# Enumeration of partitions modulo 4

Aditya Khanna

###### Abstract.

The number of standard Young tableaux possible of shape corresponding to a
partition $\lambda$ is called the dimension of the partition and is denoted by
$f^{\lambda}$. Partitions with odd dimensions were enumerated by McKay and
were further classified by Macdonald. Let $a_{i}(n)$ be the number of
partitions of $n$ with dimension congruent to $i$ modulo 4. In this paper, we
refine Macdonald’s and McKay’s results by calculating $a_{1}(n)$ and
$a_{3}(n)$ when $n$ has no consecutive 1s in its binary expansion or when the
sum of binary digits of $n$ is 2 and providing values for $a_{2}(n)$ for all
$n$. We also present similar results for irreducible representations of
alternating groups.

##  1\. Introduction

###  1.1. Foreword

A partition of $n$ is a tuple of natural numbers,
$\lambda:=(\lambda_{1},\ldots,\lambda_{k})$, with non-increasing entries that
add up to $n$. The dimension of the partition $\lambda$, $f^{\lambda}$, is the
number of standard Young tableaux of shape $\lambda$ (ref. Section 2.1) and
can be calculated by the famous hook-length formula (2). Denote by $m_{p}(n)$,
the number of partitions whose dimensions are not divisible by a given natural
number, $p$.
Macdonald, in his landmark paper [Mac71], gave a complete answer for
$m_{p}(n)$ which can be understood quite elegantly using the $p$-core tower
approach (ref. Section 6). The case for $p=2$ has a wonderful form: for
$n=2^{k_{1}}+\ldots+2^{k_{\ell}}$ with $k_{1}>\ldots>k_{\ell}$, we have,

$m_{2}(n)=2^{k_{1}+\ldots+k_{\ell}},$

which is entirely dependent on the binary expansion of $n$.
Note that $m_{2}(n)$ counts all partitions of $n$ which have odd dimension. We
call these odd partitions and will denote their count by $a(n):=m_{2}(n)$.

Recently, there has been some interest in extending Macdonald’s results. In
their unpublished notes, Amrutha P and T. Geetha [AG22] tackle the problem of
computing $m_{2^{k}}(n)$. They provide general recursive results and in the
case of “hook-partitions” of $2^{\ell}$, they find $m_{4}(2^{\ell})$ and
$m_{8}(2^{\ell})$. As we shall see, the enumeration results for $m_{4}(n)$ in
the preprint follow from our analysis of “sparse numbers”, i.e., numbers with
no consecutive 1s in their binary expansions.
This study of partitions has a representation theoretic context as partitions
of $n$ index the irreducible representations of the symmetric group, $S_{n}$.
There are results regarding the behaviour of irreducible representations
corresponding to odd partitions under restriction ([Gia+16], [APS16]). There
are results ([Pel20], [GPS20]) which show that the density of odd character
values (not just degrees) goes to zero. Although, they add to the literature
of odd partitions, none specifically focus on analysing the dimensions
further. Some papers ([OZ21], [Las08]) give explicit results for character
values of symmetric groups but the formulae do not aid us in enumeration.
We aim to fill this gap by providing explicit enumeration results which extend
the odd dimensional partition enumeration formula of Macdonald.

###  1.2. Main Results

Let $a_{i}(n)$ denote the number of partitions of $n$ with dimension congruent
to $i$ modulo 4. Then, the number of odd partitions,
$a(n)=m_{2}(n)=a_{1}(n)+a_{3}(n)$. We compute the values of $a_{1}(n)$ and
$a_{3}(n)$ for some specific values of $n$. The case where $n$ has more than
two 1s in its binary expansion and the leftmost two digits are 1 still remains
unsolved.
Define $\delta(n):=a_{1}(n)-a_{3}(n)$. In this paper, we present the following
main theorem:

######  Theorem 1.

Let $n,m,R\in\mathbb{N}$ with $R\geq 2$ and $m>0$. Suppose, $n=2^{R}+m$ with
$2^{R-1}>m$. Then, we have

$\delta(n)=\begin{cases}0,&\text{if }n\text{ is even}\\\ 4\delta(m),&\text{if
}n\text{ is odd}.\end{cases}$

Equivalently, for $k_{1}>\ldots>k_{\ell}$, we have

$a_{1}(2^{k_{1}}+\ldots+2^{k_{\ell}})=\begin{cases}2^{k_{1}+\ldots+k_{\ell}-1},&\text{if
}k_{\ell}>0\text{}\\\
4a_{1}(\sum\limits_{i=2}^{l}2^{k_{i}})+(2^{k_{1}-1}-2)2^{k_{2}+\ldots+k_{\ell}},&\text{if
}k_{\ell}=0\text{},\end{cases}$

and

$a_{3}(2^{k_{1}}+\ldots+2^{k_{\ell}})=\begin{cases}2^{k_{1}+\ldots+k_{\ell}-1},&\text{if
}k_{\ell}>0\text{}\\\
4a_{3}(\sum\limits_{i=2}^{l}2^{k_{i}})+(2^{k_{1}-1}-2)2^{k_{2}+\ldots+k_{\ell}},&\text{if
}k_{\ell}=0\text{}\end{cases}$

We call a positive integer sparse if it does not have any consecutive ones in
its binary expansion, i.e., $k_{i}>k_{i+1}+1$ in our notation. For instance,
$(42)_{10}=(101010)_{2}$ is sparse. This case has the following pleasing
corollary:

######  Corollary 1.

If $n$ is a sparse number, then

$\delta(n)=\begin{cases}2\text{,}&\text{if }n=2\\\ 0,&\text{if }n>2\text{ is
even}\\\ 4^{\nu(n)-1},&\text{if }n\text{ is odd},\end{cases}$

where $\nu(n)$ denotes the number of 1s (or equivalently the sum of digits) in
the binary expansion of $n$. For $k_{1}>\ldots>k_{\ell}\geq 0$ with
$k_{i}>k_{i+1}+1$ such that $\ell\geq 2$, we have

$a_{1}(2^{k_{1}}+\ldots+2^{k_{\ell}})=\begin{cases}2^{k_{1}+\ldots+k_{\ell}-1},&\text{if
}k_{\ell}>0\\\ 2^{k_{1}+\ldots+k_{\ell}-1}+4^{\ell-2},&\text{if
}k_{\ell}=0\text{},\end{cases}$

and

$a_{3}(2^{k_{1}}+\ldots+2^{k_{\ell}})=\begin{cases}2^{k_{1}+\ldots+k_{\ell}-1},&\text{if
}k_{\ell}>0\\\ 2^{k_{1}+\ldots+k_{\ell}-1}-4^{\ell-2},&\text{if
}k_{\ell}=0\text{},\end{cases}$

Note that we recover explicit and recursion formulae for $a_{1}(n)$ and
$a_{3}(n)$ by using $\delta(n)=a_{1}(n)-a_{3}(n)$ and
$a(n)=a_{1}(n)+a_{3}(n)$.

In the case where the binary expansion starts with $\mathtt{11}$, we have the
following partial result:

######  Theorem 2.

Let $n=2^{R}+2^{R-1}$ with $R\geq 1$, then

$\delta(n)=\begin{cases}2,&\text{if }n=3\\\ 8,&\text{if }n=6\\\
0,&else.\end{cases}$

Explicitly, we have

$a_{1}(2^{R}+2^{R-1})=\begin{cases}2,&\text{if }R=1\\\ 8,&\text{if }R=2\\\
4^{R-1},&else,\end{cases}$

and

$a_{3}(2^{R}+2^{R-1})=\begin{cases}0,&\text{if }R=1\\\ 0,&\text{if }R=2\\\
4^{R-1},&else.\end{cases}$

For $a_{2}(n)$, which counts even dimensional partitions with $f^{\lambda}$
not divisible by 4, we have recursive results for all natural numbers and
closed form results for sparse $n$:

######  Theorem 3.

Let $n=2^{R}+m$ such that $m<2^{R}$. Then we have

$a_{2}(n)=\begin{cases}2^{R}\cdot a_{2}(m)+\binom{2^{R-1}}{2}\cdot
a(m),&\text{ if }m<2^{R-1}\\\ 2^{R}\cdot
a_{2}(m)+\left(\binom{2^{R-1}}{3}+2^{R-1}\right)\cdot\displaystyle{\frac{a(m)}{2^{R-1}}},&\text{
if }2^{R-1}\leq m<2^{R}.\end{cases}$

######  Corollary 2.

When $n$ is sparse, we have

$a_{2}(n)=\begin{cases}\displaystyle{\frac{a(n)}{8}(n-2\nu(n))},&\text{if
}n\text{ is even}\\\ a_{2}(n-1),&\text{if }n\text{ is odd}.\end{cases}$

###  1.3. Structure of the paper

In Section 2, we recall the basic notions relating to partitions such as
tableaux, cores and quotients, and the hook-length formula. We present a
characterization of odd partitions due to Macdonald [Mac71] and introduce the
notion of parents. We also define a function, $\operatorname{Od}$, which
extracts the odd part of the input and returns it modulo 4. In Section 3, we
prove “the workhorse formula” that relates the $\operatorname{Od}$ values of
the cores and parents which allows us to enumerate $\delta(n)$ recursively. In
Section 4 and Section 5, we analyse the workhorse formula to obtain
enumeration results and prove 1 and 2 respectively. In Section 6, we use the
theory of 2-core towers to prove 3. In Section 8, we present some relevant
problems that remain unsolved.

##  2\. Notations and Definitions

We denote by $\mathbb{N}$ the set of natural numbers beginning at 1.

###  2.1. Partitions, Ferrers diagrams and Young Tableaux

Let
${\Lambda\subset\bigcup\limits_{i=1}^{\infty}\mathbb{N}^{i}}$ denote the set
of all tuples $\lambda=(\lambda_{1},\ldots,\lambda_{k})$ such that
$\lambda_{1}\geq\ldots\geq\lambda_{k}$.

######  Definition 1 (Partition).

An element of $\Lambda$ is known as a partition. We call $\lambda\in\Lambda$ a
partition of $n$ if $\sum\limits_{i=1}^{k}\lambda_{i}=n$.

######  Notation 1.

We use the notation $\lambda\vdash n$ to denote $\lambda$ is a partition of
$n$. Denote the size of $\lambda$ by
$|\lambda|:=\sum\limits_{i=1}^{k}\lambda_{i}=n$.

We can represent a partition in the Cartesian plane by constructing a top-left
justified array of boxes with the *i1$i^{\mathrm{st}}$
2$i^{\mathrm{nd}}$3$i^{\mathrm{rd}}$[$i^{\mathrm{th}}$] row containing
$\lambda_{i}$ many boxes.

######  Example 1.

The partition $\lambda=(4,3,3,1)$ can be represented as

4, 3, 3, 1.

This is called the Ferrers diagram of $\lambda$ and will be denoted by
$\operatorname{\text{sh}}(\lambda)$, short for shape on $\lambda$.

The boxes of the Ferrers diagram $\operatorname{\text{sh}}(\lambda)$ can be
filled with distinct numbers from $1$ to $n:=|\lambda|$. We impose the
constraint that the numbers increase from top to bottom and from left to
right. Such a filling is called a standard Young tableau (SYT) on the shape
$\operatorname{\text{sh}}(\lambda)$.

######  Example 2.

Continuing with the above example, the following filling

1 2 7 8, 3 5 10, 4 9 11, 6

is an SYT on $\operatorname{\text{sh}}((4,3,3,1))$.

We can have multiple such SYT of a given shape, and the total number of SYT on
$\operatorname{\text{sh}}(\lambda)$ is denoted by $f^{\lambda}$. We call this
number, the dimension of the partition $\lambda$.

###  2.2. Hooks and hook-lengths

We call the boxes in Ferrers diagrams cells. We label the cell in the
*i1$i^{\mathrm{st}}$ 2$i^{\mathrm{nd}}$3$i^{\mathrm{rd}}$[$i^{\mathrm{th}}$]
row from top and *j1$j^{\mathrm{st}}$
2$j^{\mathrm{nd}}$3$j^{\mathrm{rd}}$[$j^{\mathrm{th}}$] column from left by
$(i,j)$. Write $(i,j)\in\operatorname{\text{sh}}(\lambda)$ if the $(i,j)$ cell
exists.
A very useful tool in the study of partitions is the notion of a hook. For a
cell at $(i,j)$ consider all the cells to the right, to the bottom of it and
the cell itself. This constitutes the hook at $(i,j)$. The number of cells is
known as its hook-length and is denoted by $h_{i,j}$.

######  Example 3.

In the previous example, the cell at $(1,2)$ has the hook-length, $h_{1,2}=5$.
The hook is denoted in pink.

4, 3, 3, 1*[*(pink)]1 + 3, 1 +1, 1+ 1

###  2.3. Hook removal and cores

A hook of length $t$ is called a $t$-hook. If the Ferrers diagram has a
$t$-hook, then remove all the cells contained in the hook. This gives us two
(possibly empty) Ferrers diagrams. Slide the bottom diagram to the left and
then upwards to reconnect and form a Ferrers diagram.

######  Example 4.

Consider the partition $(5,5,5,4,2)$ and its Ferrers diagram

5, 5, 5, 4, 2.

In our case, let $t=5$, that is, we wish to remove a 5-hook from the diagram.
We undergo the following procedure:

boxsize = 0.3cm, aligntableaux = center [*(cyan)]5, 2, 2, 2, 2*[*(pink)]0, 0,
3 + 2, 3 + 1*[*(lightgray)]0, 2 + 3, 2 + 1, 2 + 1$\rightarrow$ [*(cyan)]5, 2,
2, 2, 2*[*(pink)]0, 0, 3 + 2, 3 + 1$\rightarrow$ , , [←] , [←], *[*(cyan)]5,
2, 2, 2, 2*[*(pink)]0, 0, 3 + 2, 3 + 1 $\rightarrow$ , , , , *[*(cyan)]5, 2,
2, 2, 2*[*(pink)]0, 0, 2+ 2, 2 + 1$\rightarrow$ , [↑][↑], , , *[*(cyan)]5, 2,
2, 2, 2*[*(pink)]0, 0, 2+ 2, 2 + 1$\rightarrow$ , , , , *[*(cyan)]5, 2, 2, 2,
2*[*(pink)]0, 2+ 2, 2 + 1 boxsize = normal, aligntableaux = top

to obtain the partition $(5,4,3,2,2)$

5, 4, 3, 2, 2.

######  Definition 2 ($t$-core of a partition).

If $\operatorname{\text{sh}}(\mu)$ does not have a $t$-hook, then we call
$\mu$ a $t$-core.

For a partition $\lambda$, if we keep applying this $t$-hook removal process
on the subsequent partitions we obtain, we will eventually reach a $t$-core.
The $t$-core of $\lambda$ is independent of the order of hook-removal and is
thus unique (cf. Example 8(c) [Mac15] p. 12). We denote it by
$\operatorname{{core}}_{t}(\lambda)$.

######  Notation 2.

The set of all $t$-cores is denoted by $\widetilde{\Lambda}_{t}$.

###  2.4. $\beta$sets

The set of first column hook-lengths of $\operatorname{\text{sh}}(\lambda)$
for $\lambda=(\lambda_{1},\ldots,\lambda_{k})$ will be denoted by

$H(\lambda)=\\{h_{i,1}\mid 1\leq i\leq k\\},$

which we conventionally order in a decreasing fashion. Also, we have the
relation, $h_{i,1}=\lambda_{i}+k-i$.

For any finite $X\subset\operatorname{\mathbb{N}\cup\\{0\\}}$, define the
$r$-shift of $X$ to be

$X^{+r}=\\{x+r\mid x\in X\\}\cup\\{0,\ldots,r-1\\}.$

We fix $X^{+0}=X$.

######  Definition 3 ($\beta$-set).

For a partition $\lambda$, all sets of the form $H(\lambda)^{+r}$ for $r\geq
0$ are known as the $\beta$-sets of $\lambda$.

We can impose the relation, $\sim_{\beta}$, on finite sets
$X,Y\subset\mathbb{N}$ such that $X\sim_{\beta}Y$ if and only if $X=Y^{+r}$ or
$Y=X^{+r}$, for some $r\in\operatorname{\mathbb{N}\cup\\{0\\}}$. This is an
equivalence relation on the set of $\beta$-sets, i.e.,
$\operatorname{\mathbb{N}\cup\\{0\\}}$.

There is a natural way to understand $t$-hooks and $t$-cores through
$\beta$-sets.

######  Proposition 1.

Let $\lambda$ be a partition and $X$ be a $\beta$-set of $\lambda$. We have
that $\operatorname{\text{sh}}(\lambda)$ contains a $t$-hook if and only if
there exists an $h$ in $X$ such that $h>t$ and $h-t$ is not in $X$.
Furthermore, if $\mu$ is the partition obtained after removing the $t$-hook,
then $H(\mu)\sim_{\beta}(H(\lambda)\cup\\{h-t\\})\backslash\\{h\\}$.

###### Proof.

The proof can be found in Example 8(a) of [Mac15] and Corollary 1.5 on page 7
of [Ols93]. ∎

######  Remark 1.

The element $h$, as above, will be referred to as the affected hook-length.
With the above notation, we write $h^{\lambda}_{\mu}:=h$.

The recipe for hook removal is quite clear. Pick an element $h$ of the
$\beta$-set, $X$, and subtract $t$ from it. If $h-t\not\in X$, then replace
$h$ by $h-t$; otherwise choose another element from $X$. If no “$x\rightarrow
x-t$” replacements can be performed, declare the partition as a $t$-core.

######  Example 5.

Let $H(\lambda)=\\{10,8,7,5,2\\}$. In this case, we have that $8\in
H(\lambda)$ but $3\not\in H(\lambda)$. Thus, there exists a 5-hook in
$\operatorname{\text{sh}}(\lambda)$ which we can remove to obtain
$\lambda^{\prime}$. Following the above rules, we get
$H(\lambda^{\prime})=\\{10,7,5,3,2\\}$. Now, we can further replace $5$ by 0
to obtain
$H(\lambda^{\prime\prime})=\\{10,7,5,3,2,0\\}\sim_{\beta}\\{9,6,4,2,1\\}$. No
further replacements can be performed. Thus,
$\lambda^{\prime\prime}=(5,3,2,1,1)$ is the 5-core of $\lambda$.

We can reconstruct the original partition from a $\beta$-set by setting
$\lambda_{i}=h_{i}+i-k$ and considering only the values where $\lambda_{i}>0$.

######  Notation 3.

If $X$ is a $\beta$-set of $\lambda$, then we define
$\operatorname{{Part}}(X)=\lambda$.

###  2.5. Odd dimensional partitions

Using the notation, $f^{\lambda}$, for the dimension of a partition $\lambda$
(refer Section 2.1), we have the following proposition due to Frobenius:

######  Proposition 2 (Hook-length formula).

Let $\lambda\in\Lambda$ and $X$ be a $\beta$-set of $\lambda$. If explicitly,
$X=\\{h_{1},\ldots,h_{k}\\}$ such that $h_{1}>\ldots>h_{k}$, then we have

$f^{\lambda}=\dfrac{|\lambda|!\prod\limits_{1\leq i<j\leq
k}(h_{i}-h_{j})}{\prod\limits_{i=1}^{k}h_{i}!}.$

######  Remark 2.

The more famous hook-length formula [FRT54]

$f^{\lambda}=\dfrac{n!}{\prod\limits_{h}h}$

has a product of all hook-lengths of $\lambda$ in the denominator. This can be
used to derive the formula in 2.

We call $\lambda$ an odd partition if $f^{\lambda}$ is odd. Macdonald [Mac71]
gave the following characterization of odd partitions which forms the backbone
of our analysis:

######  Proposition 3.

If $\lambda$ is a partition of $n=2^{R}+m$ with $m<2^{R}$, then $\lambda$ is
an odd partition if and only if $\lambda$ contains exactly one $2^{R}$-hook
and $\operatorname{{core}}_{2^{R}}(\lambda)$ is also an odd partition.

######  Definition 4.

If $\operatorname{{core}}_{2^{R}}(\lambda)=\mu$, then we call $\lambda$ a
$2^{R}$-parent of $\mu$.

We can now give an explicit form for the $2^{R}$-parents of odd partitions:

######  Proposition 4.

Let $2^{R}>m$ and $\mu$ be a partition of $m$. If $\lambda$ is a partition
such that $\operatorname{{core}}_{2^{R}}(\lambda)=\mu$, then exactly one of
the following holds:

  1. Type I.

$H(\lambda)=\left(H(\mu)\cup\\{x+2^{R}\\}\right)\backslash\\{x\\}$ for some
$x\in H(\mu)$,

  2. Type II.

$H(\lambda)=\left(H(\mu)^{+r}\cup\\{2^{R}\\}\right)\backslash\\{0\\}$ for some
$1\leq r\leq 2^{R}$ such that $2^{R}\not\in H(\mu)^{+r}$.

###### Proof.

By 1, we get
$(H(\lambda)\cup\\{h-2^{R}\\})\backslash\\{h\\}\sim_{\beta}H(\mu)$ which
implies $(H(\lambda)\cup\\{h-2^{R}\\})\backslash\\{h\\}=H(\mu)^{+r}$.
The reader can show that for non-empty finite sets $A$, $B$,
$C\subset\operatorname{\mathbb{N}\cup\\{0\\}}$, we have

  1. (1)

$A\backslash B=C\implies A=B\cup C$ if and only if $B\subset A$

  2. (2)

$A=B\cup C\implies A\backslash B=C$ if and only if $B\cap C=\varnothing$.

Using this, we get Type I by letting $r=0$, putting $h-2^{R}=x$. Similarly, we
get Type II by putting $h=2^{R}$. ∎

######  Example 6.

Let $\mu=(2,2,2)$ with $2^{R}=8$. We have $H(\mu)=\\{4,3,2\\}$. If we let
$x=3$, then we obtain a Type I parent, $\lambda$, with
$H(\lambda)=\\{11,4,2\\}$. For a Type II parent, we choose $r=2$, which gives
us $\lambda^{\prime}$ with $H(\lambda^{\prime})=\\{8,6,5,4,1\\}$. The
corresponding Ferrers diagrams for $\lambda$ (left) and $\lambda^{\prime}$
(right) with hooks added to $\mu$ are:

2,2,2*[*(pink)]2 + 7, 2 + 1 2,2,2*[*(pink)]2 + 2, 2 + 1, 2 + 1, 3, 1.

Using this, we can recover the enumeration result for odd partitions:

######  Proposition 5.

Let $n=2^{k_{1}}+\ldots+2^{k_{r}}$ with $k_{1}>\ldots>k_{r}$. Then, the number
of odd partitions of $n$ is given by $a(n):=2^{k_{1}+\ldots+k_{r}}$.

###### Proof.

Let $n=2^{k_{1}}+m$ and $\mu\vdash m$ be an odd partition. There are exactly
$2^{k_{1}}$ many $2^{k_{1}}$-parents of $\mu$ and all of them are odd
partitions. This gives us the recursion, $a(n)=2^{k_{1}}a(m)$. ∎

###  2.6. Od function

######  Notation 4.

For $n\in\mathbb{N}$, let $v_{2}(n)$ denote the largest power of 2 that
divides $n$.

######  Definition 5.

Let $\operatorname{Od}:\mathbb{N}\rightarrow\\{\pm 1\\}$ be defined as follows

$\operatorname{Od}(n)=\begin{cases}1,&\text{if }n/2^{v_{2}(n)}\equiv 1\mod
4\\\ -1,&\text{if }n/2^{v_{2}(n)}\equiv 3\mod 4.\end{cases}$

######  Lemma 1.

For all $m,n\geq 1$, we have
$\operatorname{Od}(mn)=\operatorname{Od}(m)\operatorname{Od}(n)$.

###### Proof.

The proof is elementary and follows from $o(n):=n/2^{v_{2}(n)}$ being
multiplicative. ∎

######  Notation 5 (Binary expansion).

Let $n\in\operatorname{\mathbb{N}\cup\\{0\\}}$.

  1. (1)

Let $n=\sum\limits_{i=0}^{k}b_{i}2^{i}$ such that $b_{i}=\\{0,1\\}$. We write
$n=b_{k}\ldots b_{0}$ if $b_{k}=1$ and $b_{i}=0$ for $i>k$. This is the binary
expansion of $n$.

  2. (2)

Let $\operatorname{bin}(n):=\\{i\mid b_{i}=1\\}$, i.e., the “positions” of 1s
in the binary expansion of $n$.

  3. (3)

Denote the sum of digits by $\nu(n):=\sum\limits_{i\geq 0}b_{i}$.

  4. (4)

Let $s_{2}(n):=b_{k}+b_{k-1}$, i.e., the sum of the leftmost two digits of
$n$.

######  Remark 3.

Let $n$ be as above. Denote $j:=\min(\operatorname{bin}(n))$. Then,
$\operatorname{Od}(n)\equiv b_{j+1}b_{j}\mod 4$. This corresponds to taking
the binary expansion of $n$, removing all the rightmost zeros and returning
the rightmost two digits of the newly obtained string modulo 4.

##  3\. The Workhorse Formula

In this section, we construct a relationship between partitions and their
$2^{R}$-parents, which will allow us to analyse their dimensional behaviour
modulo 4. We first define a statistic of consequence inspired by [DW89]:

######  Definition 6.

For a natural number $n$ (with binary expansion $b_{k}\ldots b_{0}$), let the
number of pairs of consecutive 1s (disjoint or overlapping) be denoted by
$D(n)$. Notationally, $D(n)$ is the number of $i$ such that the product
$b_{i}\cdot b_{i+1}=1$.

######  Example 7.

We have $D(7)=2$, $D(42)=0$ and $D(367)=4$.

We start with the following lemma:

######  Lemma 2.

For any natural number $n$, we have

$\operatorname{Od}(n!)=(-1)^{D(n)+\nu(\lfloor n/4\rfloor)},$

where $\nu(n)$ is the number of 1s in the binary expansion of $n$ as in 5 and
$\lfloor\cdot\rfloor$ denotes the integral part.

###### Proof.

As $\operatorname{Od}$ is multiplicative, we have

$\operatorname{Od}(n!)=\prod\limits_{r=1}^{n}\operatorname{Od}(r)=\prod\limits_{\begin{subarray}{c}1\leq
r\leq n\\\ r\text{
odd}\end{subarray}}\operatorname{Od}(r)\prod\limits_{\begin{subarray}{c}1\leq
r\leq n\\\ r\text{ even}\end{subarray}}\operatorname{Od}(r).$

The above rearrangement allows us to use the following two facts to obtain a
recursion. Firstly, $\operatorname{Od}(2r)=\operatorname{Od}(r)$ which gives

$\prod\limits_{\begin{subarray}{c}1\leq r\leq n\\\ r\text{
even}\end{subarray}}\operatorname{Od}(r)=\operatorname{Od}(\lfloor
n/2\rfloor!).$

We have

$\displaystyle\prod\limits_{\begin{subarray}{c}1\leq r\leq n\\\ r\text{
odd}\end{subarray}}\operatorname{Od}(r)$
$\displaystyle=\prod\limits_{\begin{subarray}{c}1\leq r\leq n\\\ r\in
4\mathbb{N}+1\end{subarray}}\operatorname{Od}(r)\prod\limits_{\begin{subarray}{c}1\leq
r\leq n\\\ r\in 4\mathbb{N}-1\end{subarray}}\operatorname{Od}(r)$
$\displaystyle=\prod\limits_{\begin{subarray}{c}1\leq r\leq n\\\ r\in
4\mathbb{N}-1\end{subarray}}(-1).$

As there are exactly $\lfloor(n+1)/4\rfloor$ terms of the arithmetic
progression $3,7,11,\ldots$ less than or equal to $n$, we get

$\operatorname{Od}(n!)=(-1)^{\lfloor(n+1)/4\rfloor}\operatorname{Od}(\lfloor
n/2\rfloor!).$

Let $n$ have the binary expansion $b_{k}\ldots b_{0}$ as in 5. By using
$\lfloor\frac{b_{\ell}\ldots b_{0}}{2}\rfloor=b_{\ell}\ldots b_{1}$, we can
rewrite our expression as

$\operatorname{Od}(n!)=(-1)^{\lfloor\frac{b_{k}\ldots
b_{0}+1}{4}\rfloor+\lfloor\frac{b_{k}\ldots
b_{1}+1}{4}\rfloor+\lfloor\frac{b_{k}\ldots
b_{2}+1}{4}\rfloor+\ldots+\lfloor\frac{b_{k}b_{k-1}+1}{4}\rfloor}\operatorname{Od}(b_{k}!).$

Notice that $\lfloor\frac{b_{k}\ldots
b_{j+1}b_{j}+1}{4}\rfloor=\lfloor\frac{b_{k}\ldots b_{j+1}b_{j}}{4}\rfloor+1$
if and only if both $b_{j}$ and $b_{j+1}$ are 1. For all other cases, we have
the equality, $\lfloor\frac{b_{k}\ldots
b_{j+1}b_{j}+1}{4}\rfloor=\lfloor\frac{b_{k}\ldots b_{j+1}b_{j}}{4}\rfloor$.
The inequality occurs exactly $D(n)$ times, giving us

$\operatorname{Od}(n!)=(-1)^{D(n)}(-1)^{\lfloor\frac{b_{k}\ldots
b_{0}}{4}\rfloor+\lfloor\frac{b_{k}\ldots
b_{1}}{4}\rfloor+\ldots+\lfloor\frac{b_{k}b_{k-1}b_{k-2}}{4}\rfloor+\lfloor\frac{b_{k}b_{k-1}}{4}\rfloor}\operatorname{Od}(b_{k}!).$

Using $\lfloor\frac{b_{\ell}\ldots b_{0}}{4}\rfloor=b_{\ell}\ldots b_{2}$, and
the fact that $b_{k}\in\\{0,1\\}$, we get

$\operatorname{Od}(n!)=(-1)^{D(n)}(-1)^{{b_{k}\ldots b_{2}}+{b_{k}\ldots
b_{3}}+\ldots+b_{k}+0}.$

When $(-1)$ is raised to a number, the parity of the number entirely decides
the result, thus $(-1)^{b_{k}\ldots b_{j}}=(-1)^{b_{j}}$. Using this, we
obtain,

$\displaystyle\operatorname{Od}(n!)$
$\displaystyle=(-1)^{D(n)+b_{2}+b_{3}+\ldots+b_{k}}$
$\displaystyle=(-1)^{D(n)+\nu(\lfloor n/4\rfloor)}.$

∎

Before stating the main formula, we define an important quantity:

######  Definition 7.

Let $\lambda$ be a $2^{R}$-parent of $\mu$ with $|\mu|<2^{R}$. Let
$h^{\lambda}_{\mu}\in H(\lambda)$ be the affected hook-length (Remark 1). We
define $\eta^{\lambda}_{\mu}\in\mathbb{Z}/2\mathbb{Z}$ as follows:

$(-1)^{\eta^{\lambda}_{\mu}}=\prod\limits_{\begin{subarray}{c}x\in
H(\lambda)\\\ x\neq
h^{\lambda}_{\mu}\end{subarray}}\dfrac{\operatorname{Od}(|h^{\lambda}_{\mu}-x|)}{\operatorname{Od}(|h^{\lambda}_{\mu}-2^{R}-x|)}.$

Now, we have the arsenal to approach the hook-length formula and apply our
$\operatorname{Od}$ function to it. This gives us the following proposition:

######  Proposition 6 (Workhorse formula).

Let $n=2^{R}+m>3$ with $m<2^{R}$. Let $\mu\vdash m$ and $\lambda$ be a
$2^{R}$-parent of $\mu$. Let $h^{\lambda}_{\mu}\in H(\lambda)$ be the affected
hook-length. Then,

$\operatorname{Od}(f^{\lambda})=(-1)^{s_{2}(n)+s_{2}(h^{\lambda}_{\mu})+\eta^{\lambda}_{\mu}}\operatorname{Od}(f^{\mu}),$

where $\eta^{\lambda}_{\mu}$ is as defined above and $s_{2}$ is as in 5.

###### Proof.

By applying $\operatorname{Od}$ on 2 for $\lambda$, we get

$\operatorname{Od}(f^{\lambda})=(-1)^{D(n)+\nu(\lfloor\frac{n}{4}\rfloor)}\prod\limits_{h\in
H(\lambda)}(-1)^{D(h)+\nu(\lfloor\frac{h}{4}\rfloor)}\prod\limits_{\begin{subarray}{c}x>y\\\
x,y\in H(\lambda)\end{subarray}}\operatorname{Od}(|x-y|).$

Suppose, $|H(\mu)^{+r}|=|H(\lambda)|$ for some $r$. Then, we obtain,

$\operatorname{Od}(f^{\mu})=(-1)^{D(m)+\nu(\lfloor\frac{m}{4}\rfloor)}\prod\limits_{h\in
H(\mu)^{+r}}(-1)^{D(h)+\nu(\lfloor\frac{h}{4}\rfloor)}\prod\limits_{\begin{subarray}{c}x>y\\\
x,y\in H(\mu)^{+r}\end{subarray}}\operatorname{Od}(|x-y|).$

Let $n$ have the binary expansion $b_{R}\ldots b_{0}$. Then,
$m=n-2^{R}=b_{R-1}\ldots b_{0}$. Clearly, $D(m)=D(n)-1$ if and only if
$b_{R-1}=1$. Also, $\nu(\lfloor m/4\rfloor)=\nu(\lfloor n/4\rfloor)-1$. These
two facts, give us

$(-1)^{D(n)+\nu(\lfloor\frac{n}{4}\rfloor)-D(m)-\nu(\lfloor\frac{m}{4}\rfloor)}=(-1)^{s_{2}(n)}.$

We have that $h^{\lambda}_{\mu}\in H(\lambda)$ and $h^{\lambda}_{\mu}-2^{R}\in
H(\mu)^{+r}$. All the other elements of the sets $H(\lambda)$, $H(\mu)^{+r}$
are same. Using the same analysis as above, we get

$\displaystyle\dfrac{\prod\limits_{h\in
H(\lambda)}(-1)^{D(h)+\nu(\lfloor\frac{h}{4}\rfloor)}}{\prod\limits_{h\in
H(\mu)^{+r}}(-1)^{D(h)+\nu(\lfloor\frac{h}{4}\rfloor)}}$
$\displaystyle=(-1)^{D(h^{\lambda}_{\mu})+\nu(\lfloor\frac{h^{\lambda}_{\mu}}{4}\rfloor)-{D(h^{\lambda}_{\mu}-2^{R})-\nu(\lfloor\frac{h^{\lambda}_{\mu}-2^{R}}{4}\rfloor)}}$
$\displaystyle=(-1)^{s_{2}(h^{\lambda}_{\mu})}.$

Extending this argument to the remaining productand using 7 gives us the
formula. ∎

Although the above recursion is quite compact, it must be untangled to aid in
enumeration. The $s_{2}$ terms are simple to deal with and so we must
investigate the $\eta^{\lambda}_{\mu}$ term.

######  Lemma 3.

For any natural number $a$, we have:

  1. (1)

for $2^{R}<a<2^{R+1}$, $\operatorname{Od}(a-2^{R})\neq\operatorname{Od}(a)\iff
a=2^{R}+2^{R-1}$,

  2. (2)

for $0<a<2^{R}$, $\operatorname{Od}(a+2^{R})\neq\operatorname{Od}(a)\iff
a=2^{R-1}$, and

  3. (3)

for $0<a<2^{R}$, $\operatorname{Od}(2^{R}-a)=\operatorname{Od}(a)\iff
a=2^{R-1}$.

###### Proof.

Let $a$ have the binary expansion $b_{R}\ldots b_{j}\ldots 0$. The results can
be shown by noticing that $\operatorname{Od}(a)\equiv b_{j+1}b_{j}\mod 4$ and
performing elementary algebraic manipulations. ∎

######  Definition 8 (Indicator function).

Let $\mathbb{I}_{X}(x)$ be equal to 1 if $x\in X$ otherwise 0. We call
$\mathbb{I}_{X}$, the indicator function on $X$.

######  Notation 6.

For a partition $\lambda$ of $n=2^{R}+m$ with $m<2^{R}$ and $h\in H(\lambda)$,
let $N_{\lambda}(h)=\\#\\{y\in H(\lambda)\mid h-2^{R}<y<h\\}$.

3 enables us to prove the following proposition which forms the crux of our
enumeration endeavour:

######  Proposition 7.

Let $\lambda$ be a $2^{R}$-parent of $\mu$ with $|\mu|<2^{R}$. Then, we have

$\eta^{\lambda}_{\mu}=N_{\lambda}(h^{\lambda}_{\mu})-\mathbb{I}_{H(\lambda)}(h^{\lambda}_{\mu}-2^{R-1})+\mathbb{I}_{H(\lambda)}(h^{\lambda}_{\mu}+2^{R-1})+\mathbb{I}_{H(\lambda)}(h^{\lambda}_{\mu}-3\cdot
2^{R-1}).$

###### Proof.

We consider the ratio

$\rho_{h}(x):=\operatorname{Od}(|h-x|)/\operatorname{Od}(|h-2^{R}-x|).$

In this proof, we will put $h=h^{\lambda}_{\mu}$ and $x\in H(\lambda)$ except
for $x=h$. The idea of this proof is to find values of $x$ such that
$\rho_{h}(x)=-1$, which allows us to compute

$(-1)^{\eta^{\lambda}_{\mu}}=\prod\limits_{\begin{subarray}{c}x\in
H(\lambda)\\\ x\neq h\end{subarray}}\rho_{h}(x).$

We split the proof into three cases:

  1. (1)

Let $x<h-2^{R}<h$. The ratio $\rho_{h}(x)$ simplifies to
$\frac{\operatorname{Od}(h-x)}{\operatorname{Od}(h-2^{R}-x)}$. Put $h-x$ as
$a$ and use (1) in 3, to give $x=h-2^{R}-2^{R-1}$. Thus,

$\rho_{h}(h-3\cdot 2^{R-1})=-1.$

  2. (2)

Let $h-2^{R}<h<x$. Similarly we use (2) from the lemma to get

$\rho_{h}(h+2^{R-1})=-1.$

  3. (3)

Let $h-2^{R}<x<h$. Now, $\rho_{h}(x)$ becomes
$\dfrac{\operatorname{Od}(h-x)}{\operatorname{Od}(-h+2^{R}+x)}$. Using (3)
from the lemma, we get that in this range, $\rho_{h}(x)=1$ if and only if
$x=h-2^{R-1}$. Thus, for exactly

${N_{\lambda}(h)-\mathbb{I}_{H(\lambda)}(h-2^{R-1})}$

many values, we have $\rho_{h}(x)=-1$.

Combining these three together gives us the proposition. ∎

We are now ready to enumerate odd dimensional partitions.

##  4\. Odd Partitions of Sparse Numbers

In this section, we prove 1 and 1 through heavy use of 6 and 7.

######  Definition 9 (Sparse number).

A natural number $n$ with the binary expansion $b_{k}\ldots b_{0}$ is called
sparse if for all $0\leq i\leq k-1$, we have the product $b_{i+1}\cdot
b_{i}=0$.

Recall that $a_{i}(n)$ denotes the number of partitions of $n$ whose
dimensions are congruent to $i$ modulo 4. Further,
$\delta(n):=a_{1}(n)-a_{3}(n)$, and the total number of odd partitions of $n$
is denoted by $a(n)=a_{1}(n)+a_{3}(n)$. We start by first defining hook
partitions and looking into their dimensions, which simplify to binomial
coefficients.

######  Definition 10.

A partition $\lambda\vdash n$ is called a hook partition if it is of the form
$(a+1,1,\ldots,1)$ for $0\leq a\leq n-2$ or $(n)$. Equivalently,
$H(\lambda)=\\{n,b,b-1,\ldots,1\\}$ or $H(\lambda)=\\{n\\}$ where $a+b+1=n$.

######  Lemma 4.

We have $\delta(1)=1$, $\delta(2)=2$ and $\delta(2^{R})=0$ for $R\geq 2$.

###### Proof.

The cases of $\delta(1)$ and $\delta(2)$ are easy to see as they have 1 and 2
partitions respectively, all of dimension 1. By 3, we deduce that all odd
partitions of $2^{R}$ are hook-partitions. Let $\lambda\vdash 2^{R}$ such that
$H(\lambda)=\\{2^{R},b,b-1,\ldots,1\\}$. Putting this into the hook-length
formula (2), we get $f^{\lambda}=\binom{2^{R}-1}{b}$. By using 8 (stated
below) and noticing that $2^{R}-1$ contains only 1s in its binary expansion,
we get the lemma. ∎

######  Proposition 8 (Davis and Webb, [DW89]).

Let $n\in\mathbb{N}$ be not sparse, then

$\\#\\{0\leq k\leq n\mid\binom{n}{k}\equiv 1\textnormal{ mod }4\\}=\\#\\{0\leq
k\leq n\mid\binom{n}{k}\equiv 3\textnormal{ mod }4\\}.$

###### Proof.

Refer to Theorem 6 of [DW89] for the proof. ∎

Let $n=2^{R}+m$ with $m<2^{R-1}<2^{R}$. We proceed with the following
strategy: consider an odd partition $\mu\vdash m$ as our $2^{R}$-core.
Consider all partitions of $n$ which are the $2^{R}$-parents of $\mu$. These
are all odd partitions by 3 and by 6, we can determine $\delta(n)$ in terms of
$\delta(m)$. Firstly, we see that 6 simplifies nicely in the $m<2^{R-1}$ case.

######  Corollary 3.

Let $n=2^{R}+m$ with $m<2^{R-1}$. For odd partitions, $\lambda\vdash n$ and
$\mu\vdash m$ such that $\lambda$ is a $2^{R}$-parent of $\mu$, we have

$\operatorname{Od}(f^{\lambda})=(-1)^{\eta^{\lambda}_{\mu}}\operatorname{Od}(f^{\mu}).$

###### Proof.

Let $n$ have the binary expansion $b_{R}b_{R-1}\ldots b_{0}$. As $n$ is
sparse, it must be that $b_{R-1}=0$ and $s_{2}(n)=b_{R}+b_{R-1}=1$. Also, we
have $2^{R}\leq h^{\lambda}_{\mu}<n<2^{R}+2^{R-1}$ as all elements of
$H(\lambda)$ are less than or equal to $n$. By the same logic as above, we get
$s_{2}(h^{\lambda}_{\mu})=1$. ∎

This makes finding $\eta^{\lambda}_{\mu}$ our primary concern. In our
particular case of $m<2^{R-1}$, even this calculation is simplified. We obtain
a corollary of 7:

######  Corollary 4.

Let $n=2^{R}+m$ with $m<2^{R-1}$. For odd partitions, $\lambda\vdash n$ and
$\mu\vdash m$ such that $\lambda$ is a $2^{R}$-parent of $\mu$, we have

$\eta^{\lambda}_{\mu}=N_{\lambda}(h^{\lambda}_{\mu})-\mathbb{I}_{H(\lambda)}(h^{\lambda}_{\mu}-2^{R-1})$

with notation as in 7.

###### Proof.

We have $2^{R}\leq h^{\lambda}_{\mu}<n<3\cdot 2^{R-1}$. Clearly,
$\mathbb{I}_{H(\lambda)}(h^{\lambda}_{\mu}-3\cdot 2^{R-1})=0$. Along similar
lines, $h^{\lambda}_{\mu}+2^{R-1}\geq 2^{R}+2^{R-1}>n$ and so it cannot be an
element of $H(\lambda)$. Thus,
$\mathbb{I}_{H(\lambda)}(h^{\lambda}_{\mu}+2^{R-1})=0$. ∎

We enumerate Type I and Type II parents (4) separately.

######  Notation 7.

When $2^{R}$ is understood, we denote the set of all $2^{R}$-parents of $\mu$
by $\operatorname{\mathcal{P}}(\mu)$. Denote the set of Type I and Type II
$2^{R}$-parents of $\mu$ by $\operatorname{\mathcal{P}}_{1}(\mu)$ and
$\operatorname{\mathcal{P}}_{2}(\mu)$ respectively.

Clearly, we have
$\operatorname{\mathcal{P}}(\mu)=\operatorname{\mathcal{P}}_{1}(\mu)\cup\operatorname{\mathcal{P}}_{2}(\mu)$.
Also, $|\operatorname{\mathcal{P}}_{1}(\mu)|=H(\lambda)$ and
$|\operatorname{\mathcal{P}}_{2}(\mu)|=2^{R}-H(\lambda)$.

######  Notation 8 (Signed-sum).

For any finite $\Lambda^{\prime}\subset\Lambda$, and a partition $\mu$, define
the signed-sum,
$S_{\Lambda^{\prime}}(\mu)=\frac{1}{\operatorname{Od}(f^{\mu})}\sum\limits_{\lambda\in\Lambda^{\prime}}\operatorname{Od}(f^{\lambda})$.

We see that $S_{\Lambda^{\prime}}(\mu)$ counts the number of partitions in
$\Lambda^{\prime}$ with same value of $\operatorname{Od}$ as $\mu$ minus the
number of partitions in $\Lambda^{\prime}$ with a different value of
$\operatorname{Od}$. For the following discussion, we will consider natural
numbers $m$ and $R$ such that $m<2^{R-1}$.
We now count Type I parents.

######  Proposition 9.

Let $\mu$ be an odd partition of $m$. Then, for
$\lambda\in\operatorname{\mathcal{P}}_{1}(\mu)$, we have
$\mathbb{I}_{H(\lambda)}(h^{\lambda}_{\mu}-2^{R-1})=0$. Further,

$S_{\operatorname{\mathcal{P}}_{1}(\mu)}(\mu)=\begin{cases}0,&\text{if
}|H(\mu)|\text{ is even}\\\ 1,&\text{if }|H(\mu)|\text{ is odd}.\\\
\end{cases}$

###### Proof.

By definition, we have
$H(\lambda)=(H(\mu)\cup\\{h^{\lambda}_{\mu}\\})\backslash\\{h^{\lambda}_{\mu}-2^{R}\\}$
where $h^{\lambda}_{\mu}>2^{R}$. Thus, if $h^{\lambda}_{\mu}-2^{R-1}\in
H(\lambda)$ then $h^{\lambda}_{\mu}-2^{R-1}\in H(\mu)$, which is not possible
as $h^{\lambda}_{\mu}-2^{R-1}>m$. This shows
$\mathbb{I}_{H(\lambda)}(h^{\lambda}_{\mu}-2^{R-1})=0$. We also have

$\displaystyle S_{\operatorname{\mathcal{P}}_{1}(\mu)}(\mu)$
$\displaystyle=\sum\limits_{\lambda\in\operatorname{\mathcal{P}}_{1}(\mu)}\frac{\operatorname{Od}(f^{\lambda})}{\operatorname{Od}(f^{\mu})}$
$\displaystyle=\sum\limits_{\lambda\in\operatorname{\mathcal{P}}_{1}(\mu)}(-1)^{\eta^{\lambda}_{\mu}}$
$\displaystyle=\sum\limits_{\lambda\in\operatorname{\mathcal{P}}_{1}(\mu)}(-1)^{N_{\lambda}(h^{\lambda}_{\mu})}.$

If, explicitly, $H(\mu)=\\{h_{1},\ldots,h_{k}\\}$ with $h_{1}>\ldots>h_{k}$,
then the map $\phi:H(\mu)\rightarrow\operatorname{\mathcal{P}}_{1}(\mu)$
defined by

$\phi(h_{i})=\operatorname{{Part}}((H(\mu)\cup\\{h_{i}+2^{R}\\})\backslash\\{h_{i}\\})$

is a bijection such that $h^{\phi(h_{i})}_{\mu}=h_{i}+2^{R}$. As all elements
of $H(\mu)$ are strictly smaller than $2^{R}$, the element
$h^{\phi(h_{i})}_{\mu}$ is the largest element in $H(\phi(h_{i}))$. Thus,

$\displaystyle N_{\phi(h_{i})}(h^{\phi(h_{i})}_{\mu})$
$\displaystyle=\\#\\{y\in H(\phi(h_{i}))\mid h_{i}+2^{R}>y>h_{i}\\}$
$\displaystyle=\\#\\{y\in H(\mu)\mid y>h_{i}\\}$ $\displaystyle=i-1.$

This gives us,

$S_{\operatorname{\mathcal{P}}_{1}(\mu)}(\mu)=\sum\limits_{i=1}^{k}(-1)^{i-1}.$

By considering the parity of $k=|H(\mu)|=|H(\lambda)|$, we get the result. ∎

The similar calculation for Type II parents is more involved. We will
eventually prove the following proposition:

######  Proposition 10.

Let $\mu$ be an odd partition of $m$ and $2^{R-1}>m$. Then,

$S_{\operatorname{\mathcal{P}}_{2}(\mu)}(\mu)=\begin{cases}2-2(-1)^{m},&\text{if
}|H(\mu)|\text{ is even}\\\ 1-2(-1)^{m},&\text{if }|H(\mu)|\text{ is odd}.\\\
\end{cases}$

Fix $\mu\vdash m$ and $n=2^{R}+m$ with $m<2^{R-1}$ as before. If $\lambda$ is
a Type II $2^{R}$-parent, i.e.,
$\lambda\in\operatorname{\mathcal{P}}_{2}(\mu)$, we have
$h^{\lambda}_{\mu}=2^{R}$. Thus, 4 simplifies to give
$\eta^{\lambda}_{\mu}=N_{\lambda}(h^{\lambda}_{\mu})-\mathbb{I}_{H(\lambda)}(2^{R-1})$.
Although it looks simpler on the surface, this calculation comes with its own
caveat that $r$-shifts do not give a $2^{R}$-parent.

######  Notation 9.

Define

$\operatorname{\mathcal{D}}:=\\{1\leq r\leq 2^{R}\mid 2^{R}\not\in
H(\mu)^{+r}\\},$

wherein we assume that $R$ and $\mu$ are understood. For a set
$X\subset[1,2^{R}]\subset\mathbb{N}$, we write
$\operatorname{\mathcal{D}}_{X}:=\operatorname{\mathcal{D}}\cap X$.
For $r\in\operatorname{\mathcal{D}}$, let

$\lambda_{[r]}=\operatorname{{Part}}\left((H(\mu)^{+r}\cup\\{2^{R}\\})\backslash\\{0\\}\right).$

We define a new quantity, which may seem arbitrary at this point, but will be
quite useful in our analysis:

######  Definition 11 (Parity gap).

For a finite $X\subset\mathbb{N}$, define the parity gap of $X$,
$\operatorname{\mathcal{G}}(X)$, to be the number of even elements of $X$
minus the number of odd elements of $X$. In notation,

$\operatorname{\mathcal{G}}(X)=\\#\\{x\in X\mid x\equiv 0\text{ mod
}2\\}-\\#\\{x\in X\mid x\equiv 1\text{ mod }2\\}.$

######  Example 8.

If $X=\\{13,12,8,5,3,1,0\\}$ then $\operatorname{\mathcal{G}}(X)=3-4=-1$.

######  Notation 10.

Let

$\operatorname{\mathcal{P}}_{2}^{\uparrow}(\mu):=\\{\lambda_{[r]}\mid
r\in\operatorname{\mathcal{D}}_{[1,2^{R-1}]}\\}$

and

$\operatorname{\mathcal{P}}_{2}^{\downarrow}(\mu):=\\{\lambda_{[r]}\mid
r\in\operatorname{\mathcal{D}}_{[2^{R-1}+1,2^{R}]}\\}.$

######  Lemma 5.

Following the above notation, we claim that $D_{[1,2^{R-1}]}=[1,2^{R-1}]$.
Further,

$S_{\operatorname{\mathcal{P}}_{2}^{\uparrow}(\mu)}(\mu)=2(-1)^{|H(\mu)|}\operatorname{\mathcal{G}}(H(\mu)).$

###### Proof.

The maximum possible element of $H(\mu)^{+r}$ is ${\max(|H(\mu)|+r)}$ which
itself is strictly smaller than $2^{R}$ as $|H(\mu)|<m<2^{R-1}$ and $r\leq
2^{R-1}$. Thus, for $1\leq r\leq 2^{R-1}$, we have $2^{R}\not\in H(\mu)^{+r}$.
For the signed-sum part, it is easy to see that
$N_{\lambda_{[r]}}(2^{R})=|H(\mu)|+r-1$. Thus, we get the following summation:

$S_{\operatorname{\mathcal{P}}_{2}^{\uparrow}(\mu)}(\mu)=\sum\limits_{r=1}^{2^{R-1}}(-1)^{|H(\mu)|+r-1}(-1)^{\mathbb{I}_{H(\lambda_{[r]})}(2^{R-1})}.$

For every $h\in H(\mu)$, we can choose an $r_{h}:=2^{R-1}-h$ which ensures
that $2^{R-1}\in H(\mu)^{+r_{h}}$. Conversely, if $2^{R-1}\in H(\mu)^{+r}$,
then as $r\leq 2^{R-1}$, we must have $h+r=2^{R-1}$ for some $h\in H(\mu)$.
Thus, there is an injective map
$H(\mu)\rightarrow\operatorname{\mathcal{P}}_{2}^{\uparrow}(\mu)$ given by
$h\mapsto\lambda_{[r_{h}]}$. Further, if $h\in H(\mu)$ and $h+r_{h}=2^{R-1}$,
then $h$ and $r_{h}$ have the same parity. We can now compute the signed-sum
as follows:

$\displaystyle S_{\operatorname{\mathcal{P}}_{2}^{\uparrow}(\mu)}(\mu)$
$\displaystyle=(-1)^{|H(\mu)|-1}\sum\limits_{r=1}^{2^{R-1}}(-1)^{r+\mathbb{I}_{H(\lambda_{[r]})}(2^{R-1})}$

We break the sum into two parts depending on whether $2^{R-1}$ belongs to
$H(\lambda_{[r]})$.

$\displaystyle S_{\operatorname{\mathcal{P}}_{2}^{\uparrow}(\mu)}(\mu)$
$\displaystyle=(-1)^{|H(\mu)|-1}\left(\sum\limits_{\begin{subarray}{c}r_{h}\\\
h\in
H(\mu)\end{subarray}}(-1)^{r_{h}+\mathbb{I}_{H(\lambda_{[r_{h}]})}(2^{R-1})}+\sum\limits_{\begin{subarray}{c}\text{other
}r\end{subarray}}(-1)^{r+\mathbb{I}_{H(\lambda_{[r]})}(2^{R-1})}\right)$
$\displaystyle=(-1)^{|H(\mu)|-1}\left(\sum\limits_{\begin{subarray}{c}h\in
H(\mu)\end{subarray}}(-1)^{r_{h}+1}+\sum\limits_{\begin{subarray}{c}\text{other
}r\end{subarray}}(-1)^{r}\right)$
$\displaystyle=(-1)^{|H(\mu)|-1}\left(\sum\limits_{\begin{subarray}{c}h\in
H(\mu)\end{subarray}}\left((-1)^{r_{h}+1}-(-1)^{r_{h}}\right)+\sum\limits_{r=1}^{2^{R-1}}(-1)^{r}\right)$
$\displaystyle=2(-1)^{|H(\mu)|-1}\left(\sum\limits_{\begin{subarray}{c}h\in
H(\mu)\end{subarray}}\left((-1)^{r_{h}+1}\right)+0\right)$
$\displaystyle=2(-1)^{|H(\mu)|}\left(\sum\limits_{\begin{subarray}{c}h\in
H(\mu)\end{subarray}}(-1)^{r_{h}}\right)$
$\displaystyle=2(-1)^{|H(\mu)|}\operatorname{\mathcal{G}}(H(\mu)).$

∎

We now hand the reader the following proposition which explicitly states what
values $\operatorname{\mathcal{G}}(X)$ can take where $X$ is a $\beta$-set of
an odd partition of $n$.

######  Lemma 6.

Let $\lambda$ be an odd partition of $n$ and $X$ be its $\beta$-set. Then, we
have

$\operatorname{\mathcal{G}}(X)=\begin{cases}1-(-1)^{n},&\text{if }|X|\text{ is
even}\\\ (-1)^{n},&\text{if }|X|\text{ is odd}.\\\ \end{cases}$

###### Proof.

Recall that if $\lambda\in\operatorname{\mathcal{P}}(\mu)$, then there exists
an $0\leq s\leq 2^{R}$ such that
$H(\lambda)=(H(\mu)^{+s}\cup\\{h^{\lambda}_{\mu}\\})\backslash\\{h^{\lambda}_{\mu}-2^{R}\\}$
for $h^{\lambda}_{\mu}\in H(\lambda)$. The case $s=0$ corresponds to Type I
parents.

Using the fact that $h^{\lambda}_{\mu}$ and $h^{\lambda}_{\mu}-2^{R}$ have the
same parity, we can deduce that
$\operatorname{\mathcal{G}}(H(\lambda))=\operatorname{\mathcal{G}}(H(\mu)^{+s})$.
With elementary computations, one can check that
$\operatorname{\mathcal{G}}(X^{+1})=1-\operatorname{\mathcal{G}}(X)$. This
shows that

$\operatorname{\mathcal{G}}(X^{+s})=\begin{cases}\operatorname{\mathcal{G}}(X),&\text{if
}s\text{ is even}\\\ 1-\operatorname{\mathcal{G}}(X),&\text{if }s\text{ is
odd.}\end{cases}$

We can now work our way down hook-by-hook and calculate
$\operatorname{\mathcal{G}}(H(\lambda))$ by finding
$\operatorname{\mathcal{G}}(H(\alpha))$ where $\alpha=\varnothing$ when $n$ is
even and $\alpha=(1)$ when $n$ is odd. Applying the above relation repeatedly,
we get,

$\operatorname{\mathcal{G}}(H(\lambda))=\begin{cases}\operatorname{\mathcal{G}}(H(\alpha)),&\text{if
}|H(\lambda)|-|H(\alpha)|\text{ is even}\\\
1-\operatorname{\mathcal{G}}(H(\alpha)),&\text{if
}|H(\lambda)|-|H(\alpha)|\text{ is odd.}\end{cases}$

In a sense, $|H(\lambda)|-|H(\alpha)|$ counts the total number of $r$-shifts
required to reach $\lambda$.
Trivially, $\operatorname{\mathcal{G}}(H(\varnothing))=0$. If $n$ is even,we
obtain

$\operatorname{\mathcal{G}}(H(\lambda))=\begin{cases}0&|H(\lambda)|\text{ is
even}\\\ 1&|H(\lambda)|\text{ is odd.}\end{cases}$

On the other hand, we have $\operatorname{\mathcal{G}}(\\{1\\})=-1$ and if $n$
is odd, then

$\operatorname{\mathcal{G}}(H(\lambda))=\begin{cases}2,&\text{if
}|H(\lambda)|\text{ is even.}\\\ -1,&\text{if }|H(\lambda)|\text{ is
odd}\par\end{cases}$

The claim follows immediately from this computation. ∎

######  Example 9.

Let $\lambda$, $\mu$, $\nu$ and $\pi$ be partitions such that

  * •

$H(\lambda)=(H(\mu^{+5})\cup\\{2^{R}\\})\backslash\\{0\\}$

  * •

$H(\mu)$ is a Type I $2^{S}$-parent of $\nu$

  * •

$H(\nu)=(H(\pi^{+7})\cup\\{2^{T}\\})\backslash\\{0\\}$

By above, we have
$\operatorname{\mathcal{G}}(H(\nu))=1-\operatorname{\mathcal{G}}(H(\pi))$.
Further,
$\operatorname{\mathcal{G}}(H(\mu))=\operatorname{\mathcal{G}}(H(\nu))$ and
$\operatorname{\mathcal{G}}(H(\lambda))=1-\operatorname{\mathcal{G}}(H(\mu))$.
Thus,
$\operatorname{\mathcal{G}}(H(\lambda))=1-\operatorname{\mathcal{G}}(H(\nu))=1-(1-\operatorname{\mathcal{G}}(H(\pi)))=\operatorname{\mathcal{G}}(H(\pi))$.

######  Lemma 7.

For $2^{R-1}+1\leq r\leq 2^{R}$, we have
$\lambda_{[r]}\in\operatorname{\mathcal{P}}_{2}^{\downarrow}(\mu)$ if and only
if $2^{R-1}\not\in H(\lambda_{[r-2^{R-1}]})$. Further,

$S_{\operatorname{\mathcal{P}}_{2}^{\downarrow}(\mu)}(\mu)=\begin{cases}0,&\text{if
}|H(\mu)|\text{ is even}\\\ 1,&\text{if }|H(\mu)|\text{ is odd}.\\\
\end{cases}$

###### Proof.

It is clear that $2^{R-1}\in\lambda_{[r-2^{R-1}]}$ if and only if $2^{R}\in
H(\mu)^{+r}$ for $2^{R-1}+1\leq r\leq 2^{R}$. In this range, the values of
$r\not\in\operatorname{\mathcal{D}}$ are exactly the values of $r$ such that
$2^{R-1}\in H(\lambda_{[r]})$. By considering these values as $r_{h}$ (refer
5), we get that
$|\operatorname{\mathcal{P}}_{2}^{\downarrow}(\mu)|=2^{R-1}-|H(\mu)|$.
For $r>2^{R-1}$, we have $2^{R-1}\in H(\lambda_{[r]})$ and thus,
$\eta^{\lambda}_{\mu}=N_{\lambda_{[r]}}(2^{R})+1$.
Suppose,
$\lambda_{[r]},\lambda_{[r+s]}\in\operatorname{\mathcal{P}}_{2}^{\downarrow}(\mu)$
and $r+1,\ldots,r+s-1\not\in\operatorname{\mathcal{D}}$. This gives us that
$H(\mu)=\\{h_{1},\ldots,h_{k},h_{k+1},\ldots,h_{k+s-1}\ldots h_{l}\\}$ where
for $1\leq i\leq s-1$, we have $h_{k+i}=h_{k+1}-i+1$ and $h_{k+i}+r+i=2^{R}$.
It follows that $h_{k}+r>2^{R}$ and $h_{k+s}+r+s<2^{R}$.

We have $N_{\lambda_{[r]}}(2^{R})=|H(\mu)|+r-1-j_{r}$ where $j_{r}$ is the
number of elements in $H(\lambda_{[r]})$ strictly greater than $2^{R}$. As
$h_{k+i}+r+s>2^{R}$ for $1\leq i\leq s-1$, we have $j_{r+s}=j_{r}+s-1$. This
gives us
$N_{\lambda_{[r+s]}}(2^{R})=|H(\mu)|+(r+s-1)-(j_{r}+s-1)=N_{\lambda_{[r]}}(2^{R})+1$.
Let $r_{1}<\ldots<r_{2^{R-1}-|H(\mu)|}$ such that
$\lambda_{[r_{1}]},\ldots,\lambda_{[r_{2^{R-1}-|H(\mu)|}]}\in\operatorname{\mathcal{P}}_{2}^{\downarrow}(\mu)$.
By the above discussion, we have
$N_{\lambda_{[r_{i}]}}(2^{R})=N_{\lambda_{[r_{1}]}}(2^{R})+i-1$. Thus,

$\displaystyle S_{\operatorname{\mathcal{P}}_{2}^{\downarrow}(\mu)}(\mu)$
$\displaystyle=\sum\limits_{i=1}^{2^{R-1}-|H(\mu)|}(-1)^{N_{\lambda_{[r_{i}]}(2^{R})}+1}$
$\displaystyle=(-1)^{N_{\lambda_{[r_{1}]}}(2^{R})+1}\sum\limits_{i=1}^{2^{R-1}-|H(\mu)|}(-1)^{i-1}$
$\displaystyle=(-1)^{|H(\mu)|+r_{1}-j_{r_{1}}}\sum\limits_{i=1}^{2^{R-1}-|H(\mu)|}(-1)^{i-1}$
$\displaystyle=(-1)^{|H(\mu)|+2^{R-1}}\sum\limits_{i=1}^{2^{R-1}-|H(\mu)|}(-1)^{i}.$

The last equality follows by noticing that if $r_{1}=2^{R-1}+x$, then
$j_{r_{1}}=x-1$ as $2^{R}\in
H(\mu)^{+(2^{R-1}+1)},\ldots,H(\mu)^{+(2^{R-1}+x-1)}$. Putting the proper
parity of $|H(\mu)|$ and evaluating the expression gives us our result. ∎

We are now ready to prove the result for Type II parents.

###### Proof of 10.

We use the value of $\operatorname{\mathcal{G}}(H(\mu))$ from 6 and apply it
to the result in 5 to obtain

$S_{\operatorname{\mathcal{P}}_{2}^{\uparrow}(\mu)}(\mu)=\begin{cases}2-2(-1)^{m},&\text{if
}|H(\mu)|\text{ is even}\\\ 2(-1)^{m+1},&\text{if }|H(\mu)|\text{ is odd}.\\\
\end{cases}$

As
$S_{\operatorname{\mathcal{P}}_{2}(\mu)}(\mu)=S_{\operatorname{\mathcal{P}}_{2}^{\uparrow}(\mu)}(\mu)+S_{\operatorname{\mathcal{P}}_{2}^{\downarrow}(\mu)}(\mu)$,
we add the result of 7, to obtain the result of 10 which is

$S_{\operatorname{\mathcal{P}}_{2}(\mu)}(\mu)=\begin{cases}2-2(-1)^{m},&\text{if
}|H(\mu)|\text{ is even}\\\ 1-2(-1)^{m},&\text{if }|H(\mu)|\text{ is odd}.\\\
\end{cases}$

∎

We now have all the tools in our hands to give a proof of 1 and 1.

###### Proof of 1.

If $\operatorname{\mathcal{P}}(\mu)$ denotes the set of $2^{R}$-parents of
$\mu$, then
$S_{\operatorname{\mathcal{P}}(\mu)}(\mu)=S_{\operatorname{\mathcal{P}}_{1}(\mu)}(\mu)+S_{\operatorname{\mathcal{P}}_{2}(\mu)}(\mu)$.
Clearly, $S_{\operatorname{\mathcal{P}}(\mu)}(\mu)=2-2(-1)^{m}$. The theorem
follows from the following computation and the observation that $n$ and $m$
have the same parity:

$\displaystyle\delta(n)$ $\displaystyle=\sum\limits_{\begin{subarray}{c}\\\
\lambda\vdash n\\\ \lambda\text{ is
odd}\end{subarray}}\operatorname{Od}(f^{\lambda})$
$\displaystyle=\sum\limits_{\begin{subarray}{c}\mu\vdash m\\\ \mu\text{ is
odd}\end{subarray}}S_{\operatorname{\mathcal{P}}(\mu)}({\mu})\operatorname{Od}(f^{\mu})$
$\displaystyle=(2-2(-1)^{m})\sum\limits_{\begin{subarray}{c}\mu\vdash m\\\
\mu\text{ is odd}\end{subarray}}\operatorname{Od}(f^{\mu})$
$\displaystyle=(2-2(-1)^{m})\delta(m).$

This gives us the result:

$\delta(n)=\begin{cases}0,&\text{if }n\text{ is even}\\\ 4\delta(m),&\text{if
}n\text{ is odd}.\end{cases}$

∎

###### Proof of 1.

Repeatedly apply 1 and use 4. The $\nu(n)-1$ comes from $\delta(1)=1$. This
gives us the result in the sparse case and we find $n=2$ separately:

$\delta(n)=\begin{cases}2\text{,}&\text{if }n=2\\\ 0,&\text{if }n>2\text{ is
even}\\\ 4^{\nu(n)-1},&\text{if }n\text{ is odd}.\end{cases}$

∎

##  5\. Odd Partitions of $n=2^{R}+2^{R-1}$

We wish to extend the results of the previous section beyond the sparse case.
Although, a general formula remains elusive; we can take confident steps
towards it. We consider all numbers whose binary expansions contain a pair of
adjacent 1s with all other bits zero. In notation, we consider $n$ with
$\nu(n)=2$ and $D(n)=1$. We now prove 2.
We use similar methods as we did in the previous section. Recall that for
$\Lambda^{\prime}\subset\Lambda$ (the set of all partitions), we have

$S_{\Lambda^{\prime}}(\mu):=\frac{1}{\operatorname{Od}(f^{\mu})}\sum\limits_{\lambda\in\Lambda^{\prime}}\operatorname{Od}(f^{\lambda}).$

We also state a corollary of 6 which will be pertinent in this case.

######  Corollary 5.

Let $n=2^{R}+2^{R-1}$ for some $R\geq 2$. Let $\lambda\vdash n$ and $\mu\vdash
2^{R-1}$ be odd partitions such that $\lambda$ is a $2^{R}$-parent of $\mu$.
Then,

$\operatorname{Od}(f^{\lambda})=(-1)^{s_{2}(h^{\lambda}_{\mu})+\eta^{\lambda}_{\mu}}\operatorname{Od}(f^{\mu}).$

The proof of the above corollary is trivial as the sum of first two digits of
$n$ is 2. Notice that we start from $R=2$ as the case of $R=1$ is slightly
different and can be handled independently.
Throughout our discussion, we will take $\mu$ to be a hook partition such that
$H(\mu)=\\{2^{R-1},b,b-1,\ldots,1\\}$, where $b=0$ implies
$H(\mu)=\\{2^{R-1}\\}$. Now, we first count the $2^{R}$-parents of Type I
which, as before, we denote by $\operatorname{\mathcal{P}}_{1}(\mu)$.

######  Lemma 8.

With the above notation, we have

$S_{\operatorname{\mathcal{P}}_{1}(\mu)}(\mu)=\begin{cases}1,&\text{if
}b\text{ is even}\\\ 2,&\text{if }b\text{ is odd}.\\\ \end{cases}$

###### Proof.

As in the proof of 9, define the bijection
$\phi:H(\mu)\rightarrow\operatorname{\mathcal{P}}_{1}(\mu)$ such that

$\phi(h)=\operatorname{{Part}}((H(\mu)\cup\\{h+2^{R}\\})\backslash\\{h\\}).$

Firstly, consider $\lambda:=\phi(2^{R-1})$. In this case,
$s_{2}(h^{\lambda}_{\mu})=2$. Further, using 7 regarding
$\eta^{\lambda}_{\mu}$, we have

$\eta^{\lambda}_{\mu}=N_{\lambda}(2^{R}+2^{R-1})-\mathbb{I_{H(\lambda)}}(2^{R})+\mathbb{I_{H(\lambda)}}(2^{R+1})+\mathbb{I_{H(\lambda)}}(0).$

As $2^{R-1}$ is the largest entry in $H(\mu)$, it is easy to check that that
$N_{\lambda}(2^{R}+2^{R-1})=0$. Further,
$H(\lambda)=\\{2^{R}+2^{R-1},b,\ldots,1\\}$ and we can see that it does not
contain 0, $2^{R}$ or $2^{R+1}$. Thus, $\eta^{\lambda}_{\mu}$ in this case is
equal to 0, which gives us $(-1)^{0}=1$.
On the other hand, for $1\leq x\leq b$, define $\lambda_{[x]}:=\phi(x)$ . In
this case, $s_{2}(h^{\lambda_{[x]}}_{\mu})=1$ and

$\eta^{\lambda_{[x]}}_{\mu}=N_{\lambda_{[x]}}(2^{R}+x)-\mathbb{I}_{H({\lambda_{[x]})}}(x+2^{R-1})+\mathbb{I}_{H(\lambda_{[x]})}(x+2^{R}+2^{R-1})+\mathbb{I}_{H(\lambda_{[x]})}(x-2^{R-1}).$

We have $H(\lambda_{[x]})=\\{x+2^{R},2^{R-1},\ldots\\}$ and
$N_{\lambda_{[x]}}(2^{R}+x)=b+1-x$. As $1\leq x<2^{R-1}$ and
$|\lambda|=2^{R}+2^{R-1}$, $H(\lambda_{[x]})$ cannot contain, $x-2^{R-1}$ or
$x+2^{R}+2^{R-1}$. From the explicit form of $H(\lambda_{[x]})$, it is easy to
see the only element greater than $2^{R-1}$ is $x+2^{R}$. Thus,
$\eta^{\lambda_{[x]}}_{\mu}=b+1-x$. This gives us,

$\displaystyle S_{\operatorname{\mathcal{P}}_{1}(\mu)}({\mu})$
$\displaystyle=1+\sum\limits_{x=1}^{b}(-1)^{b+1-x+1}$
$\displaystyle=1+\sum\limits_{i=0}^{b-1}(-1)^{x}$

which with appropriate values of $b$ simplifies to give the intended result. ∎

Notice that we are using the parity of $b$ which is opposite to the parity of
$H(\mu)$. As one gives the other, we use $b$ for convenience.
Now, we move on to the results for Type II $2^{R}$-parents.

######  Lemma 9.

With the above notation, we have

$S_{\operatorname{\mathcal{P}}_{2}(\mu)}({\mu})=\begin{cases}3,&\text{if
}b\text{ is even}\\\ 2,&\text{if }b\text{ is odd}.\\\ \end{cases}$

###### Proof.

As before, let $1\leq r\leq 2^{R}$ and $\lambda_{[r]}$ denote the
$2^{R}$-parent corresponding to that $r$-shift. In notation,

$\lambda_{[r]}=\operatorname{{Part}}\left((H(\mu)^{+r}\cup\\{2^{R}\\})\backslash\\{0\\}\right)\textnormal{
if and only if }r\in\operatorname{\mathcal{D}},$

where $\operatorname{\mathcal{D}}$ is as in 9. For
$r\in\operatorname{\mathcal{D}}$, we have
$\lambda_{[r]}\in\operatorname{\mathcal{P}}_{2}(\mu)$.
As for all $\lambda\in\operatorname{\mathcal{P}}_{2}(\mu)$,
$h^{\lambda}_{\mu}=2^{R}$, we have $s_{2}(h^{\lambda}_{\mu})=1$. Thus,

$\displaystyle S_{\operatorname{\mathcal{P}}_{2}(\mu)}({\mu})$
$\displaystyle=\sum\limits_{\lambda\in\operatorname{\mathcal{P}}_{2}(\mu)}(-1)^{\eta^{\lambda}_{\mu}+1}$
$\displaystyle=(-1)\cdot\sum\limits_{\lambda\in\operatorname{\mathcal{P}}_{2}(\mu)}(-1)^{\eta^{\lambda}_{\mu}}$

where

$\eta^{\lambda}_{\mu}=N_{\lambda}(2^{R})-\mathbb{I}_{H(\lambda)}(2^{R-1})+\mathbb{I}_{H(\lambda)}(2^{R}+2^{R-1}).$

We divide the values taken by $r$ into 6 (possibly empty, depending on $b$)
intervals, which although overkill, would help illuminate the process better.
It is recommended that the reader works out the assertions in the upcoming
discussion using the basic definitions. As before,
$H(\mu)=\\{2^{R-1},b,b-1,\ldots,1\\}$. Here, $b$ can take values from 0 to
$2^{R-1}$ where $b=0$ implies $H(\mu)=\\{2^{R-1}\\}$.

  1. (1)

$1\leq r\leq 2^{R-1}-b-1$:
In this case,
$\eta^{\lambda_{[r]}}_{\mu}=N_{\lambda_{[r]}}(2^{R})=|H(\mu)^{+r}|-1=b+r$.

  2. (2)

$2^{R-1}-b\leq r\leq 2^{R-1}-1$:
We have $\eta^{\lambda_{[r]}}_{\mu}=N_{\lambda_{[r]}}(2^{R})-1$ as $2^{R-1}\in
H(\mu)^{+r}$. Thus, $\eta^{\lambda_{[r]}}_{\mu}=b+r-1$.

  3. (3)

$r=2^{R-1}:$
As $2^{R-1}+2^{R-1}=2^{R}\in H(\mu)^{+2^{R-1}}$,
$2^{R-1}\not\in\operatorname{\mathcal{D}}$.

  4. (4)

$2^{R-1}+1\leq r\leq 2^{R}-b-1$.
For this, we have $\eta^{\lambda_{[r]}}_{\mu}=(b+r-1)-1=b+r$.

  5. (5)

$2^{R}-b\leq r\leq 2^{R}-1$.
For all these values, $r\not\in\operatorname{\mathcal{D}}$.

  6. (6)

$r=2^{R}$:
In this case,
$\eta^{\lambda_{\left[2^{R}\right]}}_{\mu}=N_{\lambda_{\left[2^{R}\right]}}(2^{R})=2^{R-1}-1$
as
$\mathbb{I}_{H(\lambda_{\left[2^{R}\right]})}(2^{R-1})=\mathbb{I}_{H(\lambda_{\left[2^{R}\right]})}(2^{R}+2^{R-1})=1$.

We can now combine the above information to give

$S_{\operatorname{\mathcal{P}}_{2}(\mu)}({\mu})=(-1)\cdot\left(\sum\limits_{r=1}^{2^{R-1}-b-1}(-1)^{b+r}+\sum\limits_{r=2^{R-1}-b}^{2^{R-1}-1}(-1)^{b+r-1}+\sum\limits_{r=2^{R-1}+1}^{2^{R}-b-1}(-1)^{b+r}+(-1)^{2^{R}-1}\right).$

With some sleight-of-hand, this can be simplified to give

$S_{\operatorname{\mathcal{P}}_{2}(\mu)}({\mu})=1+2(-1)^{b}\sum\limits_{r=1}^{2^{R-1}-b-1}(-1)^{r+1}+\sum\limits_{r=0}^{b-1}(-1)^{r}.$

When $b$ is even, we get
$S_{\operatorname{\mathcal{P}}_{2}(\mu)}({\mu})=1+2\cdot 1+0=3$. When $b$ is
odd, we get $S_{\operatorname{\mathcal{P}}_{2}(\mu)}({\mu})=1+0+1=2$. ∎

With all of these ingredients in our hands, we are ready to give a

###### Proof of 2.

Combining the results of 8 and 9, we see that
$S_{\operatorname{\mathcal{P}}(\mu)}(\mu)=4$ when $\mu\vdash 2^{R-1}$. Thus,
for $n=2^{R}+2^{R-1}$, we can do the following computation:

$\displaystyle\delta(n)$ $\displaystyle=\sum\limits_{\begin{subarray}{c}\\\
\lambda\vdash n\\\ \lambda\text{ is
odd}\end{subarray}}\operatorname{Od}(f^{\lambda})$
$\displaystyle=\sum\limits_{\begin{subarray}{c}\mu\vdash 2^{R-1}\\\ \mu\text{
is
odd}\end{subarray}}S_{\operatorname{\mathcal{P}}(\mu)}({\mu})\operatorname{Od}(f^{\mu})$
$\displaystyle=4\sum\limits_{\begin{subarray}{c}\mu\vdash 2^{R-1}\\\ \mu\text{
is odd}\end{subarray}}\operatorname{Od}(f^{\mu})$
$\displaystyle=4\delta(2^{R-1}).$

By the results of 4, we have our theorem, except for $R=1$, which can be done
by hand. ∎

##  6\. Partitions with Dimension 2 modulo 4

We now consider partitions whose dimensions are congruent to 2 modulo 4, i.e.,
$v_{2}(f^{\lambda})=1$ for all such partitions $\lambda$.
To tackle the problem of enumeration, we use the machinery of 2-core towers
which requires us to introduce the notion of a 2-quotient.
Recall that $\widetilde{\Lambda}_{t}$ is the set of all $t$-cores. It is well-
known (refer Proposition 3.7 of [Ols93] p. 20) that there exists a bijection
between the sets, $\Lambda$ and $\Lambda^{2}\times\widetilde{\Lambda}_{2}$,
given by
$\lambda\mapsto(\operatorname{quo}_{2}(\lambda),\operatorname{{core}}_{2}(\lambda))$.
Here, $\operatorname{quo}_{2}(\lambda)$ (known as the 2-quotient of $\lambda$)
is a pair $(\lambda^{(0)},\lambda^{(1)})$ with the property,

$|\lambda|=2(|\lambda^{(0)}|+|\lambda^{(1)}|)+|\operatorname{{core}}_{2}(\lambda)|.$

######  Notation 11.

We simplify the notation by writing $\lambda^{(ij)}$ for
$(\lambda^{(i)})^{(j)}$ for any binary string $i$ and $j\in\\{0,1\\}$.
Further, $\lambda^{(\varnothing)}=\lambda$ and
$(\lambda^{(\varnothing)})^{(j)}=\lambda^{(j)}$.

We recall the construction of 2-core towers as presented in [Ols93].
For $\lambda\in\Lambda$, construct an infinite rooted binary tree with nodes
labelled by 2-cores as follows:

  * •

Label the root node with
$\operatorname{{core}}_{2}(\lambda)=\operatorname{{core}}_{2}(\lambda^{(\varnothing)})$.

  * •

If the length of the unique path from the root node to a node $v$ is $i$, then
we say that the node $v$ is in the *i1$i^{\mathrm{st}}$
2$i^{\mathrm{nd}}$3$i^{\mathrm{rd}}$[$i^{\mathrm{th}}$] row.

  * •

Every node in the *i1$i^{\mathrm{st}}$
2$i^{\mathrm{nd}}$3$i^{\mathrm{rd}}$[$i^{\mathrm{th}}$] row is adjacent to two
nodes in the $(i+1)^{\text{st}}$ row. Define a recursive labelling as follows:
if the label of some node, $v$, in the *i1$i^{\mathrm{st}}$
2$i^{\mathrm{nd}}$3$i^{\mathrm{rd}}$[$i^{\mathrm{th}}$] row, is
$\operatorname{{core}}_{2}(\lambda^{(b)})$ for some binary string $b$, then
the two nodes in the $(i+1)^{\text{st}}$ row adjacent to $v$ have labels
$\operatorname{{core}}_{2}(\lambda^{(b0)})$ and
$\operatorname{{core}}_{2}(\lambda^{(b1)})$ respectively.

This tree is known as the 2-core tower of $\lambda$.

######  Example 10.

The partition $(6,5,4,2,1,1)$ has the 2-core tower:
${{(2,1)}}$${\varnothing}$${\varnothing}$${{(1)}}$${\varnothing}$${\varnothing}$${{(1)}}$${\varnothing}$${\varnothing}$${\varnothing}$${\varnothing}$${{(1)}}$${\varnothing}$${\varnothing}$${\varnothing}$
.

Every partition has a unique 2-core tower and every 2-core tower comes from a
unique partition. This bijection is inherited from the core-quotient bijection
above.
We state a well-known classification result for 2-cores which we encourage the
reader to prove on their own.

######  Lemma 10.

A partition $\lambda$ is a 2-core if and only if $\lambda=(n,n-1,\ldots 1)$
for some $n\geq 0$.

######  Notation 12.

Let $\operatorname{\mathcal{T}}_{k}(w)$ denote the number of solutions
$(\mu_{[i]})_{i=1}^{2^{k}}$ to

$\sum\limits_{i=1}^{2^{k}}|\mu_{[i]}|=w,$

where each $\mu_{[i]}$ is a 2-core.

######  Lemma 11.

With the above notation, we have,

$\operatorname{\mathcal{T}}_{k}(w)=\begin{cases}1,&\text{if }w=0\\\
2^{k},&\text{if }w=1\\\ \binom{2^{k}}{2},&\text{if }w=2\\\
\binom{2^{k}}{3}+2^{k},&\text{if }w=3.\end{cases}$

###### Proof.

When $w=0$, the only solution is $\mu_{[i]}=\varnothing$ for all $i$. Thus,
$\operatorname{\mathcal{T}}_{k}(0)=1$.
When $w=1$, choose $\mu_{[i_{0}]}=(1)$ for some $1\leq i_{0}\leq 2^{k}$ and
$\varnothing$ for the rest. As there are $2^{k}$ possible options for $i_{0}$,
we get $\operatorname{\mathcal{T}}_{k}(1)=2^{k}$.
For $w=2$, we must choose $\mu_{[i]}=\mu_{[j]}=(1)$, for some $i$ and $j$, and
the rest $\varnothing$. Thus,
$\operatorname{\mathcal{T}}_{k}(2)=\binom{2^{k}}{2}$.
There does exist a 2-core of size 3, which is $(2,1)$. If $w=3$, we have two
options, either we choose $i,j,k$ such that
$\mu_{[i]}=\mu_{[j]}=\mu_{[k]}=(1)$ or $p$ such that $\mu_{[p]}=(2,1)$. For
the former, we get $\binom{2^{k}}{3}$ ways and the latter can be done in
$2^{k}$ ways, thus giving
$\operatorname{\mathcal{T}}_{k}(3)=\binom{2^{k}}{3}+2^{k}$. ∎

######  Definition 12 (Weight in a 2-core tower).

The weight of the *k1$k^{\mathrm{st}}$
2$k^{\mathrm{nd}}$3$k^{\mathrm{rd}}$[$k^{\mathrm{th}}$] row of the 2-core
tower of a partition $\lambda$ is given by

$w_{k}(\lambda):=\sum\limits_{b\in\\{0,1\\}^{k}}|\operatorname{{core}}_{2}(\lambda^{(b)})|,$

where we define $w_{0}(\lambda)=|\operatorname{{core}}_{2}(\lambda)|$.

We state some important properties of 2-core towers and how they relate to
dimensions. For the subsequent discussion, we will let $n=\sum\limits_{i\geq
0}b_{i}2^{i}$, where $b_{i}\in\\{0,1\\}$. Further, recall that
$\operatorname{bin}(n)=\\{i\mid b_{i}=1\\}$.

######  Notation 13.

Let $\operatorname{bin}^{\prime}(n)=\\{i>0\mid b_{i}=1\\}$.

Note the strict inequality in the definition. Furthermore, we have the
relation,
$\operatorname{bin}^{\prime}(n)=\operatorname{bin}(n)\backslash\\{0\\}$.

######  Proposition 11 (Macdonald).

Let $\lambda$ be a partition of $n$. We write $w_{i}:=w_{i}(\lambda)$. In this
case, the following hold:

  1. (1)

The 2-core tower of $\operatorname{{core}}_{2^{k}}(\lambda)$ is given by
labelling the rows $0$ to $k-1$ as in the 2-core tower of $\lambda$ and by
labelling every node in *i1$i^{\mathrm{st}}$
2$i^{\mathrm{nd}}$3$i^{\mathrm{rd}}$[$i^{\mathrm{th}}$] row, for $i\geq k$,
with $\varnothing$, the empty partition.

  2. (2)

$\lambda$ is an odd partition if and only $w_{i}=b_{i}$ for all $i\geq 0$.

  3. (3)

$v_{2}(f^{\lambda})=1$ if and only if for some
$R\in\operatorname{bin}^{\prime}(n)$, we have $w_{R-1}=b_{R-1}+2$, $w_{R}=0$
and $w_{i}=b_{i}$ otherwise.

###### Proof.

The proofs of (1) and (2) are given in [Mac71]. Through section 4 of [Mac71],
we know that there exists a sequence of non-negative integers $(z_{i})_{i\geq
0}$ with $z_{0}=0$ such that

$w_{i}+z_{i}=b_{i}+2z_{i+1}.$

From Section 3 of [Mac71], we know that if $v_{2}(f^{\lambda})=1$, then we
have $\sum\limits_{i\geq 0}w_{i}=\sum\limits_{i\geq 0}b_{i}+1$. Combining
these, we get that we must have $z_{i}=1$ for exactly one $i>0$ and zero for
the rest. Putting $z_{k}=1$ gives us the equations,

$\displaystyle w_{k-1}$ $\displaystyle=b_{k-1}+2$ $\displaystyle w_{k}+1$
$\displaystyle=b_{k}.$

Notice that the second equation above tells us that $z_{k}=1$ is possible only
when $k\in\operatorname{bin}(n)$. Thus, $w_{k}=0$ and $w_{k-1}=b_{k-1}+2$. The
rest is clear as $z_{i}=z_{i+1}=0$. ∎

######  Notation 14.

For $k>0$, define a sequence of non-negative integers,
$\mathbf{w}_{k}:=(w^{k}_{i})_{i\geq 0}$ with the following properties:

  1. (1)

$w^{k}_{k-1}=b_{k-1}+2$,

  2. (2)

$w^{k}_{k}=0$, and

  3. (3)

$w^{k}_{i}=b_{i}$ for all other values of $i$.

Let $\operatorname{\mathcal{T}}(\mathbf{w}_{k}):=\prod\limits_{i\geq
0}\operatorname{\mathcal{T}}_{i}(w^{k}_{i})$.

Note that the above product is the number of ways of constructing 2-core
towers such that $w_{i}(\lambda)=w^{k}_{i}$, for some given $k$.

###### Proof of 3.

By the above discussion, we have that the number of partitions with dimensions
congruent to 2 modulo 4 is given by

$\displaystyle a_{2}(n)$
$\displaystyle=\sum\limits_{k\in\operatorname{bin}^{\prime}(n)}\operatorname{\mathcal{T}}(\mathbf{w}_{k})$
$\displaystyle=\sum\limits_{k\in\operatorname{bin}^{\prime}(n)}\prod\limits_{i\geq
0}\operatorname{\mathcal{T}}_{i}(w^{k}_{i}).$

Denote the binary expansion of $n$ as we did before. If we suppose $n=2^{R}+m$
with $m<2^{R}$, then $b_{R}=1$ and $b_{i}=0$ for $i>R$. We can break the sum
up as,

$a_{2}(n)=\left(\sum\limits_{k\in\operatorname{bin}^{\prime}(n)\backslash\\{R\\}}\prod\limits_{i\geq
0}\operatorname{\mathcal{T}}_{i}(w^{k}_{i})\right)+\operatorname{\mathcal{T}}(\mathbf{w}_{R}).$

For the term on the left of the plus sign, we have

$\displaystyle\sum\limits_{k\in\operatorname{bin}^{\prime}(n)\backslash\\{R\\}}\prod\limits_{i\geq
0}\operatorname{\mathcal{T}}_{i}(w^{k}_{i})$
$\displaystyle=\sum\limits_{k\in\operatorname{bin}^{\prime}(n)\backslash\\{R\\}}\operatorname{\mathcal{T}}_{R}(w^{k}_{R})\prod\limits_{i\neq
R}\operatorname{\mathcal{T}}_{i}(w^{k}_{i})$
$\displaystyle=2^{R}\sum\limits_{k\in\operatorname{bin}^{\prime}(m)}\prod\limits_{i\neq
R}\operatorname{\mathcal{T}}_{i}(w^{k}_{i})$
$\displaystyle=\frac{2^{R}}{\operatorname{\mathcal{T}}_{R}(0)}\sum\limits_{k\in\operatorname{bin}^{\prime}(m)}\prod\limits_{i\geq
0}\operatorname{\mathcal{T}}_{i}(w^{k}_{i})$ $\displaystyle=2^{R}a_{2}(m).$

Here, the last equality follows from point (1) in 11.
Now, we deal with the term on the right of the plus sign. We see that

$\displaystyle\operatorname{\mathcal{T}}(\mathbf{w}_{R})$
$\displaystyle=\operatorname{\mathcal{T}}_{R-1}(w^{R}_{R-1})\operatorname{\mathcal{T}}_{R}(w^{R}_{R})\prod\limits_{i\neq
R,R-1}\operatorname{\mathcal{T}}_{i}(w^{R}_{i})$
$\displaystyle=\operatorname{\mathcal{T}}_{R-1}(w^{R}_{R-1})\operatorname{\mathcal{T}}_{R}(w^{R}_{R})\prod\limits_{i\neq
R,R-1}\operatorname{\mathcal{T}}_{i}(b_{i})$
$\displaystyle=\frac{\operatorname{\mathcal{T}}_{R-1}(w^{R}_{R-1})\operatorname{\mathcal{T}}_{R}(w^{R}_{R})}{\operatorname{\mathcal{T}}_{R-1}(b_{R-1})\operatorname{\mathcal{T}}_{R}(b_{R})}a(n).$

The $a(n)$ appearing is a consequence of (2) in 11. We have $b_{R}=1$,
$w_{R}=0$ and $w_{R-1}=b_{R-1}+2$. This simplifies the above expression to

$\frac{\operatorname{\mathcal{T}}_{R-1}(b_{R-1}+2)}{2^{R}\operatorname{\mathcal{T}}_{R-1}(b_{R-1})}a(n).$

By 11 and $a(n)=2^{R}a(m)$, we can complete the proof to get

$a_{2}(n)=\begin{cases}2^{R}\cdot a_{2}(m)+\binom{2^{R-1}}{2}\cdot
a(m),&\text{ if }m<2^{R-1}\\\ 2^{R}\cdot
a_{2}(m)+\left(\binom{2^{R-1}}{3}+2^{R-1}\right)\cdot\displaystyle{\frac{a(m)}{2^{R-1}}},&\text{
if }2^{R-1}<m<2^{R}.\end{cases}$

∎

In the sparse case, this theorem takes a nice form as evident in 2. Although,
we can prove the corollary using the recursive relations in 2, we choose to do
perform direct computations as it is more illuminating.

###### Proof of 2.

Let $n$ be a sparse number, i.e., it has no consecutive 1s in its binary
expansion. The summation as before is:

$\displaystyle a_{2}(n)$
$\displaystyle=\sum\limits_{k\in\operatorname{bin}^{\prime}(n)}\operatorname{\mathcal{T}}(\mathbf{w}_{k})$
$\displaystyle=\sum\limits_{k\in\operatorname{bin}^{\prime}(n)}\frac{\operatorname{\mathcal{T}}_{k-1}(w^{k}_{k-1})\operatorname{\mathcal{T}}_{k}(w^{k}_{k})}{\operatorname{\mathcal{T}}_{k-1}(b_{k-1})\operatorname{\mathcal{T}}_{k}(b_{k})}a(n).$

As $n$ is sparse, for $k\in\operatorname{bin}^{\prime}(n)$, we have
$b_{k-1}=0$. Further, $w^{k}_{k}=0$ and $w^{k}_{k-1}=b_{k-1}+2=2$.
This gives us the following summation,

$a_{2}(n)=\sum\limits_{k\in\operatorname{bin}^{\prime}(n)}\frac{\operatorname{\mathcal{T}}_{k-1}(2)}{\operatorname{\mathcal{T}}_{k}(1)}a(n).$

By using the results from 11, this simplifies to

$\displaystyle a_{2}(n)$
$\displaystyle=a(n)\sum\limits_{k\in\operatorname{bin}^{\prime}(n)}\frac{\binom{2^{k-1}}{2}}{2^{k}}$
$\displaystyle=a(n)\sum\limits_{k\in\operatorname{bin}^{\prime}(n)}\frac{(2^{k-1})(2^{k-1}-1)}{2\cdot
2^{k}}$
$\displaystyle=\frac{a(n)}{8}\sum\limits_{k\in\operatorname{bin}^{\prime}(n)}2^{k}-2.$

If $n$ is even, then this summation becomes $n-2\nu(n)$. If $n$ is odd, the
summation becomes $(n-1)+2(\nu(n)-1)=(n-1)+2(\nu(n-1))$. This gives us the
final answer as,

$a_{2}(n)=\begin{cases}\displaystyle{\frac{a(n)}{8}(n-2\nu(n))},&\text{if
}n\text{ is even}\\\ a_{2}(n-1),&\text{if }n\text{ is odd}.\end{cases}$

∎

######  Remark 4.

The theorem above immediately leads us to the value of $m_{4}(n)$. We have
$m_{4}(n)=a_{1}(n)+a_{2}(n)+a_{3}(n)=a(n)+a_{2}(n)$ by definition.

##  7\. Irreducible Representations of Alternating Groups

We can show similar results about degrees of irreducible representations in
the case of $A_{n}$, the alternating group on $n$ letters, which is defined to
be the unique subgroup of index 2.

######  Definition 13.

Let $a_{i}^{\circ}(n)$ be the number of irreducible representations of $A_{n}$
(upto isomorphism) whose dimensions are congruent to $i$ modulo 4. Let
$\delta^{\circ}(n):=a_{1}^{\circ}(n)-a_{3}^{\circ}(n)$. Let
$a^{\circ}(n):=a_{1}^{\circ}(n)+a_{3}^{\circ}(n)$ be the number of odd
dimensional irreducible representations of $A_{n}$.

In [AG22], the value of $a^{\circ}(n)$ was computed in terms of $a(n)$ for
which we present a proof sketch here. Recall that a partition,
$\hat{\lambda}$, is called the conjugate of $\lambda$ if there is a cell in
position $(j,i)$ in $\operatorname{\text{sh}}(\hat{\lambda})$ if and only if
there is a cell in position $(i,j)$ in $\operatorname{\text{sh}}(\lambda)$. If
$\hat{\lambda}=\lambda$, then $\lambda$ is called a self-conjugate partition.

######  Remark 5.

The 2-core tower of $\hat{\lambda}$ can be obtained by flipping the 2-core
tower of $\lambda$ along the central axis of symmetry. That is, the node of
the 2-core tower of $\hat{\lambda}$ indexed by
$\operatorname{{core}}_{2}(\hat{\lambda}^{(b)})$ is indexed by
$\operatorname{{core}}_{2}(\lambda^{(a)})$ where $a$ and $b$ are complementary
binary strings.

######  Notation 15.

Let $\hat{m}_{2}(n)$ denote the number of self-conjugate partitions with
dimension congruent to 2 modulo 4.

######  Lemma 12 ([AG22]).

For $n\in\mathbb{N}$, we have

$\hat{m}_{2}(n)=\begin{cases}1,&\text{if }n=3\text{}\\\ 2^{k-2},&\text{if
}n=2^{k},2^{k}+1,k>1\text{}\\\ 0,&\text{otherwise}\text{}.\end{cases}$

###### Proof.

For a given $n$, define $w^{k}_{i}$ as in 14. By point 3 of 11, we know that
such sequences of weights of rows in the 2-core tower, $\mathbf{w}_{k}$,
correspond to partitions with dimensions congruent to 2 modulo 4.
Firstly, consider the case when the root node is labelled by $\varnothing$. We
know that $w^{k}_{i}\in\\{0,1,2,3\\}$ and for $w^{k}_{i}=1,3$, we cannot flip
the 2-core tower about the central axis to preserve symmetry (Remark 5). Thus,
$w^{k}_{i}=0$ or $2$ for all $i>0$ in this case and $w_{0}^{k}=0$ as the root
node is labelled by $\varnothing$.
Similarly, in the case when the root node is labelled by $(1)$, we have
$w^{k}_{0}=1$ and $w^{k}_{i}=0$ or $2$ for all $i>0$ and
$k\in\operatorname{bin}(n)$. In both these cases, we cannot have two rows
$j>i>0$ such that both $i$ and $j$ in $\operatorname{bin}(n)$. If this was the
case, then we will have $w^{i}_{i}=0$ but $w^{i}_{j}=1$ which violates our
condition on $w_{i}^{k}$ taking only the values 0 and 2. Thus, we must have
$\nu(n)=1$ or when $0\in\operatorname{bin}(n)$, $\nu(n)=2$. Thus, self-
conjugate partitions with dimension congruent to 2 modulo 4 exist only for
$n=2^{k},2^{k}+1$.
In these cases, we have $w^{k}_{k}=0$, $w^{k}_{k-1}=2$ and the
$(k-1)^{\text{st}}$ row is labelled by two $(1)$s arranged symmetrically about
the centre. Notationally,
$\operatorname{{core}}_{2}(\lambda^{(x)})=\operatorname{{core}}_{2}(\lambda^{(y)})=(1)$
where $x$ and $y$ are binary complementary strings in $\\{0,1\\}^{k-1}$ and
the other nodes are labelled by $\varnothing$. Clearly, there are $2^{k-2}$
many ways to achieve such an arrangement, which gives us our result. ∎

The partition $(3,3,3)\vdash 9$ has dimension congruent to 2 modulo 4 and is
self-conjugate. Its 2-core tower is the following and satisifes the symmetric
arrangement: This partition has the 2-core tower:
${{(1)}}$${\varnothing}$${\varnothing}$${{(1)}}$${\varnothing}$${\varnothing}$${{(1)}}$${\varnothing}$${\varnothing}$${\varnothing}$${\varnothing}$${\varnothing}$${\varnothing}$${\varnothing}$${\varnothing}$
.

######  Remark 6.

By Clifford theory (refer Theorem 5.12.5 in [Pra15] and Chapter 6 in [Ols93]),
we know that a self-conjugate partition, $\lambda\vdash n$, corresponds to two
irreducible representations of $A_{n}$ with dimension $f^{\lambda}/2$.
Furthermore, a conjugate pair of distinct partitions of $n$, $\mu$ and
$\hat{\mu}$, corresponds to one irreducible representation of $A_{n}$ with
dimension $f^{\mu}$.

######  Proposition 12 ([AG22]).

For $n\in\mathbb{N}$, the number of odd dimensional irreducible
representations of $A_{n}$ is given by

$a^{\circ}(n)=\begin{cases}1,&\text{if }n=1,2\text{}\\\ 3,&\text{if
}n=3\text{}\\\ 2^{k},&\text{if }n=2^{k},2^{k}+1,k>1\text{}\\\
a(n)/2,&\text{for other values of }n>4\text{}\\\ \end{cases}$

where $a(n)$ is the number of odd partitions.

###### Proof.

By Remark 6, we get that

$a^{\circ}(n)=2\hat{m}_{2}(n)+\frac{a(n)}{2}.$

Using 12 and $a(2^{k})=a(2^{k}+1)=2^{k}$, we have our result. ∎

We now wish to extend this result to find $a_{1}^{\circ}(n)$ and
$a_{3}^{\circ}(n)$. We do so by calculating $\delta^{\circ}(n)$. Pick
$\lambda$ such that $\lambda=\widehat{\lambda}$. If we want it to correspond
to an irreducible representation of $A_{n}$ with odd dimension, then
$f^{\lambda}$ must be congruent to 2 modulo 4 by Remark 6. As the dimension
gets halved, we only need to find what $\operatorname{Od}(f^{\lambda})$ is.
In order to do so, we require what $\lambda\vdash 2^{k},2^{k}+1$ with
$f^{\lambda}$ congruent to 2 modulo 4 look like. For this, we cite two
propositions from Geetha and Amruta’s paper.

######  Proposition 13 ([AG22]).

If $\lambda\vdash 2^{k}$ such that $f^{\lambda}\equiv 2$ mod 4, then

  1. (1)

$\lambda$ is a hook partition of $2^{k}$ or,

  2. (2)

$\lambda$ is of the form
$(2^{k-1}-i,2^{k-1}-j,\underbrace{2,\ldots,2}_{i\text{
times}},\underbrace{1,\ldots,1}_{j-i\text{ times}})$ where $0\leq i\leq j\leq
2^{k-1}-2$.

######  Proposition 14 ([AG22]).

If $\lambda\vdash 2^{k}+1$ such that $f^{\lambda}\equiv 2$ mod 4 and $\lambda$
is self-conjugate, then

  1. (1)

$\lambda$ has $h_{1,1}=2^{k}+1$ or

  2. (2)

$\lambda$ has $h_{1,1}=p_{1}$, $h_{2,2}=p_{2}$ and $h_{3,3}=1$ for some odd
$p_{1}>p_{2}>1$.

######  Theorem 4.

For $n\geq 3$, we have

$\delta^{\circ}(n)=\begin{cases}3,&\text{if }n=3\text{}\\\ 2^{k-1},&\text{if
}n=2^{k},k>1\text{}\\\ 2^{k-1}-2,&\text{if }n=2^{k}+1,k>1\text{}\\\
\delta(n)/2,&\text{otherwise.}\text{}\end{cases}$

###### Proof.

For $n\neq 2^{k},2^{k}+1$, there are no self-conjugate partitions
corresponding to irreducible representations of $A_{n}$ with dimension
congruent to 2 modulo 4, thus $a_{i}^{\circ}(n)=a_{i}(n)/2$ for $i=1,3$ by
Remark 6 and $\delta^{\circ}(n)=\delta(n)/2$.
Notice that $\operatorname{Od}(a)^{2}=1$. Thus, for a self conjugate partition
as we have $h_{i,j}=h_{j,i}$ for all
$(i,j)\in\operatorname{\text{sh}}(\lambda)$, it follows that

$\operatorname{Od}\left(\prod\limits_{(i,j)\in\operatorname{\text{sh}}(\lambda)}h_{i,j}\right)=\operatorname{Od}\left(\prod\limits_{\begin{subarray}{c}i\\\
(i,i)\in\operatorname{\text{sh}}(\lambda)\end{subarray}}h_{i,i}\right).$

We saw above that for only $n=2^{k},2^{k}+1$ for $k>1$, we have
$\hat{m}_{2}(n)\neq 0$. By using the formula in Remark 2 and

$\operatorname{Od}(2^{k}!)=\operatorname{Od}((2^{k}+1)!)=-1,\text{(\lx@cref{creftype~refnum}{lem:odnfac})}$

we can show that for the case when $\lambda\vdash 2^{k}$, $k>1$,and $\lambda$
is self conjugate with dimension congruent to 2 mod 4, we have

$\operatorname{Od}(f^{\lambda})=-\operatorname{Od}\left(\prod\limits_{\begin{subarray}{c}i\\\
(i,i)\in\operatorname{\text{sh}}(\lambda)\end{subarray}}h_{i,i}\right).$

To continue with our computations, we use 13. Following the notation from
[AG22], we deduce that that for $\lambda\vdash 2^{k}$ with $f^{\lambda}\equiv
2\mod 4$, we have $h_{1,1}=2^{k-1}-i+j+1$ and $h_{2,2}=2^{k-1}+i-j-1$.
Further imposing the condition that $\lambda=\hat{\lambda}$, we get
$h_{1,1}=2^{k}-(2\alpha+1)$ and $h_{2,2}=2\alpha+1$, for $0\leq\alpha\leq
2^{k-2}-1$. It is easy to see that
$\operatorname{Od}(2^{k}-(2\alpha+1))=-\operatorname{Od}(2\alpha+1)$ and thus,

$\operatorname{Od}(f^{\lambda})=-\operatorname{Od}(h_{1,1})\operatorname{Od}(h_{2,2})=-\operatorname{Od}(2\alpha+1)\cdot(-\operatorname{Od}(2\alpha+1))=1$

for this case. Thus, by Remark 6, each irreducible representation of $S_{n}$
with degree congruent to 2 modulo 4 which is indexed by a self-conjugate
representation restricts to two irreducible representations of $A_{n}$ with
degrees which are always 1 modulo 4.
An irreducible representation of $A_{n}$ has degree congruent to 1 modulo 4 if
it corresponds to a non-self conjugate pair with dimensions congruent to 1
modulo 4 or it corresponds to a self-conjugate partition with dimension
congruent to 2 modulo 4. This gives us that

$a_{1}^{\circ}(n)=2\hat{m}_{2}(n)+\frac{a_{1}(n)}{2}.$

On the other hand, $a_{3}^{\circ}(n)$ gets no contribution from
$\hat{m}_{2}(n)$. Thus, we have

  1. (1)

$a_{1}(2^{k})=a_{3}(2^{k})=2^{k-1}$ where each pair gives one irreducible
representation.

  2. (2)

There are $2^{k-2}$ possible values of $\alpha$, each of which corresponds to
two irreducible representations of $A_{n}$ with dimension congruent to 1
modulo 4 coming from representations of $S_{n}$ with dimension congruent to 2
modulo 4.

We get that $a^{\circ}_{1}(2^{k})=2^{k-1}+2^{k-2}$ and
$a_{3}^{\circ}(2^{k})=2^{k-2}$. Thus, $\delta^{\circ}(2^{k})=2^{k-1}$.

Now, we consider the case when $\lambda\vdash 2^{k}+1$. We deal with the two
cases in 14:

  1. (1)

When $\lambda$ has $h_{1,1}=2^{k}+1$, we get
$\operatorname{Od}(f^{\lambda})=-\operatorname{Od}(2^{k}+1)=-1$.

  2. (2)

In the other case, we have $p_{1}+p_{2}=2^{k}$. As $p_{1}\equiv-p_{2}\mod 4$,
we have
$\operatorname{Od}(f^{\lambda})=-\operatorname{Od}(p_{1})\operatorname{Od}(p_{2})\operatorname{Od}(1)=1$.
Note that because of the conditions $p_{1}>p_{2}$ and $p_{1}$ an odd integer,
$p_{1}$ takes exactly $2^{k-2}-1$ values.

We again use Remark 6. We add up the contributions from odd partitions of
$2^{k}+1$ and partitions of $2^{k}+1$ with dimension congruent to 2 modulo 4
which we computed above. We have

$a_{1}^{\circ}(2^{k}+1)=\frac{2^{k}-1+2}{2}+2\cdot(2^{k-2}-1)$

and

$a_{3}^{\circ}(2^{k}+1)=\frac{2^{k}-1-2}{2}+2\cdot 1.$

Thus, $\delta^{\circ}(2^{k}+1)=2^{k-1}-2$. ∎

##  8\. Closing Remarks and Open Problems

It is natural to ask why the case of $n=2^{R}+m$ for $2^{R-1}<m<2^{R}$ is so
difficult that it evades our methods. As we saw before, the values of
$S_{\operatorname{\mathcal{P}}(\mu)}(\mu)$ are constants, thus, independent of
the core $\mu$ that we are building upon. In the harder case, we do not have
this luxury. The values of $S_{\operatorname{\mathcal{P}}(\mu)}(\mu)$ vary
wildly and resist coalescing into nice summations. Along with this roadblock,
we present some relevant problems that remain unsolved.

  1. (1)

What is $\delta(n)$ in the case when $n=2^{R}+m$ for $2^{R-1}<m<2^{R}$?

  2. (2)

Is it possible to provide a reasonable bound for
$S_{\operatorname{\mathcal{P}}(\mu)}(\mu)$?

  3. (3)

Can we say something about the dimension of a partition by just looking at its
2-core tower?

  4. (4)

Is there a characterization of odd partitions modulo 4 in terms of
$\beta$-sets?

It is possible to answer (3) and (4) in the case of hook partitions as the
analysis simplifies to looking at binomial coefficients.

##  9\. Acknowledgements

This work was completed at the Indian Institute of Science Education and
Research, Pune under the supervision of my supervisor Steven Spallone who
introduced me to this fascinating problem. His crucial insights throughout the
course of this project as well as during proofreading helped me streamline my
mathematical thought and writing. My sincerest thanks goes to Jyotirmoy
Ganguly for the discussions that led to most of the proofs in the project as
well as for being a constant source of encouragement. I am also thankful to
Amritanshu Prasad for his helpful comments regarding this manuscript.

## References

  * [AG22] P. Amrutha and T. Geetha  “On the Degrees of Representations of Groups not divisible by $2^{k}$” unpublished preprint, 2022
  * [APS16] Arvind Ayyer, Amritanshu Prasad and Steven Spallone  “Odd partitions in Young’s lattice”, 2016  arXiv:1601.01776 [math.CO]
  * [DW89] K. Davis and W Webb  “Pascal’s Triangle modulo 4”  In _Fibonacci Quarterly_ 29.1, 1989, pp. 79–83
  * [FRT54] J.. Frame, G. B. Robinson and R.. Thrall  “The Hook Graphs of the Symmetric Group”  In _Canadian Journal of Mathematics_ 6 Cambridge University Press, 1954, pp. 316–324  DOI: 10.4153/CJM-1954-030-1
  * [GPS20] Jyotirmoy Ganguly, Amritanshu Prasad and Steven Spallone  “On the Divisibility of Character Values of the Symmetric Group”  In _The Electronic Journal of Combinatorics_ 27.1 The Electronic Journal of Combinatorics, 2020  DOI: 10.37236/8693
  * [Gia+16] Eugenio Giannelli, Alexander Kleshchev, Gabriel Navarro and Pham Huu Tiep  “Restriction of Odd Degree Characters and Natural Correspondences”  In _International Mathematics Research Notices_ 2017.20, 2016, pp. 6089–6118  DOI: 10.1093/imrn/rnw174
  * [Las08] Michel Lassalle  “An explicit formula for the characters of the symmetric group”  In _Mathematische Annalen_ 340.2, 2008, pp. 383–405  DOI: 10.1007/s00208-007-0156-5
  * [Mac71] I.. MacDonald  “On the Degrees of the Irreducible Representations of Symmetric Groups”  In _Bulletin of the London Mathematical Society_ 3.2, 1971, pp. 189–192  DOI: https://doi.org/10.1112/blms/3.2.189
  * [Mac15] I.. Macdonald  “Symmetric functions and hall polynomials”  Oxford University Press, 2015, pp. 12–12
  * [Ols93] Jørn B. Olsson  “Jørn B. Olsson. Combinatorics and representations of finite groups, volume 20 of Vorlesungen aus dem Fachbereich Mathematik der Universität GH Essen.”  Universit”at Essen, 1993
  * [OZ21] Rosa Orellana and Mike Zabrocki  “Symmetric group characters as symmetric functions”  In _Advances in Mathematics_ 390 Elsevier BV, 2021, pp. 107943  DOI: 10.1016/j.aim.2021.107943
  * [Pel20] Sarah Peluse  “On even entries in the character table of the symmetric group”, 2020  arXiv:2007.06652 [math.CO]
  * [Pra15] Amritanshu Prasad  “Representation Theory: A Combinatorial Viewpoint”, Cambridge Studies in Advanced Mathematics  Cambridge University Press, 2015  DOI: 10.1017/CBO9781139976824