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\begin{center}
\vskip 1cm{\LARGE\bf 
The Prime-Power Map
}
\vskip 1cm
\large
Steven\\ 
Business Mathematics Department\\
School of Applied STEM\\
Universitas Prasetiya Mulya\\
South Tangerang 15339\\
Indonesia\\
\href{mailto:steven@pmbs.ac.id}{\tt steven@pmbs.ac.id}\\
\ \\
Jonathan Hoseana\\
Department of Mathematics\\
Parahyangan Catholic University\\
Bandung 40141\\
Indonesia\\
\href{mailto:j.hoseana@unpar.ac.id}{\tt j.hoseana@unpar.ac.id} 
\end{center}

\vskip .2 in
\begin{abstract}
We introduce a modification of Pillai's prime map: the prime-power
map. This map fixes $1$, divides its argument by $p$ if it is a
prime power $p^k$, and otherwise subtracts from its argument the largest
prime power not exceeding it. We study the iteration of this map over the
positive integers, developing, firstly, results parallel to those known
for the prime map. Subsequently, we compare its dynamical properties
to those of a more manageable variant of the map under which any orbit
admits an explicit description. Finally, we present some experimental
observations, based on which we conjecture that almost every orbit of
the prime-power map contains no prime power.
\end{abstract}

\section{Introduction}\label{section:Introduction}

Let $\mathbb{P}$ be the set of primes (\seqnum{A000040} in the
{\it On-Line Encyclopedia of Integer Sequences} \cite{OEIS}). In 1930, Pillai \cite{Pillai} introduced the \textit{prime map} $x\mapsto x-r(x)$, where $r(x)$ is the largest element of $\mathbb{P}\cup\{0,1\}$ not exceeding $x$, under whose iteration every positive-integer initial condition is eventually fixed at $0$, the only fixed point of the map. Interesting results on the asymptotic behavior of the time steps $R(x)$ it takes for an initial condition $x$ to reach the fixed point were established, before subsequently improved by Luca and Thangadurai \cite{Luca} in 2009. The latter authors proved that $R(x)$ grows no faster than $\ln\ln x$ \cite[Theorem 1.1]{Luca}, and that for every $k\in\mathbb{N}$ ($\mathbb{N}$ being the set of \textit{positive} integers), the proportion of initial conditions $t\leq x$ for which $R(t)=k$ is asymptotic to $\left(\ln_k x\right)^{-1}$, where the subscript $k$ denotes $k$-fold self-composition \cite[Theorem 1.2]{Luca}.

In this paper we study a modification of the prime map, constructed essentially by letting the role of primes be taken over by prime powers (\seqnum{A000961}). A \textit{prime power}, as is well-known, is an integer of the form $p^k$, where $p\in\mathbb{P}$ and $k\in\mathbb{N}$. Letting $\mathbb{P}^{\ast}$ be the set of prime powers, we define the \textit{prime-power map} $P:\mathbb{N}\to\mathbb{N}$ by
$$P(x):=\begin{cases}
1,&\text{if }x=1;\\
x-q(x),&\text{if }x\notin\mathbb{P}^\ast;\\
\frac{x}{p},&\text{if }x\in\mathbb{P}^\ast\text{ is a power }p\in\mathbb{P},
\end{cases}$$
where $q(x)$ is the largest element of $\mathbb{P}^\ast_{1}:=\mathbb{P}^\ast\cup\{1\}$ not exceeding $x$ (see Figure \ref{fig:plotP}). By the \textit{orbit} of an \textit{initial condition} $x\in\mathbb{N}$ under $P$ we mean the sequence $\left(x_n\right)_{n=1}^\infty$ where
$$x_1=x\qquad\quad\text{and}\quad\qquad x_{n+1}=P\left(x_n\right)\quad\text{for every }n\in\mathbb{N}.$$
Clearly, every orbit under $P$ is decreasing (in the non-strict sense) and eventually reaches the sole fixed point $1$ at which it then stabilizes (becomes constant).

In the subdomain $\mathbb{P}^\ast_1\subseteq\mathbb{N}$ the dynamics of $P$ is predictable; it consists only of divisions by primes and stabilization. In its complement, by contrast, the dynamics is non-trivial and resembles that of Pillai's prime map. Such hybrid dynamical behavior naturally motivates associating to every initial condition $x$ the time steps
$$S(x):=\min\left\{n\in\mathbb{N}:x_n\in\mathbb{P}^\ast_1\right\}\qquad\text{and}\qquad T(x):=\min\left\{n\in\mathbb{N}:x_n=1\right\}$$
at which the orbit of $x$ enters the predictable subdomain $\mathbb{P}^\ast_1$, and at which it reaches $1$, respectively. These are referred to as the \textit{settling time} and \textit{transient length} of $x$, respectively. The unboundedness of the transient length function is immediate---the function diverges along the sequence of powers of any given prime---whereas that of the settling time function is less obvious and will be established in this paper. We shall also be interested in the \textit{attractor}
$$W(x):=x_{S(x)}$$
of $x$, i.e., the element of the predictable subdomain first visited by the orbit of $x$.

\begin{figure}
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\caption{The values of $P(x)$ for $x\in\{1,\ldots,250\}$.}
\label{fig:plotP}
\end{figure}

The functions $T$ and $S$ defined above constitute our primary object of
study. 

We now describe our main results, along with the structure of this paper. 
Following this introduction is Section \ref{sec:TandS},
where we establish \textit{a logarithmic upper bound for $T$} (Lemma
\ref{lemma:Bertrand2} and Theorem \ref{thm:power2}). Subsequently,
we discuss 
improved versions of Lemma \ref{lemma:Bertrand2}, derived from
improved versions of Bertrand's postulate currently known in the
literature (Propositions \ref{prop:epsilon}
and \ref{prop:Bertrandrefinement}), before we turn our attention to
the function $S$, establishing \textit{its unboundedness} (Theorem
\ref{thm:unboundednessS}) and \textit{a double-logarithmic upper bound}
(Theorem \ref{thm:orderS(x)}); these are analogues of results of Pillai
\cite[pp.~160--161]{Pillai} and
Luca and Thangadurai \cite[Theorem 1.1]{Luca}, respectively. Next,
we study the densities of some subsets of $\mathbb{N}$ induced by $P$
and prove that \textit{almost every positive integer is coprime with its
image under $P$}, i.e., that the limit of the proportion of such integers
is $1$ (Theorem \ref{thm:coprime}). At the end of the section, we connect
the map $P$ to the representations of integers as sums of prime powers;
of note is the discussion on the smallest initial condition having a
given settling time (Proposition \ref{prop:xis}).

The detailed behavior of the functions $T$ and $S$ is difficult to describe explicitly. For a comparison, in Section \ref{sec:Pp} we introduce a more manageable variant of $P$: a family of maps $P_p$, under which the analogous functions $T_p$ and $S_p$ admit explicit descriptions (Theorem \ref{thm:Fr}). We prove that these two functions are asymptotic (Corollary \ref{corollary}), which, together with some experimental observations, suggests that the original functions $T$ and $S$ exhibit an analogous asymptoticity. Motivated by these observations, we conjecture that \textit{almost every orbit of the prime-power map contains no prime power} (Conjecture \ref{finalconjecture}).

\section{Results on the map $P$}\label{sec:TandS}

Before entering into our main results, let us state some
fundamentals. First, we state the famous \textit{Bertrand's postulate}
\cite[Theorem 418]{HardyWright}, which will be used frequently.

\begin{lemma}[Bertrand's postulate]\label{lemma:Bertrand}
	 For every integer $n\geq 2$, there exists a prime $p$ such that $n <p< 2n$.
\end{lemma}

We also use standard asymptotic notation that
we now recall.
Let $f(x)$ and $g(x)$ be positive for all sufficiently large values of $x\in\mathbb{N}$. We write $f(x)\sim g(x)$ to mean $f(x)$ \textit{is asymptotic to} $g(x)$, i.e., $\frac{f(x)}{g(x)}\xrightarrow{x\to\infty}1$. In addition, $f(x)\ll g(x)$ means $f(x)$ \textit{is asymptotically bounded above by} $g(x)$, i.e., there exist $C>0$ and $x_0\in\mathbb{N}$ such that $f(x)\leq C\,g(x)$ for every integer $x\geq x_0$. Finally, $f(x)=o(g(x))$ means $\frac{f(x)}{g(x)}\xrightarrow{x\to\infty}0$.

\subsection{The transient length function}

First, we have
$$\limsup_{x\to\infty}T(x)=\infty,$$
since $T$ diverges along, e.g., the sequence of powers of two (\seqnum{A000079}). More precisely, if $x$ is a power of two, we have $T(x)= \log_2x +1$. Let us now prove that $\log_2x +1$ is an upper bound for $T$ (see Figure \ref{fig:transittime}) via the following lemma that exploits Bertrand's postulate.

\begin{lemma}\label{lemma:Bertrand2}
For every integer $x\geq 2$ we have $$P(x)\leq\frac{x}{2}.$$
\end{lemma}
\begin{proof}
Let $x\geq 2$. If $x=p^k$ for some $p\in\mathbb{P}$ and $k\in\mathbb{N}$, then $P(x)=\frac{x}{p}\leq\frac{x}{2}$. Otherwise, since Bertrand's postulate guarantees the existence of a prime between $\frac{x}{2}$ and $x$, then $q(x)\geq\frac{x}{2}$, and hence $P(x)\leq\frac{x}{2}$.
\end{proof}

\begin{theorem}\label{thm:power2}
	For every $x\in\mathbb{N}$ we have $T(x)\leq \log_2x +1$ with equality if and only if $x$ is a power of two.
\end{theorem}
\begin{proof}
	By Lemma \ref{lemma:Bertrand2}, in the orbit $\left(x_n\right)_{n=1}^\infty$ of $x\in\mathbb{N}$ we have
$$1=x_{T(x)}\leq \frac{x_{T(x)-1}}{2}\leq\frac{x_{T(x)-2}}{2^2}\leq\cdots\leq\frac{x_1}{2^{T(x)-1}}=\frac{x}{2^{T(x)-1}},$$
which implies the desired inequality. If $x$ is a power of two, we have $T(x)=\log_2 x + 1$ as previously remarked. Conversely, if $T(x)=\log_2 x + 1$, then $x=2^{T(x)-1}$.
\end{proof}

\begin{figure}
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\addplot[domain=1:250,very thick,red] {ln(x)/ln(2)+1};
\end{axis}
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\end{center}
\caption{The values of $T(x)$ for $x\in\{1,\ldots,250\}$ and the upper bound provided by Theorem \ref{thm:power2}.}
\label{fig:transittime}
\end{figure}

In the literature there have been numerous refinements of Bertrand's postulate that lead to improved versions of Lemma \ref{lemma:Bertrand2}. For instance, the prime number theorem implies that, given any $\varepsilon>0$, there exists $X_0\in\mathbb{N}$ such that for every integer $x\geq X_0$, the interval $\left[x-\varepsilon x,x\right]$ contains a prime number \cite[Section 22.19]{HardyWright}, implying that $P(x)\leq\varepsilon x$ if $x$ is not a prime power. Therefore, we have the following proposition.

\begin{proposition}\label{prop:epsilon}
For every $\varepsilon>0$ there exists $X_0\in\mathbb{N}$ such that
\begin{equation}\label{eq:epsilon}
P(x)\leq\varepsilon x
\end{equation}
for every non-prime-power $x\geq X_0$.
\end{proposition}

Notice that, since for every prime power $x=p^k$ we have $\frac{P(x)}{x}=\frac{1}{p}\leq\frac{1}{2}$ with equality if and only if $p=2$, there is no $\varepsilon<\frac{1}{2}$ such that \eqref{eq:epsilon} holds for \textit{all} sufficiently large $x$. In this sense, the non-universality of \eqref{eq:epsilon}---the fact that it holds only for non-prime-powers---is unavoidable. However, for every $\varepsilon>0$ one can always choose a sufficiently large $X_1\in\mathbb{N}$ such that \eqref{eq:epsilon} holds for all powers of all primes $p\geq X_1$.

Hoheisel \cite{Hoheisel}, as cited by Luca and Thangadurai \cite[Section 2]{Luca}, proved the following development of Bertrand's postulate.

\begin{lemma}\label{lemma:Hoheisel}
There exist $\theta\in(0,1)$ and $X_0\in\mathbb{N}$ such that for every $x\geq X_0$ the interval $\left[x-x^\theta,x\right]$ contains a prime number.
\end{lemma}

\noindent This lemma immediately implies the following proposition.

\begin{proposition}\label{prop:Bertrandrefinement}
There exist $\theta\in(0,1)$ and $X_0\in\mathbb{N}$ such that
$$P(x)\leq x^{\theta}$$
for every non-prime-power $x\geq X_0$.
\end{proposition}

\noindent Baker et al.\ \cite{Baker} have proved the existence of $X_0\in\mathbb{N}$ for which one could take $\theta=\theta_0:=0.525$.

\subsection{The settling time function}

In the first half of this subsection we follow the steps of Pillai's proof \cite[pp.~160--161]{Pillai} of the unboundedness of $R$. In the work cited, Pillai proved that
$$\limsup_{x\to\infty}R(x)=\infty$$
using a well-known result on the unboundedness of gaps between primes, i.e., that for every $n\in\mathbb{N}$, none of the $n-1$ consecutive positive integers
$$n!+2,\,\,\,n!+3,\,\,\,\ldots,\,\,\,n!+n$$
is prime, implying the existence of two consecutive primes $p_1$ and $p_2$ such that $p_2-p_1\geq n$ \cite[Lemma 1]{Pillai}. Here we first modify this argument to show the unboundedness of gaps between prime powers. Then we 
use the result to prove that $$\limsup_{x\to\infty}S(x)=\infty.$$

\begin{lemma}\label{lemma:PrimePowerGap}
	For every $n\in\mathbb{N}$ there exist two consecutive prime powers $q_1$ and $q_2$ such that $q_2-q_1\geq n$.
\end{lemma}
\begin{proof}
    Let $n\in\mathbb{N}$. If $n=1$, clearly the lemma holds. Now, suppose $n\geq 2$. Let $p_1$, \ldots, $p_m$, where $m\in\mathbb{N}$, be all prime numbers not exceeding $n$. For every $i\in\{1,\ldots,m\}$, define
    $$k_i=\max\{k\in\mathbb{N}:p_i^k\leq n\}.$$
    Letting
    $$\mathcal{N}:=\prod_{i=1}^{m} p_i^{k_i+1},$$
    we can see that none of the $n-1$ consecutive positive integers
    $$\mathcal{N}+2,\,\,\,\mathcal{N}+3,\,\,\,\ldots,\,\,\,\mathcal{N}+n$$	
	is a prime power. Therefore, letting $q_1$ be the largest prime power less than $\mathcal{N}+2$ and $q_2$ be the smallest prime power greater than $\mathcal{N}+n$, we have the lemma.
\end{proof}

\begin{theorem}\label{thm:unboundednessS}
	We have
	$$\limsup_{x\to\infty} S(x)=\infty.$$
\end{theorem}
\begin{proof}
By Lemma \ref{lemma:PrimePowerGap}, for every non-prime-power $x\in\mathbb{N}$ there exist two consecutive prime powers $q_1$ and $q_2$ such that $q_2-q_1\geq x+1$, implying $S\left(q_1+x\right)=S(x)+1$. The theorem follows.
\end{proof}

Luca and Thangadurai proved that $R(x)\ll \ln\ln x$ \cite[Theorem 1.1]{Luca}. Now, we prove that the same asymptotic upper bound also applies for the settling time function $S$, following the argument in the work cited. First, we take numbers $X_0\in\mathbb{N}$ and $\theta\in(0,1)$ that satisfy Lemma \ref{lemma:Hoheisel} (and hence Proposition \ref{prop:Bertrandrefinement}). Then we have the following lemma.

\begin{lemma}\label{lemma:orderS(x)}
Let $x\geq X_0$, and let $\left(x_n\right)_{n=1}^\infty$ be the orbit of $x$. Then for every $k\in\mathbb{N}$ the following holds:
$$\text{If }x_k\geq X_0\text{ and }S(x)\geq k+1,\text{ then }x_{k+1}\leq x^{\theta^k}.$$
\end{lemma}
\begin{proof}
We use induction on $k\in\mathbb{N}$. For $k=1$ the statement reads
$$\text{If }x_1\geq X_0\text{ and }S(x)\geq 2,\text{ then }x_{2}\leq x^{\theta},$$
which is true since, if $x_1\geq X_0$ and $S(x)\geq 2$, the latter implying that $x_1$ is not a prime power, then $$x_2=P\left(x_1\right)\leq x_1^\theta=x^\theta,$$ by Proposition \ref{prop:Bertrandrefinement}.

Now let $k\geq 2$, and suppose the statement holds for $k-1$ replacing $k$. Suppose $x_k\geq X_0$ and $S(x)\geq k+1$. The former implies that $x_{k-1}\geq X_0$ since $x_{k-1}\geq x_k$, while the latter implies that $S(x)\geq k$ and that $x_{k}$ is not a prime power. Therefore, our inductive hypothesis gives $x_k\leq x^{\theta^{k-1}}$, and so by Proposition \ref{prop:Bertrandrefinement},
$$x_{k+1}=P\left(x_k\right)\leq x_k^\theta\leq \left(x^{\theta^{k-1}}\right)^\theta=x^{\theta^k},$$
completing the induction.
\end{proof}

\begin{theorem}\label{thm:orderS(x)}
We have
$$S(x)\ll\ln\ln x.$$
\end{theorem}
\begin{proof}
Define $X_0':=\max\left\{X_0,e\right\}$. Let $x\geq X_0'$, and let $\left(x_n\right)_{n=1}^\infty$ be the orbit of $x$. Define
$$k:=\min\{S(x),\ell\}-1\in\mathbb{N}_0,\qquad\text{where}\qquad \ell:=\max\left\{\lambda\in\mathbb{N}:x_\lambda\geq X_0'\right\},$$
the existence of $\ell$ being guaranteed by the fact that $x_1=x\geq X_0'>1$.

By the definition of $k$ we have that $k\leq S(x)-1$ and $k\leq\ell-1$ (i.e., $k+1\leq\ell$), implying $S(x)\geq k+1$ and $x_k\geq x_{k+1}\geq x_{\ell}\geq X_0'\geq X_0$, respectively. Then we have the inequality
$$
x_{k+1}\leq x^{\theta^k},
$$
which is trivially true if $k=0$ and implied by Lemma \ref{lemma:orderS(x)} if $k\geq 1$. This inequality, together with the fact that $x_{k+1}\geq X_0'$, implies
$$X_0'\leq x^{\theta^k}.$$
Taking logarithms twice gives
$$\ln \ln X_0'\leq k \ln\theta + \ln\ln x.$$
Since $X_0'\geq e$, the left-hand side is non-negative, and hence so is the right-hand side, i.e.,
$$k\leq\frac{\ln\ln x}{\ln\frac{1}{\theta}}.$$
Letting $m:=\max\left\{S(1),\ldots,S\left(X_0'\right)\right\}$, by the definitions of $k$ and $\ell$ we have
$$S(x)\leq k+1+m\leq\frac{\ln\ln x}{\ln\frac{1}{\theta}}+1+m\ll \ln \ln x,$$
proving the theorem.
\end{proof}

\subsection{Densities of subsets of $\mathbb{N}$ induced by $P$} Let us now discuss the densities of some positive integer sets induced by $P$, for which the following standard terminology \cite{NassoJin,Tenenbaum} is used. Given a subset $A\subseteq\mathbb{N}$. For every $x\in\mathbb{N}$, we define
$$d_x(A):=\frac{|A\cap\{1,\ldots,x\}|}{x}.$$
The (\textit{natural}) \textit{density} of $A$ is the limit
$$d(A):=\lim_{x\to\infty}d_x(A),$$
if it exists. If $d(A)=0$, we say that $A$ is \textit{sparse}. If $d(A)=1$, we say that $A$ has \textit{full density}. If $d(A)=1$ and a statement $\mathcal{P}(x)$ holds for every $x\in A$, we say that $\mathcal{P}(x)$ holds for \textit{almost every} $x\in\mathbb{N}$.

It is fairly well-known that the sets $\mathbb{P}$ and $\mathbb{P}^\ast$ are both sparse. This can be deduced by considering the prime and prime power counting functions
$$\pi(x):=\left|\mathbb{P}\cap\{1,\ldots,\lfloor x\rfloor\}\right|\qquad\text{and}\qquad\Pi(x):=\left|\mathbb{P}^\ast\cap\{1,\ldots,\lfloor x\rfloor\}\right|$$
defined for real numbers $x>0$. Clearly, $\pi(x)\leq\Pi(x)$ for every $x>0$. The \textit{prime number theorem} \cite[page 10]{HardyWright} gives the asymptotic
$$
\pi(x)\sim\frac{x}{\ln x},
$$
from which it follows that $\mathbb{P}$ is sparse. Moreover, since
$$\Pi(x)-\pi(x)=\sum_{j=2}^{\left\lfloor\log_2 x\right\rfloor}\pi\left(x^{\frac{1}{j}}\right)\ll\frac{x^{\frac{1}{2}}}{\ln x^{\frac{1}{2}}}\left(\left\lfloor\log_2 x\right\rfloor-1\right)\ll x^{\frac{1}{2}},$$
then $\Pi(x)\sim\pi(x)$, implying that $\mathbb{P}^\ast$ is also sparse. (In fact, this holds more generally \cite{LenonettiTringali}.) Letting $\mathbb{N}_{\geq k}$ be the set of positive integers greater than or equal to $k\in\mathbb{N}$, we have the following immediate consequences.
\begin{enumerate}
\item[(i)] The set $T^{-1}\left(\mathbb{N}_{\geq 3}\right)$ has full density because $T^{-1}(\{1\})=\{1\}$ and $T^{-1}(\{2\})=\mathbb{P}\cup\left\{q+1:q\in\mathbb{P}^\ast\right\}$ are both sparse.
\item[(ii)] The set $S^{-1}\left(\mathbb{N}_{\geq 2}\right)$ has full density because $S^{-1}(\{1\})=\mathbb{P}^\ast_1$ is sparse.
\item[(iii)] \sloppy The set $\left\{x\in\mathbb{N}:P(x)\leq x^{\theta_0}\right\}$ has full density, and, consequently, the set $\left\{x\in\mathbb{N}:P(x)\leq\varepsilon x\right\}$ has full density for every $\varepsilon>0$.
\end{enumerate}
The last one follows from Propositions \ref{prop:epsilon} and \ref{prop:Bertrandrefinement}.

Every $x\in\mathbb{N}_{\geq 2}\backslash\mathbb{P}$ and its image under the \textit{prime map} is coprime; for if they have a prime common divisor then it must divide $r(x)$, and so must equal to $r(x)$, implying that $x\geq 2r(x)$ which contradicts Bertrand's postulate. Under the \textit{prime-power map}, however, the analogue does not hold. Indeed, the number $34\in\mathbb{N}_{\geq2}\backslash\mathbb{P}^\ast$ and its image $P(34)=2$ are not coprime. Nonetheless, we shall now prove that $x$ and $P(x)$ are coprime for almost every $x\in\mathbb{N}$. For this purpose, we define
$$\mathbb{G}:=\left\{x\in\mathbb{N}:\gcd(x,P(x))>1\right\},$$
and we choose numbers $X_1\in\mathbb{N}$ and $\theta\in(0,1)$ such that for every $x\geq X_1$ the interval $\left[x,x+x^\theta\right]$ contains a prime number \cite{Baker}. Our aim is to show that $d(\mathbb{G})=0$ via the following lemma.

\begin{lemma}\label{lemma:dG}
For every prime power $p^k\geq X_1$, where $p\in\mathbb{P}$ and $k\in\mathbb{N}$, we have
$$\textstyle\left|\left(\mathbb{G}\cap q^{-1}\left(\left\{p^k\right\}\right)\right)\backslash\mathbb{P}^\ast\right|\leq p^{k\theta-1}.$$
\end{lemma}
\begin{proof}
\sloppy Let $p^k\geq X_1$ be a prime power, where $p\in\mathbb{P}$ and $k\in\mathbb{N}$. For every $x\in \left(\mathbb{G}\cap q^{-1}\left(\left\{p^k\right\}\right)\right)\backslash\mathbb{P}^\ast$ we have that $x\notin\mathbb{P}^\ast$ is a preimage of $p^k$ under $q$ satisfying
$$\textstyle 1<\gcd(x,P(x))=\gcd\left(x,x-p^k\right)=\gcd\left(x,p^k\right),$$
so $x$ is a multiple of $p$, and that $x\in\left(p^k,p^k+p^{k\theta}\right]$. Therefore, the number of elements of $\left(\mathbb{G}\cap q^{-1}\left(\left\{p^k\right\}\right)\right)\backslash\mathbb{P}^\ast$ is bounded above by the number of multiples of $p$ in the interval $\left(p^k,p^k+p^{k\theta}\right]$, namely
$$\left|\left\{p^k+p,\ldots,p\left\lfloor\frac{p^k+p^{k\theta}}{p}\right\rfloor\right\}\right|=\frac{p\left\lfloor\frac{p^k+p^{k\theta}}{p}\right\rfloor-p^k}{p}\leq p^{k\theta-1},$$
proving the lemma.
\end{proof}

\begin{theorem}\label{thm:coprime}
The set $\mathbb{G}$ is sparse.
\end{theorem}
\begin{proof}
Since
$$d(\mathbb{G})=d\left(\mathbb{G}\cap\mathbb{P}^\ast\right)+d\left(\mathbb{G}\backslash\mathbb{P}^\ast\right)$$
and $d\left(\mathbb{G}\cap\mathbb{P}^\ast\right)=0$, it suffices to prove that $d\left(\mathbb{G}\backslash\mathbb{P}^\ast\right)=0$.

Notice that for every $x\in\mathbb{N}$ we have
$$d_x\left(\mathbb{G}\backslash\mathbb{P}^\ast\right)=\frac{\left|\left(\mathbb{G}\backslash\mathbb{P}^\ast\right)\cap\{1,\ldots,x\}\right|}{x}\leq \frac{1}{x}\sum_{1\leq p^k\leq x}\textstyle\left|\left(\mathbb{G}\cap q^{-1}\left(\left\{p^k\right\}\right)\right)\backslash\mathbb{P}^\ast\right|,$$
where the summation is taken over all prime powers $p^k\in\{1,\ldots,x\}$. Now, the last expression is equal to
$$\frac{1}{x}\sum_{1\leq p\leq x}\sum_{k=1}^{\left\lfloor\log_p x\right\rfloor}\textstyle\left|\left(\mathbb{G}\cap q^{-1}\left(\left\{p^k\right\}\right)\right)\backslash\mathbb{P}^\ast\right|,$$
where the outer summation is taken over all primes $p\in\{1,\ldots,x\}$. By Lemma \ref{lemma:dG}, this expression is asymptotically bounded above by
$$\frac{1}{x}\sum_{X_1\leq p\leq x}\sum_{k=1}^{\left\lfloor\log_p x\right\rfloor}p^{k\theta-1},$$
where the outer summation is taken over all primes $p\in\left\{X_1,\ldots,x\right\}$. This expression is bounded above by
$$\frac{1}{x}\sum_{j=2}^x\sum_{k=1}^{\left\lfloor\log_j x\right\rfloor}j^{k\theta-1} = \frac{1}{x}\sum_{j=2}^x \frac{j^{\theta-1}\left(\left(j^\theta\right)^{\left\lfloor\log_j x\right\rfloor}-1\right)}{j^\theta-1}\sim\frac{1}{x^{1-\theta}}\sum_{j=2}^x\frac{1}{j}\leq \frac{\ln x}{x^{1-\theta}}.$$
Since $\theta\in(0,1)$, the last expression vanishes as $x\to\infty$, so we have proved the theorem.
\end{proof}

\subsection{Connection to prime-power representations} As the \textit{prime map} is closely connected to \textit{prime representations} \cite[page 159]{Pillai}, so is the \textit{prime-power map} to \textit{prime-power representations} of positive integers. More precisely, every initial condition $x\in\mathbb{N}$ is associated with, besides its orbit, the sequence $\left(q\left(x_n\right)\right)_{n=1}^{S(x)}$ which gives a way to express $x$ as the sum of elements of $\mathbb{P}^\ast_1$:
\begin{align}\label{eq:primepowerrepresentation}
x &= x_1\nonumber\\
  &= q\left(x_1\right) + x_2\nonumber\\
  &= q\left(x_1\right) + q\left(x_2\right) + x_3\nonumber\\
  &= \cdots\nonumber\\
  &= q\left(x_1\right) + q\left(x_2\right)+\cdots+q\left(x_{S(x)-1}\right)+x_{S(x)}\nonumber\\
  &= q\left(x_1\right) + q\left(x_2\right)+\cdots+q\left(x_{S(x)-1}\right)+q\left(x_{S(x)}\right).
\end{align}
Here we have that
$$q\left(x_1\right)> q\left(x_2\right)> \cdots > q\left(x_{S(x)}\right),$$
which follows from the next two lemmas.

\begin{lemma}\label{lemma:monotonicity}
For every $x,y\in\mathbb{N}$, if $x\leq y$ then $q(x)\leq q(y)$.
\end{lemma}
\begin{proof}
This is straightforward by contradiction.
\end{proof}

\begin{lemma}\label{lemma:monotonicity2}
For every initial condition $x\in\mathbb{N}$, the sequence $\left(q\left(x_n\right)\right)_{n=1}^{S(x)}$ is strictly decreasing.
\end{lemma}
\begin{proof}
Let $x\in\mathbb{N}$. If $S(x)=1$ then there is nothing to prove. Otherwise, we have $q\left(x_1\right)\geq q\left(x_2\right)\geq \cdots \geq q\left(x_{S(x)}\right)$ by Lemma \ref{lemma:monotonicity}. Suppose for a contradiction that there exists $n\in\{1,\ldots,S(x)-1\}$ such that $q\left(x_n\right)=q\left(x_{n+1}\right)$. Then $$2q\left(x_n\right)=q\left(x_n\right)+q\left(x_{n+1}\right)=\left(x_n-x_{n+1}\right)+\left(x_{n+1}-x_{n+2}\right)\leq x_n.$$
But between $q\left(x_n\right)$ and $2q\left(x_n\right)$ there is a prime by Bertrand's postulate, contradicting the definition of $q\left(x_n\right)$.
\end{proof}

Clearly, there could be many ways to represent $x\in\mathbb{N}$ as the sum of elements of $\mathbb{P}^\ast_1$. The one given by \eqref{eq:primepowerrepresentation} is the result of applying the \textit{greedy algorithm}: for each $n\in\{1,\ldots,S(x)\}$, one chooses the largest element $q\left(x_n\right)$ of $\mathbb{P}^\ast_{1}$ not exceeding $x_n$. As in the case of the prime map \cite{Pillai,Luca}, a question one could ask is, for every $s\in\mathbb{N}$, what is the smallest positive integer which is expressed by the greedy algorithm as the sum of exactly $s$ elements of $\mathbb{P}^\ast_1$? These are precisely the numbers $\xi_s:=\min S^{-1}(\{s\})$ for every $s\in\mathbb{N}$, recently submitted to the OEIS \cite{OEIS} as sequence \seqnum{A336825} (an analogue to the Pillai sequence \seqnum{A066352}). The values of $\xi_1$, $\xi_2$, $\xi_3$, and $\xi_4$ are
$$1=1,\quad 6=5+1,\quad 95=89+5+1,\quad\text{and}\quad 360748=360653+89+5+1,$$
respectively. Anyone attempting to compute the subsequent values should take into consideration the following facts.

\begin{proposition}\label{prop:xis}
\begin{enumerate}
\leavevmode
\item[(i)] For every $s\in\mathbb{N}$ we have $P\left(\xi_{s+1}\right)=\xi_s$.
\item[(ii)] For every $s\in\mathbb{N}$ we have $\xi_{s+1}=\min P^{-1}\left(\left\{\xi_{s}\right\}\right)$.
\item[(iii)] For every integer $s\geq 3$, the number $\xi_s$ is not a prime power.
\item[(iv)] We have $$\frac{\xi_{s+1}}{\xi_s},\frac{q\left(\xi_{s+1}\right)}{q\left(\xi_s\right)}\xrightarrow{x\to\infty}\infty.$$
\end{enumerate}
\end{proposition}
\begin{proof}
First, we prove (i). Let $s\in\mathbb{N}$. Since $S\left(\xi_{s+1}\right)=s+1$, then $S\left(P\left(\xi_{s+1}\right)\right)=s$, so $\xi_s\leq P\left(\xi_{s+1}\right)$. To prove that $\xi_s=P\left(\xi_{s+1}\right)$, suppose for a contradiction that $\xi_s< P\left(\xi_{s+1}\right)$. Then,
$$q\left(\xi_{s+1}\right)<q\left(\xi_{s+1}\right)+\xi_s<q\left(\xi_{s+1}\right)+P\left(\xi_{s+1}\right)=\xi_{s+1}.$$
Since $q\left(\xi_{s+1}\right)$ is the largest prime power not exceeding $\xi_{s+1}$, this implies that $q\left(\xi_{s+1}\right)+\xi_s$ is not a prime power, and that $q\left(\xi_{s+1}\right)$ is also the largest prime power not exceeding $q\left(\xi_{s+1}\right)+\xi_s$. The latter implies that
$$P\left(q\left(\xi_{s+1}\right)+\xi_s\right)=\left(q\left(\xi_{s+1}\right)+\xi_s\right)-q\left(\xi_{s+1}\right)=\xi_s,$$
so $q\left(\xi_{s+1}\right)+\xi_s$ is a positive integer less than $\xi_{s+1}$ having settling time
$$S\left(q\left(\xi_{s+1}\right)+\xi_s\right)=S\left(\xi_s\right)+1=s+1,$$
contradicting the definition of $\xi_{s+1}$.

Both (ii) and (iii) are immediate consequences of (i). To prove (iv), notice that, by Proposition \ref{prop:Bertrandrefinement}, there exists $\theta\in(0,1)$ such that
$$\xi_{s}=P\left(\xi_{s+1}\right)\leq\xi_{s+1}^\theta,\qquad\text{i.e.},\qquad \frac{\xi_{s+1}}{\xi_s}\geq \xi_{s+1}^{1-\theta},$$
for all sufficiently large values of $s$. This implies
$$\frac{\xi_{s+1}}{\xi_s}\xrightarrow{s\to\infty}\infty\qquad\text{and}\qquad\frac{q\left(\xi_{s+1}\right)}{q\left(\xi_s\right)}=\frac{\xi_{s+1}-\xi_s}{\xi_s-\xi_{s-1}}=\frac{\frac{\xi_{s+1}}{\xi_s}-1}{1-\frac{1}{\left(\frac{\xi_s}{\xi_{s-1}}\right)}}\xrightarrow{s\to\infty}\infty,$$
as desired.
\end{proof}

Part (i) of Proposition \ref{prop:xis} also means that for every $s\in\mathbb{N}$ we have $\xi_{s+1}=\xi_s+q_1$, where $\left(q_1,q_2\right)$ is the first pair of consecutive prime powers with $q_2-q_1\geq\xi_s+1$. Therefore, to compute $\xi_5$ one must find the first pair of consecutive prime powers differing by at least $360749$.

\section{The family of maps $P_p$}\label{sec:Pp}

The unpredictability of the functions $T$ and $S$ largely originates from the fact that the prime number $p$ acting at each iteration of $P$ is not fixed. Take as an initial condition, for instance, $x_1=35$. At the first iteration, which produces $x_2=P(35)=35-2^5=3$, the acting prime is $2$. However, at the second, which produces $x_3=P(3)=\frac{3}{3}=1$, the acting prime is $3$. In order to eliminate this unpredictability, and for a comparison, let us now consider a modification of the map $P$ constructed by fixing the acting prime. This consists in a one-parameter family of maps $P_p$, where $p\in\mathbb{P}$. As we shall see, all orbits of $P_p$ are predictable via the base-$p$ representations of the iterates (Lemma \ref{transmap}), and hence so are their transient lengths and settling times (Theorem \ref{thm:Fr}).

\subsection{The maps $P_p$}
 
Fix a prime $p\in\mathbb{P}$. Define the map $P_p:\mathbb{N}\to\mathbb{N}$ by
\begin{equation}\label{eq:Fr}
P_p(x):=\begin{cases}
x,&\text{if }1\leq x\leq p-1;\\
x-q_p(x),&\text{if }x\text{ is not a power of }p;\\
\frac{x}{p},&\text{if }x\text{ is a power of }p,
\end{cases}
\end{equation}
where $q_p(x)$ is the largest power of $p$ not exceeding $x$. Notice that this map has $p-1$ distinct fixed points: $1$, \ldots, $p-1$.

As in the original map, the \textit{orbit} of $x\in\mathbb{N}$ under the map $P_p:\mathbb{N}\to\mathbb{N}$ is the sequence $\left(x_n\right)_{n=1}^\infty$ where
$$x_1=x\qquad\quad\text{and}\quad\qquad x_{n+1}=P_p\left(x_n\right)\quad\text{for every }n\in\mathbb{N}.$$
The \textit{transient length} and \textit{limit} of $x$ under $P_p$ are the positive integers
$$T_p(x):=\min\bigl\{n\in\mathbb{N}:x_n\in\{1,\ldots,p-1\}\bigr\}\qquad\text{and}\qquad L_p(x):=x_{T_p(x)},$$
respectively. Equivalently, $L_p(x)=\lim\limits_{n\to\infty} x_n$, and $T_p(x)$ is the index of the first term in the orbit having value $L_p(x)$. One immediately sees that $T_p$ is unbounded since it diverges along, e.g., the sequence of powers of $p$. The \textit{settling time} and the \textit{attractor} of $x$ under $P_p$ are the positive integers
$$S_p(x):=\min\bigl\{n\in\mathbb{N}:x_n\in\{1,\ldots,p-1\}\cup\bigl\{p^k:k\in\mathbb{N}\bigr\}\bigr\}\qquad\text{and}\qquad W_p(x):=x_{S_p(x)},$$
respectively.

Our aim is to derive explicit formulae for $T_p(x)$, $L_p(x)$, $S_p(x)$, and $W_p(x)$ in terms of $p$ and $x$. This is achieved using some number-theoretic tools \cite{HardyWright,Koshy}. First, the \textit{$p$-adic valuation} \cite[page 562]{HardyWright} of $x$ is the non-negative integer
$$\nu_p(x):=\max\left\{m\in\mathbb{N}_0:p^m\mid x\right\},$$
i.e., the exponent of the largest power of $p$ which divides $x$. Moreover, we let $x\bmod p$ denote the remainder upon dividing $x$ by $p$ \cite[page 71]{Koshy}, $\sigma_p(x)$ the sum of the coefficients in the \textit{base-$p$ representation} \cite[Section 2.2]{Koshy} of $x$, and
$$\sigma_p^\ast(x):=\sigma_p(x)-(x\bmod p).$$
The formulae are given by the following theorem.

\begin{theorem}\label{thm:Fr}
Let $p\in\mathbb{P}$. For every $x\in\mathbb{N}$ we have
\begin{align*}
T_p(x)&=
\begin{cases}
\sigma_p^\ast(x)+\nu_p(x),&\text{if }p\mid x;\\
\sigma_p^\ast(x)+1,&\text{otherwise},
\end{cases}\\
L_p(x)&=\begin{cases}
1,&\text{if }p\mid x;\\
x\bmod p,&\text{otherwise},
\end{cases}\\
S_p(x)&=\begin{cases}
\sigma_p^\ast(x),&\text{if }p\mid x;\\
\sigma_p^\ast(x)+1,&\text{otherwise},
\end{cases}\\
W_p(x)&=\begin{cases}
p^{\nu_p(x)},&\text{if }p\mid x;\\
x\bmod p,&\text{otherwise}.
\end{cases}
\end{align*}
\end{theorem}

To prove this theorem, we first study the structure of the orbits of the map $P_p$. We do this by looking at how the base-$p$ representation of $x$ changes under an application of $P_p$. More formally, we study the conjugate map $\overline{P}_p:=\Psi_p\circ P_p\circ \Psi_p^{-1}$, where $\Psi_p$ is the bijection associating to every positive integer the infinite sequence whose terms are the coefficients $a_0,\ldots,a_k\in\mathbb{Z}_p$, where $a_k\neq0$ and $k\in\mathbb{N}_0$, in its base-$p$ representation $\sum_{i=0}^k a_ip^i$ followed by infinitely many zeros denoted using the standard notation $\overline{0}$ (e.g., $\Psi_2(6)=\left(0,1,1,\overline{0}\right)$), as a self-map on the space $\Omega_p$ of all such sequences (see Figure \ref{fig:conjugation}). The action of the map $\overline{P}_p$ is described by the following lemma.

\begin{figure}
\begin{center}
\begin{tikzpicture}[scale=1.25]
\node at (-1,1) {$\mathbb{N}$};
\node at (1,1) {$\mathbb{N}$};
\node at (-1,-1) {$\Omega_p$};
\node at (1,-1) {$\Omega_p$};

\draw[->] (-0.6,1) -- (0.6,1);
\node[above] at (0,1) {$P_p$};
\draw[->] (-0.6,-1) -- (0.6,-1);
\node[below] at (0,-1) {$\overline{P}_p$};

\draw[->] (1,0.6) -- (1,-0.6);
\node[left] at (-1,0) {${\Psi_p}^{-1}$};
\draw[->] (-1,-0.6) -- (-1,0.6);
\node[right] at (1,0) {$\Psi_p$};
\end{tikzpicture}
\end{center}
\caption{The commutative diagram representing the conjugation $\overline{P}_p=\Psi_p\circ P_p\circ \Psi_p^{-1}$.}
\label{fig:conjugation}
\end{figure}

\begin{lemma}\label{transmap}
The image of $\left(a_0,\ldots,a_k,\overline{0}\right)\in\Omega_p$, where $a_k\neq0$ and $k\in\mathbb{N}_0$, under $\overline{P}_p$ is given by
\begin{equation}\label{eq:transmap}
\overline{P}_p\left(a_0,\ldots,a_k,\overline{0}\right)=\begin{cases}
\left(a_0,\ldots,a_k,\overline{0}\right),&\text{if }k=0;\\
\left(a_1,\ldots,a_k,\overline{0}\right),&\text{if }k\geq 1\text{ and }a_0=\cdots=a_{k-1}=0\text{ and }a_k=1;\\
\left(a_0,\ldots,a_k-1,\overline{0}\right),&\text{otherwise}.
\end{cases}
\end{equation}
\end{lemma}
\begin{proof}
Let $x:=\Psi_p^{-1}\left(a_0,\ldots,a_k,\overline{0}\right)\in\mathbb{N}$. This means that $\sum_{i=0}^k a_ip^i$ is the base-$p$ representation of $x$.
\begin{enumerate}
\item[(i)] If $k=0$ then $1\leq x\leq p-1$, so $P_p(x)=x$. In other words, $\overline{P}_p\left(a_0,\ldots,a_k,\overline{0}\right)=\left(a_0,\ldots,a_k,\overline{0}\right)$.
\item[(ii)] Suppose $k\geq1$, $a_0=a_1=\cdots=a_{k-1}=0$, and $a_k=1$. Then $x=p^k\geq p$, so $P_p(x)=\frac{x}{p}=p^{k-1}$. In other words, $\overline{P}_p\bigl(\underbrace{0,\ldots,0}_k,1,\overline{0}\bigr)=\bigl(\underbrace{0,\ldots,0}_{k-1},1,\overline{0}\bigr)$, i.e., $\overline{P}_p\left(a_0,\ldots,a_k,\overline{0}\right)=\left(a_1,\ldots,a_k,\overline{0}\right)$.
\item[(iii)] Suppose $k\geq1$, and suppose at least one of the following is not true: $a_0=\cdots=a_{k-1}=0$ and $a_k=1$. Then, $x$ is not a power of $p$; for otherwise, since $x>p^k$, we must have that $x=p^\ell$ for some integer $\ell>k$, meaning that $p^\ell$ is the unique base-$p$ representation of $x$, in which the leading coefficient is $1$ and all the other coefficients are $0$, contradicting our supposition. Since $p^k$ is the largest power of $p$ not exceeding $x$, then $P_p(x)=x-p^k=\sum_{i=0}^{k-1} a_ip^i + \left(a_k-1\right)p^k$. In other words, $\overline{P}_p\left(a_0,\ldots,a_k,\overline{0}\right)=\left(a_0,\ldots,a_k-1,\overline{0}\right)$.\qedhere
\end{enumerate}
\end{proof}

For every $x\in\mathbb{N}$, Lemma \ref{transmap} gives a complete description of the orbit of $\Psi_p(x)$ under $\overline{P}_p$, and hence that of the orbit of $x$ under $P_p$. For instance, a complete description of the orbit of $108$ under $P_2$,
$$108\mapsto 44\mapsto 12\mapsto 4\mapsto 2\mapsto 1,$$
can be obtained from the fact that $\Psi_2(108)=\left(0,0,1,1,0,1,1,\overline{0}\right)$ and, under $\overline{P}_2$,
$$\left(0,0,1,1,0,1,1,\overline{0}\right)\mapsto \left(0,0,1,1,0,1,\overline{0}\right)\mapsto \left(0,0,1,1,\overline{0}\right)\mapsto \left(0,0,1,\overline{0}\right)\mapsto \left(0,1,\overline{0}\right)\mapsto \left(1,\overline{0}\right).$$

Now, let $\alpha=\left(a_k,\ldots,a_0\right)$. Based on the definitions of the analogues under $P_p$, the \textit{transient length}, \textit{limit}, \textit{settling time}, and \textit{attractor} of $\alpha$ under $\overline{P}_p$ are
\begin{align*}
\overline{T}_p(\alpha)&=\min\bigl\{n\in\mathbb{N}:\text{only the first term of }\alpha_n\text{ is non-zero}\bigr\},\\
\overline{L}_p(\alpha)&=\alpha_{\overline{T}_p(\alpha)},\\
\overline{S}_p(\alpha)&=\min\bigl\{n\in\mathbb{N}:
\alpha_n\in\bigl\{ \bigl( \underbrace{0,0,\ldots,0}_t,1,\overline{0}\bigr):t\in\mathbb{N}_0\bigr\}\cup
\bigl\{\bigl(1,\overline{0}\bigr),\ldots,\bigl(p-1,\overline{0}\bigr)\bigr\}\bigr\},\\
\overline{W}_p(\alpha)&=\alpha_{\overline{S}_p(\alpha)},
\end{align*}
respectively, where $\left(\alpha_n\right)_{n=1}^\infty$ is the orbit of $\alpha$ under $\overline{P}_p$, which is determined by
$$\alpha_1=\alpha\qquad\quad\text{and}\quad\qquad\alpha_{n+1}=\overline{P}_p\left(\alpha_n\right)\quad\text{for every }n\in\mathbb{N}.$$

\noindent\textit{Proof of Theorem \ref{thm:Fr}.} 
Proving Theorem \ref{thm:Fr} is equivalent to proving that
\begin{align*}
\overline{T}_p(\alpha)&=
\begin{cases}
\nu_p\left(\Psi_p^{-1}(\alpha)\right)+a_k+\cdots+a_1,&\text{if }a_0=0;\\
1+a_k+\cdots+a_1,&\text{if }a_0\geq 1,
\end{cases}\\
\overline{L}_p(\alpha)&=\begin{cases}
\left(1,\overline{0}\right),&\text{if }a_0=0;\\
\left(a_0,\overline{0}\right),&\text{if }a_0\geq 1,
\end{cases}\\
\overline{S}_p(\alpha)&=\begin{cases}
a_k+\cdots+a_1,&\text{if }a_0=0;\\
1+a_k+\cdots+a_1,&\text{if }a_0\geq 1,
\end{cases}\\
\overline{W}_p(\alpha)&=\begin{cases}
\bigl( \underbrace{0,0,\ldots,0}_{\nu_p\left(\Psi_p^{-1}(\alpha)\right)},1,\overline{0}\bigr),&\text{if }a_0=0;\\
\left(a_0,\overline{0}\right),&\text{if }a_0\geq 1.
\end{cases}
\end{align*}
We consider each of the two cases.

\bigskip

\noindent \textit{Case 1:} $a_0\geq 1$. If $k=0$, then the orbit of $\alpha=\left(a_0,\overline{0}\right)$ is constant, so $\overline{L}_p(\alpha)=\overline{W}_p(\alpha)=\left(a_0,\overline{0}\right)$ and $\overline{T}_p(\alpha)=\overline{S}_p(\alpha)=1$, obeying the desired formulae. Now suppose $k\geq 1$. Since $a_0\geq 1$, then each iteration is prescribed by the last branch in \eqref{eq:transmap}. Since
$$\alpha_1=\left(a_0,\ldots,a_k,\overline{0}\right),$$
then
$$\alpha_{1+a_k}=\overline{P}_p^{a_k}\left(\alpha_1\right)=\left(a_0,\ldots,a_{k-1},\overline{0}\right).$$
Continuing this, we have, for every $j\in\{k-1,\ldots,1\}$,
$$\alpha_{1+a_k+\cdots+a_j}=\overline{P}_p^{a_{j}}\left(\alpha_{1+a_k+\cdots+a_{j+1}}\right)=\left(a_0,\ldots,a_{j-1},\overline{0}\right),$$
which means that
$$\overline{T}_p(\alpha)=\overline{S}_p(\alpha)=1+a_k+\cdots+a_1\qquad\text{and}\qquad\overline{L}_p(\alpha)=\overline{W}_p(\alpha)=\left(a_0,\overline{0}\right),$$
as desired.

\bigskip

\noindent \textit{Case 2:} $a_0=0$. In this case, $k\geq 1$. Let $t:=\nu_p\left(\Psi_p^{-1}(\alpha)\right)$. Then $a_0=\cdots=a_{t-1}=0$ and $a_t\geq 1$, so
$$\alpha_1=\bigl( \underbrace{0,\ldots,0}_t,a_t,\ldots,a_k,\overline{0}\bigr).$$
By the last branch in \eqref{eq:transmap}, we have
$$\alpha_{1+a_k}=\overline{P}_p^{a_k}\left(\alpha_1\right)=\bigl( \underbrace{0,\ldots,0}_t,a_t,\ldots,a_{k-1},\overline{0}\bigr),$$
and, for every $j\in\{k-1,\ldots,t+1\}$,
$$\alpha_{1+a_k+\cdots+a_{j}}=\overline{P}_p^{a_{j}}\left(\alpha_{1+a_k+\cdots+a_{j+1}}\right)=\bigl( \underbrace{0,\ldots,0}_{t},a_t,\ldots,a_{j-1},\overline{0}\bigr).$$
Next,
$$\alpha_{a_k+\cdots+a_t}=\overline{P}_p^{a_{t}-1}\left(\alpha_{1+a_k+\cdots+a_{t+1}}\right)=\bigl( \underbrace{0,\ldots,0}_t,1,\overline{0}\bigr).$$
At this point we have proved that
$$\overline{S}_p(\alpha)=a_k+\cdots+a_1\qquad\text{and}\qquad\overline{W}_p(\alpha)=\bigl( \underbrace{0,\ldots,0}_t,1,\overline{0}\bigr)=\bigl( \underbrace{0,0,\ldots,0}_{\nu_p\left(\Psi_p^{-1}(\alpha)\right)},1,\overline{0}\bigr).$$
Finally, the dynamics concludes with $t$ iterations prescribed by the second branch in \eqref{eq:transmap}:
$$\alpha_{a_k+\cdots+a_t+t}=\overline{P}_p^t\left(\alpha_{a_k+\cdots+a_t}\right)=\left(1,\overline{0}\right).$$
Therefore,
$$\overline{T}_p(\alpha)=t+a_k+\cdots+a_1=\nu_p\left(\Psi_k^{-1}(\alpha)\right)+a_k+\cdots+a_1\qquad\text{and}\qquad\overline{L}_p(\alpha)=\left(1,\overline{0}\right),$$
as desired.\hfill$\square$

\subsection{A comparison of $P_p$ and $P$}

Let $p\in\mathbb{P}$. For every $x\in\mathbb{N}$, let
$$\hat{T}_p(x):=\frac{1}{x}\sum_{t=1}^x T_p(t)\qquad\quad\text{and}\quad\qquad \hat{S}_p(x):=\frac{1}{x}\sum_{t=1}^x S_p(t)$$
be the $x$-th Ces\'aro mean of the sequences $\left(T_p(x)\right)_{x=1}^\infty$ and $\left(S_p(x)\right)_{x=1}^\infty$, respectively. From Theorem \ref{thm:Fr}, noticing that
$$\frac{\bigl|\{t\leq x:p\mid t\}\bigr|}{x}=\frac{1}{x}\left\lfloor\frac{x}{p}\right\rfloor\sim\frac{1}{p}$$
and
$$\frac{1}{x}\sum_{t=1}^x \nu_p(t)=\frac{1}{x}\sum_{t=1}^{\left\lfloor\log_p x\right\rfloor}\left\lfloor\frac{x}{p^t}\right\rfloor\sim\sum_{t=1}^{\left\lfloor\log_p x\right\rfloor}\frac{1}{p^t}\sim\frac{1}{p-1}$$
and applying
$$\frac{1}{x}\sum_{t=1}^x\sigma_p(t)\sim \frac{(p-1)\ln x}{2\ln p}$$
\cite[Theorem 1]{Bush}, one proves the following corollary.

\begin{corollary}\label{corollary}
We have
$$\hat{T}_p(x)\sim\hat{S}_p(x)\sim \frac{(p-1)\ln x}{2\ln p}.$$
\end{corollary}

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\addlegendentry{$\hat{S}_2$}
\end{axis}
\end{tikzpicture}
\end{tabular}
\end{center}
\caption{Plots of $\hat{T}(x)$ and $\hat{T}_2(x)$ versus $\ln x$ (left) and $\hat{S}(x)$ and $\hat{S}_2(x)$ versus $\ln x$ (right) for $x\in\left\{1,\ldots,1\cdot 10^4\right\}$.}
\label{fig:TSasymptotics}
\end{figure}

Thus, the growths of $\hat{T}_p(x)$ and $\hat{S}_p(x)$ with $x$ are both logarithmic. By contrast, letting
$$\hat{T}(x):=\frac{1}{x}\sum_{t=1}^x T(t)\qquad\quad\text{and}\quad\qquad \hat{S}(x):=\frac{1}{x}\sum_{t=1}^x S(t),$$
from Theorems \ref{thm:power2} and \ref{thm:orderS(x)} we know that the former grows at most logarithmically, and the latter at most double-logarithmically. In fact, Figure \ref{fig:TSasymptotics} suggests that both quantities grow at most double-logarithmically; in particular,
$$\hat{T}(x)\ll \ln \ln x.$$
This is confirmed by Figure \ref{fig:TSasymptotics2} which suggests the existence of $\lambda\geq 1$ such that
\begin{equation}\label{eq:TsimS}
\frac{\hat{T}(x)}{\hat{S}(x)}\sim\lambda.
\end{equation}

\begin{figure}
\begin{center}
\begin{tikzpicture}
\begin{axis}[
	xmin=.9198571430,
	xmax=2.202142857,
	ymin=1.336700000,
	ymax=3.296300000,
	xlabel=$\hat{S}(x)$,
	ylabel={$\hat{T}(x)$},
	width=7.25cm,
    height=6cm,
    clip=false,grid=major,legend pos=north west,reverse legend,legend cell align={left}
]
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2.823) (1.829, 2.824) (1.829, 2.825) (1.829, 2.826) (1.830, 2.820) (1.830, 2.821) (1.830, 2.823) (1.830, 2.824) (1.830, 2.825) (1.830, 2.827) (1.830, 2.828) (1.831, 2.823) (1.831, 2.824) (1.831, 2.826) (1.831, 2.827) (1.831, 2.828) (1.831, 2.830) (1.832, 2.824) (1.832, 2.825) (1.832, 2.826) (1.832, 2.828) (1.833, 2.827) (1.833, 2.829) (1.834, 2.825) (1.834, 2.826) (1.835, 2.827) (1.835, 2.828) (1.836, 2.828) (1.836, 2.830) (1.836, 2.831) (1.836, 2.832) (1.837, 2.828) (1.837, 2.831) (1.837, 2.833) (1.838, 2.831) (1.838, 2.832) (1.839, 2.833) (1.840, 2.836) (1.840, 2.837) (1.840, 2.838) (1.840, 2.840) (1.841, 2.834) (1.841, 2.836) (1.841, 2.837) (1.841, 2.838) (1.841, 2.839) (1.841, 2.840) (1.841, 2.841) (1.841, 2.842) (1.842, 2.835) (1.842, 2.836) (1.842, 2.837) (1.842, 2.838) (1.842, 2.839) (1.842, 2.840) (1.842, 2.842) (1.843, 2.835) (1.843, 2.836) (1.843, 2.837) (1.843, 2.839) (1.843, 2.840) (1.844, 2.838) (1.844, 2.839) (1.845, 2.837) (1.845, 2.840) (1.845, 2.841) (1.847, 2.842) 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2.859) (1.868, 2.860) (1.868, 2.861) (1.868, 2.862) (1.869, 2.858) (1.869, 2.859) (1.869, 2.860) (1.869, 2.861) (1.869, 2.862) (1.869, 2.863) (1.869, 2.864) (1.869, 2.865) (1.869, 2.866) (1.870, 2.858) (1.870, 2.859) (1.870, 2.860) (1.870, 2.861) (1.870, 2.862) (1.870, 2.864) (1.870, 2.865) (1.870, 2.866) (1.870, 2.867) (1.870, 2.868) (1.871, 2.860) (1.871, 2.861) (1.871, 2.862) (1.871, 2.863) (1.871, 2.865) (1.871, 2.867) (1.871, 2.868) (1.871, 2.869) (1.872, 2.862) (1.872, 2.863) (1.872, 2.868) (1.872, 2.869) (1.872, 2.870) (1.872, 2.871) (1.873, 2.871) (1.874, 2.869) (1.874, 2.870) (1.874, 2.871) (1.874, 2.872) (1.875, 2.869) (1.875, 2.870) (1.875, 2.871) (1.875, 2.872) (1.875, 2.873) (1.875, 2.874) (1.876, 2.870) (1.876, 2.871) (1.876, 2.872) (1.876, 2.873) (1.876, 2.874) (1.876, 2.875) (1.876, 2.876) (1.877, 2.874) (1.877, 2.876) (1.878, 2.874) (1.879, 2.874) (1.880, 2.875) (1.880, 2.878) (1.881, 2.878) (1.882, 2.876) (1.882, 2.877) (1.882, 2.878) (1.883, 2.877) (1.883, 2.878) (1.883, 2.879) (1.883, 2.880) (1.884, 2.877) (1.884, 2.878) (1.884, 2.879) (1.884, 2.880) (1.884, 2.881) (1.884, 2.882) (1.885, 2.880) (1.885, 2.881) (1.885, 2.882) (1.885, 2.883) (1.886, 2.881) (1.886, 2.882) (1.886, 2.883) (1.887, 2.882) (1.887, 2.883) (1.887, 2.884) (1.887, 2.885) (1.888, 2.884) (1.888, 2.885) (1.888, 2.886) (1.888, 2.887) (1.889, 2.884) (1.889, 2.885) (1.889, 2.886) (1.889, 2.887) (1.890, 2.885) (1.890, 2.886) (1.890, 2.887) (1.890, 2.889) (1.890, 2.890) (1.891, 2.886) (1.891, 2.887) (1.891, 2.888) (1.891, 2.889) (1.891, 2.890) (1.892, 2.888) (1.892, 2.889) (1.892, 2.890) (1.892, 2.891) (1.893, 2.889) (1.893, 2.891) (1.893, 2.892) (1.893, 2.893) (1.894, 2.889) (1.894, 2.890) (1.895, 2.892) (1.895, 2.893) (1.896, 2.893) (1.897, 2.893) (1.898, 2.894) (1.899, 2.894) (1.899, 2.895) (1.899, 2.896) (1.900, 2.894) (1.900, 2.895) (1.900, 2.896) (1.900, 2.897) (1.901, 2.895) (1.901, 2.896) (1.901, 2.897) (1.902, 2.897) (1.903, 2.897) (1.904, 2.897) (1.904, 2.898) (1.904, 2.899) (1.904, 2.900) (1.904, 2.901) (1.904, 2.902) (1.905, 2.898) (1.905, 2.899) (1.905, 2.900) (1.905, 2.901) (1.905, 2.902) (1.905, 2.903) (1.905, 2.904) (1.906, 2.899) (1.906, 2.900) (1.906, 2.901) (1.906, 2.902) (1.906, 2.903) (1.906, 2.905) (1.907, 2.896) (1.907, 2.897) (1.907, 2.900) (1.907, 2.901) (1.907, 2.902) (1.907, 2.903) (1.907, 2.904) (1.907, 2.905) (1.908, 2.896) (1.908, 2.897) (1.908, 2.898) (1.908, 2.899) (1.908, 2.900) (1.908, 2.901) (1.908, 2.902) (1.908, 2.903) (1.908, 2.904) (1.909, 2.899) (1.909, 2.900) (1.909, 2.901) (1.909, 2.902) (1.909, 2.903) (1.909, 2.904) (1.910, 2.900) (1.910, 2.901) (1.910, 2.902) (1.911, 2.901) (1.911, 2.902) (1.912, 2.901) (1.912, 2.902) (1.912, 2.903) (1.912, 2.904) (1.913, 2.904) (1.913, 2.905) (1.913, 2.906) (1.914, 2.905) (1.914, 2.906) (1.914, 2.907) (1.914, 2.908) (1.915, 2.907) (1.915, 2.908) (1.915, 2.909) (1.916, 2.908) (1.916, 2.909) (1.916, 2.910) (1.916, 2.911) (1.917, 2.909) (1.917, 2.910) (1.917, 2.911) (1.917, 2.912) (1.918, 2.911) (1.918, 2.912) (1.918, 2.913) (1.918, 2.914) (1.919, 2.913) (1.919, 2.914) (1.920, 2.913) (1.920, 2.915) (1.920, 2.916) (1.921, 2.914) (1.921, 2.915) (1.921, 2.916) (1.921, 2.917) (1.922, 2.916) (1.922, 2.917) (1.922, 2.918) (1.923, 2.918) (1.923, 2.919) (1.924, 2.918) (1.924, 2.919) (1.924, 2.920) (1.924, 2.921) (1.925, 2.921) (1.926, 2.920) (1.926, 2.921) (1.927, 2.921) (1.927, 2.922) (1.928, 2.922) (1.929, 2.920) (1.929, 2.921) (1.929, 2.922) (1.929, 2.923) (1.929, 2.924) (1.929, 2.927) (1.929, 2.928) (1.929, 2.929) (1.930, 2.923) (1.930, 2.924) (1.930, 2.925) (1.930, 2.928) (1.930, 2.929) (1.930, 2.930) (1.930, 2.931) (1.931, 2.924) (1.931, 2.925) (1.931, 2.928) (1.931, 2.929) (1.931, 2.930) (1.931, 2.931) (1.932, 2.930) (1.932, 2.931) (1.932, 2.932) (1.933, 2.931) (1.933, 2.933) (1.933, 2.935) (1.933, 2.936) (1.933, 2.937) (1.934, 2.932) (1.934, 2.933) (1.934, 2.935) (1.934, 2.936) (1.934, 2.937) (1.934, 2.938) (1.935, 2.935) (1.935, 2.936) (1.935, 2.937) (1.936, 2.933) (1.936, 2.934) (1.936, 2.935) (1.936, 2.937) (1.937, 2.934) (1.937, 2.935) (1.937, 2.936) (1.937, 2.937) (1.938, 2.933) (1.938, 2.934) (1.938, 2.935) (1.938, 2.936) (1.939, 2.935) (1.939, 2.936) (1.939, 2.937) (1.939, 2.938) (1.940, 2.937) (1.940, 2.938) (1.941, 2.938) (1.942, 2.939) (1.942, 2.940) (1.943, 2.939) (1.943, 2.940) (1.943, 2.941) (1.944, 2.941) (1.944, 2.942) (1.944, 2.943) (1.945, 2.942) (1.945, 2.943) (1.946, 2.943) (1.946, 2.944) (1.947, 2.942) (1.947, 2.943) (1.947, 2.944) (1.947, 2.945) (1.947, 2.946) (1.948, 2.942) (1.948, 2.943) (1.948, 2.944) (1.948, 2.945) (1.948, 2.946) (1.949, 2.943) (1.949, 2.944) (1.949, 2.945) (1.949, 2.946) (1.950, 2.944) (1.950, 2.945) (1.950, 2.946) (1.950, 2.947) (1.950, 2.948) (1.951, 2.947) (1.951, 2.948) (1.952, 2.948) (1.952, 2.949) (1.953, 2.949) (1.953, 2.950) (1.954, 2.950) (1.954, 2.951) (1.955, 2.950) (1.955, 2.951) (1.955, 2.952) (1.956, 2.952) (1.956, 2.953) (1.956, 2.954) (1.957, 2.952) (1.957, 2.953) (1.957, 2.954) (1.957, 2.955) (1.957, 2.956) (1.958, 2.956) (1.958, 2.957) (1.959, 2.956) (1.959, 2.957) (1.959, 2.958) (1.960, 2.956) (1.960, 2.957) (1.960, 2.958) (1.961, 2.957) (1.961, 2.958) (1.961, 2.959) (1.962, 2.958) (1.962, 2.959) (1.963, 2.959) (1.963, 2.960) (1.964, 2.959) (1.964, 2.960) (1.964, 2.961) (1.965, 2.960) (1.965, 2.961) (1.965, 2.962) (1.966, 2.961) (1.966, 2.962) (1.967, 2.961) (1.967, 2.962) (1.967, 2.963) (1.967, 2.964) (1.968, 2.961) (1.968, 2.962) (1.968, 2.963) (1.968, 2.964) (1.968, 2.965) (1.969, 2.963) (1.969, 2.964) (1.969, 2.965) (1.969, 2.966) (1.970, 2.965) (1.970, 2.966) (1.970, 2.967) (1.971, 2.966) (1.971, 2.967) (1.971, 2.968) (1.972, 2.967) (1.972, 2.968) (1.972, 2.969) (1.973, 2.968) (1.973, 2.969) (1.974, 2.969) (1.974, 2.970) (1.974, 2.971) (1.975, 2.969) (1.975, 2.970) (1.975, 2.971) (1.975, 2.972) (1.975, 2.973) (1.976, 2.970) (1.976, 2.971) (1.976, 2.972) (1.976, 2.973) (1.976, 2.974) (1.977, 2.973) (1.977, 2.974) (1.978, 2.974) (1.978, 2.975) (1.978, 2.976) (1.978, 2.977) (1.979, 2.975) (1.979, 2.976) (1.979, 2.977) (1.979, 2.978) (1.980, 2.978) (1.980, 2.979) (1.980, 2.980) (1.981, 2.979) (1.981, 2.980) (1.981, 2.981) (1.981, 2.982) (1.982, 2.981) (1.982, 2.982) (1.982, 2.983) (1.983, 2.982) (1.983, 2.983) (1.984, 2.983) (1.984, 2.984) (1.985, 2.983) (1.985, 2.984) (1.985, 2.985) (1.986, 2.985) (1.986, 2.986) (1.987, 2.985) (1.987, 2.986) (1.987, 2.987) (1.988, 2.987) (1.988, 2.988) (1.989, 2.987) (1.989, 2.988) (1.989, 2.989) (1.990, 2.988) (1.990, 2.989) (1.990, 2.990) (1.991, 2.990) (1.991, 2.991) (1.992, 2.991) (1.992, 2.992) (1.993, 2.992) (1.993, 2.993) (1.994, 2.992) (1.994, 2.993) (1.995, 2.992) (1.995, 2.993) (1.995, 2.994) (1.996, 2.993) (1.996, 2.994) (1.996, 2.995) (1.997, 2.994) (1.997, 2.995) (1.998, 2.994) (1.998, 2.995) (1.998, 2.996) (1.999, 2.995) (1.999, 2.996) (1.999, 2.997) (2.000, 2.996) (2.000, 2.997) (2.000, 2.998) (2.000, 2.999) (2., 2.996) (2.001, 2.997) (2.001, 2.998) (2.001, 2.999) (2.001, 3.000) (2.001, 3.) 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\end{axis}
\end{tikzpicture}
\end{center}
\caption{Plot of $\hat{T}(x)$ versus $\hat{S}(x)$ for $x\in\left\{1,\ldots,1\cdot 10^5\right\}$.}
\label{fig:TSasymptotics2}
\end{figure}

Now, since
\begin{align*}
\left(\hat{T}-\hat{S}\right)(x)&=\frac{1}{x}\sum_{t=1}^x(T-S)(t)\\
&=\frac{1}{x}\sum_{u=0}^{\max\limits_{1\leq t\leq x}(T-S)(t)}u\,\left|(T-S)^{-1}(\{u\})\cap\{1,\ldots,x\}\right|\\
&=\sum_{u=1}^{\max\limits_{1\leq t\leq x}(T-S)(t)} u\,d_x\left((T-S)^{-1}(\{u\})\right)\\
&\leq d_x\left((T-S)^{-1}(\mathbb{N})\right)\left(\max\limits_{1\leq t\leq x}(T-S)(t)\right)^2,
\end{align*}
then we have $\left(\hat{T}-\hat{S}\right)(x)\xrightarrow{x\to\infty}0$ if $d_x\left((T-S)^{-1}(\mathbb{N})\right)=d_x \left(W^{-1}\left(\mathbb{N}_{\geq 2}\right)\right)$ decays algebraically to zero as $x\to\infty$. An extensive dataset provides an evidence for the latter (Figure \ref{fig:throughprimepower}), leading to the following conjecture.

\begin{figure}
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\addplot[domain=5000:1000000,ultra thick,red] {.7839171430/x^(0.638407522714256e-2)};
\end{axis}
\end{tikzpicture}
\end{tabular}
\end{center}
\caption{Plot of $d_x\left(W^{-1}\left(\mathbb{N}_{\geq 2}\right)\right)$ versus $x$ for $x\in\{5\cdot10^4,\ldots,1\cdot 10^6\}$ in log-log scale (left) together with the least-squares line determined using its values for $x\in\{6\cdot 10^5,\ldots,1\cdot 10^6\}$ which has slope $\alpha\approx -0.006$ and ordinate intercept $\beta\approx -0.243$, and in linear-linear scale (right) together with the curve $e^{\beta}x^\alpha$.}
\label{fig:throughprimepower}
\end{figure}

\begin{conjecture}\label{finalconjecture}
There exist $c\in(0,1)$ and $\alpha\in\left(-10^{-2},-10^{-3}\right)$ such that $$d_x\left(W^{-1}\left(\mathbb{N}_{\geq2}\right)\right)\sim c x^\alpha\qquad\text{as}\qquad x\to\infty.$$
In particular, almost every orbit of $P$ contains no prime power.
\end{conjecture}

\noindent Notice that if this conjecture holds, then so does \eqref{eq:TsimS} with $\lambda=1$.

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\end{thebibliography}

\bigskip
\hrule
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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11A25;
Secondary 11A41, 37P99.

\noindent \emph{Keywords: } 
prime, prime power, prime gap.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A000040},
\seqnum{A000079},
\seqnum{A000961},
\seqnum{A066352}, and
\seqnum{A336825}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received October 12 2020;
revised versions received December 28 2020; January 6 2021.
Published in {\it Journal of Integer Sequences}, January 24 2021.

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\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{https://cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in

\end{document}
