0}$ | $\binom{3,2,1}{3,1,0}$ | $\binom{4,3,2,1}{3,2,1,0}$ | $\binom{2}{2}$ | $\binom{2,1}{2,1}$
---|---|---|---|---|---|---|---|---|---
$Z_{\rm I}$ | $\binom{3,1}{2,0}$ | $\binom{4}{0}$ | $\binom{3}{1}$ | $\binom{4}{0}$ | $\binom{4}{0}$ | $\binom{2}{0}$ | $\binom{4}{0}$ | $\binom{-}{-}$ | $\binom{-}{-}$
$|\Phi_{Z}|$ | $16$ | $4$ | $4$ | $4$ | $4$ | $4$ | $4$ | $1$ | $1$

######  Example 3.6.

Let $Z=\binom{3,1}{2,0}\in\Phi_{4,0}$. Then $\deg(Z)=2$ and we have the
following $16$ symbols $\Lambda_{M}$ in $\Phi_{Z}$:

$M$ | $\binom{-}{-}$ | $\binom{3}{-}$ | $\binom{1}{-}$ | $\binom{-}{2}$ | $\binom{-}{0}$ | $\binom{3,1}{-}$ | $\binom{-}{2,0}$ | $\binom{3}{2}$
---|---|---|---|---|---|---|---|---
$\Lambda_{M}$ | $\binom{3,1}{2,0}$ | $\binom{1}{3,2,0}$ | $\binom{3}{2,1,0}$ | $\binom{3,2,1}{0}$ | $\binom{3,1,0}{2}$ | $\binom{-}{3,2,1,0}$ | $\binom{3,2,1,0}{-}$ | $\binom{2,1}{3,0}$
$M$ | $\binom{1}{0}$ | $\binom{3}{0}$ | $\binom{1}{2}$ | $\binom{3,1}{2}$ | $\binom{3,1}{0}$ | $\binom{3}{2,0}$ | $\binom{1}{2,0}$ | $\binom{3,1}{2,0}$
$\Lambda_{M}$ | $\binom{3,0}{2,1}$ | $\binom{1,0}{3,2}$ | $\binom{3,2}{1,0}$ | $\binom{2}{3,1,0}$ | $\binom{0}{3,2,1}$ | $\binom{2,1,0}{3}$ | $\binom{3,2,0}{1}$ | $\binom{2,0}{3,1}$

###  3.3. Second proof of Theorem 1.1

######  Lemma 3.7.

Let $Z\in\Phi_{n,0}$ be a special symbol, and let $\delta=\deg(Z)$. Then

$\left|\Phi_{Z}\cap\left(\bigcup_{d\equiv
0\pmod{4}}\Phi_{n,d}\right)\right|-\left|\Phi_{Z}\cap\left(\bigcup_{d\equiv
2\pmod{4}}\Phi_{n,d}\right)\right|=\begin{cases}1,&\text{if $\delta=0$};\\\
0,&\text{if $\delta\geq 1$}.\end{cases}$

###### Proof.

Note that $|(Z_{\rm I})^{*}|+|(Z_{\rm I})_{*}|=2\delta$, and ${\rm
def}(M)\equiv|M|\pmod{2}$. Then from (3.4), we have

$\displaystyle\left|\Phi_{Z}\cap\left(\bigcup_{d\equiv
0\pmod{4}}\Phi_{n,d}\right)\right|$ $\displaystyle=|\\{\,M\subset Z_{\rm
I}\mid|M^{*}|+|M_{*}|\text{ even}\,\\}|=\sum_{k\geq 0,\text{
even}}C^{2\delta}_{k},$ $\displaystyle\left|\Phi_{Z}\cap\left(\bigcup_{d\equiv
2\pmod{4}}\Phi_{n,d}\right)\right|$ $\displaystyle=|\\{\,M\subset Z_{\rm
I}\mid|M^{*}|+|M_{*}|\text{ odd}\,\\}|=\sum_{k\geq 1,\text{
odd}}C^{2\delta}_{k}$

where $C^{l}_{m}=\frac{l!}{m!(l-m)!}$ is a binomial coefficient. Note that by
convention $C^{l}_{m}=0$ if $l,m\in{\mathbb{N}}\cup\\{0\\}$ and $l<m$. It is
well known from the binomial theorem that

$\sum_{k\geq 0,\text{ even}}C^{2\delta}_{k}-\sum_{k\geq 1,\text{
odd}}C^{2\delta}_{k}=\begin{cases}1,&\text{if $\delta=0$};\\\ 0,&\text{if
$\delta\geq 1$}.\end{cases}$

Then the lemma is proved. ∎

######  Corollary 3.8.

Let $n$ be a nonnegative integer. Then
$|\Phi_{n}^{+}|-|\Phi^{-}_{n}|=p(\tfrac{n}{2})$.

###### Proof.

From (3.2) and (3.4), we have

$\displaystyle\Phi^{+}_{n}=\bigcup_{Z\in\Phi_{n,0},\text{
special}}\left(\Phi_{Z}\cap\bigcup_{d\equiv 0\pmod{4}}\Phi_{n,d}\right),$
$\displaystyle\Phi^{-}_{n}=\bigcup_{Z\in\Phi_{n,0},\text{
special}}\left(\Phi_{Z}\cap\bigcup_{d\equiv 2\pmod{4}}\Phi_{n,d}\right).$

Then by Lemma 3.7, we have
$|\Phi_{n}^{+}|-|\Phi^{-}_{n}|=\sum_{Z\in\Phi_{n,0},\
\deg(Z)=0}1=p(\tfrac{n}{2})$. ∎

###### Second Proof of Theorem 1.1.

From (3.1), we know that
$|\Phi_{n,d}|=p_{2}\bigl{(}n-(\tfrac{d}{2})^{2}\bigr{)}$, and from (3.2), we
have

$|\Phi^{+}_{n}|=\sum_{k\in{\mathbb{Z}},\text{
even}}p_{2}\bigl{(}n-k^{2}\bigr{)},\qquad|\Phi^{-}_{n}|=\sum_{k\in{\mathbb{Z}},\text{
odd}}p_{2}\bigl{(}n-k^{2}\bigr{)}.$

Then the theorem follows from Corollary 3.8 immediately. ∎

######  Remark 3.9.

Let ${\mathbf{F}}_{q}$ denote the finite field of $q$ elements where $q$ is a
power of an odd prime. From Lusztig’s theory, for $\epsilon\in\\{+,-\\}$, the
set ${\mathcal{E}}({\rm O}^{\epsilon}_{2n}({\mathbf{F}}_{q}),1)$ of unipotent
characters of the finite even orthogonal group ${\rm
O}^{\epsilon}_{2n}({\mathbf{F}}_{q})$ is parametrized by the set
$\Phi^{\epsilon}_{n}$ (cf. [Lus77] theorem 8.2, see also [Pan24] §3.3). Hence
Corollary 3.8 means that ${\rm O}^{+}_{2n}({\mathbf{F}}_{q})$ has
$p(\frac{n}{2})$ more unipotent characters than ${\rm
O}^{-}_{2n}({\mathbf{F}}_{q})$.

######  Remark 3.10.

Another consequence of Theorem 1.1 is that the number of quadratic unipotent
conjugacy classes of ${\rm O}^{\epsilon}_{2n}({\mathbf{F}}_{q})$ is equal to
the number of quadratic unipotent characters of ${\rm
O}^{\epsilon}_{2n}({\mathbf{F}}_{q})$ for $\epsilon\in\\{+,-\\}$. The detail
will appear in a paper in preparation by the second author.

##  Appendix A A Congruence Relation for $p_{2}(n)$

The following theorem on a congruence relation for $p_{2}(n)$ has been proved
in [HL04] and [CJW06]. Below we provide another proof along the approach taken
by Michael Hirschhorn [Hir00].

######  Theorem A.1.

We have

$p_{2}(5n+2)\equiv p_{2}(5n+3)\equiv p_{2}(5n+4)\equiv 0\pmod{5}$

for any nonnegative integer $n$.

Before the proof, we first set up some notations. Let

$(a;q)=\prod_{k=0}^{\infty}(1-aq^{k}),$

and define

$R(q)=\frac{(q^{2};q^{5})(q^{3};q^{5})}{(q;q^{5})(q^{4};q^{5})}\qquad\text{and}\qquad
c=R(q^{5}).$

Then Hirschhorn obtained (see also [Aig07] p.140)

(A.2)
$\sum_{n=0}^{\infty}p(n)q^{n}=\frac{(q^{25};q^{25})^{5}}{(q^{5};q^{5})^{6}}f(q),$

where

$f(q)=c^{4}+c^{3}q+2c^{2}q^{2}+3cq^{3}+5q^{4}-3c^{-1}q^{5}+2c^{-2}q^{6}-c^{-3}q^{7}+c^{-4}q^{8}.$

From (A.2) and the fact that the expansion of $c$ only involves powers of
$q^{5}$, Hirschhorn obtained

$\sum_{n=0}^{\infty}p(5n+4)q^{5n+4}=5\frac{(q^{25};q^{25})^{5}}{(q^{5};q^{5})^{6}}q^{4}$

and concluded the famous congruence by Ramanujan: $p(5n+4)\equiv 0\pmod{5}$.

###### Proof of Theorem A.1.

We square both sides of (A.2) and reach

(A.3) $\sum_{n=0}^{\infty}p_{2}(n)q^{n}=\left(\sum_{n\geqslant
0}p(n)q^{n}\right)^{2}=\frac{(q^{25};q^{25})^{10}}{(q^{5};q^{5})^{12}}f(q)^{2},$

and

$f(q)^{2}=c^{8}+2c^{7}q+5c^{6}q^{2}+10c^{5}q^{3}+20c^{4}q^{4}+16c^{3}q^{5}\\\
+27c^{2}q^{6}+20cq^{7}+15q^{8}-20c^{-1}q^{9}+27c^{-2}q^{10}-16c^{-3}q^{11}\\\
+20c^{-4}q^{12}-10c^{-5}q^{13}+5c^{-6}q^{14}-2c^{-7}q^{15}+c^{-8}q^{16}.$

Using the fact again that the expansion of $c$ only involves powers of
$q^{5}$, we obtain

$\displaystyle\sum_{n=0}^{\infty}p_{2}(5n+2)q^{5n+2}$
$\displaystyle=5\frac{(q^{25};q^{25})^{10}}{(q^{5};q^{5})^{12}}\left(c^{6}q^{2}+4cq^{7}+4c^{-4}q^{12}\right),$
$\displaystyle\sum_{n=0}^{\infty}p_{2}(5n+3)q^{5n+3}$
$\displaystyle=5\frac{(q^{25};q^{25})^{10}}{(q^{5};q^{5})^{12}}\left(2c^{5}q^{3}+3q^{8}-2c^{-5}q^{13}\right),$
$\displaystyle\sum_{n=0}^{\infty}p_{2}(5n+4)q^{5n+4}$
$\displaystyle=5\frac{(q^{25};q^{25})^{10}}{(q^{5};q^{5})^{12}}\left(4c^{4}q^{4}-4c^{-1}q^{9}+c^{-6}q^{14}\right).$

The theorem now follows from these identities immediately. ∎

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