# Counting Restricted Partitions of Integers into Fractions: Symmetry and
Modes of the Generating Function and a Connection to $\omega(t)$

Z. Hoelscher and E. Palsson 111The second author was supported in part by
Simons Foundation Grant #360560.

###### Abstract -

Motivated by the study of integer partitions, we consider partitions of
integers into fractions of a particular form, namely with constant
denominators and distinct odd or even numerators. When numerators are odd, the
numbers of partitions for integers smaller than the denominator form symmetric
patterns. If the number of terms is restricted to $h$, then the nonzero terms
of the generating function are unimodal, with the integer $h$ having the most
partitions. Such properties can be applied to a particular class of nonlinear
Diophantine equations. We also examine partitions with even numerators. We
prove that there are $2^{\omega(t)}-2$ partitions of an integer $t$ into
fractions with the first $x$ consecutive even integers for numerators and
equal denominators of $y$, where $0<y<x<t$. We then use this to produce
corollaries such as a Dirichlet series identity and an extension of the prime
omega function to the complex plane, though this extension is not analytic
everywhere.

Keywords : Integer partitions; Restricted partitions; Partitions into
fractions

Mathematics Subject Classification (2020) : 05A17

##  1 Introduction

Integer partitions are a classic part of number thoery. In the most general,
unrestricted case, one seeks to express positive integers as the sum of
smaller positive integers. Often the function $p(n)$ is used to denote the
count of the partitions of $n$. No simple formula for this is known, though a
generating function can be written [1]. We note that a generating function is
a formal power series where the exponent of $x$ in each term refers to the
number being partitioned, and the coefficient of $x$ gives the number of
partitions.

$\sum_{n=0}^{\infty}p(n)x^{n}=\displaystyle\prod_{m\geq 1}\frac{1}{1-x^{m}}$
(1.0.1)

Ramanujan produced congruence formulas for $p(n)$ [2], while Rademacher wrote
a series expansion that allowed for asymptotic bounds [1]. Note that
$\omega(h,k)$ is the $24k^{th}$ root of unity.

$p(n)=\frac{1}{\pi\sqrt{2}}\sum_{k=1}^{\infty}\sqrt{k}\sum_{\begin{subarray}{c}0\leq
h<k\\\ (h,k)=1\end{subarray}}\omega(h,k)e^{\frac{-2\pi
inh}{k}}\frac{d}{dn}\left(\frac{\sinh\left(\frac{\pi}{k}\sqrt{\frac{2}{3}(n-\frac{1}{24})}\right)}{\sqrt{n-\frac{1}{24}}}\right)$
(1.0.2)

Much work has been done on restricted partitions. For example, we can denote
the number of partitions into distinct parts as $q(n)$, where the generating
function is given below [3].

$\sum_{n=0}^{\infty}q(n)x^{n}=\displaystyle\prod_{j=1}^{\infty}(1+x^{j})$
(1.0.3)

It is known that Rademacher-like series exist for this case as well [3]. It is
also common to find partitions into odd integers, which we count with
$p_{O}(n)$, or partitions into distinct odd terms, which we count with
$q_{O}(n)$. Both have been studied by Hagis [4], where he writes the
generating function for $q_{O}(n)$ as

$1+\sum_{n=1}^{\infty}q_{O}(n)x^{n}=\displaystyle\prod_{m=0}^{\infty}(1+x^{2m+1})$
(1.0.4)

Another common constraint is to restrict the value of the largest summand to
the integer $s$. In the general case, we can denote the number of such
partitions as $p_{s}(n)$, and the case with distinct summands as $q_{s}(n)$.
Upper bounds for $q_{s}(n)$ have been found by Bidar [3]. Alternatively, one
can restrict partitions to $h$ parts, where the number of partitions is
counted by $p_{h}(n)$, or $q_{h}(n)$, where we also require that the summands
are distinct. A recurrence relation for $q_{h}(n)$ immediately follows from
the work of Christopher, where we specify that there are $h$ different sizes
for the $h$ parts, as they are distinct [5].

$q_{h}(n)=q_{h}(n-h)+q_{h-1}(n-h)$ (1.0.5)

One can also combine many of these restrictions. We let $q_{O_{s}}(n)$ count
partitions into odd distinct summands no larger than $s$, and $q_{O_{s},h}(n)$
count partitions into $h$ distinct odd summands no larger than $s$. The latter
is the more interesting of the two. We can write the following generating
functions

$\sum_{m=0}^{j^{2}}q_{O_{2j-1}}(m)x^{m}=\displaystyle\prod_{n=1}^{j}(1+x^{2n-1})$
(1.0.6)
$\sum_{m=0}^{j^{2}}\sum_{h=0}^{j}q_{O_{2j-1},h}(m)y^{h}x^{m}=\displaystyle\prod_{n=1}^{j}(1+yx^{2n-1})$
(1.0.7)

In this paper we study partitions of integers into fractions with fixed
denominators, which is equivalent to further restrictions on the integer
partition questions given above. It is a well-known fact that
$j^{2}=\sum_{n=1}^{j}(2n-1)$, where $j$ is a positive integer. If we then
divide both sides by $j$, we find the series

$j=\frac{1}{j}+\frac{3}{j}+\cdots+\frac{2(j-1)-1}{j}+\frac{2j-1}{j}$ (1.0.8)

It is natural to ask whether there exist distinct coefficients
$a_{1},\ldots,a_{n}\in{1,3,\ldots,2j-1}$ such that any positive integer $k$,
where $0<k<j$ can be written as

$k=\frac{a_{1}}{j}+\frac{a_{2}}{j}+\ldots+\frac{a_{n-1}}{j}+\frac{a_{n}}{j}$
(1.0.9)

If $j>2$, this is, in fact, always possible. We provide a proof in Section 4.
We note that this is not always possible for other types of fractions, such as
those with even numerators. An interesting continuation of this work would be
to determine all infinite sequences $\\{n_{j}\\}$ of numerators that enable
such solutions for all $k<\lfloor G\rfloor$, where
$G=\frac{n_{1}}{j}+\frac{n_{2}}{j}+\cdots+\frac{n_{j-1}}{j}+\frac{n_{j}}{j}$,
$k,j\in\mathbb{N}$, $j>2$. For instance, one could say that $\\{n_{j}\\}$ is
the Fibonacci sequence instead of the odd integers. Note that for alternative
choices of $\\{n_{j}\\}$, $\lfloor G\rfloor$ may no longer equal $j$. If an
infinite number of such sequences exist, one could look for necessary and
sufficient conditions for a sequence to qualify. We give an example below for
the case of $j=6$:

$6=\frac{1}{6}+\frac{3}{6}+\frac{5}{6}+\frac{7}{6}+\frac{9}{6}+\frac{11}{6}$

$1=\frac{1}{6}+\frac{5}{6}\hskip 14.45377pt2=\frac{11}{6}+\frac{1}{6}\hskip
14.45377pt3=\frac{7}{6}+\frac{11}{6}\hskip
14.45377pt4=\frac{11}{6}+\frac{9}{6}+\frac{3}{6}+\frac{1}{6}\hskip
14.45377pt5=\frac{11}{6}+\frac{9}{6}+\frac{7}{6}+\frac{3}{6}$

###### Conjecture 1.0.1

The Lazy Caterer’s sequence (A000124) enables solutions for all $k<\lfloor
G\rfloor$, $j>2$, $k,j\in\mathbb{N}$. This also appears to be true for the
Cake Numbers (A000125).

It is more interesting to ask how many solutions there are, depending on the
values of $k$ and $j$. We represent this count with the function
$f_{O_{j}}(k)$.

Values for $f_{O_{j}}(k)$
---
$j$ $\hskip 5.69046pt|$ | $k$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$ | $11$
3 |  | 1 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
4 |  | 1 | 2 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
5 |  | 1 | 2 | 2 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0
6 |  | 1 | 3 | 2 | 3 | 1 | 0 | 0 | 0 | 0 | 0 | 0
7 |  | 1 | 3 | 5 | 5 | 3 | 1 | 0 | 0 | 0 | 0 | 0
8 |  | 2 | 5 | 7 | 8 | 7 | 5 | 2 | 0 | 0 | 0 | 0
9 |  | 2 | 5 | 9 | 13 | 13 | 9 | 5 | 2 | 0 | 0 | 0
10 |  | 2 | 7 | 12 | 20 | 20 | 20 | 12 | 7 | 2 | 0 | 0
11 |  | 2 | 8 | 18 | 29 | 36 | 36 | 29 | 18 | 8 | 2 | 0
12 |  | 3 | 11 | 25 | 44 | 60 | 68 | 60 | 44 | 25 | 11 | 3

It is even more interesting to ask what happens when the number of terms is
restricted to $h$. We denote the number of such solutions with
$f_{O_{j},h}(k)$. One should note that in this case, solutions need not always
exist. Below we give a number table corresponding to the case of $h=2$.

Values for $f_{O_{j},2}(k)$
---
$j$ $\hskip 5.69046pt|$ | $k$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$ | $11$ | $12$
3 |  | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
4 |  | 1 | 2 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
5 |  | 0 | 2 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
6 |  | 1 | 3 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
7 |  | 0 | 3 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
8 |  | 2 | 4 | 2 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
9 |  | 0 | 4 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
10 |  | 2 | 5 | 2 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
11 |  | 0 | 5 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
12 |  | 3 | 6 | 3 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
13 |  | 0 | 6 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0

Both of these questions are really integer partition questions in disguise, as
we are counting partitions of $kj$ into distinct odd integers no larger than
$2j-1$, and, in the more restricted case, only $h$ terms. This of course means
that $f_{O_{j}}(k)=q_{O_{2j-1}}(kj)$ and $f_{O_{j},h}(k)=q_{O_{2j-1},h}(kj)$.

The goal of this paper is to understand the behavior of the functions
$f_{O_{j}}(k)$ and $f_{O_{j},h}(k)$ beyond knowledge of their generating
functions. This is challenging, as there appears to be no work on this in the
literature. Using $q_{s}(n)$ from Bidar’s work, we begin with the following
generating function [3].

$q_{s}(x)=\displaystyle\prod_{n=1}^{s}(1+x^{n})=\sum_{m=0}^{\frac{s(s+1)}{2}}q_{s}(m)x^{m}$
(1.0.10)

Bidar writes that $q_{s}(x)$ is a symmetric, unimodal polynomial but says that
proving it is unimodal is quite difficult. Here a unimodal polynomial is a
polynomial whose coefficients strictly increase to some maximum value, then
strictly decrease. A polynomial with multiple modes then has multiple peaks in
the values of its coefficients. Bidar is not aware of an elementary proof,
with the only one he knows of requiring the use of Lie algebras. In a similar
fashion, we can consider the generating functions for $f_{O_{j}}(k)$ and
$f_{O_{j},h}(k)$. The following generating functions are polynomials in $x$,
just like $q_{s}(x)$, hence leading us to the main theorem of this paper. Note
that $0<k<j$.

$r(x)=f_{O_{j}}(1)x^{1}+f_{O_{j}}(2)x^{2}+\cdots+f_{O_{j}}(j-2)x^{j-2}+f_{O_{j}}(j-1)x^{j-1}$
(1.0.11)
$r_{h}(x)=f_{O_{j},h}(1)x^{1}+f_{O_{j},h}(2)x^{2}+\cdots+f_{O_{j},h}(j-2)x^{j-2}+f_{O_{j},h}(j-1)x^{j-1}$
(1.0.12)

###### Theorem 1.0.2

The nonzero terms of $r_{h}(x)$ are unimodal and symmetric about the term that
corresponds to $k=h$, where the maximum coefficient is $f_{O_{j},h}(h)$.

###### Proposition 1.0.3

The terms of $r(x)$ exhibit symmetry such that the coefficients follow
$f_{O_{j}}(k_{1})=f_{O_{j}}(j-k_{1})$. Note that $k_{1}\in\mathbb{N},k_{1}<j$.

We cannot make a similar statement of unimodality for $r(x)$, as the case of
$j=6$ is bimodal. This corresponds to the fourth row in the table for
$f_{O_{j}}(k)$. We also see that unlike $r_{h}(x)$, the greatest coefficient
of $r(x)$ with a given $j$ can sometimes be found for more than one
consecutive $k$-value. For example, the eighth row of the table for
$f_{O_{j}}(k)$ has three consecutive entries of 20, whereas with $r_{h}(x)$,
the greatest coefficient is only for a single value of $k$, where $k=h$. These
changes in behavior perhaps make our restrictions more interesting.

We are able to produce a simple proof of unimodality when the number of terms
is restricted to $h$, as this restriction enables a simple bijection to
another problem, where one can ultimately show that unimodality is a
consequence of the unimodality of the Gaussian binomial coefficients. This is
not possible for $r(x)$, as the number of terms per partition is not
restricted. We also note that $r(x)$ only appears to follow our conjecture
below when $k$ is an integer. If $k$ is not required to be an integer, one can
easily find instances where $r(x)$ is neither unimodal nor bimodal, such as
when $j=11$. Considering this along with the fact that the unimodality of
Gaussian binomial coefficients is difficult to prove, our following conjecture
may prove to be a fairly difficult, interesting problem.

###### Conjecture 1.0.4

$r(x)$ is always either unimodal or bimodal.

In contrast to the difficulties faced when dealing with $q_{s}(x)$, our proof
of the unimodality of nonzero terms of $r_{h}(x)$ is fairly elementary, at
least in the sense that it requires only elementary methods and previous
results that have combinatorial proofs. We prove this in Section 5. We also
explore simplifications of $r_{h}(x)$ in Section 2. In Section 3 we explore
closed-form expressions for the case of $h=2$, which we prove in Section 7. In
Section 8 we provide examples of number tables beyond the case of restriction
to two terms, and in Section 9 we include further computational results. In
Section 10 we connect the odd case to finding solutions to a certain nonlinear
Diophantine equation.

After examining such fractions with odd numerators, it is natural to
investigate a similar problem with even numerators.

$t=\frac{2}{y}+\frac{4}{y}+\cdots+\frac{2(x-1)}{y}+\frac{2x}{y}$ (1.0.13)

We observe that if one allows the denominators to be values other than $t$,
and the length of the series to vary, then there are often multiple series for
each $t$. We also note that in some cases, though not all, these series can be
partitioned into all natural numbers $k$, where $0<k<t$. Examples of such
partitions are included in Section 9. We include below a few examples of
series for $t$.

$6=\frac{2}{2}+\frac{4}{2}+\frac{6}{2}$
$10=\frac{2}{2}+\frac{4}{2}+\frac{6}{2}+\frac{8}{2}$
$10=\frac{2}{3}+\frac{4}{3}+\frac{6}{3}+\frac{8}{3}+\frac{10}{3}$
$14=\frac{2}{4}+\frac{4}{4}+\frac{6}{4}+\frac{8}{4}+\frac{10}{4}+\frac{12}{4}+\frac{14}{4}$

This case leads to a nice expression for the count of solutions, as well as a
connection to the prime omega function. We provide proofs for the following
results in Section 6.

###### Theorem 1.0.5

Let $F_{E}(t)$ be the number of partitions of the positive integer $t$ into
fractions with the first $x$ consecutive even integers as numerators and equal
denominators of $y$, such that $x$ and $y$ are positive integers, and
$0<y<x<t$. We then have $F_{E}(t)=2^{\omega(t)}-2$, where $\omega(t)$ is the
number of distinct prime factors of $t$.

We find that we can use our partition identity to produce a series identity
for $\frac{\zeta^{2}(s)}{\zeta(2s)}$ and a continuation of $\omega(t)$ to non-
integers and complex numbers $z$. This is not analytic everywhere, though it
is analytic in many places, such as $z\in\mathbb{R},n<z<n+1,n\in\mathbb{N}.$
This can be used to “assign” a quantity of distinct prime factors to numbers
that obviously have none. For example, we can define
$\omega(e)\approx-6.0963+4.5323i$. Note that we use the normalized
$\operatorname{sinc}$ function. This result is of course valid for natural
numbers as well, so it can be considered a general formula for the prime omega
function.

###### Corollary 1.0.6

$\omega(z)=\log_{2}\bigg{(}\sum_{x=1}^{\left\lceil{\Re(z)}\right\rceil}\operatorname{sinc}\bigg{(}\displaystyle\prod_{y=1}^{\left\lceil{\Re(z)}\right\rceil+1}(x^{2}+x-yz)\bigg{)}\bigg{)}$
(1.0.14)

###### Corollary 1.0.7

Let $s$ be a real number greater than two, then

$\sum_{t=1}^{\infty}\sum_{x=1}^{t}\frac{\operatorname{sinc}\bigg{(}\frac{(-t)^{t+1}\Gamma(\frac{-x^{2}}{t}-\frac{x}{t}+t+2)}{\Gamma(\frac{-x^{2}}{t}-\frac{x}{t}+1)}\bigg{)}}{t^{s}}=\frac{\zeta^{2}(s)}{\zeta(2s)}$
(1.0.15)

This is done in part by using a modification of the concept of the circle
method. These final corollaries are largely unimportant but interesting
curiosities, though the concept of a continuation of a function beyond its
usual domain is quite important in number theory. Examples include the
analytic continuation of the Riemann zeta function [6] and the gamma function
for the factorial.

##  2 Simplifying $r_{h}(x)$

We know we can write $r_{h}(x)$ as the following, where we are only interested
in values of $m$ that are integral multiples of $j$, as $m=kj$ when
$f_{O_{j},h}(k)=q_{O_{2j-1},h}(m)$, where $0<k<j$.

$\sum_{m=0}^{j^{2}}\sum_{h=0}^{j}q_{O_{2j-1},h}(m)y^{h}x^{m}=\displaystyle\prod_{n=1}^{j}(1+yx^{2n-1})$
(2.0.1)

We point out that if we remove from the sum all cases where
$q_{O_{2j-1},h}(m)=0$, we see that we can use a relation $R$ to condense the
double sum into a single sum. This relation utilizes the Rascal triangle, an
alternative to Pascal’s triangle that was published in 2010. We let the
$n^{th}$ entry, counted from left to right on the $j^{th}$ row be denoted by
$T(j,n)$. The values of this triangle are produced using the formula
$T(j,n)=\frac{(T(j-1,n-1))(T(j-1,n))+1}{T(j-2,n-1)}$, where the first and last
terms of each row are one [7]. The first few rows are

1

1, 1

1, 2, 1

1, 3, 3, 1

1, 4, 5, 4, 1

1, 5, 7, 7, 5, 1

1, 6, 9, 10, 9, 6, 1

As an example for the recurrence above, we see that for the entry $10$ in the
bottom row, we have $10=\frac{(7)(7)+1}{5}$. We see that we can write the
generating function

$\sum_{0\leq m\leq
j^{2}}q_{O_{2j-1},\lambda}(m)y^{\lambda}x^{m}=\displaystyle\prod_{n=1}^{j}(1+yx^{2n-1})$
(2.0.2)

When $q_{O_{2j-1},\lambda}(m)$ is nonzero, $R$ is a relation that takes $m$ as
its input and returns each integer $\lambda$ from $0$ to $j$ a number of times
given by the entries read from left to right on the $(j+1)^{th}$ row of the
Rascal triangle. This is then equivalent to saying that the entries in the
Rascal triangle give the number of distinct values one can find as sums of a
given number of integers removed from the set of the first $j$ odd integers.
The ability of the Rascal triangle to predict the number of distinct
restricted sums taken from consecutive integers has been noticed before, as
indicated by a comment on the OEIS page [8], though now we utilize this
property to condense a generating function. The $(j+1)^{th}$ row is needed
because we have exponents of $y$ from $0$ to $j$.

We see that $R$ does not always return its values in consecutive order and
that it can return multiple values for each $m$. $\lambda$ is even if $m$ is
even, and odd if $m$ is odd. This is because the sum of an even number of odd
integers is even, and the sum of an odd number of odd integers is odd. The
index $m$ can range from 0 to $j^{2}$, though it does not take on every value.
We define $\lambda=0$ when $m=0$ and $q_{O_{2j-1},\lambda}(m)=1$ when $m=0$
and $\lambda=0$.

As an example, we display the case of $j=6$:

$\prod_{n=1}^{6}(1+yx^{2n-1})=x^{36}y^{6}+x^{35}y^{5}+x^{33}y^{5}+x^{32}y^{4}+x^{31}y^{5}+x^{30}y^{4}+x^{29}y^{5}+2x^{28}y^{4}$
(2.0.3)
$+x^{27}y^{5}+x^{27}y^{3}+2x^{26}y^{4}+x^{25}y^{5}+x^{25}y^{3}+3x^{24}y^{4}+2x^{23}y^{3}+2x^{22}y^{4}+3x^{21}y^{3}+2x^{20}y^{4}$
$+x^{20}y^{2}+3x^{19}y^{3}+x^{18}y^{4}+x^{18}y^{2}+3x^{17}y^{3}+x^{16}y^{4}+2x^{16}y^{2}+3x^{15}y^{3}+2x^{14}y^{2}+2x^{13}y^{3}$
$+3x^{12}y^{2}+x^{11}y^{3}+x^{11}y+2x^{10}y^{2}+x^{9}y^{3}+x^{9}y+2x^{8}y^{2}+x^{7}y+x^{6}y^{2}+x^{5}y+x^{4}y^{2}+x^{3}y+xy+1$

We observe that $y^{0}$ appears one time, $y^{1}$ six times, $y^{2}$ nine
times, $y^{3}$ ten times, $y^{4}$ nine times, $y^{5}$ six times, and $y^{6}$
one time. This matches the seventh row of the Rascal triangle. We see that
some values for $m$, such as 34 and 2, do not appear. We also see that the
relation returns multiple values for some $m$, as $m=27$ returns both 5 and 3
as exponents of $y$. When the exponent of $x$ is $kj$, the coefficients of the
polynomial in this example match the entries in the tables for
$f_{O_{j},h}(k)$ with $j=6$, where the exponent of $y$ is the number of terms
per partition, $h$.

##  3 Closed form expressions for the case of $h=2$

While all tables exhibit symmetry, only certain cases can be entirely
described by readily apparent closed-form expressions. There is no obvious
closed-form expression for the coefficients of $r(x)$, though we can find
closed-form expressions for the coefficients of $r_{h}(x)$ when $h=2$. Note
that $n=j$. This notation is used to improve readability, as $j$ is too
similar to $i$, which we use to denote the imaginary unit.

For $k=1$ and $k=3:$

$f_{O_{j},2}(k)=\frac{1}{8}\bigl{(}(-1)^{1+n}-1+(-i)^{n}+i^{n}+(-1)^{n}n+n\bigr{)}$
(3.0.1)

For $k=2$:

We can write the following recurrence relation for this sequence [9].

$f_{O_{j},2}(k)=1+f_{O_{j-2},2}(k)$ (3.0.2) $\text{where}\ f_{O_{3},2}(k)=1\
\text{and}\ f_{O_{4},2}(k)=2$

Solving the recurrence relation, we find

$f_{O_{j},2}(k)=\frac{1}{4}\bigl{(}(-1)^{n}+2n-1\bigr{)}$ (3.0.3)

For $k>3:$

$f_{O_{j},2}(k)=0$ (3.0.4)

We provide a proof for this special case in Section 7.

##  4 Proof that there exists a solution for all $k$, where $k<j$ and $j>2$

###### Proposition 4.0.1

Any positive integer $j$ can be written as the sum of a set of fractions,
where the numerators are the first $j$ consecutive odd integers and the
denominators are $j$. If $j>2$, any positive integer $k$, where $k<j$, can be
written as the sum of some combination of the fractions that are summed to
produce $j$.

Proof.  We see that pairs of fractions equidistant from opposite ends of the
series for $j$ always add to two. Such pairs can be written in the form shown
below, where $m=2n-1$:

$\frac{m}{j}+\frac{2j-m}{j}=\frac{2j}{j}=2$ (4.0.1)

###  4.1 Odd value for $j$:

If $j$ is odd, then the series $\sum_{n=1}^{j}\frac{2n-1}{j}$ must have an odd
number of terms. The middle term of the series, given by
$n=\bigl{(}\frac{1}{2}(j-1)+1\bigr{)}$, is always equal to one.

$\frac{2\bigl{(}\frac{(j-1)}{2}+1\bigr{)}-1}{j}=\frac{j}{j}=1$ (4.1.1)

Any pair of terms taken from opposite ends of the series
$\sum_{n=1}^{j}\frac{2n-1}{j}$ and moving inwards sum to two, with the middle
term of the series equaling one. Any even positive integer $k$, where $k<j$,
can then be written as some multiple of pairs of fractions that add to two.
(For instance, $k=6$ would require three pairs whose terms are equidistant
from the ends of the series.) If $k$ is odd, the term from
$\sum_{n=1}^{j}\frac{2n-1}{j}$ that equals one would be added to the greatest
even integer smaller than $k$ to produce the odd integer $k$. This then proves
that any positive integer $k$, where $k<j$, can be written as the sum of some
combination of fractions from the series for $j$, where $j$ is odd.

###  4.2 Even value for $j$:

The property of pairs of terms equidistant from the ends of the series summing
to two still holds for even $j$, therefore any even positive integer $k$,
where $k<j$, can be written as the sum of some combination of the terms from
the series $\sum_{n=1}^{j}\frac{2n-1}{j}$ used to give $j$.

If $j$ is even and greater than two, the first term $\frac{1}{j}$ and the term
given by $n=\frac{j}{2}$ always sum to one.

$\frac{1}{j}+\frac{\bigl{(}2(\frac{j}{2})-1\bigr{)}}{j}=\frac{j}{j}=1$ (4.2.1)

If we know the first term and the term given by $n=\frac{j}{2}$ sum to one, we
can subtract the pair that adds to one from the total series for $j$, then
continue subtracting pairs that add to two until we are left with the odd
number of interest. This is possible because $(j-1)$ is odd when $j$ is even,
and any odd $(j-1)$ is still odd when a multiple of two is subtracted from it.

The smallest odd number that can be produced in this way is three, as two
fractions that sum to one are removed at the start. This thus removes two
possible combinations of terms that would add to two, which would leave the
smallest result as four. This is not the case because a pair adding to one was
removed from the series, thus leaving $4-1=3$ as the smallest number that can
be produced by this method. The proof is still complete; however, as we have
shown how every odd $k$ greater than or equal to three can be produced, and
have previously shown that the first term $\frac{1}{j}$ and the term given by
$n=\frac{j}{2}$ always sum to equal one. This thus proves that any positive
odd integer $k$, where $k<j$, can be found as the sum of some combination of
fractions from the series for $j$, where $j$ is even.

Example for $k=7$ where $j=10$:

$j=10=\sum_{n=1}^{10}\frac{2n-1}{10}=10=\frac{1}{10}+\frac{3}{10}+\frac{5}{10}+\frac{7}{10}+\frac{9}{10}+\frac{11}{10}+\frac{13}{10}+\frac{15}{10}+\frac{17}{10}+\frac{19}{10}$
(4.2.2)
$7=\bigl{(}\sum_{n=1}^{10}\frac{2n-1}{10}\bigr{)}-\bigl{(}(\frac{1}{10}+\frac{9}{10})+(\frac{3}{10}+\frac{17}{10})\bigr{)}=\frac{5}{10}+\frac{7}{10}+\frac{11}{10}+\frac{13}{10}+\frac{15}{10}+\frac{19}{10}$
(4.2.3)

We see that the value $k=7$ is produced when fractions adding to one and two
are subtracted from the series that gives 10.

Example for $k=3$, where $j=10$

$\bigl{(}\sum_{n=1}^{10}\frac{2n-1}{10}\bigr{)}-\bigl{(}(\frac{1}{10}+\frac{9}{10})+(\frac{3}{10}+\frac{17}{10})+(\frac{5}{10}+\frac{15}{10})+(\frac{7}{10}+\frac{13}{10})\bigr{)}=\frac{11}{10}+\frac{19}{10}$
(4.2.4)

Rewriting the above statement, we find:
$10-(1+2+2+2)=\frac{11}{10}+\frac{19}{10}=3$. These two fractions remain at
the end because they are the two fractions that would have summed to two with
$\frac{9}{10}$ and $\frac{1}{10}$ respectively, but these were removed to
produce a pair adding to one. We thus see that if $j$ is even, three is the
smallest odd integer that can be produced by subtracting pairs that add to one
and two, as only two fractions remain, which when summed equal three.

###### Remark 4.2.1

We can see that for $j=2$:

$j=\sum_{n=1}^{2}\frac{2n-1}{2}=\frac{1}{2}+\frac{3}{2}$ (4.2.5)

It is not possible to write any integer other than two as the sum of these
fractions. (The integer one cannot be found as a sum, as there is only one
pair, which when summed gives two. We also note that neither fraction is an
integer.) It is only with $j\geq 3$ that we begin to have more than a single
pair of fractions in the series, or a term in the series that equals one,
thereby enabling one to find combinations for $k<j$. This then implies we must
require $j>2$. $\Box$

##  5 Proofs relating to unimodality and symmetry

###  5.1 Proofs for relevant lemmas

Before beginning the proof, we provide necessary background on the Gaussian
binomial coefficients. These are q-analogs of the usual binomial coefficients,
hence they are polynomials in $q$ that reduce to the usual binomial
coefficients when one takes the limit as $q$ approaches one. One can define a
Gaussian binomial coefficient as the following, where we note that it is
defined as zero when $h>j$ [10].

$\binom{j}{h}_{q}=\frac{(1-q)(1-q^{2})\cdots(1-q^{j})}{(1-q)(1-q^{2})\cdots(1-q^{h})(1-q)(1-q^{2})\cdots(1-q^{j-h})}$
(5.1.1)

These appear in problems such as the counting of lattice paths. In some cases
one would wish to set $q$ to a particular value, though its value is
irrelevant in our proofs.

###### Lemma 5.1.1

The nonzero terms of $f_{O_{j},h}(k)$ are unimodal.

Proof.  There is a known generating function for partitions of an integer $m$
into $h$ parts drawn from $\\{1,2,\ldots,j\\}$, where the exponent of $q$ is
$m$. See Section 1.6 of Aigner’s textbook for a discussion of this formula
[11]. The generating function of interest is

$q^{\binom{h+1}{2}}\binom{j}{h}_{q}$ (5.1.2)

This is a polynomial in $q$ that can be read in the same way as a typical
generating function.

In our problem we partition an integer $kj$ into $h$ terms drawn from the set
of the first $j$ odd integers. If one does not require $k$ to be an integer,
one can form a bijection by mapping each summand drawn from
$\\{1,2,\ldots,j\\}$ to the corresponding odd integer. We then see that the
number of partitions of the integer $\frac{h(h+1)}{2}+c$ into $h$ parts drawn
from $\\{1,2,\ldots,j\\}$ is equal to the number of partitions of $h^{2}+2c$
into $h$ parts from $\\{1,3,5,\ldots,2j-1\\}$, where $c\in\mathbb{N}$.

Say we have a set $A$ of $h$ terms from the set $\\{1,2,3,\ldots,j-1,j\\}$.
Now suppose $A$ has the same sum as another set $B$ of $h$ terms from
$\\{1,2,3,\ldots,j-1,j\\}$. If we replace each term of the two sets with the
corresponding odd integers, forming the sets $C$ and $D$, then we see that $C$
has the same sum as $D$. This is because we have multiplied each sum by $2$
before subtracting $h$. We then see that the coefficients of the generating
function for partitions into $h$ distinct parts from
$\\{1,2,3,\ldots,j-1,j\\}$ must be the same as the coefficients of the
generating function for partitions into $h$ distinct parts from
$\\{1,3,5,\ldots,2j-1\\}$, as each partition in the former case maps to one
partition in the latter case. This indicates that there is a bijection, where
the exact rule given in the previous paragraph readily follows from well-known
formulae for the sum of the first $h$ integers and the sum of the first $h$
odd integers.

For example, if $h=3$, we see that $1+4+6=1+2+8=11$. We also see that
$1+7+11=1+3+15=19=2(11)-3$, as we have mapped $1$ to the first odd integer,
$4$ to the fourth odd integer, $6$ to the sixth odd integer, et cetera. This
bijection only works because we have restricted the number of terms to $h$. We
cannot do this for $r(x)$, which makes working with it harder.

To illustrate this bijection, one can examine the following generating
functions, which correspond to partitions into $h$ parts drawn from the first
$j$ integers and the first $j$ odd integers, respectively. Note that the
exponent of $y$ is the number of parts.

$\displaystyle\prod_{n=1}^{j}(1+yx^{n})$ (5.1.3)
$\displaystyle\prod_{n=1}^{j}(1+yx^{2n-1})$ (5.1.4)

If one examines the terms that share a given exponent $h$ of $y$, one finds
that the two polynomials have the same coefficients. All that changes are the
exponents of $x$. For example, we have:

$\displaystyle\prod_{n=1}^{6}(1+yx^{n})=x^{21}y^{6}+x^{20}y^{5}+x^{19}y^{5}+x^{18}y^{5}+x^{18}y^{4}+x^{17}y^{5}+x^{17}y^{4}+x^{16}y^{5}+2x^{16}y^{4}$
(5.1.5)
$+x^{15}y^{5}+2x^{15}y^{4}+x^{15}y^{3}+3x^{14}y^{4}+x^{14}y^{3}+2x^{13}y^{4}+2x^{13}y^{3}+2x^{12}y^{4}+3x^{12}y^{3}+x^{11}y^{4}+3x^{11}y^{3}$
$+x^{11}y^{2}+x^{10}y^{4}+3x^{10}y^{3}+x^{10}y^{2}+3x^{9}y^{3}+2x^{9}y^{2}+2x^{8}y^{3}+2x^{8}y^{2}+x^{7}y^{3}+3x^{7}y^{2}+x^{6}y^{3}+2x^{6}y^{2}$
$+x^{6}y+2x^{5}y^{2}+x^{5}y+x^{4}y^{2}+x^{4}y+x^{3}y^{2}+x^{3}y+x^{2}y+xy+1$

When we pull out the terms containing $y^{2}$, hence corresponding to two
terms per partition, we have

$x^{11}+x^{10}+2x^{9}+2x^{8}+3x^{7}+2x^{6}+2x^{5}+x^{4}+x^{3}$ (5.1.6)

Repeating the process with the odd case:

$\displaystyle\prod_{n=1}^{6}(1+yx^{2n-1})=x^{36}y^{6}+x^{35}y^{5}+x^{33}y^{5}+x^{32}y^{4}+x^{31}y^{5}+x^{30}y^{4}+x^{29}y^{5}+2x^{28}y^{4}+x^{27}y^{5}$
(5.1.7)
$x^{27}y^{3}+2x^{26}y^{4}+x^{25}y^{5}+x^{25}y^{3}+3x^{24}y^{4}+2x^{23}y^{3}+2x^{22}y^{4}+3x^{21}y^{3}+2x^{20}y^{4}+x^{20}y^{2}+3x^{19}y^{3}$
$+x^{18}y^{4}+x^{18}y^{2}+3x^{17}y^{3}+x^{16}y^{4}+2x^{16}y^{2}+3x^{15}y^{3}+2x^{14}y^{2}+2x^{13}y^{3}+3x^{12}y^{2}+x^{11}y^{3}+x^{11}y$
$+2x^{10}y^{2}+x^{9}y^{3}+x^{9}y+2x^{8}y^{2}+x^{7}y+x^{6}y^{2}+x^{5}y+x^{4}y^{2}+x^{3}y+xy+1$

Pulling out terms containing $y^{2}$:

$x^{20}+x^{18}+2x^{16}+2x^{14}+3x^{12}+2x^{10}+2x^{8}+x^{6}+x^{4}$ (5.1.8)

Both sequences of coefficients follow those of

$q^{\binom{2+1}{2}}\binom{6}{2}_{q}=q^{11}+q^{10}+2q^{9}+2q^{8}+3q^{7}+2q^{6}+2q^{5}+q^{4}+q^{3}$
(5.1.9)

When we require $k$ to be an integer, we require each exponent of $x$ to be an
integral multiple of $6$, as $j=6$. After pulling those terms out from

$x^{20}+x^{18}+2x^{16}+2x^{14}+3x^{12}+2x^{10}+2x^{8}+x^{6}+x^{4}$ (5.1.10)

We have

$x^{18}+3x^{12}+x^{6}$ (5.1.11)

As expected, this is unimodal.

It is known that the Gaussian binomial coefficients are unimodal. A
constructive combinatorial proof was produced by O’Hara [11]. If one knows
that $\binom{j}{h}_{q}$ is unimodal, one sees that
$q^{\binom{h+1}{2}}\binom{j}{h}_{q}$ is unimodal as well, as multiplying each
term by the same power of $q$ does not affect unimodality. As we have a
bijection to our problem when $k$ is not necessarily an integer, we see we
have unimodality if $k$ is not required to be an integer. This generating
function corresponds to the terms of the following polynomial, where we only
examine terms with the same exponent of $y$.

$\displaystyle\prod_{n=1}^{j}(1+yx^{2n-1})$ (5.1.12)

The nonzero terms of $r_{h}(x)$ are the terms from the polynomial above where
the exponent of $x$ is $kj$, $k\in\mathbb{N}$. If one does not change the
order of the terms relative to one another, any polynomial composed of terms
taken from a unimodal polynomial must also be unimodal. We thus see that the
nonzero coefficients of $r_{h}(x)$ are unimodal. $\Box$

###### Lemma 5.1.2

The coefficients of $r_{h}(x)$ are symmetric about the coefficient
corresponding to $k=h$.

Proof.  Picture a group of $h$ terms that sum to some value $k_{1}$. If we
measure the distance of each term from the left end of the series for $j$,
then replace that term with a term located that many units from the right end
of the series, we find a group of terms that sum to $(2h-k_{1})$, hence
corresponding to $k=2h-k_{1}$. We observe that the values $k_{1}$ and
$2h-k_{1}$ are equidistant from $k=h$.

$\frac{2(n_{1})-1}{j}+\frac{2(n_{2})-1}{j}+\cdots+\frac{2(n_{h-1})-1}{j}+\frac{2(n_{h})-1}{j}=\frac{2(n_{1}+n_{2}+\cdots+n_{h})-h}{j}=k_{1}$
$\frac{2(j-n_{1}+1)-1}{j}+\frac{2(j-n_{2}+1)-1}{j}+\cdots+\frac{2(j-n_{h-1}+1)-1}{j}+\frac{2(j-n_{h}+1)-1}{j}$
$=\frac{2hj+h-2(n_{1}+n_{2}+\cdots+n_{h-1}+n_{h})}{j}=2h-k_{1}$ (5.1.13)

We observe that we are really reflecting the group of selected boxes shown
below over a vertical line located in the center of the series. Any
combination for $k_{1}$ can then be reflected about a central vertical line to
find a new combination for $2h-k_{1}$. As $k_{1}$ and $2h-k_{1}$ are
equidistant from $k=h$, we can then see that we have symmetry about $k=h$. We
write an example below, where the odd numerators are listed out. Boxes are
drawn around terms used in a sum. Here $k=5$, $j=8$, and $h=4$:

1 3 5 7 9 11 13 15

We see that $3+9+13+15=40=(5)(8)$. After reflecting over a central axis, we
have:

1 3 5 7 9 11 13 15

Note that $1+3+7+13=24=(3)(8)$, hence corresponding to $k=3$.

As we pointed out in the proof for unimodality that this problem is connected
to the Gaussian binomial coefficients, it appears that this proof for symmetry
can be combined with the bijection in the proof for unimodality to potentially
serve as an excessively long alternative proof that $\binom{j}{h}_{q}$ is
symmetric. The symmetry of Gaussian binomial coefficients is a well-known
result, though it is usually proven in a different, more direct way. $\Box$

###### Lemma 5.1.3

A group of $h$ consecutive terms from the center of the series for $j$ sums to
$k=h$. The first $\frac{h}{2}$ consecutive terms and the last $\frac{h}{2}$
consecutive terms together sum to $k=h$.

Proof.

$\frac{2(\frac{j-h}{2}+1)-1}{j}+\frac{2(\frac{j-h}{2}+2)-1}{j}+\cdots+\frac{2(\frac{j-h}{2}+h-1)-1}{j}+\frac{2(\frac{j-h}{2}+h)-1}{j}$
$=\frac{h(j-h)+h(h+1)-h}{j}=\frac{hj}{j}=h$ (5.1.14)

We thus see that a group of $h$ terms centered in the series for $j$ sums to
$k=h$.

$\frac{2(1)-1}{j}+\frac{2(2)-1}{j}+\cdots+\frac{2(\frac{h}{2})-1}{j}+\frac{2(j)-1}{j}+\frac{2(j-1)-1}{j}+\cdots+\frac{2j+1-h}{j}$
$=\frac{\frac{h}{2}(\frac{h}{2}+1)-\frac{h}{2}}{j}+\frac{2(j+(j-1)+(j-2)+\cdots+(j-\frac{h}{2}+1))-\frac{h}{2}}{j}$
$=\frac{\frac{h}{2}(\frac{h}{2}+1)-\frac{h}{2}}{j}+\frac{hj-(\frac{h}{2}-1)\frac{h}{2}-\frac{h}{2}}{j}=\frac{hj}{j}=h$
(5.1.15)

This then indicates that the first $\frac{h}{2}$ terms and the last
$\frac{h}{2}$ terms together sum to $k=h$. $\Box$

###  5.2 Proof for Theorem 1.0.2

Proof.  By Lemmas 5.1.1 and 5.1.2 the nonzero coefficients of $r_{h}(x)$ are
unimodal and symmetric about the coefficient corresponding to $k=h$. These
results alone do not prove that the maximum nonzero coefficient of $r_{h}(x)$
is $f_{O_{j},h}(h)$, though, as one must dismiss the scenario where
$f_{O_{j},h}(h)=0$ but $f_{O_{j},h}(k)$ is nonzero for values of $k$
equidistant from $h$.

We see that there is at least one partition for $k=h$ when both $h$ and $j$
are even, or both $h$ and $j$ are odd, as in these cases it is possible to
find a centered group of $h$ consecutive terms from the series for $j$, which
by Lemma 5.1.3 sum to $h$. If $h$ is even and $j$ is odd, by Lemma 5.1.3 one
can take the first $\frac{h}{2}$ terms and the last $\frac{h}{2}$ terms from
the series for $j$, which also sum to $k=h$. If $h$ is odd and $j$ is even,
one cannot find a centered group of $h$ terms or two groups of $\frac{h}{2}$
terms. This is not a problem, though, as no $k$ can be found as the sum of an
odd number of terms when $j$ is even. The sum of an odd number of odd integers
is odd, though $j$ is even, hence an odd sum divided by an even $j$ cannot
give an integral $k$. We thus have symmetry about $k=h$, see that
$f_{O_{j},h}(k)$ is unimodal, and know that we have at least one partition for
$k=h$ when there are partitions for any integer $k$, where $0<k<j$. This then
indicates that $k=h$ must have the most partitions with a given $j$, or none
at all, in which case every value of $f_{O_{j},h}(k)$ is zero with that given
$j$. $\Box$

###  5.3 Proof for proposition 1.0.3

Proof.  Suppose we consider two $k$-values of $k_{1}$ and $(j-k_{1})$, where
we note that $k_{1}$ and $(j-k_{1})$ are equidistant from the ends of the
sequence $\\{1,2,3,\ldots,j-1\\}$. We know that each combination for $k_{1}$
is a portion of the larger series for $j$, thus any combination for the
$k$-value of $(j-k_{1})$ is what is left over whenever each possible
combination for $k_{1}$ is removed from the larger series for $j$. This then
necessitates that $k_{1}$ and $(j-k_{1})$ must have the same number of
combinations, as if there are $M$ combinations for $k_{1}$, then there are
only $M$ ways to subtract $k_{1}$ from $j$, thus causing there to also be $M$
combinations for $(j-k_{1})$. We then have
$f_{O_{j}}(k_{1})=f_{O_{j}}(j-k_{1})\forall k_{1}<j$. $\Box$

##  6 Even numerators

###  6.1 Proof for Theorem 1.0.5

We note that the sum of the first $x$ even integers can be given by

$\sum_{n=1}^{x}2n=x^{2}+x$ (6.1.1)

We can thus transform our problem into counting solutions for a given $t$ to
the following Diophantine equation, where $x,y,t\in\mathbb{N}$ and $0<y<x<t$.
Note that the restriction $0<y<x<t$ serves to prevent one from being able to
find obvious solutions that hold for all $t$.

$\frac{x^{2}+x}{y}=t$ (6.1.2)

For matters of convenience that will later become clear, it is useful to alter
our restrictions within the majority of our proofs, temporarily inserting two
trivial solutions. In fact, if we alter our restrictions to $0<x\leq t$ and
$0<y\leq t+1$, we only add the trivial solutions of $x=t$ and $x=t-1$, which
are solutions for every $t$.

###### Lemma 6.1.1

There exist exactly two integral solutions to $t=\frac{x^{2}+x}{y}$ that
satisfy $0<x\leq t$ and $0<y\leq(t+1)$ but not $0<y<x<t$.

Proof.  Let $y>(x+1)$, thus $\frac{x+1}{y}<1$. We then see
$\frac{x(x+1)}{y}=t<x$, which is not allowed by our new restriction $0<x\leq
t$. We see that $y$ can only equal $x+1$ when $t=x$, which is the trivial
solution of $y=t+1$. The variable $y$ can only equal $x$ when $t$ is $x+1$,
which is the trivial solution $y=t-1$. $y$ cannot equal $t$, as this would
imply that $t^{2}=x(x+1)$. For a natural number to be a perfect square, its
prime factors must all appear an even number of times. Consecutive natural
numbers share no prime factors, thus both $x$ and $x+1$ would have to be
perfect squares, which is impossible. We then see that all remaining solutions
fit the restriction $0<y<x<t$, hence the two trivial solutions are the only
points added by the restriction change. $\Box$

###### Lemma 6.1.2

The number of solutions for a given $t$ is equal to the product of the number
of solutions for each $p_{i}^{j}$, where we have the prime factorization
$t=p_{i_{1}}^{j_{1}}p_{i_{2}}^{j_{2}}p_{i_{3}}^{j_{3}}\cdots$.

Proof.  We see that we can rewrite our problem as

$x^{2}+x\equiv 0\bmod t$ (6.1.3)

It is known that if one has a polynomial $f(x)$, where

$f(x)\equiv 0\bmod t$ (6.1.4)

the number of solutions for a given $t$ can be given by the product of the
number of solutions to each

$f(x)\equiv 0\bmod p_{i}^{j}$ (6.1.5)

where we have the prime factorization
$t=p_{i_{1}}^{j_{1}}p_{i_{2}}^{j_{2}}p_{i_{3}}^{j_{3}}\cdots$. This is a
consequence of the ring isomorphism of the Chinese Remainder Theorem. $\Box$

###### Lemma 6.1.3

Let $x$ be a positive integer, where $x\neq t-1$ and $x$ is a solution to
$\frac{x^{2}+x}{y}=t$, subject to the restrictions $0<x\leq t$ and
$0<y\leq(t+1)$. $x$ must then be a multiple of at least one prime factor of
$t$.

Proof.  We observe that $x$ and $x+1$ are relatively prime, thus they share no
prime factors. If $x$ and $t$ share no prime factors, $y$ must cancel all
prime factors of $x$. We then have $y=Ax$, where $A$ is a natural number. If
$A=1$, $x=t-1$, which is a trivial solution.

Let $A>1$:

$t=\frac{x(x+1)}{Ax}=\frac{x+1}{A}$ (6.1.6)

We thus see that $t<x$, which is not allowed by our restriction $0<x\leq t$.
This then indicates that the only way for $t$ and $x$ to share no prime
factors is if $x=t-1$, hence indicating that all other $x$ one can find as
solutions must share at least one prime factor with $t$. $\Box$

###### Lemma 6.1.4

If $t$ is prime, then there are only two solutions.

Proof.  We know that $x$ and $t$ must share at least one prime factor, or none
at all, where if they share none, then $x=t-1$. If $t$ is prime, it only has
one prime factor, so $x$ must be $t-1$ or an integral multiple of $t$. Let
$a\in\mathbb{Z}$, $a>0$, where $x=at$. If $a=1$, we have $x=t$ and $y=t+1$,
which is one of our two trivial solutions.

$\frac{x^{2}+x}{y}=\frac{t^{2}+t}{t+1}=t$ (6.1.7)

If $a>1$, we see that $y$ must increase, though that would mean $y>t+1$, which
our restrictions $0<x\leq t$ and $0<y\leq t+1$ forbid. We thus have only two
solutions when $t$ is prime, where these two solutions are the trivial
solutions. $\Box$

###### Lemma 6.1.5

Let $p$ be a prime number, and $n$ a natural number. The number of solutions
for $p$ is the same as the number of solutions for $p^{n}$.

Proof.  We can split $x^{2}+x$ into its factors, hence enabling us to look at
the following two subproblems:

$g(x)=x\equiv 0\bmod p$ (6.1.8) $h(x)=x+1\equiv 0\bmod p$ (6.1.9)

The integer one is not a prime, thus we see that

$\frac{dg}{dx}=1\not\equiv 0\bmod p$ (6.1.10) $\frac{dh}{dx}=1\not\equiv
0\bmod p$ (6.1.11)

By Hensel’s Lemma, there are unique solutions to $x\equiv 0\bmod p$ and
$x+1\equiv 0\bmod p$. The polynomial $x^{2}+x$ has no repeated factors. We
also note that $x$ and $x+1$ are relatively prime, hence this indicates that
there are only two solutions for $x^{2}+x\equiv 0\bmod p^{n}$, where we see
those solutions are $x=p^{n}$ and $x=p^{n}-1$. In other words, we have the
same number of solutions for $t=p$ and $t=p^{n}$. $\Box$

We are now in a position to combine our lemmas to prove the theorem.

Proof.  We know that the number of solutions to

$x^{2}+x\equiv 0\bmod t$ (6.1.12)

is equal to the product of the number of solutions to each

$x^{2}+x\equiv 0\bmod p_{i}^{j}$ (6.1.13)

where the prime factorization of $t$ is
$t=p_{i_{1}}^{j_{1}}p_{i_{2}}^{j_{2}}p_{i_{3}}^{j_{3}}\cdots$. As there are
two solutions for $t=p^{n}$, we see that there are $2^{\omega(t)}$ solutions
for

$\frac{x^{2}+x}{y}=t$ (6.1.14)

where $0<x\leq t$ and $0<y\leq t+1$, $x,y,t\in\mathbb{N}$. As this restriction
change adds two trivial solutions, we thus see that for the restriction
$0<y<x<t$, we have $2^{\omega(t)}-2$ solutions, hence
$F_{E}(t)=2^{\omega(t)}-2$. $\Box$

###  6.2 Proof for corollary 1.0.6

Proof.  In this section we provide a proof for the continuation we give in
Corollary 1.0.6. We apply a slightly different take on the circle method and
the concept of a generating function to achieve this. In the following
subsection, we use part of this process as a stepping stone to produce the
series identity in Corollary 1.0.7. If we examine our Diophantine equation, we
see that if we have found a solution, we have

$\bigg{(}\frac{x^{2}+x}{y}-t\bigg{)}=0$ (6.2.1)

Shifting our restrictions from $0<y<x<t$ to $0<x\leq t$ and $0<y\leq t+1$ is
convenient in part because it removes the need for $y$ to always be less than
$x$. We can then eliminate $y$ by taking a product over its range of values.
We see that whenever we find a solution, one of the factors equals zero, thus
causing the entire polynomial to equal zero.

$\displaystyle\prod_{y=1}^{t+1}\bigg{(}\frac{x^{2}+x}{y}-t\bigg{)}=0$ (6.2.2)

After multiplying both sides by $(t+1)!$, we can simplify to

$\Psi_{t}=\displaystyle\prod_{y=1}^{t+1}(x^{2}+x-yt)=0$ (6.2.3)

This is a family of polynomials in $x$, as $y$ is an index within the product
and $t$ becomes some positive index that is constant within this product. We
thus have a different polynomial $\Psi_{t}$ for each value of $t$. In a
partition problem, one would normally work with the coefficients of the
generating function, which is a polynomial, though instead we accomplish the
same sort of thing by counting the positive integral roots of a different
polynomial, which we call $\Psi_{t}$. This is a bit more natural for our
partition problem than a typical, non-Dirichlet generating function. For
comparison, we write the generating function for this problem below.

$F_{E}(t_{1})x^{t_{1}}+F_{E}(t_{1})x^{t_{2}}+F_{E}(t_{3})x^{t_{3}}+F_{E}(t_{4})x^{t_{4}}+\cdots$
(6.2.4)

We have a different function $\Psi_{t}$ associated with each $t$, where
$\Psi_{t}$ is a function of $x$ alone, hence the number of positive integral
roots of $\Psi_{t}$ is a function of $t$. We count the number of positive
integral roots with the function $R(t)$. As we have added two trivial
solutions, we see that $R(t)=F_{E}(t)+2$. Each coefficient in the generating
function is then two less than the number of positive integral roots of
$\Psi_{t}$, where one uses the appropriate value of $t$. For example, when
$t=6$, we have

$\Psi_{6}=x^{14}+7x^{13}-147x^{12}-973x^{11}+9107x^{10}+54621x^{9}-309953x^{8}$
(6.2.5)
$-1578527x^{7}+6290256x^{6}+24636024x^{5}-76219920x^{4}-195456240x^{3}$
$+507586176x^{2}+609700608x-1410877440$

We see that the positive integral roots are $x=2$, $x=3$, $x=5$, and $x=6$.
These are the values of $x$ that are solutions when $t=6$, where we include
the trivial solutions $x=5$ and $x=6$.

By Theorem 1.0.5, we know that $R(t)=2^{\omega(t)}$. We now examine the
following integral. We place $\Psi_{t}$ in the exponent of $z$, where the
coefficient of $z$ is one. This may seem odd, though it becomes convenient
later. Note that $\gamma$ is the unit circle in the complex plane, oriented
counterclockwise.

$y,t,x\in\mathbb{Z}\therefore\frac{1}{2\pi
i}\int_{\gamma}z^{\Psi_{t}-1}dz\in\\{0,1\\}$ (6.2.6)

We have a solution when $\Psi_{t}=0$. This leaves the exponent of $z$ as
negative one, hence causing the integral to evaluate as one. As the
coefficient of $z$ is always exactly one, we have no convergence issues from
the integral itself, as it is always identically zero or one. Rather than
divide the circle into major and minor arcs, we instead repeatedly integrate
over the undivided circle. As mentioned previously, we refer to the number of
roots of $\Psi_{t}$ as $R(t)$.

$R(t)=\sum_{x=1}^{t}\frac{1}{2\pi i}\int_{\gamma}z^{\Psi_{t}-1}dz$ (6.2.7)

We observe that we can rewrite the integral, where $\operatorname{sinc}$ is
the normalized $\operatorname{sinc}$ function:

$\sum_{x=1}^{t}\frac{1}{2\pi
i}\int_{\gamma}z^{\Psi_{t}-1}dz=\sum_{x=1}^{t}\operatorname{sinc}(\Psi_{t})$
(6.2.8)

This works because $\operatorname{sinc}(m)=1$ when $m=0$, and zero when $m$ is
a nonzero integer. We observe that since $\Psi_{t}$ is really a product of
polynomials, where $x$, $y$, and $t$ are integers, $\Psi_{t}$ must output an
integer for every value in our problem.

Figure 1: Plot of the normalized function $\operatorname{sinc}(x)$

We thus have:

$2^{\omega(t)}=\bigg{(}\sum_{x=1}^{t}\operatorname{sinc}\bigg{(}\displaystyle\prod_{y=1}^{t+1}(x^{2}+x-yt)\bigg{)}\bigg{)}$
(6.2.9)

To extend the definition of this function to non-integral inputs and inputs
with an imaginary part, we alter the definition, where we see that the
expression still correctly counts the number of distinct prime factors of the
positive integers. We rename $t$ to $z$ to highlight that in general, the
input of this function is complex.

$\omega(z)=\log_{2}\bigg{(}\sum_{x=1}^{\left\lceil{\Re(z)}\right\rceil}\operatorname{sinc}\bigg{(}\displaystyle\prod_{y=1}^{\left\lceil{\Re(z)}\right\rceil+1}(x^{2}+x-yz)\bigg{)}\bigg{)}$
(6.2.10)

We thus have a continuation of the prime omega function, though we note that
the inclusion of the ceiling function prevents it from being analytic
everywhere. This enables us to define $\omega(z)$ for inputs that are not
positive integers. This is somewhat similar to the practice of using the gamma
function to assign a value to expressions such as the factorial of
$\frac{1}{2}$, which otherwise would be absurd. For example, we can “count”
the number of distinct prime factors of numbers such as $\pi$ and $e$. Such
numbers obviously have no prime factors, though it is interesting to see that
we can assign them a quantity of such factors by extending $\omega(z)$ to the
complex plane.

$\omega(\pi)\approx-9.9287$ (6.2.11) $\omega(e)\approx-6.0963+4.5323i$
(6.2.12) $\omega(4+i)\approx 181729.6967-0.0798i$ (6.2.13)

$\Box$

###  6.3 Proof for corollary 1.0.7

Proof.  We observe the following, where the expression
$\bigg{(}1-\frac{f(x)}{t}\bigg{)}_{t+1}$ is a pochhammer symbol.

$\displaystyle\prod_{n=1}^{n+1}(f(x)-nt)=(-t)^{t+1}\bigg{(}1-\frac{f(x)}{t}\bigg{)}_{t+1}$
(6.3.1) $(x)_{n}=\frac{\Gamma(x+n)}{\Gamma(x)}$ (6.3.2)

Hence when $x$ and $t$ are integers, we have

$\Psi_{t}=\displaystyle\prod_{y=1}^{t+1}(x^{2}+x-yt)=\frac{(-t)^{t+1}\Gamma(\frac{-x^{2}}{t}-\frac{x}{t}+t+2)}{\Gamma(\frac{-x^{2}}{t}-\frac{x}{t}+1)}$
(6.3.3)

We now write a Dirichlet series $D(s)$:

$D(s)=\sum_{t=1}^{\infty}\frac{a_{t}}{t^{s}}$ (6.3.4)
$a_{t}=\sum_{x=1}^{t}(\operatorname{sinc}(\Psi_{t}))$ (6.3.5)

We specify that $s$ is a real number greater than two. Since real numbers are
complex numbers with an imaginary part of zero, $s$ is still complex.

$D(s)=\sum_{t=1}^{\infty}\sum_{x=1}^{t}\frac{1}{2\pi
i}\int_{\gamma}\frac{z^{\Psi_{t}-1}}{t^{s}}dz=\sum_{t=1}^{\infty}\sum_{x=1}^{t}\frac{\operatorname{sinc}(\Psi_{t})}{t^{s}}$
(6.3.6)

The coefficient $a_{t}$ increases by one every time we find a positive
integral root of $\Psi_{t}$ using that value of $t$. $\Psi_{t}$ is a
polynomial of degree $2t+2$, thus it cannot have more than $2t+2$ such roots
for a given $t$. In our case, we cannot have more than one solution for each
value of $x$ with a given $t$, where $0<x\leq t$, hence we have at most $t$
solutions. We thus have the following comparison:

$\sum_{t=1}^{\infty}\sum_{x=1}^{t}\frac{\operatorname{sinc}(\Psi_{t})}{t^{s}}\leq\sum_{t=1}^{\infty}\frac{t}{t^{s}}$
(6.3.7)

As $s$ is a real number greater than two, we see that the series on the right
is a convergent p-series.

$\sum_{t=1}^{\infty}\frac{t}{t^{s}}=\sum_{t=1}^{\infty}\frac{1}{t^{s-1}}$
(6.3.8)

As $D(s)$ is always less than or equal to a convergent series, it must also
converge.

The following Dirichlet series is known [12].

$\sum_{n=1}^{\infty}\frac{2^{\omega(n)}}{n^{s}}=\frac{\zeta^{2}(s)}{\zeta(2s)}$
(6.3.9)

We thus have

$\sum_{t=1}^{\infty}\sum_{x=1}^{t}\frac{\operatorname{sinc}(\Psi_{t})}{t^{s}}=\frac{\zeta^{2}(s)}{\zeta(2s)}$
(6.3.10)

Hence

$\sum_{t=1}^{\infty}\sum_{x=1}^{t}\frac{\operatorname{sinc}\bigg{(}\frac{(-t)^{t+1}\Gamma(\frac{-x^{2}}{t}-\frac{x}{t}+t+2)}{\Gamma(\frac{-x^{2}}{t}-\frac{x}{t}+1)}\bigg{)}}{t^{s}}=\frac{\zeta^{2}(s)}{\zeta(2s)}$
(6.3.11)

$\Box$

##  7 Proof for the special case of restriction to two terms

We divide this proof into sections, focusing on each value of $k$ in turn. We
prove that the terms of the table corresponding to $h=2$ (given on page 4)
must always follow certain simple patterns, which then lead to the closed-form
expressions given in Section 3.

###### Lemma 7.0.1

When $k=1$, we have
$f_{O_{j},2}(k)=\frac{1}{8}\bigl{(}(-1)^{1+n}-1+(-i)^{n}+i^{n}+(-1)^{n}n+n\bigr{)}$,
where $n=j$.

Proof.  An odd value of $j$ cannot have any combination of two terms that add
to $k=1$, as the sum of any two odd numbers is even, and thus unequal to the
odd denominator. (Any pairs of numerators chosen must sum to the denominator
to produce $k=1$.) This then results in the alternating zeros found with the
column for $k=1$. We know that any even value of $j$ has at least one pair of
terms that sum to $k=1$, as we proved in Section 4 that the terms given by
$n=1$ and $n=\frac{j}{2}$ sum to this value. If two is subtracted from one of
the resulting numerators, then added to the other, the sum does not change.
The numerators in the series for $j$ are consecutive odd integers, thus they
differ from their neighbors by two. We thus find that the number of pairs that
can sum to one for an even $j$ is equal to the number of pairs equidistant
from $n=1$ and $n=\frac{j}{2}$, as we can move inwards from those terms to
produce more combinations that work. This results in the pattern
$\\{0,1,0,1,0,2,0,2,0,3,0,\ldots\\}$ seen for $k=1$.

Example for $j=8$:

$8=\frac{1}{8}+\frac{3}{8}+\frac{5}{8}+\frac{7}{8}+\frac{9}{8}+\frac{11}{8}+\frac{13}{8}+\frac{15}{8}$
(7.0.1)

Pulling out the section of the series from $n=1$ to $n=\frac{j}{2}$ we have:

$\frac{1}{8}+\frac{3}{8}+\frac{5}{8}+\frac{7}{8}$ (7.0.2)

We see that there are two pairs equidistant from the ends of this section of
the series. Each of these sum to one.

$1=\frac{1}{8}+\frac{7}{8}$ (7.0.3) $1=\frac{3}{8}+\frac{5}{8}$ (7.0.4)

The numerator $3=1+2$, and the numerator $5=7-2$. The number of these pairs
increases by one for every other even $j$, therefore we get the pattern seen
in the table. Through an argument nearly identical to that used to prove Lemma
7.0.2, one can see that only pairs equidistant from the terms given by $n=1$
and $n=\frac{j}{2}$ can sum to one.

We thus know that for $k=1$, $f_{O_{j},2}(k)$ must always follow the simple
pattern seen in the first column of the table. We can see that this pattern is
generated by the formula
$\frac{1}{8}\bigl{(}(-1)^{1+n}-1+(-i)^{n}+i^{n}+(-1)^{n}n+n\bigr{)}$, hence
$f_{O_{j},2}(k)$ has this closed-form expression when $k=1$. $\Box$

###### Lemma 7.0.2

Only pairs of terms equidistant from the ends of the series for $j$ can sum to
two.

Proof.  Suppose there are two positive odd integers, $a$ and $b$, where

$a+b=kj$ (7.0.5) $a<b\leq(2j-1)$ (7.0.6)

This thus requires $a$ and $b$ to be numerators within the series for $j$,
where the fractions in question sum to $k$, as the denominators are $j$.
Suppose there are two other positive odd integers, $c$ and $d$, where:

$c+d=kj$ (7.0.7) $c<d<(2j-1)$ (7.0.8)

For $c+d$ to equal $a+b$, there must exist an integer $Q$ such that:

$c=a+Q$ (7.0.9) $d=b-Q$ (7.0.10)

If we begin with a pair of integers, $a$ and $b$, where $a$ and $b$ are
numerators equidistant from the ends of the series for $j$, we know they sum
to $2j$, as we proved in Section 4 that the respective fractions these
numerators are a part of sum to two. Any other pair, $c$ and $d$, where
$c=a+Q$ and $d=b-Q$ must also be equidistant from the ends of the series for
$j$. We then know that such pairs are the only pairs that can sum to $kj$,
where $k=2$. $\Box$

###### Lemma 7.0.3

When $k=2$ we have $f_{O_{j},2}(k)=\frac{1}{4}\bigl{(}(-1)^{n}+2n-1\bigr{)}$,
where $n=j$.

Proof.

We proved in Section 4 that pairs of terms equidistant from the ends of the
series for $j$ always sum to two. By Lemma 7.0.2, we also know that such pairs
are the only pairs that can sum to two. When $j$ is even, there are
$\frac{j}{2}$ pairs equidistant from the ends of the series for $j$, though
when $j$ is odd we can find $\frac{(j-1)}{2}$ equidistant pairs. We then see
that an even $j$ and the following odd $j$ have the same number of pairs that
sum to two. This produces the pattern seen where each number of combinations
is displayed twice.

We thus know that for $k=2$, $f_{O_{j},2}(k)$ always follows the pattern
$\\{1,2,2,3,3,4,4,\ldots\\}$ seen in the second column of the table. We can
see that this pattern is generated by the formula
$f_{O_{j},2}(k)=\frac{1}{4}\bigl{(}(-1)^{n}+2n-1\bigr{)}$, hence
$f_{O_{j},2}(k)$ has this closed-form expression when $k=2$. $\Box$

###### Lemma 7.0.4

When $k=3$, we have
$f_{O_{j},2}(k)=\frac{1}{8}\bigl{(}(-1)^{1+n}-1+(-i)^{n}+i^{n}+(-1)^{n}n+n\bigr{)}$,
where $n=j$.

Proof.

By Lemma 5.1.2, symmetry implies that $f_{O_{j},2}(1)=f_{O_{j},2}(3)$. $\Box$

###### Lemma 7.0.5

If $k>3$, $f_{O_{j},2}(k)=0$.

Proof.

Suppose it is possible to find $h$ terms from the series for $j$ that sum to a
$k$-value greater than or equal to $2h$, where $h>1$. This then implies the
following statement, where $q\geq 0$.

$\left(\frac{2n_{1}-1}{j}+\frac{2n_{2}-1}{j}+\cdots+\frac{2n_{h-1}-1}{j}+\frac{2n_{h}-1}{j}\right)=(2h+q)$
(7.0.11) $2\bigl{(}n_{1}+n_{2}+\cdots+n_{h-1}+n_{h}\bigr{)}=(2h+q)j+h$
(7.0.12)

We know that $n_{1}$, $n_{2}$, …, $n_{h-1}$, $n_{h}$ are all positive integers
less than or equal to $j$, where at most one is equal to $j$, thus:

$n_{1}=j-a_{1},\ n_{2}=j-a_{2},\ \cdots n_{h-1}=j-a_{h-1},\ n_{h}=j-a_{h}$
(7.0.13)

The variables $a_{1}$, $a_{2}$, …, $a_{h-1}$, $a_{h}$ are positive integers
with at most one equaling zero.

$a_{1},a_{2},\cdots,a_{h-1},$ and $a_{h}$ are distinct.

These constants are distinct to prevent any term from being repeated. No term
can be repeated because no term in the series for $j$ is repeated, thus no
portion of the larger series can contain repeated terms. Summing the
$n$-values, we then find

$hj-\bigl{(}a_{1}+a_{2}+\cdots+a_{h-1}+a_{h}\bigr{)}=\left(h+\frac{q}{2}\right)j+\frac{1}{2}h$
(7.0.14) $h>1\ \therefore\ \bigl{(}a_{1}+a_{2}+\cdots+a_{h-1}+a_{h}\bigr{)}>0$
(7.0.15) $hj-\bigl{(}a_{1}+a_{2}+\cdots+a_{h-1}+a_{h}\bigr{)}<hj$ (7.0.16)
$\left(\left(h+\frac{q}{2}\right)j+\frac{1}{2}h\right)>hj$ (7.0.17)
$hj-\bigl{(}a_{1}+a_{2}+\cdots+a_{h-1}+a_{h}\bigr{)}\neq\left(\left(h+\frac{q}{2}\right)j+\frac{1}{2}h\right)$
(7.0.18)
$\left(\frac{2n_{1}-1}{j}+\frac{2n_{2}-1}{j}+\cdots+\frac{2n_{h-1}-1}{j}+\frac{2n_{h}-1}{j}\right)\neq(2h+q)$
(7.0.19)

We can conclude that if combinations are restricted to $h$ terms, where $h>1$,
then it is impossible to find any combination for a value of $k$ greater than
or equal to $2h$. If $k>3$ where $h=2$, then $k\geq 2h$, thus there cannot be
any combination of $2$ terms for $k>3$. We have thus proven that the case of
$h=2$ must always follow the simple patterns visible in its number table.
These patterns can then be produced by the closed-form expressions we give.
$\Box$

We now see that we can combine these lemmas to prove our result.

Proof.  Combining Lemma 7.0.1, Lemma 7.0.2, Lemma 7.0.3, Lemma 7.0.4, and
Lemma 7.0.5 we see that $f_{O_{j},2}(k)$ follows the closed-form expressions
given in Section 3. $\Box$

##  8 More examples of number tables

To generate the number table for $r(x)$, the MATLAB script prompts the user
for the maximum $j$-value, then creates a counter to keep track of what row it
is on, where each row corresponds to a different $j$-value. The script finds
all $k$-values for each $j$-value, then loads these into a vector. It finds
the first $j$ odd integers, then applies the built-in nchoosek function to
find every possible combination of $q$ integers from set of the first $j$
consecutive odd numbers, where $q<j$. The script loads the combinations into a
matrix. We repeat this process for every $q$ smaller than $j$, where the
script checks each combination of $q$ odd integers to see if they sum to the
product of $k$ and $j$. This is because the first $j$ consecutive odd integers
are numerators from the series for $j$, thus if the combination is valid, it
will sum to the product of $k$ and $j$. Each time the script finds a
combination that works, it increases a counter by one, then prints the total
number of combinations for each $k$-value in the appropriate position in a
number table. All loops are while loops. We apply a similar process to produce
tables for $r_{h}(x)$, except the user is asked to input $h$, where $h$ is the
fixed number of terms per partition. The program then only allows values of
$q$ equal to this number.

Note that in the tables below, nonzero terms are written in bold to make the
patterns easier to read. Values for $j$ are given in the column on the far
left; values for $k$ are given across the first row.

###  8.1 Restriction to three terms:

Values for $f_{O_{j},3}(k)$
---
$j$ $\hskip 5.69046pt|$ | $k$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$ | $11$ | $12$ | $13$ | $14$
3 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
4 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
5 |  | 0 | 0 | 2 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
6 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
7 |  | 0 | 0 | 5 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
8 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
9 |  | 1 | 0 | 8 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
10 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
11 |  | 1 | 0 | 13 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
12 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
13 |  | 2 | 0 | 18 | 0 | 2 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
14 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
15 |  | 3 | 0 | 25 | 0 | 3 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0

###  8.2 Restriction to four terms:

Values for $f_{O_{j},4}(k)$
---
$j$ $\hskip 5.69046pt|$ | $k$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$ | $11$ | $12$ | $13$ | $14$
3 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
4 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
5 |  | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
6 |  | 0 | 0 | 1 | 3 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
7 |  | 0 | 0 | 0 | 5 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
8 |  | 0 | 1 | 5 | 8 | 5 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
9 |  | 0 | 1 | 0 | 12 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
10 |  | 0 | 2 | 10 | 18 | 10 | 2 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
11 |  | 0 | 3 | 0 | 24 | 0 | 3 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
12 |  | 0 | 5 | 21 | 33 | 21 | 5 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
13 |  | 0 | 6 | 0 | 43 | 0 | 6 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
14 |  | 0 | 9 | 35 | 55 | 35 | 9 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
15 |  | 0 | 11 | 0 | 69 | 0 | 11 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0

###  8.3 Restriction to five terms:

Values for $f_{O_{j},5}(k)$
---
$j$ $\hskip 5.69046pt|$ | $k$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$ | $11$ | $12$ | $13$ | $14$
3 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
4 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
5 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
6 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
7 |  | 0 | 0 | 0 | 0 | 3 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
8 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
9 |  | 0 | 0 | 1 | 0 | 12 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0
10 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
11 |  | 0 | 0 | 5 | 0 | 32 | 0 | 5 | 0 | 0 | 0 | 0 | 0 | 0 | 0
12 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
13 |  | 0 | 0 | 13 | 0 | 73 | 0 | 13 | 0 | 0 | 0 | 0 | 0 | 0 | 0
14 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
15 |  | 0 | 0 | 30 | 0 | 141 | 0 | 30 | 0 | 0 | 0 | 0 | 0 | 0 | 0

###  8.4 Restriction to six terms:

Values for $f_{O_{j},6}(k)$
---
$j$ $\hskip 5.69046pt|$ | $k$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$ | $11$ | $12$ | $13$ | $14$
3 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
4 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
5 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
6 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
7 |  | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
8 |  | 0 | 0 | 0 | 0 | 2 | 4 | 2 | 0 | 0 | 0 | 0 | 0 | 0 | 0
9 |  | 0 | 0 | 0 | 1 | 0 | 8 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0
10 |  | 0 | 0 | 0 | 2 | 10 | 18 | 10 | 2 | 0 | 0 | 0 | 0 | 0 | 0
11 |  | 0 | 0 | 0 | 5 | 0 | 32 | 0 | 5 | 0 | 0 | 0 | 0 | 0 | 0
12 |  | 0 | 0 | 1 | 11 | 39 | 58 | 39 | 11 | 1 | 0 | 0 | 0 | 0 | 0
13 |  | 0 | 0 | 0 | 19 | 0 | 94 | 0 | 19 | 0 | 0 | 0 | 0 | 0 | 0
14 |  | 0 | 0 | 3 | 33 | 103 | 151 | 103 | 33 | 3 | 0 | 0 | 0 | 0 | 0
15 |  | 0 | 0 | 0 | 54 | 0 | 227 | 0 | 54 | 0 | 0 | 0 | 0 | 0 | 0

###  8.5 Restriction to seven terms:

Values for $f_{O_{j},7}(k)$
---
$j$ $\hskip 5.69046pt|$ | $k$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$ | $11$ | $12$ | $13$ | $14$
3 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
4 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
5 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
6 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
7 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
8 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
9 |  | 0 | 0 | 0 | 0 | 0 | 0 | 4 | 0 | 0 | 0 | 0 | 0 | 0 | 0
10 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
11 |  | 0 | 0 | 0 | 0 | 3 | 0 | 24 | 0 | 3 | 0 | 0 | 0 | 0 | 0
12 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
13 |  | 0 | 0 | 0 | 0 | 19 | 0 | 94 | 0 | 19 | 0 | 0 | 0 | 0 | 0
14 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
15 |  | 0 | 0 | 0 | 0 | 70 | 0 | 289 | 0 | 70 | 0 | 0 | 0 | 0 | 0

###  8.6 Restriction to eight terms:

Values for $f_{O_{j},8}(k)$
---
$j$ $\hskip 5.69046pt|$ | $k$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ | $9$ | $10$ | $11$ | $12$ | $13$ | $14$
3 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
4 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
5 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
6 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
7 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
8 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0
9 |  | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 0
10 |  | 0 | 0 | 0 | 0 | 0 | 0 | 2 | 5 | 2 | 0 | 0 | 0 | 0 | 0
11 |  | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 13 | 0 | 1 | 0 | 0 | 0 | 0
12 |  | 0 | 0 | 0 | 0 | 0 | 5 | 21 | 33 | 21 | 5 | 0 | 0 | 0 | 0
13 |  | 0 | 0 | 0 | 0 | 0 | 13 | 0 | 73 | 0 | 13 | 0 | 0 | 0 | 0
14 |  | 0 | 0 | 0 | 0 | 3 | 33 | 103 | 151 | 103 | 33 | 3 | 0 | 0 | 0
15 |  | 0 | 0 | 0 | 0 | 0 | 70 | 0 | 289 | 0 | 70 | 0 | 0 | 0 | 0

Upon inspection, we see that sequences such as the third column for
restriction to four terms and the fifth column for restriction to five terms
are not listed in the OEIS.

##  9 Further computational results:

###  9.1 Partitioning series with even numerators into integers

Using a MATLAB script, we studied the series corresponding to integral points
on the surface $t=\frac{x^{2}+x}{y}$ to find what integers said series can be
partitioned into. Not all can be partitioned into every integer smaller than
$t$, though this is frequently the case, especially for longer series. Often
there are multiple combinations for a given integer. We include the results
for only three series below, as most others would require too much space to be
practical to show.

Series:

$6=\frac{2}{2}+\frac{4}{2}+\frac{6}{2}$

Partitions:

$1=\frac{2}{2}\hskip 14.22636pt2=\frac{4}{2}\hskip
14.22636pt3=\frac{6}{2}\hskip 14.22636pt3=\frac{2}{2}+\frac{4}{2}\hskip
14.22636pt4=\frac{2}{2}+\frac{6}{2}\hskip 14.22636pt5=\frac{4}{2}+\frac{6}{2}$

Series:

$10=\frac{2}{2}+\frac{4}{2}+\frac{6}{2}+\frac{8}{2}$

Partitions:

$1=\frac{2}{2}\hskip 14.22636pt2=\frac{4}{2}\hskip
14.22636pt3=\frac{6}{2}\hskip 14.22636pt3=\frac{2}{2}+\frac{4}{2}\hskip
14.22636pt4=\frac{8}{2}\hskip 14.22636pt4=\frac{2}{2}+\frac{6}{2}\hskip
14.22636pt5=\frac{2}{2}+\frac{8}{2}\hskip 14.22636pt5=\frac{4}{2}+\frac{6}{2}$
$6=\frac{4}{2}+\frac{8}{2}\hskip
14.22636pt6=\frac{2}{2}+\frac{4}{2}+\frac{6}{2}\hskip
14.22636pt7=\frac{6}{2}+\frac{8}{2}\hskip
14.22636pt7=\frac{2}{2}+\frac{4}{2}+\frac{8}{2}\hskip
14.22636pt8=\frac{2}{2}+\frac{6}{2}+\frac{8}{2}\hskip
14.22636pt9=\frac{4}{2}+\frac{6}{2}+\frac{8}{2}$

Series:

$10=\frac{2}{3}+\frac{4}{3}+\frac{6}{3}+\frac{8}{3}+\frac{10}{3}$

Partitions:

$2=\frac{6}{3}\hskip 14.22636pt2=\frac{2}{3}+\frac{4}{3}\hskip
14.22636pt4=\frac{2}{3}+\frac{10}{3}\hskip
14.22636pt4=\frac{4}{3}+\frac{8}{3}\hskip
14.22636pt4=\frac{2}{3}+\frac{4}{3}+\frac{6}{3}\hskip
14.22636pt6=\frac{8}{3}+\frac{10}{3}$
$6=\frac{2}{3}+\frac{6}{3}+\frac{10}{3}\hskip
14.22636pt6=\frac{4}{3}+\frac{6}{3}+\frac{8}{3}\hskip
14.22636pt8=\frac{6}{3}+\frac{8}{3}+\frac{10}{3}\hskip
14.22636pt8=\frac{2}{3}+\frac{4}{3}+\frac{8}{3}+\frac{10}{3}$

##  10 Connection to Diophantine Equations

Diophantine equations have proved interesting in part due to the difficulty in
predicting when integral solutions occur, and if so, how many exist.
Matiyasevich proved that there is no general algorithm to predict when these
solutions occur [13], though that does not mean we cannot work with specific
cases. Andrew Wiles’ proof of Fermat’s Last Theorem is a famous example of
this fact [14, 15]. Just as even numerator series can be connected to finding
integral points on a surface that is really a nonlinear Diophantine equation,
odd numerator series can also be connected to such an equation. The case for
odd numerators is different though, as it contains a variable number of
unknowns.

We find that it is possible to predict whether the equation
$2(n_{1}+n_{2}+\cdots+n_{h-1}+n_{h})-(kj+h)=0$ can have integral solutions. We
see that in certain situations we can also predict how many there are. This
particular class of equations is interesting not only because it is nonlinear,
but also because the total number of unknowns is dependent on the value of the
variable $h$.

###### Proposition 10.0.1

Suppose we have a Diophantine equation of the form

$2(n_{1}+n_{2}+\cdots+n_{h-1}+n_{h})-(kj+h)=0$ (10.0.1)

where $n_{1}$, $n_{2},\ldots,n_{h-1}$, $n_{h}$, $j$, $k$, and $h$ are
variables with positive integral values,

$k<j$

$n_{1},n_{2},\cdots,n_{h-1},$ and $n_{h}$ are distinct

$n_{1},n_{2},\ldots,n_{h-1},n_{h}\leq j$

There is never a solution if $j<3$ or $k\geq 2h$. If $0<k<2h$ and $j>2$, there
can be a solution, but it is not guaranteed. If there are $M$ solutions when
using a given $k_{1}$ and $j$, where $k_{1}<h<j$, then there are also $M$
solutions for $k_{2}=2h-k_{1}$, where $j$ is fixed.

Proof.  When we fix the number of terms per combination to $h$, we are really
working with a Diophantine equation of the form:

$\left(\frac{2n_{1}-1}{j}+\frac{2n_{2}-1}{j}+\cdots+\frac{2n_{h-1}-1}{j}+\frac{2n_{h}-1}{j}\right)=k$
(10.0.2) $\frac{2\bigr{(}n_{1}+n_{2}+\cdots+n_{h-1}+n_{h}\bigr{)}-h}{j}=k$
(10.0.3)

$n_{1},n_{2},\cdots,n_{h-1},$ and $n_{h}$ are distinct.

$n_{1},n_{2},\ \ldots,\ n_{h-1},n_{h}\leq j$ $k<j,\ j>2$

We can rewrite as:

$2(n_{1}+n_{2}+\cdots+n_{h-1}+n_{h})-(kj+h)=0$ (10.0.4)

This equation is then nonlinear because it includes the product of two
variables.

Proposition 4.0.1 stipulates that we must have $j>2$ and $k<j$ for there to be
any combinations for $k$, regardless of the number of terms used. In Section 7
we proved that there is never a solution if $k\geq 2h$, where $h>1$. This is
also true when $h=1$. All numerators are less than or equal to $2j-1$, hence
no $k\geq 2$ can be found using one term. We know that $k$ must be positive,
as all fractions in the series for $j$ are positive. We thus know that for
there to be a possibility of a solution, $0<k<2h$. In our proof for symmetry,
we found that if there are $M$ solutions for $k_{1}$, where $k_{1}<h<j$, then
there are also $M$ solutions for $k_{2}=2h-k_{1}$. Using this result, we then
have $f_{O_{j},h}(k_{1})=f_{O_{j},h}(k_{2})$. $\Box$

## References

  * [1] A. Sills, Rademacher-type formulas for restricted partition and overpartition functions, Ramanujan J., 23 (2010), 253–264, available online at the URL: https://doi.org/10.1007/s11139-009-9184-y
  * [2] S. Ramanujan, Some properties of p(n), the number of partitions of n, Math. Proc. Cambridge Philos. Soc., 19 (1919), 207–210.
  * [3] M. Bidar, Partition of an integer into distinct bounded parts, identities and bounds, Integers, 12 (2012), available online at the URL: http://math.colgate.edu/ integers/m8/m8.pdf
  * [4] P. Hagis, Jr, Partitions into odd and unequal parts, Amer. J. Math., 86 (1964), 317–324, available online at the URL: https://www.jstor.com/stable/2373167
  * [5] D. Christopher, Partitions with fixed number of sizes, J. Integer Seq., 18 (2015), available online at the URL: https://cs.uwaterloo.ca/journals/JIS/VOL18/Christopher/chris7.pdf
  * [6] B. Riemann, Über die anzahl der primzahlen unter einer gegebenen grösse, Monatsberichte der Berliner Akademie, 7 (1859), 136–144.
  * [7] A. Anggoro, E. Liu, and A. Tulloch, The rascal triangle, College Math. J., 41 (2010), 393–395, available online at the URL: https://www.maa.org/sites/default/files/Anggoro2010.pdf
  * [8] N.J.A. Sloane, Sequence A077028, The On-Line Encyclopedia of Integer Sequences, available online at the URL: https://oeis.org
  * [9] N.J.A. Sloane, Sequence A004526, The On-Line Encyclopedia of Integer Sequences, available online at the URL: https://oeis.org
  * [10] K. O’Hara, Unimodality of Gaussian coefficients: A constructive proof, _J. Combin. Theory Ser. A_ , 53 (1990), 29-52, available online at the URL: https://doi.org/10.1016/0097-3165(90)90018-R
  * [11] M. Aigner, _Graduate Texts in Mathematics: A Course in Enumeration_ , Springer-Verlag, (2007).
  * [12] L. Tóth, Counting solutions of quadratic congruences in several variables revisited, J. of Integer Seq., 17 (2014), available online at the URL: https://cs.uwaterloo.ca/journals/JIS/VOL17/Toth/toth12.html
  * [13] Y. Matiyasevich, Hilbert’s Tenth Problem, MIT Press, (1993).
  * [14] A. Wiles, Modular elliptic curves and Fermat’s Last Theorem, Ann. of Math., 141 (1995), 443–551, available online at the URL: https://www.jstor.org/stable/2118559
  * [15] A. Wiles and R. Taylor, Ring-theoretic properties of certain Hecke algebras, Ann. of Math., 141 (1995), 553–572, available online at the URL: https://www.jstor.org/stable/2118560

Zachary Hoelscher
Virginia Polytechnic Institute and State University
Blacksburg, Virginia 24061
E-mail<EMAIL_ADDRESS>

Eyvindur Palsson
Virginia Polytechnic Institute and State University
Blacksburg, Virginia 24061
E-mail<EMAIL_ADDRESS>