# Syracuse Random Variables and the Periodic Points of Collatz-type maps

Maxwell C. Siegel University of Southern California, Dornsife School of
Letters & Sciences, Department of Mathematics. Department of Mathematics 3620
S. Vermont Ave., KAP 104 Los Angeles, CA 90089-2532. E-mail<EMAIL_ADDRESS>Declarations of interest: none

(Monday, July 13, 2020)

###### Abstract

Let $p$ be an odd prime, and consider the map $H_{p}$ which sends an integer
$x$ to either $\frac{x}{2}$ or $\frac{px+1}{2}$ depending on whether $x$ is
even or odd. The values at $x=0$ of arbitrary composition sequences of the
maps $\frac{x}{2}$ and $\frac{px+1}{2}$ can be parameterized over the $2$-adic
integers ($\mathbb{Z}_{2}$) leading to a continuous function from
$\mathbb{Z}_{2}$ to $\mathbb{Z}_{p}$ which the author calls the “numen” of
$H_{p}$, denoted $\chi_{p}$; the $p=3$ case turns out to be an alternative
version of the “Syracuse Random Variables” constructed by Tao[10]. This paper
establishes the “Correspondence Theorem”, which shows that an odd integer
$\omega$ is a periodic point of $H_{p}$ if and only if
$\omega=\chi_{p}\left(n\right)/\left(1-r_{p}\left(n\right)\right)$ for some
integer $n\geq 1$, where
$r_{p}\left(n\right)=p^{\\#_{1}\left(n\right)}/2^{\lambda\left(n\right)}$,
where $\\#_{1}\left(n\right)$ is the number of $1$s digits in the binary
expansion of $n$ and $\lambda\left(n\right)$ is the number of digits in the
binary expansion of $n$. Using this fact, the bulk of the paper is devoted to
examining the Dirichlet series associated to $\chi_{p}$ and $r_{p}$, which are
used along with Perron’s Formula to reformulate the Correspondence Theorem in
terms of contour integrals, to which Residue calculus is applied so as to
obtain asymptotic formulae for the quantities therein. A sampling of other
minor results on $\chi_{p}$ are also discussed in the paper’s final section.

Keywords: Collatz Conjecture, $p$-adic analysis, $2$-adic integers, Mellin
transform, parity vector, contour integration, Pontryagin duality, $px+1$ map,
$3x+1$ map, $5x+1$ map, Bayesian inference.

###### Contents

  1. 1 The Numen $\chi_{p}$ and the Correspondence Theorem
    1. 1.1 Construction and Characterization of $\chi_{p}$
    2. 1.2 The Correspondence Theorem
  2. 2 Our Toolbox
    1. 2.1 Complex-Analytic Methods
    2. 2.2 Rising-continuity and functions from $\mathbb{Z}_{p}$ to $\mathbb{Z}_{q}$
  3. 3 Asymptotics, Blancmanges, and Contour Integrals
    1. 3.1 Blancmange Curves, $\sum_{n=1}^{N}r_{p}\left(n\right)$, & $\sum_{n=1}^{N}\chi_{p}\left(n\right)$
      1. 3.1.1 Pictures at an Exhibition
      2. 3.1.2 The Function $\zeta_{p}$
      3. 3.1.3 The summatory function of $\chi_{p}$
    2. 3.2 Integral Criteria for Periodic Points
  4. 4 Miscellaneous Results
    1. 4.1 A Lipschitz-type estimate for $\chi_{p}$
    2. 4.2 The Probabilistic Approach and Bayesian Inference
    3. 4.3 An Archimedean Upper Bound for $\chi_{p}$ on certain subsets of $\mathbb{Z}_{2}$

###### List of Figures

  1. 1 The blancmange curve $T_{1/2}\left(x\right)$ on $\left[-1,1\right]$
  2. 2 $\textrm{Bl}\left(x\right)$, the $\\#_{1}$-summatory blancmange
  3. 3 $\textrm{Bl}_{3}\left(x\right)$, the $r_{3}$-summatory blancmange
  4. 4 Graph of a segment of $\tilde{\textrm{Bl}}_{3}\left(x\right)$, with blue regions showing the values of $x$ for which $r_{3}\left(x\right)<1$.
  5. 5 Log-scale graphs of $\left|\tilde{\textrm{Bl}}_{3}\left(x\right)\right|$ (red), $\left|\tilde{\textrm{Bl}}_{5}\left(x\right)\right|$ (blue), and $\left|\tilde{\textrm{Bl}}_{7}\left(x\right)\right|$ (green)

This research did not receive any specific grant from funding agencies in the
public, commercial, or not-for-profit sectors.

## Introduction

Let $p$ be an odd prime, and consider the map
$H_{p}:\mathbb{N}_{0}\rightarrow\mathbb{N}_{0}$ (where
$\mathbb{N}_{k}\overset{\textrm{def}}{=}\left\\{n\in\mathbb{Z}:x\geq
k\right\\}$) defined by:

$H_{p}\left(x\right)\overset{\textrm{def}}{=}\begin{cases}\frac{x}{2}&\textrm{if
}n\overset{2}{\equiv}0\\\ \frac{px+1}{2}&\textrm{if
}n\overset{2}{\equiv}1\end{cases}$ (1)

where $a\overset{r}{\equiv}b$ denotes the congruence of $a$ and $b$ mod $r$.
In 1978, Böhm and Sontacchi [11] proved what has since become a well-known
result in the Collatz literature: the fact that every integer periodic point
of the Collatz map _must_ be of the form:

$\sum_{k=1}^{n}\frac{2^{m-b_{k}-1}3^{k-1}}{2^{m}-3^{n}}$ (2)

for some integers $m,n\geq 1$ and some strictly increasing sequence of
positive integers $b_{1}<b_{2}<\cdots<b_{n}$. The use of probabilistic
techniques to study the Collatz map’s dynamics is by no means new. Tao’s 2019
paper on the subject, for examples, attack 2 by treating the $b_{k}$s as
random variables. A second probabilistic approach is to study the parity
vector of a given input $x$ under an $H_{p}$ map; this is the sequence
$\left[x\right]_{2},\left[H_{p}\left(x\right)\right]_{2},\left[H_{p}^{\circ
2}\left(x\right)\right]_{2},\ldots$ of the values mod $2$ of the iterates of
$x$ under $H_{p}$. [10]

The present paper came about as a result of the author’s idea to subordinate
the input $x$ to its parity vector. In other words, rather than asking “given
$x$, what is $x$’s parity vector under $H_{p}$”, we consider the question
“given a parity vector, what $x$ could have generated it?” Doing so quickly
leads to a method of parameterizing $x$ over the space of all possible parity
vectors, which we identify with $\mathbb{Z}_{2}$, the space of $2$-adic
integers.

Our method is as follows. First, we define the maps:

$\displaystyle h_{0}\left(x\right)$ $\displaystyle\overset{\textrm{def}}{=}$
$\displaystyle\frac{x}{2}$ $\displaystyle h_{1}\left(x\right)$
$\displaystyle\overset{\textrm{def}}{=}$ $\displaystyle\frac{px+1}{2}$

(the branches of $H_{p}$). Next, for each $n\in\mathbb{N}_{1}$, let
$\left\\{0,1\right\\}^{n}$ denote the set of all $n$-tuples whose entries are
$0$s and $1$s. We denote an arbitrary such tuple as
$\mathbf{j}=\left(j_{1},\ldots,j_{n}\right)$, and we say $\mathbf{j}$ is non-
zero if $\mathbf{j}$ contains at least one $1$. Given any such tuple, we write
$\left|\mathbf{j}\right|$ to denote the length of $\mathbf{j}$, so that any
tuple can be written as
$\mathbf{j}=\left(j_{1},\ldots,j_{\left|\mathbf{j}\right|}\right)$. For any
integer $m\geq 1$, we write $\mathbf{j}^{m}$ to denote the tuple obtained by
concatenating $m$ copies of $\mathbf{j}$.

Writing:

$\mathcal{J}\overset{\textrm{def}}{=}\bigcup_{n=1}^{\infty}\left\\{0,1\right\\}^{n}$
(3)

to denote the set of all tuples of $0$s and $1$s of finite length, we then
consider the composition sequence $h_{\mathbf{j}}$ defined by:

$h_{\mathbf{j}}=h_{j_{1}}\circ\cdots\circ
h_{j_{\left|\mathbf{j}\right|}},\textrm{ }\forall\mathbf{j}\in\mathcal{J}$ (4)

Note the near-equivalence of $\mathbf{j}$ with parity vectors. Indeed, given
any $x,n\in\mathbb{N}_{1}$, there exists a unique $\mathbf{j}\in\mathcal{J}$
of length $n$ so that $H_{p}^{\circ
n}\left(x\right)=h_{\mathbf{j}}\left(x\right)$. Said $\mathbf{j}$ is the
parity vector for the first $n$ iterates of $x$ under $H_{p}$, albeit written
in reverse order.

Since the $h_{j}$s are affine linear maps (that is, of the form $ax+b$), so
any composition sequence $h_{\mathbf{j}}$. As such, for any
$\mathbf{j}\in\mathcal{J}$, we have:

$h_{\mathbf{j}}\left(x\right)=h_{\mathbf{j}}^{\prime}\left(0\right)x+h_{\mathbf{j}}\left(0\right)=\frac{p^{\\#_{1}\left(\mathbf{j}\right)}}{2^{\left|\mathbf{j}\right|}}x+h_{\mathbf{j}}\left(0\right),\textrm{
}\forall\mathbf{j}\in\mathcal{J}$ (5)

where:

$\\#_{1}\left(\mathbf{j}\right)\overset{\textrm{def}}{=}\textrm{number of
}1\textrm{s in }\mathbf{j}$ (6)

Observing that the quantity
$p^{\\#_{1}\left(\mathbf{j}\right)}/2^{\left|\mathbf{j}\right|}$ has a
$p$-adic magnitude of $p^{-\\#_{1}\left(\mathbf{j}\right)}$—which tends to
zero in $\mathbb{Z}_{p}$ whenever
$\\#_{1}\left(\mathbf{j}\right)\rightarrow\infty$, we then have that:

$\lim_{\\#_{1}\left(\mathbf{j}\right)\rightarrow\infty}h_{\mathbf{j}}\left(x\right)\overset{\mathbb{Z}_{p}}{=}\lim_{\\#_{1}\left(\mathbf{j}\right)\rightarrow\infty}h_{\mathbf{j}}\left(0\right),\textrm{
}\forall x\in\mathbb{Z}$

In letting the number of $1$s in $\mathbf{j}$ tends to infinity, we can
identify $\mathbf{j}$ with the $2$-adic integer $\mathfrak{z}$ whose $2$-adic
digits are precisely the entries of $\mathbf{j}$. Since the above limit is
independent of the value of $x$, this allows us to realize the map
$\mathbf{j}\mapsto h_{\mathbf{j}}\left(0\right)$ as a function
$\chi_{p}:\mathbb{Z}_{2}\rightarrow\mathbb{Z}_{p}$ by letting the length of
$\mathbf{j}$ tend to infinity; the author calls $\chi_{p}$ the numen of
$H_{p}$.

It is worth noting that this approach has the peculiar distinction of being
neither novel nor well-known. While the author independently formulated this
approach in October of 2019, at around the same time, Tao published a paper
[14] on the Collatz Conjecture, focusing on the analysis of what he called
“Syracuse random variables”. Tao’s Syracuse random variables and the author’s
numen are essentially one and the same. Indeed, equation 1.3 on page 5 of
Tao’s paper is equivalent to 5. Tao mentions the fact that his Syracuse random
variables can be unified into a single $3$-adic-valued random variable—the
author’s $\chi_{3}$—but his paper chooses not to adopt that point of view.
Another noteworthy difference in our approaches is that whereas Tao builds the
Syracuse random variables using a purely probabilistic approach, treating them
as collections of geometrically distributed random variables, the present
paper treats the unified Syracuse random variable (and its corresponding
generalizations for the other $px+1$ maps) explicitly parameterizes these
functions in terms of the properties of the binary representations of
integers. There are other similarities in the two approaches, to be discussed
in Section 4.

While Tao’s work already establishes the importance of the Syracuse random
variables—and hence, of $\chi_{p}$—to understanding the Collatz Conjecture and
its relatives, the driving force behind the present paper’s analysis is based
off the following relation which, to all appearances, seems to have gone
unnoticed. $\chi_{p}$ is intimately related to Böhm and Sontacchi’s formula 2
for the periodic points of $H_{3}$. As established in Section 1 of this paper
(Lemma 3), 2 turns out to be equivalent to the functional equation$\chi_{3}$:

$\chi_{p}\left(B\left(t\right)\right)\overset{\mathbb{Z}_{p}}{=}\frac{\chi_{p}\left(t\right)}{1-r_{p}\left(t\right)},\textrm{
}\forall t\in\mathbb{N}_{1}$ (7)

where:

$\displaystyle B\left(t\right)$
$\displaystyle=\frac{t}{1-2^{\lambda\left(t\right)}}$ (8) $\displaystyle
r_{p}\left(t\right)$
$\displaystyle=\frac{p^{\\#_{1}\left(t\right)}}{2^{\lambda\left(t\right)}}$
(9) $\displaystyle\\#_{1}\left(t\right)$ $\displaystyle=\textrm{number of
}1\textrm{s in the binary expansion of }t\in\mathbb{N}_{1}$ (10)
$\displaystyle\lambda\left(t\right)$ $\displaystyle=\textrm{total number of
digits in the binary expansion of }t\in\mathbb{N}_{1}$ (11)

The principal result of Section 1 (which the author terms the Correspondence
Theorem) is that:

$\chi_{p}\left(\mathbb{Z}_{2}\right)\cap\mathbb{Z}=\left\\{\chi_{p}\left(B\left(t\right)\right):t\in\mathbb{N}_{0}\right\\}\cap\mathbb{Z}=\left\\{\textrm{periodic
points of }H_{p}\textrm{ in }\mathbb{Z}\right\\}$ (12)

As such, $\chi_{p}$ presents us with a means of attacking the problem of
determining the periodic points of the $H_{p}$ maps. Beyond this fundamental
result, the numen itself (along with related objects, especially $r_{p}$) has
the advantage of providing a unified foundation—both archimedean
($\mathbb{N}_{0}$, $\mathbb{Z}$, viewed as subsets of $\mathbb{R}$) and non
($\chi_{p}\left(B\left(\mathbb{N}_{0}\right)\right)\subseteq\mathbb{Q}$ viewed
as a subset of $\mathbb{Z}_{p}$) for studying both the $H_{p}$ maps’ dynamics
and the more general number-theoretic properties implicated therein.

While the author has obtained a smorgasbord of various interesting minor
results (which shall be detailed in Section 4), the principal aim of this
paper—Section 3—is to use classical complex-analytic Mellin transform methods
to obtain an “exact formula” (a.k.a., generalized Fourier series) for 7. The
essence of this approach is to use the Dirichlet series associated to the
functions $\chi_{3}$ and $r_{3}$ to restate the Correspondence Theorem in
terms of a contour integral 135; namely, an odd integer $\omega$ is a periodic
point of the Collatz map if and only if there is an integer $n\geq 1$ so that:

$\frac{1}{2\pi
i}\int_{2-i\infty}^{2+i\infty}\frac{\kappa_{n}\left(s\right)\mathcal{C}_{3,\omega}\left(s\right)}{s\left(s+1\right)^{2}}ds=\frac{1}{2}$

where:

$\displaystyle C_{3,\omega}\left(s\right)$
$\displaystyle=\mathcal{B}_{3}\left(s\right)-\frac{1}{\omega}\Xi_{3}\left(s\right)$
$\displaystyle\mathcal{B}_{3}\left(s\right)$
$\displaystyle=\sum_{n=0}^{\infty}\frac{\frac{3}{2}-r_{3}\left(n\right)}{\left(n+1\right)^{s}}$
$\displaystyle\Xi_{3}\left(s\right)$
$\displaystyle=\sum_{n=0}^{\infty}\frac{\chi_{3}\left(n\right)}{\left(n+1\right)^{s}}$

and where:

$\displaystyle\kappa_{n}\left(s\right)$
$\displaystyle=n\left(\left(n+2\right)^{s+1}-3\left(n+1\right)^{s+1}+3n^{s+1}-\left(n-1\right)^{s+1}\right)$
$\displaystyle+\left(n+2\right)^{s+1}-\left(n+1\right)^{s+1}-\left(n^{s+1}-\left(n-1\right)^{s+1}\right)$

By shifting the contour of integration in 135 from
$\textrm{Re}\left(s\right)=2$ to $\textrm{Re}\left(s\right)=-1/4$, we obtain
an expression $R_{3}\left(\omega,n\right)$ 143 so that:

$\frac{1}{2\pi
i}\int_{2-i\infty}^{2+i\infty}\frac{\kappa_{n}\left(s\right)C_{3,\omega}\left(s\right)}{s\left(s+1\right)^{2}}ds=R_{3}\left(\omega,n\right)+O\left(n^{3/4}\right)$

_Note_ : here, we adopt the admitted abuse of notation by writing $\mathbf{j}$
and $\mathbf{j}^{m}$ in place of $\beta\left(\mathbf{j}\right)$ and
$\beta\left(\mathbf{j}^{m}\right)$; so, $\chi_{3}\left(\mathbf{j}\right)$, for
instance, means $\chi_{3}$ evaluated at the integer
$\beta\left(\mathbf{j}\right)$ whose sequence of binary digits are exactly the
entries of $\mathbf{j}$.

Definition: Let $X_{3}^{+}$ denote the set of all positive integers $n$ for
which $r_{3}\left(n\right)>1$; equivalently, we say that a tuple
$\mathbf{j}\in\mathcal{J}$ is in $X_{3}^{+}$ if and only if
$r_{3}\left(\mathbf{j}\right)\overset{\textrm{def}}{=}r_{3}\left(\beta\left(\mathbf{j}\right)\right)>1$.

Theorem I (Contour integral criteria for periodic points of the Collatz Map):
An odd integer $\omega$ is a periodic point of the Collatz map if and only if
there is a non-zero $\mathbf{j}\in\mathcal{J}$ so that:

$\frac{1}{2}=\frac{1}{2\pi
i}\int_{2-i\infty}^{2+i\infty}\frac{\kappa_{\mathbf{j}^{m}}\left(s\right)C_{3,\omega}\left(s\right)}{s\left(s+1\right)^{2}}ds,\textrm{
}\forall m\in\mathbb{N}_{1}$ (13)

Shifting the integral from $\textrm{Re}\left(s\right)=2$ to
$\textrm{Re}\left(s\right)=-1/4$ and taking residues into account gives the
equivalent condition:

$\frac{1}{2}=R_{3}\left(\omega,\mathbf{j}^{m}\right)+\frac{1}{2\pi
i}\int_{-\frac{1}{4}-i\infty}^{-\frac{1}{4}+i\infty}\frac{\kappa_{\mathbf{j}^{m}}\left(s\right)C_{3,\omega}\left(s\right)}{s\left(s+1\right)^{2}}ds,\textrm{
}\forall m\in\mathbb{N}_{1}$ (14)

Theorem II (Asymptotic and limit formulae : For all
$\omega\in\mathbb{C}\backslash\left\\{0\right\\}$, all
$\mathbf{j}\in\mathcal{J}$, and all $m\in\mathbb{N}_{1}$:

$\chi_{3}\left(B\left(\mathbf{j}\right)\right)=\frac{3\omega}{2}+\left(\chi_{3}\left(B\left(\mathbf{j}\right)\right)-\omega\right)r_{3}^{m}\left(\mathbf{j}\right)-\frac{\omega}{2\pi
i}\int_{2-i\infty}^{2+i\infty}\frac{\kappa_{\mathbf{j}^{m}}\left(s\right)C_{3,\omega}\left(s\right)}{s\left(s+1\right)^{2}}ds$
(15)

Shifting the integral from $\textrm{Re}\left(s\right)=2$ to
$\textrm{Re}\left(s\right)=-1/4$ and taking residues into account gives:

$\chi_{3}\left(B\left(\mathbf{j}\right)\right)=\frac{3\omega}{2}+\left(\chi_{3}\left(B\left(\mathbf{j}\right)\right)-\omega\right)r_{3}^{m}\left(\mathbf{j}\right)-\omega
R_{3}\left(\omega,\mathbf{j}^{m}\right)-\frac{\omega}{2\pi
i}\int_{-\frac{1}{4}-i\infty}^{-\frac{1}{4}+i\infty}\frac{\kappa_{\mathbf{j}^{m}}\left(s\right)C_{3,\omega}\left(s\right)}{s\left(s+1\right)^{2}}ds$

In particular, in both these formulae, note that since
$r_{3}\left(\mathbf{j}\right)<1$ if and only if $\mathbf{j}\in X_{3}^{+}$, for
such $\mathbf{j}$, we can take the limit as $m\rightarrow\infty$ to get:

$\chi_{3}\left(B\left(\mathbf{j}\right)\right)=\frac{3\omega}{2}-\omega\lim_{m\rightarrow\infty}\left(R_{3}\left(\omega,\mathbf{j}^{m}\right)+\frac{1}{2\pi
i}\int_{-\frac{1}{4}-i\infty}^{-\frac{1}{4}+i\infty}\frac{\kappa_{\mathbf{j}^{m}}\left(s\right)C_{3,\omega}\left(s\right)}{s\left(s+1\right)^{2}}ds\right)$

In all of these formulae, we have the asymptotic estimate:

$\left|\int_{-\frac{1}{4}-i\infty}^{-\frac{1}{4}+i\infty}\frac{\kappa_{\mathbf{j}^{m}}\left(s\right)C_{3,\omega}\left(s\right)}{s\left(s+1\right)^{2}}ds\right|\ll\left(\beta\left(\mathbf{j}^{m}\right)\right)^{3/4}$
(16)

where the constant depends only on $\omega$.

Theorem III (Explicit Formula for $R_{3}$): The following formula holds for
all $\omega\in\mathbb{C}\backslash\left\\{0\right\\}$ for all
$n\in\mathbb{N}_{2}$:

$\displaystyle R_{3}\left(\omega,n\right)$
$\displaystyle=-\frac{1}{\omega}\frac{1}{\ln 2}\sum_{k\neq
0}\frac{\zeta\left(1+\frac{2k\pi i}{\ln 2}\right)+F\left(1+\frac{2k\pi i}{\ln
2}\right)}{\left(1+\frac{2k\pi i}{\ln 2}\right)\left(2+\frac{2k\pi i}{\ln
2}\right)^{2}}\underbrace{\kappa_{n}\left(1+\frac{2k\pi i}{\ln
2}\right)}_{O\left(n\right)}$ (17) $\displaystyle-\frac{1}{4\ln 2}\sum_{k\neq
0}\frac{G\left(1+\frac{2k\pi i}{\ln 2}\right)}{\left(1+\frac{2k\pi i}{\ln
2}\right)\left(2+\frac{2k\pi i}{\ln
2}\right)^{2}}\underbrace{\kappa_{n}\left(1+\frac{2k\pi i}{\ln
2}\right)}_{O\left(n\right)}$
$\displaystyle+\frac{3}{2}-\frac{G\left(1\right)}{4\ln
2}+\frac{1}{\omega}\frac{1}{\ln 2}\left(2-\gamma+\frac{\ln 2}{2}+\frac{4+\ln
2}{2}F\left(1\right)-F^{\prime}\left(1\right)\right)$
$\displaystyle-\frac{1}{\omega}\frac{1+F\left(1\right)}{4\ln
2}\underbrace{\kappa_{n}^{\prime}\left(1\right)}_{O\left(\ln
n\right)}+\frac{1}{\omega}\frac{2}{\ln
2}\underbrace{\kappa_{n}^{\prime}\left(0\right)}_{O\left(1/n\right)}$
$\displaystyle+\frac{1}{\omega}\frac{2}{\ln 2}\sum_{k\neq
0}\frac{\zeta\left(1+\frac{2k\pi i}{\ln 2}\right)+F\left(1+\frac{2k\pi i}{\ln
2}\right)}{\left(1+\frac{2k\pi i}{\ln
2}\right)^{2}}\underbrace{\kappa_{n}\left(\frac{2k\pi i}{\ln
2}\right)}_{O\left(1\right)}$ $\displaystyle+\frac{1}{16\ln 2}\sum_{k\neq
0}\frac{G\left(1+\frac{2k\pi i}{\ln 2}\right)}{\left(1+\frac{2k\pi i}{\ln
2}\right)^{2}}\underbrace{\kappa_{n}\left(\frac{2k\pi i}{\ln
2}\right)}_{O\left(1\right)}$

Here:

$F\left(s\right)\overset{\textrm{def}}{=}\sum_{n=1}^{\infty}\left(\frac{1}{2}\right)^{n}\binom{s+n-1}{n}\Xi_{3}\left(s+n\right)$

$G\left(s\right)\overset{\textrm{def}}{=}\sum_{n=1}^{\infty}\left(\frac{1}{2}\right)^{n}\binom{s+n-1}{n}\zeta_{3}\left(s+n\right)$

Outline of the Paper:

Section 1 constructs the numen $\chi_{p}$ and proves a variety of its
properties, culminating with the Correspondence Theorem.

Section 2 consists of a review of the pertinent tools from analytical number
theory (Perron’s formula, etc.) as well as a handful of technical results, all
to be utilized in Sections 3 and 4. This includes some bits of $p$-adic
analysis and $p$-adic Fourier analysis, particularly as regards the van der
Put basis for continuous functions from $\mathbb{Z}_{p}$ to $\mathbb{Z}_{q}$
where $p$ and $q$ are distinct primes.

Section 3 analyzes the summatory functions of $r_{p}$ and $\chi_{p}$, as well
as the ensuing Dirichlet series ($\zeta_{3}$ and $\Xi_{3}$, respectively).
Asymptotics for the summatory functions are established, functional equations
and analytic continuations are derived for the aforementioned Dirichlet
series, and the Correspondence Theorem from Section 1 is reformulated as a
contour integral. Using standard estimates on the growth rate of the Riemann
Zeta Function (RZF) along vertical lines in the complex plane, similar growth
rates are established for $\zeta_{3}$ and $\Xi_{3}$. Using these asymptotics,
we can justify shifting the contour of integration in our reformulation of the
Correspondence Theorem and use residues to compute an asymptotic formula for
the integral in question, which we give in full.

Section 4 features an array of different minor results obtained by the author.
First, a Lipschitz-type estimate for the $p$-adic absolute value of
$\chi_{p}$. Second, the author gives his version of the probabilistic
formulation given by Tao in [14]. This is accompanied by a crude but hopefully
interesting inequality providing a necessary condition for an odd integer
$\omega$ to be a periodic point of $H_{p}$; namely, roughly speaking, such a
periodic point has the property that the probability that $\chi_{p}$ is
congruent to $\omega$ modulo $p^{n}$ cannot decrease too quickly as
$n\rightarrow\infty$. Finally, using a sorcerous intermixing of the topologies
of $\mathbb{C}$ and $\mathbb{C}_{p}$, $p$-adic Fourier analysis is used to
provide a method (“the $L^{1}$-method”) of bounding the archimedean absolute
values of $\mathbb{C}_{p}$-valued functions when those functions are
restricted to appropriate subsets of their domains on which they take values
in $\mathbb{Q}\subseteq\mathbb{C}_{p}$. This is then used to obtain upper
bounds on the absolute value of $\chi_{p}$ over subsets of $\mathbb{Z}_{2}$
consisting of all those $2$-adic integers in whose $2$-adic expansions there
are at least $\left\lceil\log_{2}p\right\rceil$ $0$s between any two
consecutive $1$s.

##  1 The Numen $\chi_{p}$ and the Correspondence Theorem

###  1.1 Construction and Characterization of $\chi_{p}$

As Lagarias and others have done, we begin by reinterpreting $\mathbf{j}$ in
terms of binary expansions of integers [10, 5]. In writing a tuple
$\mathbf{j}=\left(j_{1},\ldots,j_{\left|\mathbf{j}\right|}\right)\in\mathcal{J}$,
we say a tuple $\mathbf{i}\in\mathcal{J}$ was obtained by “adding $m$ terminal
zeroes to $\mathbf{j}$” whenever:

$\mathbf{i}=\left(j_{1},\ldots,j_{\left|\mathbf{j}\right|},\underbrace{0,\ldots,0}_{m\textrm{
times}}\right)$

Definition 1.1.1:

I. We define an equivalence relation $\sim$ on $\mathcal{J}$ by the property
that, for all $\mathbf{i},\mathbf{j}\in\mathcal{J}$,
$\mathbf{i}\sim\mathbf{j}$ if and only if one of $\mathbf{i}$ or $\mathbf{j}$
can be obtained by adding finitely many terminal zeroes to the other.

Observe that the value of $h_{\mathbf{j}}\left(0\right)$ is independent of the
number of terminal zeroes in $\mathbf{j}$:

$h_{\mathbf{i}}\left(0\right)=h_{\mathbf{j}}\left(0\right),\textrm{
}\forall\mathbf{i},\mathbf{j}\in\mathcal{J}\textrm{ such that
}\mathbf{i}\sim\mathbf{j}$

II. Let $\mathcal{J}/\sim$ be the set of equivalence classes of $\mathcal{J}$
under the equivalence relation $\sim$ defined above. Then, we define the map
$\beta:\mathcal{J}/\sim\rightarrow\mathbb{N}_{0}$ by:

$\beta\left(\mathbf{j}\right)\overset{\textrm{def}}{=}\sum_{\ell=1}^{\left|\mathbf{j}\right|}j_{\ell}2^{\ell-1}$
(18)

Note:

i. $\beta\left(\mathbf{j}\right)$ is well-defined with respect to $\sim$,
since adding terminal zeroes to $\mathbf{j}$ does not change the value of
$\beta\left(\mathbf{j}\right)$.

ii. Observe that $\beta\left(\mathbf{i}\right)=\beta\left(\mathbf{j}\right)$
occurs if and only if $\mathbf{i}\sim\mathbf{j}$.

iii. The map $\beta$ is well-known in Collatz literature, where—following
Lagarias’ convention—it is usually denoted by $Q$. [10]

III. Our first definition of the numen of $H_{p}$ is the function
$\chi_{p}:\mathbb{N}_{0}\rightarrow\mathbb{Q}$ defined by the rule:

$\chi_{p}\left(t\right)\overset{\textrm{def}}{=}h_{\beta^{-1}\left(t\right)}\left(0\right),\textrm{
}\forall t\in\mathbb{N}_{0}$ (19)

where $\beta^{-1}\left(t\right)$ is any element $\mathbf{j}$ of $\mathcal{J}$
so that $\beta\left(\mathbf{j}\right)=t$.

$\chi_{p}$ is well-defined, since the choice of a pre-image of $t$ under
$\beta$ does not affect the value of $\chi_{p}\left(t\right)$.

Definition 1.1.2:

I. For all $t\in\mathbb{N}_{0}$, let $\\#_{1}\left(t\right)$ denote the number
of $1$s in the binary/$2$-adic expansion of $t$. We extend $\\#_{1}$ to
$\mathbb{Z}_{2}$ by defining $\\#_{1}\left(\mathfrak{z}\right)=\infty$ for all
$\mathfrak{z}\in\mathbb{Z}_{2}\backslash\mathbb{N}_{0}$.

II. For all $t\in\mathbb{N}_{0}$, let $\lambda\left(t\right)$ denote the
number of digits ($0$s and $1$s) in the binary/$2$-adic expansion of $t$.
Using the convention that $\lambda\left(0\right)=0$, it follows that:

$\lambda\left(t\right)=\left\lceil\log_{2}\left(t+1\right)\right\rceil,\textrm{
}\forall t\in\mathbb{N}_{0}$ (20)

We extend $\lambda$ to $\mathbb{Z}_{2}$ by defining
$\lambda\left(\mathfrak{z}\right)=\infty$ for all
$\mathfrak{z}\in\mathbb{Z}_{2}\backslash\mathbb{N}_{0}$. Note also that
$\lambda\left(t\right)=\left\lfloor\log_{2}t\right\rfloor+1$ for all
$t\in\mathbb{N}_{1}$.

Note that $\\#_{1}$ and $\lambda$ satisfy the functional equations:

$\displaystyle\\#_{1}\left(2^{m}t+k\right)$
$\displaystyle=\\#_{1}\left(t\right)+\\#_{1}\left(k\right),\textrm{ }\forall
t\in\mathbb{N}_{0},\textrm{ }\forall m\in\mathbb{N}_{1},\textrm{ }\forall
k\in\left\\{0,\ldots,2^{m}-1\right\\}$ (21)
$\displaystyle\lambda\left(2^{m}t+k\right)$
$\displaystyle=\lambda\left(t\right)+m,\textrm{ }\forall
t\in\mathbb{N}_{1},\textrm{ }\forall m\in\mathbb{N}_{1},\textrm{ }\forall
k\in\left\\{0,\ldots,2^{m}-1\right\\}$ (22)

In particular:

$\displaystyle\\#_{1}\left(2t\right)$ $\displaystyle=\\#_{1}\left(t\right)$
$\displaystyle\\#_{1}\left(2t+1\right)$
$\displaystyle=\\#_{1}\left(t\right)+1$

for all $t\in\mathbb{N}_{0}$, and:

$\lambda\left(2t\right)=\lambda\left(2t+1\right)=\lambda\left(t\right)+1$

for all $t\in\mathbb{N}_{1}$.

Letting $t\in\mathbb{N}_{1}$, and writing $\beta^{-1}\left(t\right)$ to denote
the shortest tuple $\mathbf{j}$ for which $\beta\left(t\right)=\mathbf{j}$, we
then have the identities:

$\displaystyle\\#_{1}\left(\beta^{-1}\left(t\right)\right)$
$\displaystyle=\\#_{1}\left(t\right)$
$\displaystyle\left|\beta^{-1}\left(t\right)\right|$
$\displaystyle=\lambda\left(t\right)$

The first identity tells us that, given a positive integer $t$, the number of
$1$s in the string $\mathbf{j}=\beta^{-1}\left(t\right)$
($\\#_{1}\left(\beta^{-1}\left(t\right)\right)$) is equal to the number of
$1$s in the $2$-adic expansion of $t$ ($\\#_{1}\left(t\right)$). The second
identity tells us that, given a positive integer $t$, the length of the
shortest tuple $\mathbf{j}$ for which $\beta\left(\mathbf{j}\right)=t$
($\left|\beta^{-1}\left(t\right)\right|$) is equal to the number of digits in
the $2$-adic expansion of $t$ ($\lambda\left(t\right)$).

Letting $t\in\mathbb{N}_{1}$, we can write 5 using $\beta^{-t}\left(t\right)$
in lieu of $\mathbf{j}$. This gives us:

$h_{\beta^{-1}\left(t\right)}\left(x\right)=\frac{p^{\\#_{1}\left(\beta^{-1}\left(t\right)\right)}}{2^{\left|\beta^{-1}\left(t\right)\right|}}x+h_{\beta^{-1}\left(t\right)}\left(0\right),\textrm{
}\forall x\in\mathbb{Z},\textrm{ }\forall t\in\mathbb{N}_{1}$ (23)

Applying Remark 1.1.5, using $\chi_{p}\left(t\right)$ to denote
$h_{\beta^{-1}\left(t\right)}\left(0\right)$, we then obtain:

$h_{\beta^{-1}\left(t\right)}\left(x\right)=\frac{p^{\\#_{1}\left(t\right)}}{2^{\lambda\left(t\right)}}x+\chi_{p}\left(t\right),\textrm{
}\forall x\in\mathbb{Z},\textrm{ }\forall t\in\mathbb{N}_{1}$ (24)

Next, as is well known, every positive integer $t$ has a unique binary
expansion of the form:

$t=\sum_{k=1}^{\\#_{1}\left(t\right)}2^{n_{k}}$

for some strictly increasing sequence of non-negative integers $0\leq
n_{1}<n_{2}<\cdots<n_{K}<\infty$. We can extend this notation to include $0$
by defining the sum to be $0$ when $t=0$. Using this notation, we can give our
first somewhat closed-form expression for $\chi_{p}\left(t\right)$.

Proposition 1.1.1: Let $t\in\mathbb{N}_{0}$, and write the $2$-adic expansion
of $t$ as $\sum_{k=1}^{\\#_{1}\left(t\right)}2^{n_{k}}$. Then:

$\chi_{p}\left(\sum_{k=1}^{\\#_{1}\left(t\right)}2^{n_{k}}\right)=\sum_{k=1}^{\\#_{1}\left(t\right)}\frac{p^{k-1}}{2^{n_{k}+1}}$
(25)

where the sum on the right is defined to be $0$ whenever $t=0$.

_Remark_ 1.1.6: This formula is well-known in Collatz literature; see [14] for
but one example. Indeed, as remarked by Tao, it follows by induction that, as
defined, $\chi_{p}$ is injective (Ibid). However, this injectivity does not
necessarily hold when we consider the value of $\chi_{p}$ modulo $p^{N}$ for
some $N\in\mathbb{N}_{1}$. This line of thinking leads to the aforementioned
Lipschitz-type estimates that instigated the researches detailed in this
paper.

Proof: Since $\chi_{p}\left(0\right)$ is defined to be $0$, formula 25 holds
for $t=0$. So, fix $K\in\mathbb{N}_{1}$ and let $t$ be any positive integer
for which $\\#_{1}\left(t\right)=K$. We write $t=\sum_{k=1}^{K}2^{n_{k}}$, and
proceed by induction on $K$.

When $K=1$,
$\chi\left(2^{n_{1}}\right)=h_{\beta^{-1}\left(2^{n_{1}}\right)}\left(0\right)$.
Note that the $n_{1}+1$ tuple
$\mathbf{j}=\left(\underbrace{0,\ldots,0}_{n_{1}\textrm{ times}},1\right)$
then satisfies:

$\beta\left(\mathbf{j}\right)=\sum_{\ell=1}^{n_{1}+1}j_{\ell}2^{\ell-1}=\underbrace{0\cdot
2^{1-1}+0\cdot 2^{2-1}+\cdots+0\cdot 2^{n_{1}-1}}_{n_{1}\textrm{
terms}}+1\cdot 2^{\left(n_{1}+1\right)-1}=2^{n_{1}}$

and so:

$\displaystyle\chi_{p}\left(2^{n_{1}}\right)$
$\displaystyle=h_{\beta^{-1}\left(2^{n_{1}}\right)}\left(0\right)$
$\displaystyle=h_{\left(\underbrace{0,\ldots,0}_{n_{1}\textrm{
times}},1\right)}\left(0\right)=h_{0}^{\circ
n_{1}}\left(h_{1}\left(0\right)\right)=\frac{1}{2^{n_{1}}}\frac{p\cdot
0+1}{2}=\frac{1}{2^{n_{1}+1}}=\sum_{k=1}^{1}\frac{p^{k-1}}{2^{n_{k}+1}}$

Thus, the formula holds for $K=1$.

Now, proceeding to the inductive step, suppose the formula 25 holds for an
arbitrary $K\geq 1$. So, let $t$ be any non-negative integer with at most $K$
non-zero binary digits. Without loss of generality, suppose $t$ has $K$ non-
zero binary digits, with $t=\sum_{k=1}^{K}2^{n_{k}}$ for some $n_{k}$. Note
that $\lambda\left(t\right)=n_{K}+1$ (that is,
$2^{\lambda\left(t\right)-1}=2^{\left\lfloor\log_{2}t\right\rfloor}$ is the
largest power of $2$ which is less than or equal to $2$). Next, let $n_{K+1}$
be any integer greater than $n_{K}$. Then:

$\sum_{k=1}^{K+1}2^{n_{k}}=t+2^{n_{K+1}}=t+\underbrace{0\cdot
2^{n_{K}+1}+\cdots+0\cdot
2^{n_{K}+\left(n_{K+1}-n_{K}-1\right)}}_{n_{K+1}-n_{K}-1\textrm{
terms}}+2^{n_{K+1}}$

Hence:

$\displaystyle\chi_{p}\left(\sum_{k=1}^{K+1}2^{n_{k}}\right)$ $\displaystyle=$
$\displaystyle\chi_{p}\left(t+2^{n_{K+1}}\right)$ $\displaystyle=$
$\displaystyle\left(h_{\beta^{-1}\left(t\right)}\circ h_{0}^{\circ
n_{K+1}-n_{K}-1}\circ h_{1}\right)\left(0\right)$ $\displaystyle=$
$\displaystyle h_{\beta^{-1}\left(t\right)}\left(\frac{1}{2\cdot
2^{n_{K+1}-n_{K}-1}}\right)$ $\displaystyle=$ $\displaystyle
h_{\beta^{-1}\left(t\right)}\left(\frac{1}{2^{n_{K+1}-n_{K}}}\right)$
$\displaystyle\left(\textrm{use }\ref{eq:Affine map formula for
Chi_p}\right);$ $\displaystyle=$
$\displaystyle\frac{p^{\\#_{1}\left(t\right)}}{2^{\lambda\left(t\right)}}\frac{1}{2^{n_{K+1}-n_{K}}}+\chi_{p}\left(t\right)$
$\displaystyle\left(n_{K}=\lambda\left(t\right)-1;\chi_{p}\left(t\right)=\sum_{k=1}^{K}\frac{p^{k-1}}{2^{n_{k}+1}};\\#_{1}\left(t\right)=K\right);$
$\displaystyle=$
$\displaystyle\frac{p^{K}}{2^{n_{K+1}+1}}+\sum_{k=1}^{K}\frac{p^{k-1}}{2^{n_{k}+1}}$
$\displaystyle=$ $\displaystyle\sum_{k=1}^{K+1}\frac{p^{k-1}}{2^{n_{k}+1}}$

Thus, the $K$th case implies the $\left(K+1\right)$th case. As such, by
induction, the formula holds for all $K\in\mathbb{N}_{0}$.

Q.E.D.

Next, we $2$-adically interpolate $\chi_{p}$ into a function
$\chi_{p}:\mathbb{Z}_{2}\rightarrow\mathbb{Z}_{p}$. To do this, we note that,
in 25, as $\\#_{1}\left(t\right)\rightarrow\infty$, the series
$\sum_{k=1}^{\\#_{1}\left(t\right)}\frac{p^{k-1}}{2^{n_{k}+1}}$ converges to a
limit in $\mathbb{Z}_{p}$. As such:

Definition 1.1.3: We $2$-adically interpolate $\chi_{p}$ to a function
$\chi_{p}:\mathbb{Z}_{2}\rightarrow\mathbb{Z}_{p}$ by way of the rule:

$\chi_{p}\left(\sum_{k=1}^{\infty}2^{n_{k}}\right)\overset{\mathbb{Z}_{p}}{=}\sum_{k=1}^{\infty}\frac{p^{k-1}}{2^{n_{k}+1}}$
(26)

where $\left\\{n_{k}\right\\}_{k\in\mathbb{N}_{1}}$ is any strictly increasing
sequence of non-negative integers. We write $\overset{\mathbb{Z}_{p}}{=}$ to
emphasize that, in the above equality, the series’ convergence occurs in
$\mathbb{Z}_{p}$, rather than in $\mathbb{R}$.

While this gives us a valid $2$-adic interpolation of $\chi_{p}$, this
interpolation is _not_ continuous on all of $\mathbb{Z}_{2}$. As an example,
the sequence $\left\\{2^{n}\right\\}_{n\in\mathbb{N}_{1}}$ converges to $0$ in
$\mathbb{Z}_{2}$, yet, using 25, we have that
$\chi_{p}\left(2^{n}\right)=2^{-n-1}$. Were
$\chi_{p}:\mathbb{Z}_{2}\rightarrow\mathbb{Z}_{p}$ continuous,
$\chi_{p}\left(2^{n}\right)=2^{-n-1}$ would converge in $\mathbb{Z}_{p}$ to
$\chi_{p}\left(0\right)=0$, but $2^{-n-1}$ has unit magnitude in
$\mathbb{Z}_{p}$ for all $n\in\mathbb{N}_{1}$. Nevertheless, we _do_ have the
following:

Lemma 1.1.1: $\chi_{p}:\mathbb{Z}_{2}\rightarrow\mathbb{Z}_{p}$ is continuous
as a function from $\mathbb{Z}_{2}\backslash\mathbb{N}_{0}$ to
$\mathbb{Z}_{p}$. (That is to say, in the terminology of Section 2.2,
$\chi_{p}$ is “rising-continuous”.) In particular:

$\chi_{p}\left(\mathfrak{z}\right)\overset{\mathbb{Z}_{p}}{=}\lim_{n\rightarrow\infty}\chi_{p}\left(\left[\mathfrak{z}\right]_{2^{n}}\right)$
(27)

Proof: Let $\mathfrak{z}\in\mathbb{Z}_{2}\backslash\mathbb{N}_{0}$ be
arbitrary, and write $\mathfrak{z}=\sum_{n=0}^{\infty}a_{n}2^{n}$, where the
$a_{n}$s are the $2$-adic digits of $\mathfrak{z}$. Then, let
$\left\\{\mathfrak{z}_{m}\right\\}_{m\in\mathbb{N}_{1}}\subseteq\mathbb{Z}_{2}\backslash\mathbb{N}_{0}$
be any sequence converging to $\mathfrak{z}$ in $2$-adic absolute value. or
each $m$, we write $\mathfrak{z}_{m}=\sum_{n=0}^{\infty}a_{m,n}2^{m}$. This
convergence is equivalent to saying that, for every $N\in\mathbb{N}_{1}$,
there is an $m_{N}\in\mathbb{N}_{1}$ so that
$\mathfrak{z}_{m}\overset{2^{N}}{\equiv}\mathfrak{z}$ for all $m\geq m_{N}$,
and hence, that $a_{m,n}=a_{n}$ for all $n\in\left\\{0,\ldots,N-1\right\\}$
whenever $m\geq m_{N}$. As such:

$\mathfrak{z}_{m}\overset{\mathbb{Z}_{2}}{=}\sum_{n=0}^{N-1}a_{n}2^{n}+\sum_{n=N}^{\infty}a_{m,n}2^{n},\textrm{
}\forall m\geq m_{N}$

Next, we can write the rational integer $\sum_{n=0}^{N-1}a_{n}2^{n}$ as:

$\sum_{n=0}^{N-1}a_{n}2^{n}=\sum_{k=1}^{K_{N}}2^{n_{k}}$

for some $K_{N}\in\mathbb{N}_{0}$ (where $\sum_{n=0}^{N-1}a_{n}2^{n}=0$
whenever $K_{N}=0$). In terms of $N$, $K_{N}$ is the number of $1$s present in
the $2$-adic expansion of
$\left[\mathfrak{z}\right]_{2^{N}}=\sum_{n=0}^{N-1}a_{n}2^{n}$.

Anyhow, using 26, we then have that:

$\chi_{p}\left(\mathfrak{z}_{m}\right)\overset{\mathbb{Z}_{p}}{=}\chi_{p}\left(\sum_{k=1}^{K_{N}}2^{n_{k}}+\sum_{n=N}^{\infty}a_{m,n}2^{n}\right)\overset{\mathbb{Z}_{p}}{=}\sum_{k=1}^{K_{N}}\frac{p^{k-1}}{2^{n_{k}+1}}+O\left(p^{K_{N}}\right)$

Indeed, here, $\sum_{n=N}^{\infty}a_{m,n}2^{n}$—the error between
$\mathfrak{z}$ and $\mathfrak{z}_{m}$—contains any and all $1$s in the
$2$-adic expansion of $\mathfrak{z}_{m}$ that occur after the $K_{N}$th $1$ in
said $2$-adic expansion. 26 tells us that $2^{n_{k}}$—the term of the $2$-adic
expansion of $\mathfrak{z}_{m}$ corresponding to the $n_{k}$th $1$ in said
$2$-adic expansion—yields the term $\frac{p^{k-1}}{2^{n_{k}+1}}$ in
$\chi_{p}\left(\mathfrak{z}_{m}\right)$. As such, since
$\sum_{n=N}^{\infty}a_{m,n}2^{n}$ contains all the $1$s after the $K_{N}$th
$1$, it must be that all of the corresponding terms of
$\chi_{p}\left(\mathfrak{z}\right)$ have powers of $p$ which are $\geq
p^{K_{N}+1-1}=p^{K_{N}}$; hence the $O\left(p^{K_{N}}\right)$ term. Similarly:

$\chi_{p}\left(\mathfrak{z}\right)\overset{\mathbb{Z}_{p}}{=}\chi_{p}\left(\sum_{k=1}^{K_{N}}2^{n_{k}}+\sum_{n=N}^{\infty}a_{n}2^{n}\right)\overset{\mathbb{Z}_{p}}{=}\sum_{k=1}^{K_{N}}\frac{p^{k-1}}{2^{n_{k}+1}}+O\left(p^{K_{N}}\right)$

As such, for any $N\in\mathbb{N}_{1}$,
$\left|\mathfrak{z}-\mathfrak{z}_{m}\right|_{2}\leq p^{N}$ implies that there
are positive integers $K_{N}$ and $m_{N}$ so that:

$\left|\chi_{p}\left(\mathfrak{z}\right)-\chi_{p}\left(\mathfrak{z}_{m}\right)\right|_{p}=O\left(p^{-K_{N}}\right),\textrm{
}\forall m\geq m_{N}$

Now, the clincher: since $\mathfrak{z}$ is a $2$-adic integer which is _not_
in $\mathbb{N}_{0}$, the $2$-adic expansion of $\mathfrak{z}$ contains
infinitely many $1$s. Thus, as $N\rightarrow\infty$, the number of $1$s digits
in the $2$-adic expansion of the integer
$\left[\mathfrak{z}\right]_{2^{N}}=\sum_{n=0}^{N-1}a_{n}2^{n}$ will tend to
$\infty$. “The number of $1$s digits in the $2$-adic expansion of the integer
$\left[\mathfrak{z}\right]_{2^{N}}$” is, of course, exactly the quantity we
denoted by $K_{N}$. Thus, $\mathfrak{z}$ being an element of
$\mathbb{Z}_{2}\backslash\mathbb{N}_{0}$ guarantees that the magnitude of
$\left|\chi_{p}\left(\mathfrak{z}\right)-\chi_{p}\left(\mathfrak{z}_{m}\right)\right|_{p}$
tends to $0$ in $\mathbb{R}$ as $N\rightarrow\infty$. As such, we have that
$\mathfrak{z}_{m}\rightarrow\mathfrak{z}$ in $\mathbb{Z}_{2}$ forces
$\chi_{p}\left(\mathfrak{z}_{m}\right)$ to converge to
$\chi_{p}\left(\mathfrak{z}\right)$ in $\mathbb{Z}_{p}$ whenever
$\mathfrak{z}\in\mathbb{Z}_{2}\backslash\mathbb{N}_{0}$, which proves the
continuity of
$\chi_{p}:\mathbb{Z}_{2}\backslash\mathbb{N}_{0}\rightarrow\mathbb{Z}_{p}$.

Q.E.D.

Using some extra bits of terminology, we can give $\chi_{p}$ a more closed-
form expression.

Definition 1.1.4: Let $\mathfrak{z}\in\mathbb{Z}_{2}$. Then, we write
$\left\\{\beta_{k}\left(\mathfrak{z}\right)\right\\}_{k\in\left\\{1,\ldots,\\#_{1}\left(\mathfrak{z}\right)\right\\}}$
to denote the unique, strictly increasing sequence of non-negative integers so
that the $2$-adic expansion of $\mathfrak{z}$ is given by:

$\mathfrak{z}=\sum_{k=1}^{\\#_{1}\left(\mathfrak{z}\right)}2^{\beta_{k}\left(\mathfrak{z}\right)}$
(28)

The above sum is defined to be $0$ whenever $\mathfrak{z}=0$.

Theorem 1.1.1 ($\chi_{p}$ functional equations): $\chi_{p}$ is the unique
function from $\mathbb{Z}_{2}$ to $\mathbb{Z}_{p}$ satisfying the functional
equations:

$\displaystyle\chi_{p}\left(2\mathfrak{z}\right)$
$\displaystyle=\frac{1}{2}\chi_{p}\left(\mathfrak{z}\right)$ (29)
$\displaystyle\chi_{p}\left(2\mathfrak{z}+1\right)$
$\displaystyle=\frac{p\chi_{p}\left(\mathfrak{z}\right)+1}{2}$

for all $\mathfrak{z}\in\mathbb{Z}_{2}$.

Proof: Let $t=\sum_{k=1}^{K}2^{n_{k}}\in\mathbb{N}_{0}$. Then 25implies:

$\chi_{p}\left(2\sum_{k=1}^{K}2^{n_{k}}\right)=\chi_{p}\left(\sum_{k=1}^{K}2^{n_{k}+1}\right)=\sum_{k=1}^{K}\frac{p^{k-1}}{2^{n_{k}+2}}=\frac{1}{2}\sum_{k=1}^{K}\frac{p^{k-1}}{2^{n_{k}+1}}=\chi_{p}\left(\sum_{k=1}^{K}2^{n_{k}}\right)$

Similarly:

$\displaystyle\chi_{p}\left(2\left(\sum_{k=1}^{K}2^{n_{k}}\right)+1\right)$
$\displaystyle=\chi_{p}\left(1+2^{n_{1}+1}+2^{n_{2}+1}+\cdots+2^{n_{K}+1}\right)$
$\displaystyle=\frac{1}{2^{0+1}}+\frac{p}{2^{n_{1}+2}}+\frac{p^{2}}{2^{n_{2}+2}}+\cdots+\frac{p^{K}}{2^{n_{K}+2}}$
$\displaystyle=\frac{1}{2}\left(1+p\left(\frac{1}{2^{n_{1}+1}}+\frac{p}{2^{n_{2}+1}}+\cdots+\frac{p^{K-1}}{2^{n_{K}+1}}\right)\right)$
$\displaystyle=\frac{1}{2}\left(1+p\chi_{p}\left(\sum_{k=1}^{K}2^{n_{k}}\right)\right)$

Hence 29 hold true for all $\mathfrak{z}\in\mathbb{N}_{0}$. By the $2$-adic
interpolability of $\chi_{p}$, we have that $\sum_{k=1}^{K}2^{n_{k}}$ tends to
an arbitrary element of $\mathbb{Z}_{2}\backslash\mathbb{N}_{0}$ as
$K\rightarrow\infty$, which then shows that $\chi_{p}$ satisfies the
functional equations for all
$\mathfrak{z}\in\mathbb{Z}_{2}\backslash\mathbb{N}_{0}$. Since $\chi_{p}$ also
satisfies these functional equations for all $\mathfrak{z}\in\mathbb{N}_{0}$,
it then follows that it satisfies them for all
$\mathfrak{z}\in\mathbb{Z}_{2}$.

Finally, let $f:\mathbb{Z}_{2}\rightarrow\mathbb{Z}_{p}$ be any function
satisfying:

$\displaystyle f\left(2\mathfrak{z}\right)$
$\displaystyle=\frac{1}{2}f\left(\mathfrak{z}\right)$ $\displaystyle
f\left(2\mathfrak{z}+1\right)$
$\displaystyle=\frac{pf\left(\mathfrak{z}\right)+1}{2}$

for all $\mathfrak{z}\in\mathbb{Z}_{2}$. Setting $\mathfrak{t}=0$ gives
$f\left(0\right)=\frac{1}{2}f\left(0\right)$, which forces
$f\left(0\right)=0$. Consequently, $f\left(1\right)=f\left(2\cdot
0+1\right)=\frac{pf\left(0\right)+1}{2}=\frac{1}{2}$, and so
$f\left(2^{n_{1}}\right)=\frac{1}{2^{n_{1}}}f\left(\frac{1}{2}\right)=\frac{1}{2^{n_{1}+1}}$.

Then, for any $n_{2}>n_{1}$:

$\displaystyle f\left(2^{n_{1}}+2^{n_{2}}\right)$
$\displaystyle=f\left(2^{n_{1}}\left(2^{n_{2}-n_{1}}+1\right)\right)$
$\displaystyle=\frac{1}{2^{n_{1}}}f\left(2^{n_{2}-n_{1}}+1\right)$
$\displaystyle=\frac{1}{2^{n_{1}}}\frac{pf\left(2^{n_{2}-n_{1}-1}\right)+1}{2}$
$\displaystyle=\frac{1}{2^{n_{1}+1}}+\frac{p}{2^{n_{1}+1}}f\left(2^{n_{2}-n_{1}-1}\right)$
$\displaystyle=\frac{1}{2^{n_{1}+1}}+\frac{p}{2^{n_{1}+1}}\frac{1}{2^{n_{2}-n_{1}-1}}f\left(1\right)$
$\displaystyle=\frac{1}{2^{n_{1}+1}}+\frac{p}{2^{n_{2}+1}}$

Continuing in this manner, by induction, it then follows that:

$f\left(\sum_{k=1}^{K}2^{n_{K}}\right)=\sum_{k=1}^{K}\frac{p^{k-1}}{2^{n_{k}+1}}=\chi_{p}\left(\sum_{k=1}^{K}2^{n_{K}}\right)$

and thus, that $f\left(t\right)=\chi_{p}\left(t\right)$ holds for all
$t\in\mathbb{N}_{0}$. Applying the argument from part (I) then shows that
$f\left(t\right)=\chi_{p}\left(t\right)$ on $\mathbb{N}_{0}$ then determines
the values of $f$ at every other $2$-adic integer, forcing
$f\left(\mathfrak{z}\right)=\chi_{p}\left(\mathfrak{z}\right)$ for all
$\mathfrak{t}\in\mathbb{Z}_{2}$.

Q.E.D.

###  1.2 The Correspondence Theorem

In this subsection we prove the titular Correspondence Theorem (Theorem
1.2.1): an odd integer $\omega$ is a periodic point of $H_{p}$ if and only if
$\omega=\chi_{p}\left(\mathfrak{z}\right)$ for some
$\mathfrak{z}\in\mathbb{Z}_{2}$. To do this, we begin with a nearly
tautological observation.

Proposition 1.2.1: Let $\omega\in\mathbb{Z}$ be a periodic point of $H_{p}$
(i.e., $H_{p}^{\circ n}\left(\omega\right)=\omega$ for some
$n\in\mathbb{N}_{1}$). There exists a $t\in\mathbb{N}_{0}$ so that:

$\chi_{p}\left(t\right)\overset{\mathbb{R}}{=}\left(1-\frac{p^{\\#_{1}\left(t\right)}}{2^{\lambda\left(t\right)}}\right)\omega$
(30)

Additionally, for any $\mathbf{j}\in\mathcal{J}$ for which
$h_{\mathbf{j}}\left(\omega\right)=\omega$, it must be that:

$\chi_{p}\left(\beta\left(\mathbf{j}\right)\right)\overset{\mathbb{R}}{=}\left(1-\frac{p^{\\#_{1}\left(\beta\left(\mathbf{j}\right)\right)}}{2^{\lambda\left(\beta\left(\mathbf{j}\right)\right)}}\right)\omega$
(31)

Proof: If $\omega\in\mathbb{N}_{1}$ is a periodic point of $H_{p}$ (say, of
period $n$), there is then a unique $\mathbf{j}\in\mathcal{J}$ of length $n$
so that $h_{\mathbf{j}}\left(\omega\right)=H_{p}^{\circ
n}\left(\omega\right)=\omega$. As such, for $t=\beta\left(\mathbf{j}\right)$:

$\displaystyle\omega$ $\displaystyle=$ $\displaystyle
h_{\mathbf{j}}\left(\omega\right)h_{\beta^{-1}\left(t\right)}\left(\omega\right)=\frac{p^{\\#_{1}\left(t\right)}\omega}{2^{\lambda\left(t\right)}}+\chi_{p}\left(t\right)$
$\displaystyle\Updownarrow$
$\displaystyle\left(1-\frac{p^{\\#_{1}\left(t\right)}}{2^{\lambda\left(t\right)}}\right)\omega$
$\displaystyle=$ $\displaystyle\chi_{p}\left(t\right)$

By the same reasoning, for any $\mathbf{j}\in\mathcal{J}$ for which
$h_{\mathbf{j}}\left(\omega\right)=\omega$, it must be that:

$\chi_{p}\left(\beta\left(\mathbf{j}\right)\right)=\left(1-\frac{p^{\\#_{1}\left(\beta\left(\mathbf{j}\right)\right)}}{2^{\lambda\left(\beta\left(\mathbf{j}\right)\right)}}\right)\omega$

holds true as an equality in $\mathbb{R}$, as desired.

Q.E.D.

Lemma 1.2.1: Every periodic point of $H_{p}$ in $\mathbb{N}_{1}$ is of the
form:

$\chi_{p}\left(\frac{t}{1-2^{\lambda\left(t\right)}}\right)$ (32)

for some $t\in\mathbb{N}_{1}$. In particular, if $\omega$ is the periodic
point of $H_{p}$, then:

$\omega=\chi_{p}\left(\frac{\beta\left(\mathbf{j}\right)}{1-2^{\lambda\left(\beta\left(\mathbf{j}\right)\right)}}\right)$
(33)

where $\mathbf{j}$ is any element of $\mathcal{J}$ such that
$h_{\mathbf{j}}\left(\omega\right)=\omega$.

Proof: For a periodic point $\omega$, let $\mathbf{j}\in\mathcal{J}$ be the
shortest tuple for which $h_{\mathbf{j}}\left(\omega\right)=\omega$, and let
$\mathbf{j}^{n}$ denote the concatenation of $n$ copies of $\mathbf{j}$. No
matter how many copies of $\mathbf{j}$ we concatenate, the associated
composition sequence leaves $\omega$ fixed:

$h_{\mathbf{j}^{n}}\left(\omega\right)=h_{\mathbf{j}}^{\circ
n}\left(\omega\right)=h_{\mathbf{j}}^{\circ
n-1}\left(h_{\mathbf{j}}\left(\omega\right)\right)=h_{\mathbf{j}}^{\circ
n-1}\left(\omega\right)=\cdots=\omega$

As such, 31 yields:

$\chi_{p}\left(\beta\left(\mathbf{j}^{n}\right)\right)=\left(1-\frac{p^{\\#_{1}\left(\beta\left(\mathbf{j}^{n}\right)\right)}}{2^{\lambda\left(\beta\left(\mathbf{j}^{n}\right)\right)}}\right)\omega$

Now, observe that:

$\beta\left(\mathbf{j}^{n}\right)=\sum_{m=0}^{n-1}2^{\left|\mathbf{j}\right|m}\underbrace{\left(j_{1}+j_{2}2+\cdots+j_{\left|\mathbf{j}\right|}2^{\left|\mathbf{j}\right|-1}\right)}_{\beta\left(\mathbf{j}\right)}=\frac{1-2^{\left|\mathbf{j}\right|n}}{1-2^{\left|\mathbf{j}\right|}}\beta\left(\mathbf{j}\right)$

Letting $t$ denote $\beta\left(\mathbf{j}\right)$, we write:

$t_{n}\overset{\textrm{def}}{=}\beta\left(\mathbf{j}^{n}\right),\textrm{
}\forall n\in\mathbb{N}_{1}$ (34)

Since $\left|\mathbf{j}\right|=\lambda\left(t\right)$, we can then write:

$t_{n}=\frac{1-2^{n\lambda\left(t\right)}}{1-2^{\lambda\left(t\right)}}t$ (35)

Additionally, since $\mathbf{j}^{n}$ is the concatenation of $n$ copies of
$\mathbf{j}$, we have that:

$\\#_{1}\left(\beta\left(\mathbf{j}^{n}\right)\right)=\\#_{1}\left(\mathbf{j}^{n}\right)=\\#_{1}\left(\mathbf{j}\right)n=\\#_{1}\left(t\right)n$
$\lambda\left(\beta\left(\mathbf{j}^{n}\right)\right)=\left|\mathbf{j}^{n}\right|-1=n\left|\mathbf{j}\right|-1=n\lambda\left(t\right)-1$

As such:

$\displaystyle\chi_{p}\left(\beta\left(\mathbf{j}^{n}\right)\right)$
$\displaystyle=$
$\displaystyle\left(1-\frac{p^{\\#_{1}\left(\beta\left(\mathbf{j}^{n}\right)\right)}}{2^{\lambda\left(\beta\left(\mathbf{j}^{n}\right)\right)}}\right)\omega$
$\displaystyle\Updownarrow$
$\displaystyle\chi_{p}\left(\frac{1-2^{n\lambda\left(t\right)}}{1-2^{\lambda\left(t\right)}}t\right)$
$\displaystyle=$
$\displaystyle\left(1-\left(\frac{p^{\\#_{1}\left(t\right)}}{2^{\lambda\left(t\right)}}\right)^{n}\right)\omega$
(36)

Finally, since:

$\lim_{n\rightarrow\infty}\frac{1-2^{n\lambda\left(t\right)}}{1-2^{\lambda\left(t\right)}}t\overset{\mathbb{Z}_{2}}{=}\frac{t}{1-2^{\lambda\left(t\right)}},\textrm{
}\forall t\in\mathbb{N}_{0}$

and since $\frac{t}{1-2^{\lambda\left(t\right)}}\in\mathbb{N}_{0}$ if and only
if $t=0$, the rising continuity of
$\chi_{p}:\mathbb{Z}_{2}\rightarrow\mathbb{Z}_{p}$ guarantees that:

$\displaystyle\chi_{p}\left(\frac{t}{1-2^{\lambda\left(t\right)}}\right)$
$\displaystyle\overset{\mathbb{Z}_{p}}{=}\lim_{n\rightarrow\infty}\chi_{p}\left(\left[\frac{t}{1-2^{\lambda\left(t\right)}}\right]_{2^{n\lambda\left(t\right)}}\right)$
$\displaystyle=\lim_{n\rightarrow\infty}\chi_{p}\left(\frac{1-2^{n\lambda\left(t\right)}}{1-2^{\lambda\left(t\right)}}t\right)$
$\displaystyle\overset{\mathbb{Z}_{p}}{=}\lim_{n\rightarrow\infty}\left(1-\left(\frac{p^{\\#_{1}\left(t\right)}}{2^{\lambda\left(t\right)}}\right)^{n}\right)\omega$
$\displaystyle\overset{\mathbb{Z}_{p}}{=}\omega$

This proves every periodic point of $H_{p}$ in $\mathbb{Z}$ is of the form 32
for some $t\in\mathbb{N}_{1}$; in particular,
$t=\beta\left(\mathbf{j}\right)$, where $\mathbf{j}$ is any element of
$\mathcal{J}$ satisfying $h_{\mathbf{j}}\left(\omega\right)=\omega$, which
gives us 33.

Q.E.D.

Definition 1.2.1: Let $B:\mathbb{N}_{0}\rightarrow\mathbb{Z}_{2}$ be defined
by:

$B\left(t\right)\overset{\textrm{def}}{=}\frac{t}{1-2^{\lambda\left(t\right)}}=\frac{t}{1-2^{\left\lceil\log_{2}\left(t+1\right)\right\rceil}},\textrm{
}\forall t\in\mathbb{N}_{0}$ (37)

Proposition 1.2.2: $B$ extends to a continuous function from $\mathbb{Z}_{2}$
to $\mathbb{Z}_{2}$. In particular:

I. The restriction of $B$ to $\mathbb{Z}_{2}\backslash\mathbb{N}_{0}$ is the
identity map: $B\left(\mathfrak{t}\right)=\mathfrak{t}$ for all
$\mathfrak{t}\in\mathbb{Z}_{2}\backslash\mathbb{N}_{0}$. (Hence, with respect
to the Haar probability measure of $\mathbb{Z}_{2}$, $B$ is equal to the
identity map almost everywhere on $\mathbb{Z}_{2}$).

II. $B\left(2^{n}-1\right)=-1$ for all $n\in\mathbb{N}_{1}$.

Proof: As a sequence $\left\\{t_{n}\right\\}_{n\in\mathbb{N}_{1}}$ of positive
integers converges $2$-adically to a limit $\mathfrak{t}\in\mathbb{Z}_{2}$,
one of two things can happen. If $\lambda\left(t_{n}\right)$ remains bounded
as $n\rightarrow\infty$, then $\mathfrak{t}$ must be an element of
$\mathbb{N}_{0}$, and so,
$\lim_{n\rightarrow\infty}B\left(t_{n}\right)=B\left(\mathfrak{t}\right)$. The
other possibility is that $\lambda\left(t_{n}\right)\rightarrow\infty$ as
$n\rightarrow\infty$, in which case, $2^{\lambda\left(t_{n}\right)}$ tends to
$0$ in $\mathbb{Z}_{2}$, and hence:

$\lim_{n\rightarrow\infty}B\left(t_{n}\right)=\lim_{n\rightarrow\infty}\frac{t_{n}}{1-2^{\lambda\left(t_{n}\right)}}\overset{\mathbb{Z}_{2}}{=}\lim_{n\rightarrow\infty}\frac{t_{n}}{1-0}=\mathfrak{t}$

which shows that $B\left(\mathfrak{t}\right)=\mathfrak{t}$ for all
$\mathfrak{t}\in\mathbb{Z}_{2}\backslash\mathbb{N}_{0}$. Thus, we can
continuously (and, therefore, uniquely) extend $B$ to a function on
$\mathbb{Z}_{2}$, and this extension equals the identity map for all
$\mathfrak{t}\in\mathbb{Z}_{2}\backslash\mathbb{N}_{1}$, a set of full measure
in $\mathbb{Z}_{2}$.

Finally, for (II), note that
$2^{n}-1=\frac{2^{n}-1}{2-1}=\sum_{k=0}^{n-1}2^{k}$, and so,
$\lambda\left(2^{n}-1\right)=n$, and so:

$B\left(2^{n}-1\right)=\frac{2^{n}-1}{1-2^{\lambda\left(2^{n}-1\right)}}=\frac{2^{n}-1}{1-2^{n}}=\frac{2^{n}-1}{1-2^{n}}=-1$

as claimed.

Q.E.D.

Using $B\left(t\right)$, we can re-interpret 33 as a functional equation
relating $\chi_{p}\left(B\left(t\right)\right)$ and $\chi_{p}\left(t\right)$.

Lemma 1.2.2 ($\chi_{p}\circ B$ functional equation):

$\chi_{p}\left(B\left(t\right)\right)\overset{\mathbb{Z}_{p}}{=}\frac{\chi_{p}\left(t\right)}{1-r_{p}\left(t\right)},\textrm{
}\forall t\in\mathbb{N}_{1}$ (38)

Proof: Letting $t\in\mathbb{N}_{1}$, we have that:

$\displaystyle\chi_{p}\left(B\left(t\right)\right)$
$\displaystyle\overset{\mathbb{Z}_{p}}{=}$
$\displaystyle\chi_{p}\left(\sum_{m=0}^{\infty}\sum_{k=1}^{\\#_{1}\left(t\right)}2^{m\lambda\left(t\right)+\beta_{k}\left(t\right)+1}\right)$
$\displaystyle=$
$\displaystyle\sum_{k=1}^{\\#_{1}\left(t\right)}\frac{p^{k-1}}{2^{\beta_{k}\left(t\right)+1}}+\sum_{k=1}^{\\#_{1}\left(t\right)}\frac{p^{k+\\#_{1}\left(t\right)-1}}{2^{\lambda\left(t\right)+\beta_{k}\left(t\right)+1}}+\sum_{k=1}^{\\#_{1}\left(t\right)}\frac{p^{k+2\\#_{1}\left(t\right)-1}}{2^{2\lambda\left(t\right)+\beta_{k}\left(t\right)+1}}+\cdots$
$\displaystyle=$
$\displaystyle\sum_{m=0}^{\infty}\sum_{k=1}^{\\#_{1}\left(t\right)}\frac{p^{k+m\\#_{1}\left(t\right)-1}}{2^{m\lambda\left(t\right)+\beta_{k}\left(t\right)+1}}$
$\displaystyle=$
$\displaystyle\underbrace{\sum_{m=0}^{\infty}\left(\frac{p^{\\#_{1}\left(t\right)}}{2^{\lambda\left(t\right)}}\right)^{m}}_{\textrm{converges
if }\\#_{1}\left(t\right)\geq
1}\times\underbrace{\sum_{k=1}^{\\#_{1}\left(t\right)}\frac{p^{k-1}}{2^{\beta_{k}\left(t\right)+1}}}_{\chi_{p}\left(t\right)}$
$\displaystyle\left(\textrm{if }t\neq 0\right);$
$\displaystyle\overset{\mathbb{Z}_{p}}{=}$
$\displaystyle\frac{\chi_{p}\left(t\right)}{1-\frac{p^{\\#_{1}\left(t\right)}}{2^{\lambda\left(t\right)}}}$
$\displaystyle=$
$\displaystyle\frac{\chi_{p}\left(t\right)}{1-r_{p}\left(t\right)}$

Thus, the identity holds for all $t\in\mathbb{N}_{1}$.

Q.E.D.

__Lemma 1.2.2 shows that $B\left(\mathbb{N}_{1}\right)$ is a subset of
$\mathbb{Q}\cap\mathbb{Z}_{2}$ mapped into $\mathbb{Q}$ by $\chi_{p}$. It
would be of interest to know if the “converse” holds true; that is, is
$B\left(\mathbb{N}_{1}\right)$ the pre-image of $\mathbb{Q}\cap\mathbb{Z}_{p}$
under $\chi_{p}$?

Proposition 1.2.3: 38 allows us to view $\chi_{p}\circ B$ as a function from
$\mathbb{N}_{1}$ to $\mathbb{R}$ defined by the rule:

$\chi_{p}\left(B\left(t\right)\right)\overset{\mathbb{R}}{=}\frac{\chi_{p}\left(t\right)}{1-r_{p}\left(t\right)}$
(39)

Proof: Because the series defining $\chi_{p}\left(B\left(t\right)\right)$
simplifies in $\mathbb{Z}_{p}$ to the product of a rational number
($\chi_{p}\left(t\right)$) and a $p$-adically convergent geometric series
whose common ratio is a rational number, the universality of the geometric
series formula allows us to conclude that for every $t\in\mathbb{N}_{0}$, the
value of $\chi_{p}\left(B\left(t\right)\right)$ is an honest-to-goodness
rational number. Consequently, using the right-hand side of 38, we can define
$\chi_{p}\circ B$ as a function from $\mathbb{N}_{1}$ to $\mathbb{R}$.

Q.E.D.

Finally, we have the main result of this section, which shows why the numen is
of interest:

Theorem 1.2.1 (The Correspondence Theorem): An odd integer $\omega$ is a
periodic point of $H_{p}$ if and only if:

$\omega\overset{\mathbb{R}}{=}\chi_{p}\left(B\left(t\right)\right)\overset{\mathbb{R}}{=}\frac{\chi_{p}\left(t\right)}{1-\frac{p^{\\#_{1}\left(t\right)}}{2^{\lambda\left(t\right)}}}=\sum_{k=1}^{\\#_{1}\left(t\right)}\frac{2^{\lambda\left(t\right)-\beta_{k}\left(t\right)-1}p^{k-1}}{2^{\lambda\left(t\right)}-p^{\\#_{1}\left(t\right)}}$
(40)

for some $t\in\mathbb{N}_{0}$.

Proof: Let $\omega$ be an odd non-zero periodic point of $H_{p}$. Then, by
Lemma 1.2.1, there is a $t\in\mathbb{N}_{1}$ so that
$\chi_{p}\left(t\right)=\left(1-\frac{p^{\\#_{1}\left(t\right)}}{2^{\lambda\left(t\right)}}\right)\omega$.
Hence:

$\omega=\frac{\chi_{p}\left(t\right)}{1-\frac{p^{\\#_{1}\left(t\right)}}{2^{\lambda\left(t\right)}}}=\chi_{p}\left(B\left(t\right)\right)$

as desired.

II. Let $t\in\mathbb{N}_{1}$, and suppose that
$\chi_{p}\left(B\left(t\right)\right)\in\mathbb{Z}$. Letting $\omega$ denote
$\chi_{p}\left(B\left(t\right)\right)$—and noting that we then have
$\omega=\frac{\chi_{p}\left(t\right)}{1-\frac{p^{\\#_{1}\left(t\right)}}{2^{\lambda\left(t\right)}}}$—by
23, we can write:

$h_{\beta^{-1}\left(t\right)}\left(\omega\right)=\frac{p^{\\#_{1}\left(t\right)}\omega}{2^{\lambda\left(t\right)}}+\chi_{p}\left(t\right)=\frac{p^{\\#_{1}\left(t\right)}\omega}{2^{\lambda\left(t\right)}}+\omega\left(1-\frac{p^{\\#_{1}\left(t\right)}}{2^{\lambda\left(t\right)}}\right)=\omega$

Letting $\mathbf{j}\in\beta^{-1}\left(t\right)$, it then follows that
$h_{\mathbf{j}}\left(\omega\right)=\omega$.

Claim: Let $t\in\mathbb{N}_{1}$, and suppose there is a
$\mathbf{j}\in\beta^{-1}\left(t\right)$ so that
$h_{\mathbf{j}}\left(\omega\right)=\omega$. Then, $\omega$ is a periodic point
of $H_{p}$.

Proof of Claim: Let $t$ and $\mathbf{j}$ be as given, By way of contradiction,
suppose that $\omega$ was _not_ a periodic point of $H_{p}$. Since the
composition sequence $h_{\mathbf{j}}$ fixes $\omega$, our counterfactual
assumption on $\omega$ forces there to be an
$n\in\left\\{1,\ldots,\left|\mathbf{j}\right|\right\\}$ so that:

$H^{\circ n}\left(\omega\right)=h_{j_{1},\ldots,j_{n}}\left(\omega\right)$

but for which:

$H^{\circ n+1}\left(\omega\right)\neq
h_{j_{1},\ldots,j_{n},j_{n+1}}\left(\omega\right)$

(Note that $t\in\mathbb{N}_{1}$ then guarantees that
$\left|\mathbf{j}\right|\geq 1$.) So, at some point in applying to $\omega$
the maps in the composition sequence defining $h_{\mathbf{j}}$, we must have
applied the “wrong” map; that is, $h_{j_{1},\ldots,j_{n}}\left(\omega\right)$
was odd (resp. even), but $j_{n+1}=0$ (resp. $j_{n+1}=1$), causing the
composition sequence $h_{\mathbf{j}}$ to diverge from the “natural” path
corresponding to the motions of $\omega$ under iterations of $H$.

Noting that for any $m\in\mathbb{N}_{0}$:

$\displaystyle h_{0}\left(2m+1\right)$ $\displaystyle=$
$\displaystyle\frac{2m+1}{2}\in\mathbb{Z}\left[\frac{1}{2}\right]\backslash\mathbb{Z}$
$\displaystyle h_{1}\left(2m\right)$ $\displaystyle=$
$\displaystyle\frac{2pm+1}{2}\in\mathbb{Z}\left[\frac{1}{2}\right]\backslash\mathbb{Z}$

we see that applying the “wrong” map at any step in the composition sequence
sends us from $\mathbb{Z}$ to an element of
$\mathbb{Z}\left[\frac{1}{2}\right]$ (the dyadic rational numbers) of the form
$\frac{a}{2}$, where $a$ is an odd integer. So, letting $m\in\mathbb{N}_{1}$
and $a\in 2\mathbb{Z}+1$ be arbitrary, we have that:

$\left|h_{0}\left(\frac{a}{2^{m}}\right)\right|_{2}=\left|\frac{a}{2^{m+1}}\right|_{2}=2\left|\frac{a}{2^{m}}\right|_{2}$

$\left|h_{1}\left(\frac{a}{2^{m}}\right)\right|_{2}=\left|\frac{p\frac{a}{2^{m}}+1}{2}\right|_{2}=\left|\frac{pa+2^{m+1}}{2^{m+1}}\right|_{2}\overset{\textrm{St}\Delta}{=}\frac{\left|pa\right|_{2}}{\left|2^{m+1}\right|_{2}}=2\left|\frac{a}{2^{m}}\right|_{2}$

where $\textrm{St}\Delta$ denotes an application of the strong triangle
inequality, using the facts that $a$ and $p$ are odd and $m\geq 1$.

Since a dyadic rational number $q$ is a non-integer if and only if
$q=\frac{a}{2^{m}}$ for some $a\in 2\mathbb{Z}+1$ and $m\in\mathbb{N}_{1}$,
this shows that, for any such $q$,
$\left|h_{0}\left(q\right)\right|_{2}=\left|h_{1}\left(q\right)\right|_{2}=2\left|q\right|_{2}$.
So,
$\left|h_{\mathbf{i}}\left(q\right)\right|_{2}=2^{\left|\mathbf{i}\right|}\left|q\right|_{2}$
for all such $q$ and any $\mathbf{i}\in\mathcal{J}$. Hence:

$h_{\mathbf{i}}\left(q\right)\notin\mathbb{Z},\textrm{
}\forall\mathbf{i}\in\mathcal{J},\forall
q\in\mathbb{Z}\left[\frac{1}{2}\right]\backslash\mathbb{Z}$ (41)

In other words, _once a composition sequence of $h_{0}$ and $h_{1}$ outputs a
non-integer dyadic rational, there is no way to apply composition sequences of
$h_{0}$ and $h_{1}$ to arrive back at an integer output_.

In summary, our assumption that $\omega\in\mathbb{Z}$ was not a periodic point
of $H_{p}$ forces $h_{j_{1},\ldots,j_{n},j_{n+1}}\left(\omega\right)\neq
H^{\circ n+1}\left(\omega\right)$, and hence, forces
$h_{j_{1},\ldots,j_{n},j_{n+1}}\left(\omega\right)$ to be a non-integer dyadic
rational number. But then, $\mathbb{Z}$ contains:

$\omega=h_{\mathbf{j}}\left(\omega\right)=h_{j_{n+2},\ldots,j_{\left|\mathbf{j}\right|}}\left(h_{j_{1},\ldots,j_{n},j_{n+1}}\left(\omega\right)\right)\overset{\ref{eq:Composition
Sequence of No Return}}{\notin}\mathbb{Z}$

which is _not_ an integer. This is impossible. Consequently, for our
$\mathbf{j}$, $h_{\mathbf{j}}\left(\omega\right)=\omega$ forces $\omega$ to be
a periodic point of $H_{p}$. This proves the Claim. $\checkmark$

Since $t$ was an arbitrary positive-integer for which
$\chi_{p}\left(B\left(t\right)\right)\in\mathbb{Z}$, this shows that
$\chi_{p}\left(B\left(t\right)\right)$ is a periodic point of $H_{p}$ in
$\mathbb{Z}$ for any $t\in\mathbb{N}_{1}$ for which
$\chi_{p}\left(B\left(t\right)\right)\in\mathbb{Z}$. Finally, we need to deal
with the fact that our periodic points must be odd. To see why this must be
so, taking the $2$-adic absolute value of the expression in Lemma 1.2.2 for
any $t\in\mathbb{N}_{1}$, we have that:

$\left|\chi_{p}\left(B\left(t\right)\right)\right|_{2}=\left|\frac{\chi_{p}\left(t\right)}{1-\frac{p^{\\#_{1}\left(t\right)}}{2^{\lambda\left(t\right)}}}\right|_{2}=\left|\frac{2^{\lambda\left(t\right)}\chi_{p}\left(t\right)}{\underbrace{2^{\lambda\left(t\right)}-p^{\\#_{1}\left(t\right)}}_{\textrm{odd
for
}t\in\mathbb{N}_{1}}}\right|_{2}=\left|2^{\lambda\left(t\right)}\chi_{p}\left(t\right)\right|$

Since $t\in\mathbb{N}_{1}$ implies $\chi_{p}\left(t\right)\in\mathbb{Q}$, and
since $\lambda\left(t\right)=\beta_{\\#_{1}\left(t\right)}\left(t\right)+1$,
we have that:

$\left|2^{\lambda\left(t\right)}\chi_{p}\left(t\right)\right|_{2}=\left|\sum_{k=1}^{\\#_{1}\left(t\right)}2^{\lambda\left(t\right)-\beta_{k}\left(t\right)-1}p^{k-1}\right|_{2}=\left|2C+p^{\\#_{1}\left(t\right)-1}\right|_{2}$

where $C$ is some non-negative integer depending on $p$ and $t$. Since
$p^{\\#_{1}\left(t\right)-1}$ is always an odd integer, this shows that:

$\left|2^{\lambda\left(t\right)}\chi_{p}\left(t\right)\right|_{2}=\left|2C+p^{\\#_{1}\left(t\right)-1}\right|_{2}=1$

and hence:

$\left|\chi_{p}\left(B\left(t\right)\right)\right|_{2}=\left|\frac{2^{\lambda\left(t\right)}\chi_{p}\left(t\right)}{2^{\lambda\left(t\right)}-p^{\\#_{1}\left(t\right)}}\right|_{2}=\left|2^{\lambda\left(t\right)}\chi_{p}\left(t\right)\right|=1$

which shows that, for any $t\in\mathbb{N}_{1}$, both the numerator and
denominator of the rational number $\chi_{p}\left(B\left(t\right)\right)$ are
co-prime to $2$. Consequently, when $\chi_{p}\left(B\left(t\right)\right)$ is
an integer for $t\in\mathbb{N}_{1}$, it is necessarily an odd integer, as
desired.

This proves the Correspondence Theorem.

Q.E.D.

The following table displays the first few values of the various functions
considered thus far. As per Proposition 1.2.3, note that the integer values
attained by $\chi_{3}\left(B\left(t\right)\right)$ and
$\chi_{5}\left(B\left(t\right)\right)$ are all periodic points of the
shortened collatz map $H_{3}$—this includes fixed points at negative integers,
as well.

$t$ | $\\#_{1}\left(t\right)$ | $\lambda\left(t\right)$ | $\chi_{p}\left(t\right)$ | $\chi_{p}\left(B\left(t\right)\right)$ | $\chi_{3}\left(B\left(t\right)\right)$ | $\chi_{5}\left(B\left(t\right)\right)$
---|---|---|---|---|---|---
$0$ | $0$ | $0$ | $0$ | $0$ | $0$ | $0$
$1$ | $1$ | $1$ | $\frac{1}{2}$ | $\frac{1}{2-p}$ | $-1$ | $-\frac{1}{3}$
$2$ | $1$ | $2$ | $\frac{1}{4}$ | $\frac{1}{4-p}$ | $1$ | $-1$
$3$ | $2$ | $2$ | $\frac{2+p}{4}$ | $\frac{2+p}{4-p^{2}}$ | $-1$ | $-\frac{1}{3}$
$4$ | $1$ | $3$ | $\frac{1}{8}$ | $\frac{1}{8-p}$ | $\frac{1}{5}$ | $\frac{1}{3}$
$5$ | $2$ | $3$ | $\frac{4+p}{8}$ | $\frac{4+p}{8-p^{2}}$ | $-7$ | $-\frac{9}{17}$
$6$ | $2$ | $3$ | $\frac{2+p}{8}$ | $\frac{2+p}{8-p^{2}}$ | $-5$ | $-\frac{7}{17}$
$7$ | $3$ | $3$ | $\frac{4+2p+p^{2}}{8}$ | $\frac{4+2p+p^{2}}{8-p^{3}}$ | $-1$ | $-\frac{1}{3}$
$8$ | $1$ | $4$ | $\frac{1}{16}$ | $\frac{1}{16-p}$ | $\frac{1}{13}$ | $\frac{1}{11}$
$9$ | $2$ | $4$ | $\frac{8+p}{16}$ | $\frac{8+p}{16-p^{2}}$ | $\frac{11}{7}$ | $-\frac{13}{9}$
$10$ | $2$ | $4$ | $\frac{4+p}{16}$ | $\frac{4+p}{16-p^{2}}$ | $1$ | $-1$
$11$ | $3$ | $4$ | $\frac{8+4p+p^{2}}{16}$ | $\frac{8+4p+p^{2}}{16-p^{3}}$ | $-\frac{29}{11}$ | $-\frac{53}{109}$
$12$ | $2$ | $4$ | $\frac{2+p}{16}$ | $\frac{2+p}{16-p^{2}}$ | $\frac{5}{7}$ | $-\frac{7}{9}$
$13$ | $3$ | $4$ | $\frac{8+2p+p^{2}}{16}$ | $\frac{8+2p+p^{2}}{16-p^{3}}$ | $-\frac{23}{11}$ | $-\frac{43}{109}$
$14$ | $3$ | $4$ | $\frac{4+2p+p^{2}}{16}$ | $\frac{4+2p+p^{2}}{16-p^{3}}$ | $\frac{19}{7}$ | $-\frac{39}{109}$
Table 1: $\chi_{p}$ and related functions

By looking at $\chi_{p}\left(B\left(t\right)\right)$, we can see definite
patterns emerge, such as the fact that:

$\chi_{p}\left(B\left(2^{n}-1\right)\right)=\frac{1}{2-p},\textrm{ }\forall
n\in\mathbb{N}_{1}$

and, more significant, that $\chi_{3}\left(B\left(t\right)\right)$ is more
often positive than negative, but that the opposite holds for $p=5$ (and,
heuristically, for all odd integers $p\geq 5$). Most significant, in the
author’s opinion, however, is the example of:

$\chi_{p}\left(B\left(10\right)\right)=\frac{4+p}{16-p^{2}}=\frac{1}{4-p}$

In showing that every integer periodic point $x$ of $H_{p}$ is of the form:

$\omega=\chi_{p}\left(B\left(t\right)\right)=\frac{\chi_{p}\left(t\right)}{1-\frac{p^{\\#_{1}\left(t\right)}}{2^{\lambda\left(t\right)}}}=\sum_{k=1}^{\\#_{1}\left(t\right)}\frac{2^{\lambda\left(t\right)-\beta_{k}\left(t\right)-1}p^{k-1}}{2^{\lambda\left(t\right)}-p^{\\#_{1}\left(t\right)}}$

for some $t\in\mathbb{N}_{0}$, Proposition 1.2.3 demonstrates that
$\left|2^{\lambda\left(t\right)}-p^{\\#_{1}\left(t\right)}\right|=1$ is a
_sufficient condition_ for $\chi_{p}\left(B\left(t\right)\right)$ to be a
periodic point of $H_{p}$. However, as the $t=10$ case shows, this is _not_ a
_necessary_ condition, because there could be values of $t$ where the
numerator and denominator of $\chi_{p}\left(B\left(t\right)\right)$ share a
common divisor that, when cancelled out, reduce
$\chi_{p}\left(B\left(t\right)\right)$ to an integer, despite the potentially
large absolute value of $2^{\lambda\left(t\right)}-p^{\\#_{1}\left(t\right)}$
for that value of $t$. In fact, thanks to P. Mihăilescu’s resolution of
Catalan’s Conjecture, it would seem that estimates on the _archimedean_ size
of $2^{\lambda\left(t\right)}-p^{\\#_{1}\left(t\right)}$ do not help us
understand $\chi_{p}\left(B\left(t\right)\right)$ a whit better!

Mihăilescu’s Theorem [2]: The only choice of $x,y\in\mathbb{N}_{1}$ and
$m,n\in\mathbb{N}_{2}$ for which:

$x^{m}-y^{n}=1$

are $x=3$, $m=2$, $y=2$, $n=3$ (that is, $3^{2}-2^{3}=1$).

With Mihăilescu’s Theorem, it is easy to see that, for any odd integer $p\geq
3$, $\left|2^{\lambda\left(t\right)}-p^{\\#_{1}\left(t\right)}\right|$ will
_never_ be equal to $1$ _for any_ $t\geq 3$. Consequently, for all prime
$p\geq 3$, any rational integer value of
$\chi_{p}\left(B\left(t\right)\right)$ (and hence, any periodic point of
$H_{p}$) of the form $\chi_{p}\left(B\left(t\right)\right)$, where $t\geq 3$,
must occur as a result of
$\sum_{k=1}^{\\#_{1}\left(t\right)}2^{\lambda\left(t\right)-\beta_{k}\left(t\right)-1}p^{k-1}$
being a multiple of
$2^{\lambda\left(t\right)}-p^{\\#_{1}\left(t\right)}\in\mathbb{Z}\backslash\left\\{-1,1\right\\}$.
This would suggest that classical techniques of transcendence theory (ex:
Baker’s Method) or additive number theory (ex: the Circle Method) need not
bother to show up if and when the call is ever sent out to gather the
miraculous mathematical army needed to conquer the Collatz Conjecture.
Instead, it seems that the key—at least for proving the (non-)existence of
periodic points—is to understand the roots of the polynomials:

$\sum_{k=1}^{\\#_{1}\left(t\right)}2^{\lambda\left(t\right)-\beta_{k}\left(t\right)-1}x^{k-1},\sum_{k=1}^{\\#_{1}\left(t\right)}p^{k-1}x^{\lambda\left(t\right)-\beta_{k}\left(t\right)-1},2^{\lambda\left(t\right)}-x^{\\#_{1}\left(t\right)},x^{\lambda\left(t\right)}-y^{\\#_{1}\left(t\right)},\textrm{
etc.}$

over the finite fields ($\mathbb{Z}/q\mathbb{Z}$, for prime $q$).

##  2 Our Toolbox

This section is the paper’s principal mathematical toolbox, consisting of all
the lemmata and other assorted knick-knacks we shall need.

###  2.1 Complex-Analytic Methods

We need a slight generalization of Perron’s Formula:

Theorem 2.1.1 Generalized Perron Formula [3]: Consider a sequence
$\left\\{a_{k}\right\\}_{k\in\mathbb{N}_{1}}\subseteq\mathbb{C}$, and let:

$D_{A}\left(s\right)\overset{\textrm{def}}{=}\sum_{n=1}^{\infty}\frac{a_{n}}{n^{s}}$
(42)

Writing $\sigma_{a}\left(D_{A}\right)$ to denote the abscissa of absolute
convergence of $D_{A}$, for any $m\in\mathbb{N}_{0}$ and any
$b>\max\left\\{0,\sigma_{a}\left(D_{A}\right)\right\\}$, we have that:

$\sum_{k=1}^{\left\lfloor x\right\rfloor-1}a_{k}\left(1-\frac{k}{\left\lfloor
x\right\rfloor}\right)^{m}=\frac{m!}{2\pi
i}\int_{b-i\infty}^{b+i\infty}\frac{x^{s}}{s\left(s+1\right)\cdots\left(s+m\right)}D_{A}\left(s\right)ds,\textrm{
}\forall n\in\mathbb{N}_{2}$ (43)

Proposition 2.1.1: Letting $D_{A}$ and $b$ be as given in 43, define
$A:\left[1,\infty\right)\rightarrow\mathbb{C}$ by:

$A\left(x\right)\overset{\textrm{def}}{=}\sum_{k=1}^{\left\lfloor
x\right\rfloor}a_{k}$ (44)

Then:

$\sum_{k=1}^{\left\lfloor x\right\rfloor}A\left(k\right)=\frac{1}{2\pi
i}\int_{b-i\infty}^{b+i\infty}\frac{\left(\left\lfloor
x\right\rfloor+1\right)\left(x+1\right)^{s}}{s\left(s+1\right)}D_{A}\left(s\right)ds,\textrm{
}\forall x\in\mathbb{R}\geq 1$ (45)

$a_{\left\lfloor x\right\rfloor}=\frac{1}{2\pi
i}\int_{b-i\infty}^{b+i\infty}\frac{\left(\left\lfloor
x\right\rfloor+1\right)\left(x+1\right)^{s}-2\left\lfloor x\right\rfloor
x^{s}+\left\lfloor
x-1\right\rfloor\left(x-1\right)^{s}}{s\left(s+1\right)}D_{A}\left(s\right)ds,\textrm{
}\forall x\in\mathbb{R}\geq 1$ (46)

Proof: We apply summation by parts:

$\displaystyle\sum_{k=1}^{\left\lfloor
x\right\rfloor-1}a_{k}\left(1-\frac{k}{\left\lfloor x\right\rfloor}\right)$
$\displaystyle=\left(1-\frac{\left\lfloor x\right\rfloor-1}{\left\lfloor
x\right\rfloor}\right)\sum_{k=1}^{\left\lfloor
x\right\rfloor-1}a_{k}-\sum_{k=1}^{\left\lfloor
x\right\rfloor-2}\left(\left(1-\frac{k+1}{\left\lfloor
x\right\rfloor}\right)-\left(1-\frac{k}{\left\lfloor
x\right\rfloor}\right)\right)\sum_{j=1}^{k}a_{j}$
$\displaystyle=\frac{1}{\left\lfloor x\right\rfloor}A\left(\left\lfloor
x\right\rfloor-1\right)-\sum_{k=1}^{\left\lfloor
x\right\rfloor-2}\left(-\frac{1}{\left\lfloor
x\right\rfloor}\right)A\left(k\right)$ $\displaystyle=\frac{1}{\left\lfloor
x\right\rfloor}\sum_{k=1}^{\left\lfloor x\right\rfloor-1}A\left(k\right)$

Setting $m=1$ in 43, replace the left-hand side with $\frac{1}{\left\lfloor
x\right\rfloor}\sum_{k=1}^{\left\lfloor x\right\rfloor-1}A\left(k\right)$,
multiply both sides by $\left\lfloor x\right\rfloor$, and then replace $x$
with $x+1$; this gives 45.

Applying the backward difference operator:

$\nabla\left\\{f\right\\}\left(x\right)\overset{\textrm{def}}{=}f\left(x\right)-f\left(x-1\right)$

to 45 twice with respect to $x$ then yields 46

Q.E.D.

While the formula for $a_{\left\lfloor x\right\rfloor}$ is good, it can be
made better. At present, the denominator of the integrand introduces a decay
of $O\left(\left|s\right|^{-2}\right)$. Our purposes, however, will require
$O\left(\left|s\right|^{-3}\right)$ decay. This can be done by integrating
with respect to $x$; the fact that $a_{\left\lfloor x\right\rfloor}$ is
constant over the interval $x\in\left(n,n+1\right)$ will then accrue one more
backward difference into our integral.

Proposition 2.1.2: Letting $D_{A}$ and $b$ be as given in 43, we have that:

$a_{n}=\frac{1}{2\pi
i}\int_{b-i\infty}^{b+i\infty}\frac{\kappa_{n}\left(s\right)D_{A}\left(s\right)}{s\left(s+1\right)^{2}}ds,\textrm{
}\forall n\in\mathbb{N}_{1}$ (47)

where:

$\displaystyle\kappa_{n}\left(s\right)$
$\displaystyle=n\left(\left(n+2\right)^{s+1}-3\left(n+1\right)^{s+1}+3n^{s+1}-\left(n-1\right)^{s+1}\right)$
(48)
$\displaystyle+\left(n+2\right)^{s+1}-\left(n+1\right)^{s+1}-\left(n^{s+1}-\left(n-1\right)^{s+1}\right)$
$\displaystyle=n\nabla^{3}\left(\left(n+2\right)^{s+1}\right)+\nabla\left(\left(n+2\right)^{s+1}\right)-\nabla\left(n^{s+1}\right)$

where:

$\nabla^{k}\left(f\left(n\right)\right)\overset{\textrm{def}}{=}\nabla^{k}\left\\{f\right\\}\left(n\right)$

Proof: We note that:

$\int_{n}^{n+1}a_{\left\lfloor x\right\rfloor}dx=a_{n}\int_{n}^{n+1}dx=a_{n}$
$\displaystyle\int_{n}^{n+1}\left(\left\lfloor
x\right\rfloor+c\right)\left(x+c\right)^{s}dx$
$\displaystyle=\left(n+c\right)\int_{n}^{n+1}\left(x+c\right)^{s}dx$
$\displaystyle=\frac{n+c}{s+1}\left(\left(n+c+1\right)^{s+1}-\left(n+c\right)^{s+1}\right)$

As such:

$\int_{n}^{n+1}\left(\left(\left\lfloor
x\right\rfloor+1\right)\left(x+1\right)^{s}-2\left\lfloor x\right\rfloor
x^{s}+\left\lfloor x-1\right\rfloor\left(x-1\right)^{s}\right)dx$

becomes:

$\displaystyle\frac{n}{s+1}\left(\left(n+2\right)^{s+1}-3\left(n+1\right)^{s+1}+3n^{s+1}-\left(n-1\right)^{s+1}\right)$
$\displaystyle+\frac{1}{s+1}\left(\left(n+2\right)^{s+1}-\left(n+1\right)^{s+1}-n^{s+1}+\left(n-1\right)^{s+1}\right)$

which is:

$\frac{n\nabla^{3}\left(\left(n+2\right)^{s+1}\right)}{s+1}+\frac{\nabla\left(\left(n+2\right)^{s+1}\right)-\nabla\left(n^{s+1}\right)}{s+1}$

Consequently:

$\displaystyle a_{n}$ $\displaystyle=\frac{1}{2\pi
i}\int_{b-i\infty}^{b+i\infty}\frac{n\nabla^{3}\left(\left(n+2\right)^{s+1}\right)+\nabla\left(\left(n+2\right)^{s+1}\right)-\nabla\left(n^{s+1}\right)}{s\left(s+1\right)^{2}}D_{A}\left(s\right)ds$
$\displaystyle=\frac{1}{2\pi
i}\int_{b-i\infty}^{b+i\infty}\frac{\kappa_{n}\left(s\right)D_{A}\left(s\right)}{s\left(s+1\right)^{2}}ds$

as desired.

Proposition 2.1.3 (Behavior and Asymptotics of $\kappa_{n}\left(s\right)$):

I. Let $n\in\mathbb{N}_{1}$ be arbitrary. Then, as a function of the complex
variable $s$, $\kappa_{n}\left(s\right)$ is holomorphic for
$\textrm{Re}\left(s\right)>-1$ (and is an entire function whenever $n\geq 2$)
with a simple zero at $s=0$; that is, $\kappa_{n}\left(0\right)=0$ for al
$n\in\mathbb{N}_{1}$. Also, $\kappa_{n}\left(1\right)=4$, for all
$n\in\mathbb{N}_{1}$.

II. For all $s\in\mathbb{C}$, with $\textrm{Re}\left(s\right)=\sigma$:

$\left|\kappa_{n}\left(s\right)\right|\leq\left|s+1\right|\left(\frac{\left|\sigma\left(\sigma-1\right)\right|}{3}+2\right)\left(n+2\right)^{\sigma},\textrm{
}\forall n\in\mathbb{N}_{1}$ (49)

III.

$\kappa_{n}^{\prime}\left(s\right)=n\nabla^{3}\left(\left(n+2\right)^{s+1}\ln\left(n+2\right)\right)+\nabla\left(\left(n+2\right)^{s+1}\ln\left(n+2\right)\right)-\nabla\left(n^{s+1}\ln
n\right)$ (50)

with:

$\displaystyle\left|\kappa_{n}^{\prime}\left(s\right)\right|$
$\displaystyle\leq
n\sup_{x\in\left[n-1,n+2\right]}x^{\sigma-2}\left|3s^{2}-1+s\left(s^{2}-1\right)\ln
x\right|$ (51)
$\displaystyle+2\sup_{x\in\left[n,n+2\right]}\left|\left(s+1\right)\nabla\left(x^{s}\ln
x\right)+\nabla\left(x^{s}\right)\right|$

for all $n\geq 2$. In particular:

$\displaystyle\left|\kappa_{n}^{\prime}\left(0\right)\right|$
$\displaystyle\leq\frac{n}{\left(n-1\right)^{2}}+2\ln\left(1+\frac{1}{n-1}\right)=O\left(\frac{1}{n}\right)$
(52) $\left|\kappa_{n}^{\prime}\left(1\right)\right|\leq
4\ln\left(n+3+\frac{1}{n+1}\right)+4n\ln\left(1+\frac{1}{n+1}\right)+4+\frac{2}{n-1}=O\left(\ln
n\right)$ (53)

Proof: Using:

$\displaystyle\kappa_{n}\left(s\right)$
$\displaystyle=n\left(\left(n+2\right)^{s+1}-3\left(n+1\right)^{s+1}+3n^{s+1}-\left(n-1\right)^{s+1}\right)$
$\displaystyle+\left(\left(n+2\right)^{s+1}-\left(n+1\right)^{s+1}-n^{s+1}+\left(n-1\right)^{s+1}\right)$

we see that for $n\geq 2$, $\kappa_{n}\left(s\right)$ is a linear combination
of functions of the form $a\cdot b^{s}$ for real numbers $a\neq 0$ and $b>0$,
forcing $\kappa_{n}\left(s\right)$ to be entire. When $n=1$, we have
$0^{s+1}$, which is zero for all $\textrm{Re}\left(s\right)>-1$ but is
otherwise undefined or singular for $\textrm{Re}\left(s\right)\leq-1$, showing
that for all $n\geq 1$, $\kappa_{n}\left(s\right)$ is holomorphic for
$\textrm{Re}\left(s\right)>-1$.

Setting $s=1$, the formula for $\kappa_{n}\left(1\right)$ ends up simplifying
to $4$. Setting $s=0$, the formula for $\kappa_{n}\left(0\right)$ ends up
simplifying to $0$, indicating that $\kappa_{n}\left(s\right)$ has a zero at
$s=0$. Differentiating with respect to $s$ gives:

$\kappa_{n}^{\prime}\left(s\right)=n\nabla^{3}\left(\left(n+2\right)^{s+1}\ln\left(n+2\right)\right)+\nabla\left(\left(n+2\right)^{s+1}\ln\left(n+2\right)\right)-\nabla\left(n^{s+1}\ln
n\right)$

(which proves 50 for (III)). Setting $s=0$, we obtain:

$\displaystyle\kappa_{n}^{\prime}\left(0\right)$
$\displaystyle=n\nabla^{3}\left(\left(n+2\right)\ln\left(n+2\right)\right)+\nabla\left(\left(n+2\right)\ln\left(n+2\right)\right)-\nabla\left(n\ln
n\right)$
$\displaystyle=n\left(n+2\right)\ln\left(n+2\right)-3n\left(n+1\right)\ln\left(n+1\right)+3n^{2}\ln
n-n\left(n-1\right)\ln\left(n-1\right)$
$\displaystyle+\left(n+2\right)\ln\left(n+2\right)-\left(n+1\right)\ln\left(n+1\right)$
$\displaystyle-n\ln n+\left(n-1\right)\ln\left(n-1\right)$
$\displaystyle=\left(n+1\right)\left(n+2\right)\ln\left(n+2\right)-\left(3n+1\right)\left(n+1\right)\ln\left(n+1\right)$
$\displaystyle+n\left(3n-1\right)\ln
n-\left(n-1\right)^{2}\ln\left(n-1\right)$

Now, $\kappa_{n}\left(s\right)$ vanishes at $s=0$ with order $1$ if and only
if $\kappa_{n}^{\prime}\left(0\right)\neq 0$. Setting the above expression
equal to $0$ gives:

$1=\frac{\left(n+2\right)^{\left(n+1\right)\left(n+2\right)}n^{n\left(3n-1\right)}}{\left(n+1\right)^{\left(3n+1\right)\left(n+1\right)}\left(n-1\right)^{2\left(n-1\right)}}$

Clearly $n$ cannot be $1$. Next, note that if $n\geq 2$ is even (resp. odd),
the fraction on the right is of the form $\textrm{even}/\textrm{odd}$ (resp.
$\textrm{odd}/\textrm{even}$), and such a fraction can never be equal to $1$;
any fraction $a/b$ which reduces to $1$ must be of the form
$\textrm{even}/\textrm{even}$ or $\textrm{odd}/\textrm{odd}$. Thus,
$\kappa_{n}^{\prime}\left(0\right)\neq 0$ for any $n\in\mathbb{N}_{1}$, which
shows that $\kappa_{n}$’s zero at $s=0$ is indeed of order $1$.

For the estimates on $\kappa_{n}$, we note that:

$x^{s}-\left(x-1\right)^{s}=s\int_{x-1}^{x}t^{s+1}dt$

implies:

$\left|\nabla\left(x^{s}\right)\right|=\left|x^{s}-\left(x-1\right)^{s}\right|\leq\left|s\right|x^{\sigma-1},\textrm{
}\forall x\in\mathbb{R}\geq 1,\textrm{ }\forall
s\in\mathbb{C}\backslash\left\\{0\right\\}$ (54)

Thus, by induction:

$\left|\nabla^{k}\left(x^{s}\right)\right|\leq\left|s\right|\left|\frac{\left(\sigma-1\right)!}{\left(\sigma-k\right)!}\right|x^{\sigma-k},\textrm{
}\forall x\in\mathbb{R}\geq 1,\textrm{ }\forall
s\in\mathbb{C}\backslash\left\\{0\right\\},\textrm{ }\forall
k\in\mathbb{N}_{1}$ (55)

where in both formulae, $\sigma=\textrm{Re}\left(s\right)$.

So:

$\kappa_{n}\left(s\right)=n\nabla^{3}\left(\left(n+2\right)^{s+1}\right)+\nabla\left(\left(n+2\right)^{s+1}\right)-\nabla\left(n^{s+1}\right)$

implies:

$\displaystyle\left|\kappa_{n}\left(s\right)\right|$
$\displaystyle\leq\sigma\left(\sigma-1\right)\left|s+1\right|n\left(n+2\right)^{\sigma-2}+\left|s+1\right|\left(n+2\right)^{\sigma}+\left|s+1\right|n^{\sigma}$
$\displaystyle\leq\left|s+1\right|\left(\frac{\left|\sigma\left(\sigma-1\right)\right|}{n+2}+2\right)\left(n+2\right)^{\sigma}$
$\displaystyle\left(n\geq 1\right);$
$\displaystyle\leq\left|s+1\right|\left(\frac{\left|\sigma\left(\sigma-1\right)\right|}{3}+2\right)\left(n+2\right)^{\sigma}$

which proves 49.

Finally, using

$\kappa_{n}^{\prime}\left(s\right)=n\nabla^{3}\left(\left(n+2\right)^{s+1}\ln\left(n+2\right)\right)+\nabla\left(\left(n+2\right)^{s+1}\ln\left(n+2\right)\right)-\nabla\left(n^{s+1}\ln
n\right)$

we have:

$\displaystyle\nabla\left(\left(n+2\right)^{s+1}\ln\left(n+2\right)\right)-\nabla\left(n^{s+1}\ln
n\right)$
$\displaystyle=\int_{n}^{n+2}\frac{d}{dx}\left\\{\nabla\left\\{x^{s+1}\ln
x\right\\}\left(x\right)\right\\}dx$
$\displaystyle=\int_{n}^{n+2}\left(\left(s+1\right)\nabla\left(x^{s}\ln
x\right)+\nabla\left(x^{s}\right)\right)dx$

and so:

$\displaystyle\left|\nabla\left(\left(n+2\right)^{s+1}\ln\left(n+2\right)\right)-\nabla\left(n^{s+1}\ln
n\right)\right|$ $\displaystyle\leq
2\sup_{x\in\left[n,n+2\right]}\left|\left(s+1\right)\nabla\left(x^{s}\ln
x\right)+\nabla\left(x^{s}\right)\right|$
$\displaystyle=2\sup_{x\in\left[n,n+2\right]}\left|\left(s+1\right)\nabla\left(x^{s}\ln
x\right)+\nabla\left(x^{s}\right)\right|$

For the third-order difference, we proceed inductively:

$\displaystyle\nabla\left(n^{s+1}\ln n\right)$
$\displaystyle=\int_{n-1}^{n}t^{s}\left(1+\left(s+1\right)\ln t\right)dt$
$\displaystyle\Downarrow$ $\displaystyle\nabla^{2}\left(n^{s+1}\ln n\right)$
$\displaystyle=\int_{n-1}^{n}t^{s}\left(1+\left(s+1\right)\ln
t\right)dt-\int_{n-2}^{n-1}t^{s}\left(1+\left(s+1\right)\ln t\right)dt$
$\displaystyle=\int_{n-1}^{n}\left(t^{s}\left(1+\left(s+1\right)\ln
t\right)-\left(t-1\right)^{s}\left(1+\left(s+1\right)\ln\left(t-1\right)\right)\right)dt$
$\displaystyle=\int_{n-1}^{n}\left(\int_{t-1}^{t}\frac{d}{dx}\left\\{x^{s}\left(1+\left(s+1\right)\ln
x\right)\right\\}dx\right)dt$
$\displaystyle=\int_{n-1}^{n}\int_{t-1}^{t}x^{s-1}\left(2s+1+s\left(s+1\right)\ln
x\right)dxdt$ $\displaystyle\Downarrow$
$\displaystyle\nabla^{3}\left(n^{s+1}\ln n\right)$
$\displaystyle=\int_{n-1}^{n}\int_{t-1}^{t}\int_{x-1}^{x}y^{s-2}\left(3s^{2}-1+s\left(s^{2}-1\right)\ln
y\right)dydxdt$ $\displaystyle\Downarrow$
$\displaystyle\left|\nabla^{3}\left(n^{s+1}\ln n\right)\right|$
$\displaystyle\leq\sup_{y\in\left[n-3,n\right]}y^{\sigma-2}\left|3s^{2}-1+s\left(s^{2}-1\right)\ln
y\right|$

and so:

$\left|\nabla^{2}\left(\left(n+2\right)^{s+1}\ln\left(n+2\right)\right)\right|\leq\sup_{x\in\left[n-1,n+2\right]}x^{\sigma-2}\left|3s^{2}-1+s\left(s^{2}-1\right)\ln
x\right|$

Consequently:

$\displaystyle\left|\kappa_{n}^{\prime}\left(s\right)\right|$
$\displaystyle\leq
n\sup_{x\in\left[n-1,n+2\right]}x^{\sigma-2}\left|3s^{2}-1+s\left(s^{2}-1\right)\ln
x\right|$
$\displaystyle+2\sup_{x\in\left[n,n+2\right]}\left|\left(s+1\right)\nabla\left(x^{s}\ln
x\right)+\nabla\left(x^{s}\right)\right|$

which is 51. Plugging in $s=0$ and $s=1$ gives 52 and 53, respectively.

Q.E.D.

Proposition 2.1.4: Let $\left\\{a_{n}\right\\}_{n\geq 1}$ be a sequence of
complex numbers, and suppose that there is a real number $c>0$ so that the
$a_{n}$s’ summatory function, $A\left(N\right)$, satisfies:

$\left|A\left(N\right)\right|\ll N^{c}\ln N\textrm{ as }N\rightarrow\infty$
(56)

Then:

$\sup_{t\in\mathbb{R}}\left|D_{A}\left(\sigma+it\right)\right|<\infty,\textrm{
}\forall\sigma>c$

Proof: Writing $s=\sigma+it$, summing by parts gives:

$\displaystyle\sum_{n=1}^{\infty}\frac{a_{n}}{n^{s}}$
$\displaystyle=\lim_{N\rightarrow\infty}\left(\frac{A\left(N\right)}{N^{s}}+\sum_{n=1}^{N-1}\left(\frac{1}{n^{s}}-\frac{1}{\left(n+1\right)^{s}}\right)A\left(n\right)\right)$
$\displaystyle\left(\textrm{if }\sigma>c\right);$
$\displaystyle=\sum_{n=1}^{\infty}\left(\frac{1}{n^{s}}-\frac{1}{\left(n+1\right)^{s}}\right)A\left(n\right)$
$\displaystyle=\left(1-2^{-s}\right)a_{1}+\sum_{n=2}^{\infty}\frac{1}{n^{s}}\left(1-\frac{1}{\left(1+\frac{1}{n}\right)^{s}}\right)A\left(n\right)$
$\displaystyle\left(k\textrm{-}\Sigma\textrm{ conv. unif. in }n\right);$
$\displaystyle=\left(1-2^{-s}\right)a_{1}-\sum_{k=1}^{\infty}\binom{-s}{k}\sum_{n=2}^{\infty}\frac{A\left(n\right)}{n^{s+k}}$

Since $\left(1-2^{-s}\right)a_{1}$ is uniformly bounded for
$\textrm{Re}\left(s\right)=\sigma$, we have that $\sigma>c$ implies:

$\displaystyle\left|D_{A}\left(\sigma+it\right)\right|$ $\displaystyle\leq
C_{\sigma}+\sum_{k=1}^{\infty}\left|\binom{-s}{k}\sum_{n=2}^{\infty}\frac{A\left(n\right)}{n^{\sigma+it+k}}\right|$
$\displaystyle\ll
C_{\sigma,A}\sum_{k=1}^{\infty}\left|\binom{-s}{k}\sum_{n=2}^{\infty}\frac{\ln
n}{n^{\sigma-c+k}}\right|$
$\displaystyle=C_{\sigma,A}\sum_{k=1}^{\infty}\left|\binom{-s}{k}\zeta^{\prime}\left(\sigma-c+k\right)\right|$

Since:

$\displaystyle\lim_{k\rightarrow\infty}\left|\frac{\binom{-s}{k+1}\zeta^{\prime}\left(\sigma-c+k+1\right)}{\binom{-s}{k}\zeta^{\prime}\left(\sigma-c+k\right)}\right|$
$\displaystyle=\lim_{k\rightarrow\infty}\frac{\left|s+k\right|}{k+1}\frac{\zeta^{\prime}\left(\sigma-c+k+1\right)}{\zeta^{\prime}\left(\sigma-c+k\right)}$
$\displaystyle=\lim_{k\rightarrow\infty}\frac{\zeta^{\prime}\left(\sigma-c+k+1\right)}{\zeta^{\prime}\left(\sigma-c+k\right)}$
$\displaystyle=\lim_{k\rightarrow\infty}\frac{\zeta^{\prime}\left(k+1\right)}{\zeta^{\prime}\left(k\right)}$
$\displaystyle=\lim_{k\rightarrow\infty}\frac{\frac{\ln 2}{2^{k+1}}+\frac{\ln
3}{3^{k+1}}+\cdots}{\frac{\ln 2}{2^{k}}+\frac{\ln 3}{3^{k}}+\cdots}$
$\displaystyle=\lim_{k\rightarrow\infty}\frac{\frac{\ln
2}{2}+\left(\frac{2}{3}\right)^{k}\frac{\ln
3}{3}+\left(\frac{2}{4}\right)^{k}\frac{\ln 4}{4}+\cdots}{\ln
2+\left(\frac{2}{3}\right)^{k}\ln 3+\left(\frac{2}{4}\right)^{k}\ln 4+\cdots}$
$\displaystyle=\lim_{k\rightarrow\infty}\frac{\frac{\ln 2}{2}}{\ln 2}$
$\displaystyle=\frac{1}{2}$

we have that, by the ratio test, when $\sigma>c$, the series:

$\sum_{k=1}^{\infty}\binom{-s}{k}\zeta^{\prime}\left(\sigma-c+k\right)$

converges absolutely for each $s\in\mathbb{C}$; note that this convergence
occurs compactly with respect to $s$. Consequently, when $\sigma>c$,
$\sum_{k=1}^{\infty}\binom{-s}{k}\zeta^{\prime}\left(\sigma-c+k\right)$
defines an entire function of the complex variable $s$. Moreover, it can be
shown that:

$\sigma>c\Rightarrow\sup_{t\in\mathbb{R}}\left|\sum_{k=1}^{\infty}\binom{-\sigma-
it}{k}\zeta^{\prime}\left(\sigma-c+k\right)\right|<\infty$

from which it follows that:

$\sup_{t\in\mathbb{R}}\left|D_{A}\left(\sigma+it\right)\right|\ll
C_{\sigma,A}\sup_{t\in\mathbb{R}}\sum_{k=1}^{\infty}\left|\binom{-\sigma-
it}{k}\zeta^{\prime}\left(\sigma-c+k\right)\right|<\infty$

as desired.

Q.E.D.

Lemma 2.1.1: Let $\left\\{a_{n}\right\\}_{n\geq 1}$ be a sequence of complex
numbers, and suppose that there is a real number $c>0$ so that the $a_{n}$s’
summatory function, $A\left(N\right)$, satisfies:

$\left|A\left(N\right)\right|\ll N^{c}\ln N\textrm{ as }N\rightarrow\infty$

Then, for all $b>c$, the integral:

$\int_{b-i\infty}^{b+i\infty}\frac{\kappa_{n}\left(s\right)D_{A}\left(s\right)}{s\left(s+1\right)^{2}}ds$

is absolutely convergent for all $n\in\mathbb{N}_{1}$.

Proof: For brevity, let us write:

$K\left(x,s\right)\overset{\textrm{def}}{=}\left(\left\lfloor
x\right\rfloor+1\right)\left(x+1\right)^{s}-2\left\lfloor x\right\rfloor
x^{s}+\left\lfloor x-1\right\rfloor\left(x-1\right)^{s}$ (57)

so that 46 becomes:

$a_{\left\lfloor x\right\rfloor}=\frac{1}{2\pi
i}\int_{b-i\infty}^{b+i\infty}\frac{K\left(x,s\right)}{s\left(s+1\right)}D_{A}\left(s\right)ds$
(58)

The growth condition on $\left|A\left(N\right)\right|$ guarantees that
$C_{A,b}\overset{\textrm{def}}{=}\sup_{t\in\mathbb{R}}\left|D_{A}\left(b+it\right)\right|<\infty$
for all $b>c$, and hence, for all $x>1$:

$\displaystyle\int_{b-i\infty}^{b+i\infty}\left|\frac{K\left(x,s\right)}{s\left(s+1\right)}D_{A}\left(s\right)\right|ds$
$\displaystyle\leq\left(\left\lfloor
x\right\rfloor+1\right)\int_{b-i\infty}^{b+i\infty}\left|\frac{\left(x+1\right)^{s}}{s\left(s+1\right)}D_{A}\left(s\right)\right|ds$
$\displaystyle+2\left\lfloor
x\right\rfloor\int_{b-i\infty}^{b+i\infty}\left|\frac{x^{s}}{s\left(s+1\right)}D_{A}\left(s\right)\right|ds$
$\displaystyle+\left\lfloor
x-1\right\rfloor\int_{b-i\infty}^{b+i\infty}\left|\frac{\left(x-1\right)^{s}}{s\left(s+1\right)}D_{A}\left(s\right)\right|ds$
$\displaystyle\leq C_{A,b}^{\prime}\left(\left(\left\lfloor
x\right\rfloor+1\right)\left(x+1\right)^{b}+2\left\lfloor x\right\rfloor
x^{b}+\left\lfloor x-1\right\rfloor\left(x-1\right)^{b}\right)$

where:

$C_{A,b}^{\prime}=C_{A,b}\int_{b-i\infty}^{b+i\infty}\frac{ds}{\left|s\left(s+1\right)\right|}<\infty$

Thus, 58 converges absolutely for all $x>1$, with:

$a_{\left\lfloor x\right\rfloor}\ll\left(\left\lfloor
x\right\rfloor+1\right)\left(x+1\right)^{b}+2\left\lfloor x\right\rfloor
x^{b}+\left\lfloor
x-1\right\rfloor\left(x-1\right)^{b}\ll\left(x+1\right)^{b+1}$

Since we obtained 47 by integrating 46 with respect to
$x\in\left(n,n+1\right)$ we have that:

$\displaystyle\int_{b-i\infty}^{b+i\infty}\left|\frac{\kappa_{n}\left(s\right)D_{A}\left(s\right)}{s\left(s+1\right)^{2}}\right|ds$
$\displaystyle=\int_{b-i\infty}^{b+i\infty}\left|\int_{n}^{n+1}\frac{K\left(x,s\right)D_{A}\left(s\right)}{s\left(s+1\right)}dx\right|ds$
$\displaystyle\leq\int_{b-i\infty}^{b+i\infty}\left(\int_{n}^{n+1}\left|\frac{K\left(x,s\right)D_{A}\left(s\right)}{s\left(s+1\right)}\right|dx\right)ds$
$\displaystyle\left(\textrm{Fubini}\right);$
$\displaystyle\leq\int_{n}^{n+1}\left(\int_{b-i\infty}^{b+i\infty}\left|\frac{K\left(x,s\right)D_{A}\left(s\right)}{s\left(s+1\right)}\right|ds\right)dx$
$\displaystyle\ll\int_{n}^{n+1}\left(x+1\right)^{b+1}dx$
$\displaystyle\leq\sup_{x\in\left[n,n+1\right]}\left(x+1\right)^{b+1}$

which shows that our integral is indeed absolutely convergent for all $n\geq
1$.

Q.E.D.

While we have shown that the integral will converge, we need stronger
estimates on $\kappa_{n}\left(s\right)$ to be able to get what we want.

Lemma 2.1.2: For all $s=\sigma+it\in\mathbb{C}$ and all $n\in\mathbb{N}_{2}$:

$\left|\kappa_{n}\left(s\right)\right|\ll\left|s\right|n^{\sigma}\textrm{ as
}\left|\textrm{Im}\left(s\right)\right|\rightarrow\infty$ (59)

where the constant of proportionality is independent of both
$\textrm{Im}\left(s\right)$ and $n$.

Proof: Let:

$M_{n}\left(s\right)\overset{\textrm{def}}{=}\frac{\kappa_{n}\left(s\right)}{n^{s}}$
(60)

Noting that for any real number $r$:

$\int_{1}^{\infty}\frac{x^{r}}{x^{s+1}}dx=\frac{1}{s-r},\textrm{
}\forall\textrm{Re}\left(s\right)>r$ (61)

we have for:

$\sigma_{n,b}\overset{\textrm{def}}{=}\inf\left\\{\sigma\in\mathbb{R}:\int_{1}^{\infty}\left|\frac{M_{n}\left(b+ix\right)}{x^{\sigma+1}}\right|dx<\infty\right\\}$
(62)

$M_{n}$ satisfies111Technically, 61 only implies
$\left|M_{n}\left(b+it\right)\right|\ll\left|t\right|^{\sigma_{n,b}}\textrm{
as }t\rightarrow\infty$. However, by the Reflection Principle:
$M_{n}\left(b-it\right)=\overline{M_{n}\left(b+it\right)}$ and hence, the
bound for $t\rightarrow\infty$ also holds as $t\rightarrow-\infty$.:

$\left|M_{n}\left(b+it\right)\right|\ll\left|t\right|^{\sigma}\textrm{ as
}\left|t\right|\rightarrow\infty,\textrm{ }\forall\sigma\geq\sigma_{n,b}$ (63)

Now:

$\displaystyle\left|\int_{1}^{\infty}\frac{\kappa_{n}\left(b+ix\right)}{n^{b+ix}}\frac{1}{x^{s+1}}dx\right|$
$\displaystyle=\left|\frac{1}{n^{b}}\int_{1}^{\infty}\frac{\kappa_{n}\left(b+ix\right)e^{-i\left(\ln
n\right)x}}{x^{s+1}}dx\right|$ $\displaystyle\left(\textrm{use
}\ref{eq:kappa_n asymptotics}\right);$
$\displaystyle\leq\frac{1}{n^{b}}\int_{1}^{\infty}\frac{\left|b+1+ix\right|\left(\frac{\left|b\left(b-1\right)\right|}{3}+2\right)\left(n+2\right)^{b}}{x^{\sigma+1}}dx$
$\displaystyle=\left(\frac{\left|b\left(b-1\right)\right|}{3}+2\right)\frac{\left(n+2\right)^{b}}{n^{b}}\int_{1}^{\infty}\frac{\sqrt{x^{2}+\left(b+1\right)^{2}}}{x^{\sigma+1}}dx$
$\displaystyle\ll\left(\frac{\left|b\left(b-1\right)\right|}{3}+2\right)\underbrace{\frac{\left(n+2\right)^{b}}{n^{b}}}_{O\left(1\right)}\int_{1}^{\infty}\frac{x}{x^{\sigma+1}}dx$
$\displaystyle\ll\frac{1}{\sigma-1}$

Thus, $\sigma=1$ must be greater than or equal to $\sigma_{n,b}$, and so:

$\left|M_{n}\left(b+it\right)\right|\ll\left|t\right|\textrm{ as
}\left|t\right|\rightarrow\infty$

where the constant of proportionality is independent of $n$. Consequently:

$\left|\kappa_{n}\left(\sigma+it\right)\right|=\left|n^{s}M_{n}\left(s\right)\right|\ll\left|t\right|n^{\sigma}\ll\left|s\right|n^{\sigma}\textrm{
as }\left|t\right|=\left|\textrm{Im}\left(s\right)\right|\rightarrow\infty$

as desired.

Q.E.D.

Our final proposition is a computation of an upper bound on an integral which
occurs when estimating the integral in Lemma 2.1.1.

Proposition 2.1.5:

$\left|\int_{b-i\infty}^{b+i\infty}\frac{dz}{\left|z\right|\left|z+1\right|^{2}}\right|=\frac{2}{b^{2}}\frac{\textrm{arccosh}\left(\frac{b^{2}+4b+2}{b^{2}}\right)}{\sqrt{\left(\frac{b^{2}+4b+2}{b^{2}}\right)^{2}-1}},\textrm{
}\forall b\in\mathbb{R}\backslash\left\\{0,-1\right\\}$

Proof: Letting $x=\frac{z-b}{i}$, $dx=\frac{1}{i}dz$

$\displaystyle\left|\frac{1}{i}\int_{b-i\infty}^{b+i\infty}\frac{dz}{\left|z\right|\left|z+1\right|^{2}}\right|$
$\displaystyle=\int_{-\infty}^{\infty}\frac{dx}{\left|b+ix\right|\left|b+1+ix\right|^{2}}$
$\displaystyle=\int_{-\infty}^{\infty}\frac{dx}{\sqrt{x^{2}+b^{2}}\left(x^{2}+\left(b+1\right)^{2}\right)}$
$\displaystyle=2\int_{0}^{\infty}\frac{dx}{\sqrt{x^{2}+b^{2}}\left(x^{2}+\left(b+1\right)^{2}\right)}$

Letting:

$x=\sqrt{u-\left(b+1\right)^{2}}$
$dx=\frac{du}{2\sqrt{u-\left(b+1\right)^{2}}}$

we have that:

$\displaystyle\frac{1}{i}\int_{b-i\infty}^{b+i\infty}\frac{dz}{\left|z\right|\left|z+1\right|^{2}}$
$\displaystyle=2\int_{0}^{\infty}\frac{dx}{\sqrt{x^{2}+b^{2}}\left(x^{2}+\left(b+1\right)^{2}\right)}$
$\displaystyle=\int_{\left(b+1\right)^{2}}^{\infty}\frac{dx}{u\sqrt{\left(u+b^{2}-\left(b+1\right)^{2}\right)\left(u-\left(b+1\right)^{2}\right)}}$
$\displaystyle=\int_{\left(b+1\right)^{2}}^{\infty}\frac{du}{u\sqrt{\left(u^{2}-\left(b^{2}+4b+2\right)u+\left(2b+1\right)\left(b+1\right)^{2}\right)}}$
$\displaystyle=\int_{\left(b+1\right)^{2}}^{\infty}\frac{du}{u\sqrt{\left(u-\frac{b^{2}+4b+2}{2}\right)^{2}-\frac{b^{4}}{4}}}$
$\displaystyle=\int_{\frac{b^{2}}{2}}^{\infty}\frac{du}{\left(u+\frac{b^{2}+4b+2}{2}\right)\sqrt{u^{2}-\frac{b^{4}}{4}}}$
$\displaystyle=\int_{\frac{b^{2}}{2}}^{\infty}\frac{\left(2/b^{2}\right)du}{\left(u+\frac{b^{2}+4b+2}{2}\right)\sqrt{\left(2u/b^{2}\right)^{2}-1}}$
$\displaystyle\left(t=\frac{2u}{b^{2}}\right);$
$\displaystyle=\frac{2}{b^{2}}\int_{1}^{\infty}\frac{1}{t+\frac{b^{2}+4b+2}{b^{2}}}\frac{dt}{\sqrt{t^{2}-1}}$
$\displaystyle\left(\textrm{Wolfram Alpha}\right);$
$\displaystyle=\frac{2}{b^{2}}\frac{\textrm{arccosh}\left(\frac{b^{2}+4b+2}{b^{2}}\right)}{\sqrt{\left(\frac{b^{2}+4b+2}{b^{2}}\right)^{2}-1}}$

Q.E.D.

Finally, because it will appear _everywhere_ , we adopt the notation:

$\sigma_{p}\overset{\textrm{def}}{=}\log_{2}\left(\frac{p+1}{2}\right)$ (64)

###  2.2 Rising-continuity and functions from $\mathbb{Z}_{p}$ to
$\mathbb{Z}_{q}$

In dealing with $\chi_{p}$, we will vary the particular domain and range/co-
domain being used to make sense of the function. Sometimes we will work
$p$-adically; other times, we will work in $\mathbb{C}$. Intermixing
archimedean and non-archimedean convergence in this manner is infamously
dangerous. In general, if a series of rational numbers
$\sum_{n=0}^{\infty}a_{n}$ converges to a sum $s$ in $\mathbb{R}$ and to a sum
$s_{p}$ in $\mathbb{Z}_{p}$, it _cannot_ be assumed that $s=s_{p}$. This
mistake is as old as the $p$-adic numbers themselves, going all the way back
to Kurt Hensel himself.[16] However, there are two special cases where $s$ and
$s_{p}$ must be equal, and these are all the cases we need:

Case 1: All but finitely many of the $a_{n}$s are $0$. Here, the sum
$\sum_{n=0}^{\infty}a_{n}$ reduces to a sum of finitely many non-zero rational
numbers, which is necessarily convergent in any completion of $\mathbb{Q}$.

Case 2: All but finitely many of the $a_{n}$s form a geometric sequence. That
is, there is a rational number $r$ and an $n_{0}\in\mathbb{N}_{0}$ so that
$a_{n_{0}+k}=a_{n_{0}}r^{k}$ for all $k\in\mathbb{N}_{0}$. The reason for this
is due to the _universality_ of the geometric series formula:

$\sum_{n=0}^{\infty}r^{n}=\frac{1}{1-r}$

which is to say, if a geometric series of rational numbers converges in both
$\mathbb{R}$ and $\mathbb{Z}_{p}$, then the sum of the series in $\mathbb{R}$
is the same as the sum of the series in $\mathbb{Z}_{p}$.

We start with van der Put’s basis for continuous functions on
$\mathbb{Z}_{p}$.

Theorem 2.2.1 (The van der Put Theorem (vdPT))[2]: Letting $p$ be a prime,
number, define the following222The author writes $\mathbb{N}_{\kappa}$ to
denote the set of all integers $\geq\kappa$:

I. Let $\lambda_{p}:\mathbb{N}_{0}\rightarrow\mathbb{N}_{0}$ be the function
so that $\lambda_{p}\left(n\right)$ denotes the number of digits in the base
$p$ expansion of $n$; we define $\lambda_{p}\left(0\right)$ to be $0$. (Note
that $\lambda_{p}\left(n\right)=\left\lfloor\log_{p}n\right\rfloor+1$ for all
$n\in\mathbb{N}_{0}$, and
$\lambda_{p}\left(n\right)=\left\lceil\log_{p}\left(n+1\right)\right\rceil$
for all $n\in\mathbb{N}_{0}$.)

II. We call the indicator functions
$\left[\mathfrak{z}\overset{p^{\lambda_{p}\left(j\right)}}{\equiv}j\right]$
the van der Put functions and refer to
$\left\\{\left[\mathfrak{z}\overset{p^{\lambda_{p}\left(j\right)}}{\equiv}j\right]\right\\}_{j\in\mathbb{N}_{0}}$
as the van der Put basis.

III. Given an integer $n\geq 1$, let $n_{-}$ denote the integer obtained by
deleting the right-most/terminal non-zero $p$-adic digit of $n$. That is:

$n_{-}=n-n_{\lambda_{p}\left(n\right)-1}p^{\lambda_{p}\left(n\right)-1}$ (65)

where $n_{\lambda_{p}\left(n\right)-1}$ is the coefficient of the largest
power of $p$ present in the $p$-adic expansion of $n$.

With this terminology, let $K$ be any complete field extension of
$\mathbb{Z}_{p}$, and let $C\left(\mathbb{Z}_{p},K\right)$ be the $K$-vector
space of continuous functions $f:\mathbb{Z}_{p}\rightarrow K$. Then:

I. For any choice of constants
$\left\\{c_{n}\right\\}_{n\in\mathbb{N}_{0}}\subseteq K$, the van der Put
series:

$c_{0}+\sum_{n=0}^{\infty}c_{n}\left[\mathfrak{z}\overset{p^{\lambda_{p}\left(n\right)}}{\equiv}n\right]$

converges in $K$ uniformly for $\mathfrak{z}\in\mathbb{Z}_{p}$ if and only if
$\left|c_{n}\right|_{p}\rightarrow 0$ as $n\rightarrow\infty$.

II. Let $f:\mathbb{Z}_{p}\rightarrow K$ be arbitrary, and set:

$c_{n}\left(f\right)\overset{\textrm{def}}{=}\begin{cases}f\left(0\right)&\textrm{if
}n=0\\\ f\left(n\right)-f\left(n_{-}\right)&\textrm{if }n\geq
1\end{cases},\textrm{ }\forall n\in\mathbb{N}_{0}$ (66)

We call $c_{n}\left(f\right)$ the $n$th van der Put coefficient of $f$. Then,
$\left|c_{n}\left(f\right)\right|_{p}\rightarrow 0$ as $n\rightarrow\infty$ if
and only if $f\in C\left(\mathbb{Z}_{p},K\right)$. When $f\in
C\left(\mathbb{Z}_{p},K\right)$, $f$ has the van der Put series
representation:

$f\left(\mathfrak{z}\right)\overset{K}{=}c_{0}\left(f\right)+\sum_{n=0}^{\infty}c_{n}\left(f\right)\left[\mathfrak{z}\overset{p^{\lambda_{p}\left(n\right)}}{\equiv}n\right],\textrm{
}\forall\mathfrak{z}\in\mathbb{Z}_{p}$ (67)

which is uniformly convergent333We write $\overset{K}{=}$ to make clear that
the equality in question occurs in $K$ and that all notions of convergence
utilized on either side of the equality are those belonging to $K$ and its
$p$-adic absolute value. over $\mathbb{Z}_{p}$.

III. Every $f\in C\left(\mathbb{Z}_{p},K\right)$ is uniquely determined by its
the van der Put coefficients,
$\left\\{\text{$c_{n}\left(f\right)$}\right\\}_{n\in\mathbb{N}_{0}}$.

This construction extends to a more general class of functions, such as
$C\left(\mathbb{Z}_{p},\mathbb{Z}_{q}\right)$, the space of continuous
$q$-adic integer valued functions of a $p$-adic integer variable. In
particular, the author found it useful to coin the following bits of
terminology:

Definition 2.2.1: Let $p$ and $q$ be primes, possibly distinct.

I. We write $\mathbb{Z}_{p}^{\prime}$ to denote the set:

$\mathbb{Z}_{p}^{\prime}\overset{\textrm{def}}{=}\mathbb{Z}_{p}\backslash\mathbb{N}_{0}=\left\\{\mathfrak{z}\in\mathbb{Z}_{p}:\mathfrak{z}\textrm{
is not a non-negative real integer}\right\\}$ (68)

II. A sequence
$\left\\{\mathfrak{z}_{n}\right\\}_{n\in\mathbb{N}_{0}}\subseteq\mathbb{Z}_{p}$
is said to be (_p_ -adically) rising (or is called a (_p_ -adically)__ rising
sequence) whenever it is convergent and its limit is an element of
$\mathbb{Z}_{p}^{\prime}$.

_Remark_ : The author chose the term “rising” to refer to such sequences to
reflect the fact that a sequence
$\left\\{n_{j}\right\\}_{j\in\mathbb{N}_{0}}\subseteq\mathbb{N}_{0}$ of non-
negative integers is $p$-adically rising if and only if the number of non-zero
$p$-adic digits in each $n_{j}$ “rises” to $\infty$ as $j\rightarrow\infty$.

III. Let $K$ be a complete ring extension of $\mathbb{Z}_{q}$, and let
$f:\mathbb{Z}_{p}\rightarrow K$ be a function. We say $f$ is rising-continuous
whenever
$\lim_{n\rightarrow\infty}f\left(\mathfrak{z}_{n}\right)\overset{K}{=}f\left(\lim_{n\rightarrow\infty}\mathfrak{z}_{n}\right)$
holds for every rising sequence
$\left\\{\mathfrak{z}_{n}\right\\}_{n\in\mathbb{N}_{0}}\subseteq\mathbb{Z}_{p}$.

Proposition 2.2.1: Let $K$ be a complete ring extension of $\mathbb{Z}_{q}$,
and let $f:\mathbb{Z}_{p}\rightarrow K$ be a function. Then, $f$ is rising-
continuous if and only if $f$ both the following conditions hold true:

I. $f$ is continuous on $\mathbb{Z}_{p}^{\prime}$.

II.
$\lim_{j\rightarrow\infty}f\left(n_{j}\right)\overset{K}{=}f\left(\lim_{j\rightarrow\infty}n_{j}\right)$
holds true for all sequences of non-negative integers $\left\\{n_{j}\right\\}$
which are $p$-adically rising.

_Remark_ :__ As the author discovered when reading through [1], exercise 62.B
on page 192 of that text contains half of the characterization of rising-
continuity; the requirement that (II) occur for all rising sequences in
$\mathbb{Z}_{p}$ (rather than arbitrary convergent sequences in
$\mathbb{Z}_{p}$) is what is needed to guarantee to make things work.

The utility of the notion of “rising continuity” occurs in the light of the
interpolation of a function $f:\mathbb{N}_{0}\rightarrow\mathbb{Q}$ to a
function $\tilde{f}:\mathbb{Z}_{p}\rightarrow\mathbb{Z}_{q}$. First, however,
a useful identity:

Proposition 2.2.2: Consider a function
$f:\mathbb{Z}_{p}\rightarrow\mathbb{F}$, where $\mathbb{F}$ is a valued field
of characteristic zero which is complete as a metric space with respect to the
absolute value induced by $\mathbb{F}$’s valuation. Then, we have the
identity:

$\sum_{n=0}^{\infty}c_{n}\left(f\right)\left[\mathfrak{z}\overset{p^{\lambda_{p}\left(n\right)}}{\equiv}n\right]\overset{\mathbb{F}}{=}\lim_{k\rightarrow\infty}f\left(\left[\mathfrak{z}\right]_{p^{k}}\right),\textrm{
}\forall\mathfrak{z}\in\mathbb{Z}_{p}$ (69)

where $\left[\mathfrak{z}\right]_{p^{k}}$ denotes the unique integer in
$\left\\{0,\ldots,p^{k}-1\right\\}$ congruent to $\mathfrak{z}$ modulo
$p^{k}$. We call this the van der Put identity.

Definition 2.2.2: Let $p$ and $q$ be primes, let $K$ be a complete ring
extension of $\mathbb{Z}_{q}$, and let $f:\mathbb{N}_{0}\rightarrow K$ be a
function. We say $f$ has/admits a rising-continuation to $\mathbb{Z}_{p}$ / is
(_p_ -adically) rising-continuable to $\mathbb{Z}_{p}$ whenever there exists a
rising-continuous function $g:\mathbb{Z}_{p}\rightarrow K$ so that
$g\left(n\right)=f\left(n\right)$ for all $n\in\mathbb{N}_{0}$. We call such a
$g$ a (_p_ -adic)__ rising-continuation of $f$ (to $\mathbb{Z}_{p}$).

Lemma 2.2.1: Let $f:\mathbb{N}_{0}\rightarrow K$ be a function admitting a
rising-continuation to $\mathbb{Z}_{p}$. Then:

I. The rising-continuation of $f$ is unique.

II. Letting $\tilde{f}$ denote the rising-continuation of $f$, we have that:

$\tilde{f}\left(\mathfrak{z}\right)=\lim_{j\rightarrow\infty}f\left(\left[\mathfrak{z}\right]_{p^{j}}\right),\textrm{
}\forall\mathfrak{z}\in\mathbb{Z}_{p}$

that is, we make sense of $f\left(\mathfrak{z}\right)$ for
$\mathfrak{z}\in\mathbb{Z}_{p}^{\prime}$ by using the above limit procedure
(the convergence here is point-wise with respect to $\mathfrak{z}$). In fact,
more generally, we have that:

$\tilde{f}\left(\mathfrak{z}\right)=\lim_{j\rightarrow\infty}f\left(n_{j}\right),\textrm{
}\forall\mathfrak{z}\in\mathbb{Z}_{p}^{\prime}$

for any rising sequence
$\left\\{n_{j}\right\\}_{j\in\mathbb{N}_{1}}\subseteq\mathbb{N}_{0}$
converging to $\mathfrak{z}\in\mathbb{Z}_{p}^{\prime}$.

Rising-continuations always admit van der Put series representations.

Theorem 2.2.1: Let $p$ and $q$ be primes, let $K$ be a complete ring extension
of $\mathbb{Z}_{q}$, let $f:\mathbb{N}_{0}\rightarrow K$ be a rising-
continuable function, and let $\tilde{f}$ be the rising-continuation of $f$ to
$\mathbb{Z}_{p}$. Then, $\tilde{f}$ is given by:

$\tilde{f}\left(\mathfrak{z}\right)\overset{K}{=}\sum_{n=0}^{\infty}c_{n}\left(f\right)\left[\mathfrak{z}\overset{p^{\lambda_{p}\left(n\right)}}{\equiv}n\right]$
(70)

where the series converges _point-wise_(to $\tilde{f}$) for all
$\mathfrak{z}\in\mathbb{Z}_{p}$.

Proof: Use 69 and Lemma 2.2.1.

Q.E.D.

Next, we show that, for certain values of $\mathfrak{z}$, under the
appropriate circumstances, it is possible for the van der Put series of
$\tilde{f}$ to converge to $\tilde{f}\left(\mathfrak{z}\right)$ not only in
the non-archimidean topology of $K$, but in the archimedean topology of
$\mathbb{R}$ and/or $\mathbb{C}$, as well! This result is best described as an
interpolation theorem for real or complex-valued functions on the non-negative
integers that happen to admit a rising-continuation to a rising-continuous
function from a $p$-adic space to a $q$-adic one.

Theorem 2.2.2: Let $p$ and $q$ be a primes, let $K$ be a complete ring
extension of $\mathbb{Z}_{q}$, suppose that
$K^{\prime}\overset{\textrm{def}}{=}\mathbb{C}\cap K$ is non-empty, and let
$f:\mathbb{N}_{0}\rightarrow K$ be a function whose image (that is,
$f\left(\mathbb{N}_{0}\right)$) lies in $K^{\prime}$. If $f$ has a rising-
continuation $\tilde{f}:\mathbb{Z}_{p}\rightarrow K$, then:

$\tilde{f}\left(\mathfrak{z}\right)\overset{\mathbb{C},K}{=}\sum_{n=0}^{\infty}c_{n}\left(f\right)\left[\mathfrak{z}\overset{p^{\lambda_{p}\left(n\right)}}{\equiv}n\right]$
(71)

(meaning that the van der put series converges in both $\mathbb{C}$ _and_ $K$
to the true value of $\tilde{f}\left(\mathfrak{z}\right)$) occurs for all
$\mathfrak{z}\in\mathbb{Z}_{p}$ for which either of the following conditions
hold true:

I.
$c_{n}\left(f\right)\left[\mathfrak{z}\overset{p^{\lambda_{p}\left(n\right)}}{\equiv}n\right]=0$
for all but at most finitely many $n\in\mathbb{N}_{0}$.

II. There exist constants $a,r\in K^{\prime}$ so that:

$\sum_{n=0}^{\infty}c_{n}\left(f\right)\left[\mathfrak{z}\overset{p^{\lambda_{p}\left(n\right)}}{\equiv}n\right]\overset{K^{\prime}}{=}a\sum_{n=0}^{\infty}r^{n}$

where $\max\left\\{\left|r\right|_{\mathbb{C}},\left|r\right|_{q}\right\\}<1$,
where $\left|\cdot\right|_{\mathbb{C}}$ is the absolute value on $\mathbb{C}$.
In particular, for such a $\mathfrak{z}$, we have that:

$\tilde{f}\left(\mathfrak{z}\right)\overset{\mathbb{C}}{=}\frac{a}{1-r}$

We write $D_{p,q}\left(f\right)$ or $D_{p,q}\left(\tilde{f}\right)$ to denote
the set of all $\mathfrak{z}\in\mathbb{Z}_{p}$ for which 71 holds true. We
call $D_{p,q}\left(f\right)$ or $D_{p,q}\left(\tilde{f}\right)$ $f$’s (or
$\tilde{f}$’s) domain of double-convergence.

Proof: Let everything be as given.

I. Suppose
$c_{n}\left(f\right)\left[\mathfrak{z}\overset{p^{\lambda_{p}\left(n\right)}}{\equiv}n\right]=0$
for all but at most finitely many $n\in\mathbb{N}_{0}$. Since
$c_{n}\left(f\right)\in K^{\prime}\subseteq\mathbb{C}$ for all
$n\in\mathbb{N}_{0}$, we then have that 71 reduces to a sum of finitely many
terms in $K^{\prime}$. Being finite, this sum necessarily converges, and does
so in both $K$ and $\mathbb{C}$, and its sum is an element of $K^{\prime}$. By
Theorem 2.2.1, the convergence of the finite sum in $K^{\prime}\subseteq K$
tells us that the value of the sum is equal to the true value of
$\tilde{f}\left(\mathfrak{z}\right)$. On the other hand, since $K^{\prime}$ is
also a subring of $\mathbb{C}$, this tells us that
$\tilde{f}\left(\mathfrak{z}\right)\in\mathbb{C}$, and hence, that the value
to which the series converges in $\mathbb{C}$ is the same as the value to
which it converges in $K$.

II. Suppose there are $a,r\in K^{\prime}$, with
$\max\left\\{\left|r\right|_{\mathbb{C}},\left|r\right|_{q}\right\\}<1$, so
that

$\sum_{n=0}^{\infty}c_{n}\left(f\right)\left[\mathfrak{z}\overset{p^{\lambda_{p}\left(n\right)}}{\equiv}n\right]\overset{K^{\prime}}{=}a\sum_{n=0}^{\infty}r^{n}$

Since $\left|r\right|_{q}<1$ and $\left|r\right|_{\mathbb{C}}<1$ we have that:

$\sum_{n=0}^{\infty}r^{n}\overset{K}{=}\frac{1}{1-r}$
$\sum_{n=0}^{\infty}r^{n}\overset{\mathbb{C}}{=}\frac{1}{1-r}$

as such, by the universality of the geometric series formula, it follows that
the sum of this geometric series in $K$ is equal to (as an element of
$K^{\prime}$) its sum in $\mathbb{C}$. Consequently:

$\sum_{n=0}^{\infty}c_{n}\left(f\right)\left[\mathfrak{z}\overset{p^{\lambda_{p}\left(n\right)}}{\equiv}n\right]\overset{K^{\prime}}{=}a\sum_{n=0}^{\infty}r^{n}\overset{K^{\prime}}{=}a\times\frac{1}{1-r}$

Since $a$ and $\frac{1}{1-r}$ are both elements of $K^{\prime}$, their product
is also in $K^{\prime}$, and is thus simultaneously in $K$ and in
$\mathbb{C}$. Thus:

$\tilde{f}\left(\mathfrak{z}\right)=\sum_{n=0}^{\infty}c_{n}\left(f\right)\left[\mathfrak{z}\overset{p^{\lambda_{p}\left(n\right)}}{\equiv}n\right]\overset{\mathbb{C}}{=}\frac{a}{1-r}\overset{K}{=}\frac{a}{1-r}$

as desired.

Q.E.D.

Using the fact that for each $\mathfrak{z}\in\mathbb{N}_{0}$, all but finitely
many terms of $\tilde{f}\left(\mathfrak{z}\right)$’s van der Put series
vanish, we can truncate the van der Put series to obtain a closed-form
expression for $f:\mathbb{N}_{0}\rightarrow\mathbb{Q}$ which converges for all
$z\in\mathbb{N}_{0}$. Similarly, we can truncate
$\tilde{f}\left(\mathfrak{z}\right)$ to obtain obtain a closed-form expression
for $f$, albeit over $\mathbb{Z}_{p}$, which converges to the true value of
$f$ for all sufficiently small $x\in\mathbb{N}_{0}$

Corollary 2.2.1: Let $p$ and $q$ be a primes, let $K$ be a complete ring
extension of $\mathbb{Z}_{q}$, suppose that
$K^{\prime}\overset{\textrm{def}}{=}\mathbb{C}\cap K$ is non-empty, and let
$f:\mathbb{N}_{0}\rightarrow K$ be a $p$-adically rising-continuable function
whose image (that is, $f\left(\mathbb{N}_{0}\right)$) lies in $K^{\prime}$.
Then:

I.

$f\left(x\right)\overset{\mathbb{C}}{=}c_{x}\left(f\right)+\sum_{n=0}^{p^{\lambda_{p}\left(x\right)-1}-1}c_{n}\left(f\right)\left[n\overset{p^{\lambda_{p}\left(n\right)}}{\equiv}x\right],\textrm{
}\forall x\in\mathbb{N}_{0}$ (72)

where the sum is defined to be $0$ when $x=0$. This can also be written as:

$f\left(x\right)\overset{\mathbb{C}}{=}c_{0}\left(f\right)+c_{x}\left(f\right)+\sum_{m=1}^{\lambda_{p}\left(x\right)-1}\sum_{n=p^{m-1}}^{p^{m}-1}c_{n}\left(f\right)\left[n\overset{p^{m}}{\equiv}x\right],\textrm{
}\forall x\in\mathbb{N}_{1}$ (73)

II. If we define $\tilde{f}_{N}:\mathbb{Z}_{p}\rightarrow\mathbb{C}$ by:

$\tilde{f}_{N}\left(\mathfrak{z}\right)\overset{\textrm{def}}{=}\sum_{n=0}^{p^{N}-1}\left[\mathfrak{z}\overset{p^{N}}{\equiv}n\right]f\left(n\right)$

Then $\tilde{f}_{N}$ is continuous (in particular, locally-constant), and
satisfies:

$\tilde{f}_{N}\left(n\right)=f\left(n\right),\textrm{ }\forall
n\in\left\\{0,\ldots,p^{N}-1\right\\}$

Finally, we need to note the formula for the Fourier Coefficients of a
$q$-adic complex-valued function of a $p$-adic integer variable. We can
formally derive this by using the identity:

$\left[\mathfrak{z}\overset{p^{m}}{\equiv}n\right]=\frac{1}{p^{m}}\sum_{k=0}^{p^{m}-1}e^{2\pi
i\left\\{k\frac{\mathfrak{z}-n}{p^{m}}\right\\}_{p}}$ (74)

to obtain a formula for a formal $q$-adic van der Put series
$f\left(\mathfrak{z}\right)$, which is an expression of the form:

$f\left(\mathfrak{z}\right)=\sum_{n=0}^{\infty}c_{n}\left(f\right)\left[\mathfrak{z}\overset{p^{\lambda_{p}\left(n\right)}}{\equiv}n\right]$
(75)

where the $c_{n}\left(f\right)$s are constants in $\mathbb{C}_{q}$. In 74,
$\left\\{\cdot\right\\}_{p}$ is the $p$-adic fractional part. 74 holds both
over $\mathbb{C}$ _and_ $\mathbb{C}_{q}$ for any prime $q$ (including $q=p$).
In the non-archimedean case, we fix an embedding of the complex roots of unity
$\left\\{e^{2\pi i/p^{n}}\right\\}_{n\in\mathbb{N}_{1}}$ in $\mathbb{C}_{q}$
and, for all $\mathfrak{y}\in\mathbb{Q}_{p}$, writing
$\mathfrak{y}=\mathfrak{z}/p^{n}$, where $\mathfrak{z}\in\mathbb{Z}_{p}$ and
$n\in\mathbb{N}_{1}$, we abuse notation slightly to write $e^{2\pi
i\left\\{\mathfrak{y}\right\\}_{p}}$ to denote the
$\left[\mathfrak{z}\right]_{p^{n}}$th power of the image of $e^{2\pi i/p^{n}}$
under our chosen embedding. Applying 74 to 75, we obtain the formal
identities:

Theorem 2.2.3: Let $p$ and $q$ be distinct primes, and let $f$ be a formal van
der Put series of the form 75. Then, the formal $\left(p,q\right)$-adic
Fourier coefficient of $f$, denoted $\hat{f}\left(t\right)$, are the formal
sums so that:

$f\left(\mathfrak{z}\right)=\sum_{t\in\hat{\mathbb{Z}}_{2}}\hat{f}\left(t\right)e^{2\pi
i\left\\{t\mathfrak{z}\right\\}_{p}}$ (76)

and are given by:

$\hat{f}\left(t\right)=\sum_{n=\frac{1}{p}\left|t\right|_{p}}^{\infty}\frac{c_{n}\left(f\right)}{p^{\lambda_{p}\left(n\right)}}e^{-2n\pi
it},\textrm{ }\forall t\in\hat{\mathbb{Z}}_{2}$ (77)

In particular, the following are equivalent:

i. $\hat{f}\left(t\right)$ converges in $\mathbb{C}_{q}$ for all
$t\in\hat{\mathbb{Z}}_{2}$.

ii. $\lim_{n\rightarrow\infty}c_{n}\left(f\right)\overset{\mathbb{C}_{q}}{=}0$

iii. $f$ is a continuous $\mathbb{C}_{q}$-valued function on $\mathbb{Z}_{p}$.

iv. 76 converges uniformly in $\mathbb{C}_{q}$ for all
$\mathfrak{z}\in\mathbb{Z}_{p}$.

_Remark_ : If $f:\mathbb{Z}_{p}\rightarrow\mathbb{C}$ has a van der Put series
expression so that 77 converges in $\mathbb{C}$ for all
$t\in\hat{\mathbb{Z}}_{p}$, and $\hat{f}\in
L^{1}\left(\hat{\mathbb{Z}}_{p}\right)$, then 76 converges absolutely to $f$
in $\mathbb{C}$ for all $\mathfrak{z}\in\mathbb{Z}_{p}$. If $\hat{f}\in
L^{2}\left(\hat{\mathbb{Z}}_{p}\right)$, then 76 converges to $f$ in
$\mathbb{C}$ for almost every $\mathfrak{z}\in\mathbb{Z}_{p}$.

##  3 Asymptotics, Blancmanges, and Contour Integrals

###  3.1 Blancmange Curves, $\sum_{n=1}^{N}r_{p}\left(n\right)$, &
$\sum_{n=1}^{N}\chi_{p}\left(n\right)$

####  3.1.1 Pictures at an Exhibition

As it turns out, like in the episode of Monty Python’s Flying Circus where
people everywhere are being mysteriously transformed into Scotsmen, behind the
asymptotics of $\chi_{p}$ and $r_{p}$ lurks a blancmange—only of the
mathematical type, rather than an earthly dessert from outer space. The
mathematical kind of blancmanges are Blancmange curves, named after the
dessert they so happen to resemble, also known as the Takagi Function, after
their discoverer. These are a family of fractal curves created by repeated
midpoint subdivision. There is a great deal to be said about these curves; the
exposition here is liberally borrowed from Lagarias’ excellent article [16] on
the subject. The Takagi-Landsberg function
$T_{w}:\left[0,1\right]\rightarrow\mathbb{R}$ can be written as:

$T_{w}\left(x\right)=\sum_{n=0}^{\infty}w^{n}s\left(2^{n}x\right)=\frac{1}{4}\frac{1}{1-w}-\frac{2}{\pi^{2}}\sum_{n=0}^{\infty}\sum_{k=0}^{\infty}w^{n}\frac{\cos\left(2^{n+1}\left(2k+1\right)\pi
x\right)}{\left(2k+1\right)^{2}}$ (78)

where $w$ is a real parameter with $\left|w\right|<1$, and $s\left(x\right)$
is the triangle wave function:

$s\left(x\right)\overset{\textrm{def}}{=}\min_{n\in\mathbb{Z}}\left|x-n\right|=\left|x-\left\lfloor
x+\frac{1}{2}\right\rfloor\right|$ (79)

which gives the distance from $x$ to the nearest integer. Takagi’s original
curve (_the_ “blancmange curve”) comes from setting $w=1/2$:

$T\left(x\right)\overset{\textrm{def}}{=}T_{1/2}\left(x\right)$ (80)

A parabola occurs for $w=1/4$, a result of Archimedes.

Figure 1: The blancmange curve $T_{1/2}\left(x\right)$ on $\left[-1,1\right]$

The author’s first direct encounter with Blancmange curves came from examining
the summatory function of $\\#_{1}\left(m\right)$. By applying logarithmic
differentiation with respect to the parameter $a$ in the identity:

$\prod_{m=0}^{n-1}\left(1+a\cdot
b^{2^{m}}\right)=1+\sum_{k=1}^{2^{n}-1}a^{\\#_{1}\left(k\right)}b^{k},\textrm{
}\forall n\in\mathbb{N}_{1},\textrm{ }\forall a,b\in\mathbb{C}$ (81)

one obtains:

$\sum_{k=1}^{2^{n}-1}\\#_{1}\left(k\right)a^{\\#_{1}\left(k\right)-1}b^{k}=\sum_{k=0}^{n-1}\frac{b^{2^{k}}}{1+a\cdot
b^{2^{k}}}\prod_{m=0}^{n-1}\left(1+a\cdot b^{2^{m}}\right)$ (82)

which, after setting $a=b=1$ gives:

$\sum_{k=1}^{2^{n}-1}\\#_{1}\left(k\right)=\sum_{k=0}^{n-1}\frac{1}{2}\prod_{m=0}^{n-1}2=\sum_{k=0}^{n-1}2^{n-1}=n2^{n-1}$

Replacing $n$ with $\log_{2}\left(n+1\right)$ suggests that:

$\sum_{m=1}^{n}\\#_{1}\left(m\right)\approx\frac{n+1}{2}\log_{2}\left(n+1\right)\textrm{
as }n\rightarrow\infty$ (83)

Plotting:

$\textrm{Bl}\left(x\right)\overset{\textrm{def}}{=}\frac{\left\lfloor
x\right\rfloor+1}{2}\log_{2}\left(\left\lfloor
x\right\rfloor+1\right)-\sum_{m=1}^{\left\lfloor
x\right\rfloor}\\#_{1}\left(m\right)$ (84)

reveals the blancmange:

Figure 2: $\textrm{Bl}\left(x\right)$, the $\\#_{1}$-summatory blancmange

In essence, our naïve substitution of $\log_{2}\left(n+1\right)$ into the
summatory function of $\\#_{1}\left(n\right)$ produced the dominant growth
term; subtracting off that term reveals the blancmange batrachion hidden
beneath it. Delightfully, this naïveté is fully and rigorously justified. In
1968, Trollope provided a closed-form expression demonstrating exactly the
relation between $\sum_{m=1}^{n}\\#_{1}\left(m\right)$ and the Takagi function
$T\left(x\right)$:

Theorem 3.1.1.1 (Trollope 1968) [16]:

$\sum_{m=1}^{n}\\#_{1}\left(m\right)=\frac{\left(n+1\right)\left(\lambda\left(n\right)+1\right)-2^{\lambda\left(n\right)}}{2}-2^{\lambda\left(n\right)-2}T\left(\frac{n+1}{2^{\lambda\left(n\right)-1}}-1\right),\textrm{
}\forall n\in\mathbb{N}_{1}$ (85)

Tellingly, this formula can also be obtained using the Perron’s Formula Mellin
transform methods detailed in Section 2.2 [3]. The methods used there are what
inspired the present approach.

Using the identity 97, we can do exactly the same for $r_{p}$ and $\chi_{p}$.
For $r_{p}$, the resulting formula is:

$\sum_{m=1}^{2^{n}-1}r_{p}\left(m\right)=\frac{p}{p-1}\left(\left(\frac{p+1}{2}\right)^{n}-1\right)$
(86)

Replacing $n$ with $\log_{2}\left(n+1\right)$ gives the suggestion:

$\sum_{m=1}^{n}r_{p}\left(m\right)\approx\frac{p}{p-1}\left(\left(n+1\right)^{\sigma_{p}}-1\right)=O\left(n^{\sigma_{p}}\right)$
(87)

which we now make rigorous.

Lemma 3.1.1.1 (Replacing $n$ with $\log_{2}\left(n+1\right)$ to obtain
dominant asymptotics is justified): Let
$\left\\{a_{m}\right\\}_{m\in\mathbb{N}_{1}}$ be a sequence of non-negative
real numbers for which there is a continuously differentiable
$\mathcal{A}:\left(0,\infty\right)\rightarrow\left[0,\infty\right)$, with
$\mathcal{A}\left(0\right)=0$ and with
$\sup_{x>0}\mathcal{A}^{\prime}\left(x\right)<\infty$ so that:

$\sum_{m=1}^{2^{n}-1}a_{m}=\mathcal{A}\left(n\right),\textrm{ }\forall
n\in\mathbb{N}_{1}$

Then:

$\sum_{m=1}^{n}a_{m}\leq\mathcal{A}\left(\log_{2}\left(n+1\right)\right)+\sup_{x>0}\left|\mathcal{A}^{\prime}\left(x\right)\right|,\textrm{
}\forall n\in\mathbb{N}_{1}$

Proof: Let $A\left(n\right)=\sum_{m=1}^{n}a_{m}$. Then:

$\mathcal{A}\left(n\right)=\sum_{m=1}^{2^{n}-1}a_{m}=A\left(2^{n}-1\right)$

Hence:

$\displaystyle\sum_{m=1}^{2^{\left\lfloor
x\right\rfloor}-1}a_{m}=A\left(2^{\left\lfloor x\right\rfloor}-1\right)$
$\displaystyle=\mathcal{A}\left(\left\lfloor x\right\rfloor\right)$
$\displaystyle\Updownarrow$ $\displaystyle
A\left(2^{\left\lfloor\log_{2}\left(x\right)+1\right\rfloor}-1\right)$
$\displaystyle=\mathcal{A}\left(\left\lfloor\log_{2}\left(x\right)+1\right\rfloor\right)$
$\displaystyle\Updownarrow$ $\displaystyle
A\left(2^{\lambda\left(x\right)}-1\right)$
$\displaystyle=\mathcal{A}\left(\left\lfloor\log_{2}\left(x\right)+1\right\rfloor\right)$
$\displaystyle=\mathcal{A}\left(\left\lceil\log_{2}\left(x+1\right)\right\rceil\right)$

Since $\mathcal{A}$ is continuously differentiable, and since
$\mathcal{A}\left(0\right)=0$:

$\displaystyle\mathcal{A}\left(\left\lceil\log_{2}\left(x+1\right)\right\rceil\right)$
$\displaystyle=\mathcal{A}\left(\left\lceil\log_{2}\left(x+1\right)\right\rceil\right)-\mathcal{A}\left(0\right)$
$\displaystyle=\int_{0}^{\left\lceil\log_{2}\left(x+1\right)\right\rceil}\mathcal{A}^{\prime}\left(t\right)dt$
$\displaystyle=\int_{0}^{\log_{2}\left(x+1\right)}\mathcal{A}^{\prime}\left(t\right)dt+\int_{\log_{2}\left(x+1\right)}^{\left\lceil\log_{2}\left(x+1\right)\right\rceil}\mathcal{A}^{\prime}\left(t\right)dt$
$\displaystyle=\mathcal{A}\left(\log_{2}\left(x+1\right)\right)-\underbrace{\mathcal{A}\left(0\right)}+\int_{\log_{2}\left(x+1\right)}^{\left\lceil\log_{2}\left(x+1\right)\right\rceil}\mathcal{A}^{\prime}\left(t\right)dt$
$\displaystyle=\mathcal{A}\left(\log_{2}\left(x+1\right)\right)+\int_{\log_{2}\left(x+1\right)}^{\left\lceil\log_{2}\left(x+1\right)\right\rceil}\mathcal{A}^{\prime}\left(t\right)dt$

By the mean value theorem, for all $x>0$, there is an
$\eta_{x}\in\left[\log_{2}\left(x+1\right),\left\lceil\log_{2}\left(x+1\right)\right\rceil\right]$
so that:

$\int_{\log_{2}\left(x+1\right)}^{\left\lceil\log_{2}\left(x+1\right)\right\rceil}\mathcal{A}^{\prime}\left(t\right)dt=\mathcal{A}^{\prime}\left(\eta_{x}\right)\leq\sup_{t>0}\mathcal{A}^{\prime}\left(t\right)<\infty$

and so, for all $n\in\mathbb{N}_{1}$:

$\displaystyle\sum_{m=1}^{n}a_{m}$
$\displaystyle=\sum_{m=1}^{2^{\log_{2}\left(n+1\right)}-1}a_{m}$
$\displaystyle\leq\sum_{m=1}^{2^{\left\lceil\log_{2}\left(n+1\right)\right\rceil}-1}a_{m}$
$\displaystyle=\mathcal{A}\left(\left\lceil\log_{2}\left(n+1\right)\right\rceil\right)$
$\displaystyle=\mathcal{A}\left(\log_{2}\left(n+1\right)\right)+\mathcal{A}^{\prime}\left(\eta_{n}\right)$
$\displaystyle\leq\mathcal{A}\left(\log_{2}\left(n+1\right)\right)+\sup_{x>0}\mathcal{A}^{\prime}\left(x\right)$

as desired.

Q.E.D.

Applying this to 86 gives:

Proposition 3.1.1.1 (Dominant asymptotics of
$\sum_{m=1}^{n}r_{p}\left(m\right)$):

$\sum_{m=1}^{n}r_{p}\left(m\right)\leq\frac{p}{p-1}\left(\left(n+1\right)^{\sigma_{p}}-1\right)$
(88)

Since $\sigma_{3}=1$, setting $p=3$ yields $\frac{3n}{2}$. Thus, the case of
the Collatz map ($p=3$) is distinguished as the only one where the summatory
function of $r_{p}$ grows _linearly_ with $n$ when $p=3$. For all larger
values of $p$, however, we instead get power-law type growth, with polynomial
growth occurring precisely when $p+1$ is a power of $2$.

Now, like with $\\#_{1}$, let us subtract off our asymptotic estimate 87 and
define:

Definition 3.1.1.1: We write
$\textrm{Bl}_{p}:\left[0,\infty\right)\rightarrow\mathbb{R}$ to denote the
function:

$\textrm{Bl}_{p}\left(x\right)\overset{\textrm{def}}{=}\frac{p}{p-1}\left(\left(\left\lfloor
x\right\rfloor+1\right)^{\sigma_{p}}-1\right)-\sum_{m=1}^{\left\lfloor
x\right\rfloor}r_{p}\left(m\right)=\sum_{m=1}^{\left\lfloor
x\right\rfloor}\left(\frac{p}{p-1}\Delta\left(m^{\sigma_{p}}\right)-r_{p}\left(m\right)\right)$
(89)

where the sum is defined to be $0$ when $\left\lfloor x\right\rfloor=0$, and
where:

$\Delta\left(m^{\sigma_{p}}\right)=\left(m+1\right)^{\sigma_{p}}-m^{\sigma_{p}}$
(90)

where $\Delta$ is the forward difference operator. When $p=3$, this reduces
to:

$\textrm{Bl}_{3}\left(x\right)=\sum_{m=1}^{\left\lfloor
x\right\rfloor}\left(\frac{3}{2}-r_{3}\left(m\right)\right)$ (91)

Figure 3: $\textrm{Bl}_{3}\left(x\right)$, the $r_{3}$-summatory blancmange

While $\textrm{Bl}_{p}$ still retains blancmange structure for $p\geq 5$, the
curve is much more diffusively spaced, requiring one to pass to logarithmic
scaling on the $y$-axis in order for the behavior to become apparent. Thus,
the presence of blancmange structure is not the unique purview of the $p=3$
case. Nevertheless, there is a decisive distinguishing factor: _the size of
$r_{p}\left(n\right)$ relative to $1$_.

Define:

$\tilde{\textrm{Bl}}_{p}\left(x\right)\overset{\textrm{def}}{=}\sum_{m=1}^{\left\lfloor
x\right\rfloor}\left(1-r_{p}\left(m\right)\right)$ (92)

As constructed, note that:

$\tilde{\textrm{Bl}}_{p}\left(x+1\right)=1-r_{p}\left(\left\lfloor
x\right\rfloor\right)+\tilde{\textrm{Bl}}_{p}\left(x\right)$

and thus, that $r_{p}\left(x\right)<1$ if and only if
$\tilde{\textrm{Bl}}_{p}\left(x+1\right)>\tilde{\textrm{Bl}}_{p}\left(x\right)$.
Figure 4 shows a section of the graph of
$\tilde{\textrm{Bl}}_{3}\left(x\right)$, with the blue-shaded regions denoting
the values of $x$ for which $r_{3}\left(x\right)<1$.

Figure 4: Graph of a segment of $\tilde{\textrm{Bl}}_{3}\left(x\right)$, with
blue regions showing the values of $x$ for which $r_{3}\left(x\right)<1$.

When we plot $\tilde{\textrm{Bl}}_{p}\left(x\right)$ for values of $p$ other
than $3$, a key difference emerges: when $p>3$, there is significantly less
“wiggle” in the curve. Figure 5 shows
$\left|\tilde{\textrm{Bl}}_{p}\left(x\right)\right|$ graphed logarithmically;
the places where the curves _decrease_ corresponds to the values of $x$ where
$r_{p}\left(x\right)<1$.

Figure 5: Log-scale graphs of
$\left|\tilde{\textrm{Bl}}_{3}\left(x\right)\right|$ (red),
$\left|\tilde{\textrm{Bl}}_{5}\left(x\right)\right|$ (blue), and
$\left|\tilde{\textrm{Bl}}_{7}\left(x\right)\right|$ (green)

Defining:

$\displaystyle X_{p}^{+}$
$\displaystyle\overset{\textrm{def}}{=}\left\\{n\in\mathbb{N}_{1}:r_{p}\left(n\right)<1\right\\}$
$\displaystyle X_{p}^{-}$
$\displaystyle\overset{\textrm{def}}{=}\left\\{n\in\mathbb{N}_{1}:r_{p}\left(n\right)>1\right\\}$

conjecturally, it appears that for odd $p\geq 3$:

$\limsup_{N\rightarrow\infty}\frac{\left|X_{p}^{+}\cap\left\\{1,\ldots,N\right\\}\right|}{N}>\limsup_{N\rightarrow\infty}\frac{\left|X_{p}^{-}\cap\left\\{1,\ldots,N\right\\}\right|}{N}\Leftrightarrow
p=3$

with the preponderance of local minima in the
$\left|\tilde{\textrm{Bl}}_{3}\left(x\right)\right|$ curve in Figure 5 and the
absence thereof in the $\left|\tilde{\textrm{Bl}}_{5}\left(x\right)\right|$
and $\left|\tilde{\textrm{Bl}}_{7}\left(x\right)\right|$ curves being
computational evidence in support of this conjecture.

####  3.1.2 The Function $\zeta_{p}$

We begin with some useful notations, as well as a generating function
identity:

Definition 3.1.2.1: Let $p$ be an odd integer $\geq 3$.

I. We define $\zeta_{p}\left(s\right)$ as the Dirichlet series associated to
$r_{p}$:

$\zeta_{p}\left(s\right)\overset{\textrm{def}}{=}\sum_{n=0}^{\infty}\frac{r_{p}\left(n\right)}{\left(n+1\right)^{s}}$
(93)

II. We define $\mathcal{B}_{p}\left(s\right)$ as the Dirichlet Series
associated to $\nabla\left\\{\textrm{Bl}_{p}\right\\}$:

$\mathcal{B}_{p}\left(s\right)=\sum_{n=0}^{\infty}\frac{\frac{p}{p-1}\Delta\left(n^{\sigma_{p}}\right)-r_{p}\left(n\right)}{\left(n+1\right)^{s}}$
(94)

_Remark_ : expanding $\Delta\left(n^{\sigma_{p}}\right)$ (which, note, is $1$
when $p=3$), we get:

$\mathcal{B}_{p}\left(s\right)=\begin{cases}\frac{3}{2}\zeta\left(s\right)-\zeta_{3}\left(s\right)&\textrm{if
}p=3\\\
\frac{p}{p-1}\zeta\left(s-\sigma_{p}\right)-\sum_{n=1}^{\infty}\frac{n^{\sigma_{p}}}{\left(n+1\right)^{s}}-\zeta_{p}\left(s\right)&\textrm{if
}p\geq 5\end{cases}$ (95)

Proposition 3.1.2.1: Let $\mathbb{F}$ be any valued field of characteristic
$0$, and let $a$ and $z$ be any elements of $\mathbb{F}$. Then, we have the
identities:

$\sum_{n=0}^{2^{m}-1}a^{\\#_{1}\left(n\right)}z^{n}=\prod_{j=0}^{m-1}\left(1+az^{2^{j}}\right),\textrm{
}\forall m\in\mathbb{N}_{0}$ (96)
$\sum_{n=2^{m-1}}^{2^{m}-1}a^{\\#_{1}\left(n\right)}z^{n}=az^{2^{m-1}}\prod_{j=0}^{m-2}\left(1+az^{2^{j}}\right),\textrm{
}\forall m\in\mathbb{N}_{1}$ (97)

where the products are defined to be $1$ whenever $m$ is $0$ and $1$,
respectively.

Proof:

I. Letting:

$C_{a}\left(z\right)\overset{\textrm{def}}{=}\sum_{n=0}^{\infty}a^{\\#_{1}\left(n\right)}z^{n}$
(98)

Note that:

$\frac{C_{a}\left(z\right)+C_{a}\left(-z\right)}{2}=\sum_{n=0}^{\infty}a^{\\#_{1}\left(2n\right)}z^{2n}=\sum_{n=0}^{\infty}a^{\\#_{1}\left(n\right)}z^{2n}=C_{a}\left(z^{2}\right)$

and:

$\frac{C_{a}\left(z\right)-C_{a}\left(-z\right)}{2}=\sum_{n=0}^{\infty}a^{\\#_{1}\left(2n+1\right)}z^{2n+1}=az\sum_{n=0}^{\infty}a^{\\#_{1}\left(n\right)}z^{2n}=azC_{a}\left(z^{2}\right)$

Adding these two identities yields:

$C_{a}\left(z\right)=\left(1+az\right)C_{a}\left(z^{2}\right)$ (99)

Hence:

$\displaystyle C_{a}\left(z\right)$
$\displaystyle=\left(1+az\right)C_{a}\left(z^{2}\right)$
$\displaystyle=\left(1+az\right)\left(1+az^{2}\right)C_{a}\left(z^{2^{2}}\right)$
$\displaystyle\vdots$
$\displaystyle=\lim_{N\rightarrow\infty}C_{a}\left(x^{2^{N}}\right)\prod_{n=0}^{N-1}\left(1+az^{2^{n}}\right)$
$\displaystyle\left(\forall\left|z\right|<1\right);$
$\displaystyle=C_{a}\left(0\right)\prod_{n=0}^{\infty}\left(1+az^{2^{n}}\right)$
$\displaystyle\left(C_{a}\left(0\right)=a^{\\#_{1}\left(0\right)}=a^{0}=1\right);$
$\displaystyle=\prod_{n=0}^{\infty}\left(1+az^{2^{n}}\right)$

and so:

$C_{a}\left(z\right)=\prod_{n=0}^{\infty}\left(1+az^{2^{n}}\right),\textrm{
}\forall\left|z\right|<1$ (100)

Since the partial product:

$\prod_{j=0}^{m-1}\left(1+az^{2^{j}}\right)$

contains all terms of $C_{a}$’s power series of degree:
$\leq\sum_{j=0}^{m-1}2^{j}=2^{m}-1$, it then follows that:

$\prod_{j=0}^{m-1}\left(1+az^{2^{j}}\right)=\sum_{n=0}^{2^{m}-1}a^{\\#_{1}\left(n\right)}z^{n}$