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\begin{center}
\vskip 1cm{\Large\bf On a Family of Solutions to Arithmetic \\
\vskip .1in
Differential Equations Involving the Collatz Map
}
\vskip 1cm
\large
Zachary P. Bradshaw\\
Advanced Processing Branch\\
Naval Surface Warfare Center\\
Panama City, FL 32407\\
USA \\
\href{mailto:email}{\tt zbradshaw@tulane.edu} \\
\end{center}

\vskip .2 in
\begin{abstract}
The arithmetic derivative is a nonlinear derivation on the
positive integers which forms a natural analog of the conventional
derivative. While exploring solutions to arithmetic differential
equations, we stumbled across a curious pattern in the positive integers
for which the arithmetic derivative and the Collatz map commute. Here
we report on these empirical findings, and prove several analytical results
on the form of such numbers. Among these findings is the existence of
a family of semiprime numbers which are mapped by the Collatz function
to another semiprime having a sum of prime factors which is half of the
original semiprime's. We show that this family of semiprimes solves the
commutation problem and that the sum of their reciprocals converges.
\end{abstract}

\section{Introduction} 

The arithmetic derivative is a distinctive analogue to the conventional derivative from calculus. Unlike its continuous counterpart, the arithmetic derivative is tailored specifically for positive integers and encodes information about a number's prime factorization, offering a novel perspective on the structural properties of numbers. It was introduced initially by Shelly in 1911 \cite{shelly1911} and subsequently refined by various mathematicians, including Barbeau \cite{barbeau1961}, who extended it to the rationals, showing that the arithmetic derivative satisfies an analog of the familiar quotient rule. Later, this operation was extended to the set of irrational numbers which can be written as the product of primes raised to rational powers by Ufnarovski and Ahlander \cite{ufnarovski2003}. Central to the arithmetic derivative is its adherence to a product rule akin to that of traditional calculus, facilitating computations that reveal relationships between integers. This operation has garnered attention not only for its theoretical elegance but also for its connections to conjectures in number theory, such as the Goldbach and twin prime conjectures \cite{ufnarovski2003}, and these links hint at deeper insights into the distribution and behavior of prime numbers.

Meanwhile, the Collatz conjecture has been studied at least as far back as 1937, and it remains a surprisingly challenging open problem. It has been attacked from many angles, including studies of the number of steps needed to reach 1 (the total stopping time) \cite{lagarias2023}, continuous extensions \cite{chamberland1996}, and the determination of bounds on the size of a nontrivial cycle \cite{eliahou1993}. In 1972, Conway showed that a natural generalization of the Collatz conjecture was undecidable \cite{conway1972} and later built on this work to construct a method for universal computation known as FRACTRAN \cite{conway1987}. Interestingly, the Collatz iterates were also related to Benford's Law \cite{Kontorovich_2005,lagarias2006}. More recently, it was shown by Tao that almost all orbits of the Collatz map attain almost bounded values in the sense of logarithmic density \cite{tao2019}, and Barina has confirmed that positive integers as high as $2^{68}$ obey the conjecture \cite{barina2020}. Despite decades of scrutiny and extensive computational exploration, the Collatz conjecture remains unproven, captivating mathematicians with its deceptively simple nature.

The original motivation for this work was to investigate numbers for which the arithmetic derivative and compositions of the Collatz map commute. It can be shown that arithmetic functions which satisfy this property for some integer $n$ in a cycle $\Omega$ necessarily send $n$ into a cycle of length dividing $\lvert\Omega\rvert$. Indeed, if $D(C^{\lvert\Omega\rvert}(n))=C^{\lvert\Omega\rvert}(D(n))$ and $C^{\lvert\Omega\rvert}(n)=n$, then $D(n)=C^{\lvert\Omega\rvert}(D(n))$, so that $D(n)$ generates a cycle. Note that the choice of $D$ was arbitrary here, and the arithmetic derivative was chosen merely out of curiosity while exploring concrete examples. In the course of our exploration, we find that for $\lvert\Omega\rvert=1$, these numbers satisfy a puzzling property. Indeed, all but one example are congruent to 9 modulo 10 and can be written as the product of two distinct primes, one which ends in 1, while the other ends in 9. The exception is the number 606938385, which is the product of the three distinct primes $3, 5$, and $40462559$.

In this work, we introduce a class of semiprime numbers which we call \textit{compatible} (\seqnum{A376275} in the OEIS \cite{oeis}), showing that they satisfy the arithmetic differential equation $D\bigl(\frac{3n+1}{2}\bigr)=\frac{D(n)}{2}$, thereby solving the commutation problem $D(C(n))=C(D(n))$. Naturally, we ask whether there are infinitely many such numbers. Just as in Brun's work on twin primes \cite{brun1919}, we show that the sum of the reciprocals of the compatible semiprimes converges, so that an easy answer to this question does not seem likely. However, we show that all known solutions to the commutation problem are semiprimes of this form except for the case 606938385, which instead solves the arithmetic differential equation $D\bigl(\frac{3n+1}{2}\bigr)=\frac{3D(n)+1}{2}$. Moreover, we explain the observation that all solutions (again, except for 606938385) are congruent to 9 modulo 10 by showing that this is the case for all compatible semiprimes. This result does not rule out the possibility that further solutions which break this congruence exist, but merely explains what is seen empirically.

The rest of this paper is organized as follows. In Section~\ref{sec:arithmetic}, we review the arithmetic derivative, re-deriving several known results, and we define the Collatz map. In Section~\ref{sec:results}, we present several analytical results about the numbers for which these maps commute. We show that all such numbers are odd, and motivated by our empirical results, attempt to characterize the distinct primes appearing in these numbers by showing that they cannot have the forms of several classes of prime numbers. We then give a bound on the difference between these primes, showing that it grows at least linearly with the larger prime. We also produce a class of semiprime numbers which belong to this sequence of numbers and show that all known elements of the sequence belong to this class, with the exception of 606938385. In Section~\ref{sec:sum}, we show that the sum of the reciprocals of the numbers in this class converges. Finally, in Section~\ref{sec:conclusion}, we give concluding remarks and put forth several conjectures, including the assertion that there are infinitely many numbers for which the arithmetic derivative and Collatz map commute.

\section{Background}\label{sec:arithmetic}

The arithmetic derivative is a natural analog of the conventional derivative from calculus, at least algebraically. We define the arithmetic derivative to be a non-linear derivation $D:\mathbb{N}\to\mathbb{N}$ on the set of natural numbers with the property that $D(1)=D(0)=0$ and $D(p)=1$ for all primes $p$. Explicitly, we define $D$ so that
\begin{equation}\label{eq:leibniz}
    D(mn)=D(m)n+mD(n)
\end{equation}
for every $m,n\in\mathbb{N}$. This requirement already demands $D(1)=D(1)+D(1)$, so that $D(1)=0$ by construction. Moreover, if $n=pm$ for some $m\in\mathbb{N}$ and some prime $p$, it follows from \eqref{eq:leibniz} that $D(n)=D(p)m+pD(m)$. If $m$ is composite, we may proceed inductively until the only derivatives appearing on the right side of the equality are derivatives of primes. Thus, the arithmetic derivative (and indeed every arithmetic function that satisfies \eqref{eq:leibniz}) is completely determined by its action on prime numbers. The choice of $D(p)=1$ serves to treat all prime numbers with equal weighting. Another consequence of \eqref{eq:leibniz} is that the arithmetic derivative satisfies a power rule. Indeed, we have
\begin{equation*}
    D(n^k)=kn^{k-1}D(n),
\end{equation*}
and so it follows that if $n=p_1^{\nu_1}\cdots p_k^{\nu_k}$ is the prime factorization of $n$, then
\begin{equation}\label{eq:general_deriv}
    D(n)=n\sum_{j=1}^k\frac{\nu_j}{p_j}D(p_j)=n\sum_{j=1}^k\frac{\nu_j}{p_j}.
\end{equation}
Notice the logarithmic derivative flavor of \eqref{eq:general_deriv}. If we divide by $n$ and define the logarithmic arithmetic derivative $\textnormal{ld}(n):=D(n)/n$, then \eqref{eq:general_deriv} becomes
\begin{equation*}
    \textnormal{ld}(n)=\sum_{j=1}^k\frac{\nu_j}{p_j}.
\end{equation*}

Interestingly, the arithmetic derivative can be linked to both the Goldbach and twin primes conjectures rather easily. Indeed, if $p,p+2$ are twin primes, then $D(2p)=D(2)p+2D(p)=p+2$. Now applying the arithmetic derivative again, we find that $D^2(2p)=1$. Thus, if the twin prime conjecture holds, then there are infinitely many $n$ for which $D^2(n)=1$. As for the Goldbach conjecture, observe that if $2n=p+q$ for some $p,q$ prime, then $D(pq)=p+q=2n$. Thus, if the Goldbach conjecture holds, then for every even integer $2n$, there is another integer $k$ such that $D(k)=2n$.

We make use of inequalities for the arithmetic derivative derived by Barbeau \cite{barbeau1961} and then strengthened by Dahl, Olsson, and Loiko \cite{dahl2011}. In particular, we make use of the following lemma featuring the prime omega function (the number of prime factors in $n$).
\begin{lemma}\label{lemma:inequality}
    Let $\Omega$ denote the prime omega function and let $p$ be the least prime in $n$. Then
    \begin{equation*}
        \Omega(n)n^{\frac{\Omega(n)-1}{\Omega(n)}}\le D(n)\le\frac{n\log_p(n)}{p}.
    \end{equation*}
    Moreover, equality holds if and only if $n$ is a prime power.
\end{lemma}
\begin{proof}
    Let $n=p_1^{\nu_1}\cdots p_k^{\nu_k}$. Then
    \begin{equation*}
        D(n)=n\sum_{j=1}^k\frac{\nu_j}{p_j}.
    \end{equation*}
    Let $p$ be the least prime of $p_1,\ldots,p_k$. Then
    \begin{equation*}
        D(n)\le n\sum_{j=1}^k\frac{\nu_j}{p}\le n\sum_{j=1}^k\frac{\nu_j\log_p(p_j)}{p}=\frac{n\log_p(n)}{p}
    \end{equation*}
    and equality holds whenever $p_1=\cdots=p_k=p$. For the lower bound, note that $\frac{1}{\Omega(n)}D(n)=\frac{n}{\Omega(n)}\sum_{j=1}^k\frac{\nu_j}{p_j}$ is an arithmetic mean and apply the AM-GM inequality. This produces
    \begin{equation*}
        \frac{D(n)}{\Omega(n)}\ge n\bigg(\prod_{j=1}^k\frac{1}{p_j^{\nu_j}}\bigg)^{1/\Omega(n)}=n^{\frac{\Omega(n)-1}{\Omega(n)}},
    \end{equation*}
    and equality holds in the AM-GM inequality whenever $p_1=\cdots=p_k$.
\end{proof}

The Collatz mapping is the arithmetic function defined by
\begin{equation*}
    C(n)=\begin{cases}
        3n+1, &amp; \text{ if $n$ odd;}\\
        \frac{n}{2}, &amp; \text{ if $n$ even.}
    \end{cases}
\end{equation*}
A longstanding open problem in mathematics is the so-called Collatz conjecture, which states that the discrete dynamical system with trajectories defined by $C$ always converges to the $\{4,2,1\}$ cycle. Equivalently, it says that for all $n\in\mathbb{N}$, there exists a $k\in\mathbb{N}$ such that $C^k(n)=1$. This seemingly simple problem has gone unproven for more than eighty years, and those who study it deeply often comment that the problem is completely intractable and outside the scope of the mathematical tools available today. While this may be the case, it can be interesting to explore certain aspects of this problem in an attempt to gain even the smallest insight. We forgo a study of what is known about the Collatz conjecture and instead refer the reader to the excellent survey by Lagarias \cite{lagarias2023}. We, however, point out that when $n$ is odd, the next iterate in the trajectory of $n$ is even. It is therefore common to redefine the Collatz map as
\begin{equation*}
    C(n)=\begin{cases}
        \frac{3n+1}{2}, &amp; \text{ if $n$ odd;}\\
        \frac{n}{2}, &amp; \text{ if $n$ even,}
    \end{cases}
\end{equation*}
and we take this as our definition throughout the rest of our work.

\section{The commutation problem and compatible semiprimes}\label{sec:results}

We let $(a_n)_{n=1}^\infty$ denote the sequence of numbers for which the arithmetic derivative and Collatz map commute. In our empirical investigation, we have found the first 30 numbers of this sequence and they are listed along with their prime factorization in Table~\ref{tab:an}. We note that all but one of these numbers are congruent to 9 mod 10 and can be written as a product of two distinct primes, one of which is congruent to 1 while the other is congruent to 9. All numbers with the exclusion of the exceptional case $n=12$ are also congruent to 2 modulo 3. For $n=12$, we see that $a_n$ is divisible by 3. We note also that in the case of $n=12$, the largest prime is at least an order of magnitude larger than all of the remaining primes in the table.

\begin{table}
    \centering
    \begin{tabular}{|cccccc|}
        \hline
         $n$ &amp; $a_n$ &amp; Prime Factorization &amp; $D(a_n)$ &amp; $C(a_n)$ &amp; Prime Factorization\\
         \hline\hline
         1 &amp; 114239 &amp; $71\times1609$ &amp; 1680 &amp; 171359 &amp; $349\times491$\\
         \hline
         2 &amp; 144059 &amp; $71\times2029$ &amp; 2100 &amp; 216089 &amp; $281\times769$\\
         \hline
         3 &amp; 933899 &amp; $131\times7129$ &amp; 7260 &amp; 1400849 &amp; $439\times3191$ \\
         \hline
         4 &amp; 1918199 &amp; $79\times24281$ &amp; 24360 &amp; 2877299 &amp; $241\times11939$\\ 
         \hline
         5 &amp; 25054499 &amp; $149\times168151$ &amp; 168300 &amp; 37581749 &amp; $449\times83701$\\
         \hline
         6 &amp; 30495419 &amp; $1129\times27011$ &amp; 28140 &amp; 45743129 &amp; $5099\times8971$\\
         \hline
         7 &amp; 33065159 &amp; $569\times58111$ &amp; 58680 &amp; 49597739 &amp; $1801\times27539$\\
         \hline
         8 &amp; 72602039 &amp; $1511\times48049$ &amp; 49560 &amp; 108903059 &amp; $5711\times19069$\\
         \hline
         9 &amp; 255442559 &amp; $809\times315751$ &amp; 316560 &amp; 383163839 &amp; $2459\times155821$\\
         \hline
         10 &amp; 353104079 &amp; $1511\times233689$ &amp; 235200 &amp; 529656119 &amp; $4691\times112909$\\
         \hline
         11 &amp; 575473559 &amp; $3631\times158489$ &amp; 162120 &amp; 863210339 &amp; $12611\times68449$\\
         \hline
         12 &amp; 606938385 &amp; $3\times5\times40462559$ &amp; 323700487 &amp; 910407578 &amp; $2\times47\times83\times116689$\\
         \hline
         13 &amp; 808589879 &amp; $2801\times288679$ &amp; 291480 &amp; 1212884819 &amp; $8861\times136879$\\
         \hline
         14 &amp; 846509819 &amp; $2861\times295879$ &amp; 298740 &amp; 1269764729 &amp; $9049\times140321$\\
         \hline
         15 &amp; 1042804799 &amp; $6871\times151769$ &amp; 158640 &amp; 1564207199 &amp; $36709\times42611$\\
         \hline
         16 &amp; 1055710979 &amp; $2999\times352021$ &amp; 355020 &amp; 1583566469 &amp; $9421\times168089$\\
         \hline
         17 &amp; 1059728279 &amp; $4079\times259801$ &amp; 263880 &amp; 1589592419 &amp; $13411\times118529$\\
         \hline
         18 &amp; 1184657879 &amp; $2281\times519359$ &amp; 521640 &amp; 1776986819 &amp; $7001\times253819$\\
         \hline
         19 &amp; 1247085239 &amp; $4751\times262489$ &amp; 267240 &amp; 1870627859 &amp; $15889\times117731$\\
         \hline
         20 &amp; 1791627599 &amp; $8609\times208111$ &amp; 216720 &amp; 2687441399 &amp; $38431\times69929$\\
         \hline
         21 &amp; 2196997739 &amp; $9059\times242521$ &amp; 251580 &amp; 3295496609 &amp; $37199\times88591$\\
         \hline
         22 &amp; 2323221179 &amp; $7741\times300119$ &amp; 307860 &amp; 3484831769 &amp; $27581\times126349$\\
         \hline
         23 &amp; 2372469179 &amp; $9091\times260969$ &amp; 270060 &amp; 3558703769 &amp; $35899\times99131$\\
         \hline
         24 &amp; 2591327159 &amp; $10369\times249911$ &amp; 260280 &amp; 3886990739 &amp; $46439\times83701$\\
         \hline
         25 &amp; 3063507719 &amp; $8191\times374009$ &amp; 382200 &amp; 4595261579 &amp; $28211\times162889$\\
         \hline
         26 &amp; 3276652079 &amp; $5881\times557159$ &amp; 563040 &amp; 4914978119 &amp; $18701\times262819$\\
         \hline
         27 &amp; 4021840859 &amp; $2909\times1382551$ &amp; 1385460 &amp; 6032761289 &amp; $8821\times683909$\\
         \hline
         28 &amp; 5489857619 &amp; $2309\times2377591$ &amp; 2379900 &amp; 8234786429 &amp; $6961\times1182989$\\
         \hline
         29 &amp; 5716553879 &amp; $3881\times1472959$ &amp; 1476840 &amp; 8574830819 &amp; $11801\times726619$\\
         \hline
         30 &amp; 6022735799 &amp; $929\times6483031$ &amp; 6483960 &amp; 9034103699 &amp; $2789\times3239191$\\
         \hline
    \end{tabular}
    \caption{First 30 elements of $a_n$.}
    \label{tab:an}
\end{table}

Our empirical results seem to indicate several constraints on the form of $a_n$, some of which can be proven by elementary methods. Take, for example, the fact that all numbers in the table are odd. We can easily show that this is always the case.

\begin{proposition}\label{prop:odd}
    Let $a_n$ be the sequence of positive integers for which $D(C(a_n))=C(D(a_n))$. Then $a_n$ is odd for all $n$.
\end{proposition}
\begin{proof}
    Suppose $a_n$ were even for some $n$. Then there exists an $m\in\mathbb{N}$ such that $a_n=2m$, and we must have $D(C(a_n))=D(m)$ while $C(D(a_n))=C(m+2D(m))$. If $m+2D(m)$ is even, it follows that $2D(m)=m+2D(m)$, so that $m=0$, a contradiction (since $a_n&gt;0$ by assumption). On the other hand, if $m+2D(m)$ is odd, it follows that $D(m)=(3m+6D(m)+1)/2$, so that $4D(m)=-3m-1$, which is again a contradiction. Thus, $a_n$ is odd for all $n$. 
\end{proof}

Our data suggests that $a_n$ must be a product of distinct primes (an almost-prime), and we notice that no single prime appears in the table. This latter fact is indeed the case in general, as the next proposition shows.

\begin{proposition}
    Let $a_n$ be the sequence of positive integers for which $D(C(a_n))=C(D(a_n))$. Then $a_n$ is a composite number for all $n$.
\end{proposition}
\begin{proof}
    Suppose $a_n=p$ were prime for some $n$ and note that $p\ne2$ by Proposition~\ref{prop:odd}. Then $D(C(a_n))=D\bigl(\frac{3p+1}{2}\bigr)$, while $C(D(a_n))=C(1)=2$. It follows that $D\bigl(\frac{3p+1}{2}\bigr)=2$. If $\Omega(\frac{3p+1}{2})=1$, then $\frac{3p+1}{2}$ is prime and so $D\bigl(\frac{3p+1}{2}\bigr)=1$, a contradiction. Then by Lemma~\ref{lemma:inequality}, we have
    \begin{equation*}
        2=D\bigg(\frac{3p+1}{2}\bigg)\ge2\sqrt{\frac{3p+1}{2}},
    \end{equation*}
    which produces a contradiction. Thus, $a_n$ is composite for all $n$.
\end{proof}

It is worth noting that a weak connection between the arithmetic derivative and the Collatz conjecture can be constructed. Indeed, a tetration of a prime belongs to the sequence $a_n$ if and only if its image under the Collatz map is a tetration of a prime. 
\begin{proposition}\label{prop:tetration}
    Let $p^p$ be a tetration of a prime. Then $a_n=p^p$ for some $n$ if and only if $C(p^p)=q^q$ for some prime $q$.
\end{proposition}
\begin{proof}
    If $C(p^p)=q^q$ for some prime $q$, then
    \begin{equation*}
        D(C(p^p))=D(q^q)=q^q=C(p^p)=C(D(p^p)),
    \end{equation*}
    and we are done. Conversely, suppose $p^p$ belongs to the sequence $a_n$. From Proposition~\ref{prop:odd}, we know that $p\ne2$. Then we have
    \begin{equation}\label{eq:tetration}
        D\bigg(\frac{3p^p+1}{2}\bigg)=D(C(p^p))=C(D(p^p))=\frac{3p^p+1}{2}.
    \end{equation}
    We claim that the only positive fixed points of the arithmetic derivative are tetrations of primes. To see this, assume that $m=p_1^{\nu_1}\cdots p_k^{\nu_k}$ and let $D(m)=m$. It follows from \eqref{eq:general_deriv} that
    \begin{equation*}
        \sum_{j=1}^k\frac{\nu_j}{p_j}=1,
    \end{equation*}
    and since all terms are non-negative, this tells us that $\nu_j\le p_j$ for all $j$. Multiplying both sides by $p_1\cdots p_k$ then produces
    \begin{equation*}
        \sum_{j=1}^k\nu_j\prod_{i\ne j}p_i=p_1\cdots p_k,
    \end{equation*}
    from which it follows that $p_j$ divides $\nu_j$. But since $\nu_j\le p_j$, it follows that $\nu_j=0$ or $\nu_j=p_j$. Of course, the latter cannot be true for more than one choice of $j$. Thus, the only positive fixed points are of the form $q^q$ for some prime $q$, and it now follows from \eqref{eq:tetration} that $C(p^p)=q^q$ for some prime $q$.
\end{proof}

Note that it is not known whether a prime tetration of the form in Proposition~\ref{prop:tetration} exists, only that if it does exist, the connection outlined therein holds. Let us now dive deeper into our empirical results. By examining Table~\ref{tab:an}, we are led to believe that most values of $a_n$ are squarefree semiprimes. It is therefore natural to wonder how the primes making up $a_n$ are related to each other. In the next proposition, we give a bound for the difference between these two primes.

\begin{proposition}\label{prop:gapbound}
    Let $a_n$ be the sequence of positive integers for which $D(C(a_n))=C(D(a_n))$. If $a_n=pq$ for some primes $q&lt;p$, then
    \begin{equation*}
        p-q\ge2\sqrt{30p^2+2}-10p,
    \end{equation*}
    that is, the difference between the two primes grows at least linearly with the larger prime.
\end{proposition}
\begin{proof}
    The proof is a simple calculation using Lemma~\ref{lemma:inequality}. Note that neither $p$ nor $q$ is equal to 2, as $n$ would then be even, contradicting Proposition~\ref{prop:odd}. We can therefore assume that $m:=p-q$ is even. Observe that
    \begin{align*}
        D(C(p(p-m)))=D\bigg(\frac{3p(p-m)+1}{2}\bigg)
    \end{align*}
    and that
    \begin{align*}
        C(D(p(p-m)))=C(2p-m)=p-\frac{m}{2},
    \end{align*}
    where we have used the fact that $m$ is even in the last equality. If $\frac{3p(p-m)+1}{2}$ is prime, it follows that $m=2(p-1)$. But this implies that $a_n=p(2-p)$, a contradiction. It therefore follows from Lemma~\ref{lemma:inequality} that
    \begin{equation*}
        p-\frac{m}{2}\ge 2\sqrt{\frac{3p(p-m)+1}{2}}.
    \end{equation*}
    Squaring and rearranging the terms produces the inequality $20pm+m^2-8\ge20p^2$, and now factoring the left side gives
    \begin{equation*}
        (m+10p)^2\ge 120p^2+8,
    \end{equation*}
    from which it follows that
    \begin{equation*}
        m\ge2\sqrt{30p^2+2}-10p.
    \end{equation*}
\end{proof}

Proposition~\ref{prop:gapbound} eliminates several commonly paired primes. As every prime is obviously a distance zero from itself, squares of primes (\seqnum{A001248}) cannot belong to $(a_n)_{n=1}^\infty$. The twin primes (\seqnum{A001359}) differ from each other by two, and so their products are also eliminated, as are the products of the so-called cousin primes (\seqnum{A046132}) which differ by four. Similarly, we are able to rule out products of Sophie Germain primes (\seqnum{A005384}) with their corresponding safe prime; that is, products of the form $p(2p+1)$ with both $p$ and $2p+1$ prime.

\begin{corollary}
    There does not exist an $n$ such that $a_n$ is a square of a prime, a product of twin primes, a product of cousin primes, or a product of a Sophie Germain prime and its corresponding safe prime.
\end{corollary}

From Table~\ref{tab:an}, we see that when $D(a_n)$ is even, the known $a_n$ are a product of two distinct primes, as are the corresponding $C(a_n)$. Moreover, when this is the case, we have $\frac{p+q}{2}=C(D(a_n))=D(C(a_n))=r+s$. The entries in the table are exhaustive in the sense that there are no additional elements of $(a_n)_{n=1}^\infty$ less than $a_{30}=6022735799$. Using the observations just outlined, we can extend this list in a non-exhaustive manner with the following proposition.

\begin{proposition}
    Let $p,q$ be primes such that both
    \begin{equation*}
        r_\pm=\frac{p+q\pm\sqrt{p^2-22pq+q^2-8}}{4}
    \end{equation*}
    are also prime. Then $C(pq)=r_+r_-$ and $pq$ is an element of $(a_n)_{n=1}^\infty$.
\end{proposition}
\begin{proof}
    Observe that 
    \begin{equation*}
        \frac{3pq+1}{2}=\bigg(\frac{p+q+\sqrt{p^2-22pq+q^2-8}}{4}\bigg)\bigg(\frac{p+q-\sqrt{p^2-22pq+q^2-8}}{4}\bigg)
    \end{equation*}
    so that $C(pq)=r_+r_-$. Moreover, $r_++r_-=\frac{p+q}{2}$, and it follows that $D(C(pq))=D(r_+r_-)=r_++r_-=\frac{p+q}{2}=C(D(pq))$.
\end{proof}

We call such semiprimes $pq$ and $r_+r_-$ \textit{compatible}. Equivalently, $pq$ and $rs$ are compatible if $r$ and $s$ are the roots of the quadratic equation $2x^2-(p+q)x+3pq+1=0$. In a computer search for compatible semiprimes, we uncovered the elements of $(a_n)_{n=1}^\infty$ listed in Table~\ref{tab:conjugate_pairs}, which is so long that we have banished it to the appendix. Note that the position in the sequence $n$ is not labeled since this search is non-exhaustive. To facilitate a better understanding of the solutions to the arithmetic differential equation $D\bigl(\frac{3n+1}{2}\bigr)=\frac{D(n)}{2}$ for $n$ odd, let us now derive several modular restrictions on compatible semiprimes.

\begin{proposition}\label{prop:mod3}
    Suppose $pq$ and $rs$ are compatible semiprimes. Then $pq\equiv2\equiv \modd{rs} {3}$. It follows that either $p\equiv1$ and $q\equiv2$ or $p\equiv2$ and $q\equiv\modd{1} {3}$, and likewise for $r,s$.
\end{proposition}
\begin{proof}
    Observe that $rs=\frac{3pq+1}{2}\equiv \modd{2} {3}$, independent of $p,q$. Without loss of generality, take $r\equiv1$ and $s\equiv2$. Then $p+q=2(r+s)\equiv0$. Since $p,q$ are prime, this congruence can only be satisfied by $p\equiv1$ and $q\equiv2$ or the reverse.
\end{proof}
We make use of the following lemma.
\begin{lemma}\label{lemma:rs_odd}
    Suppose $pq$ and $rs$ are compatible semiprimes. Then $p,q,r,s$ are all odd.
\end{lemma}
\begin{proof}
    That $p$ and $q$ are odd follows from Proposition~\ref{prop:odd}. Without loss of generality, suppose $s=2$. Then $\frac{p+q}{2}=2+r=2+\frac{3pq+1}{4}$, from which it follows that $2p+2q=3pq+9&gt;3pq$. Without loss of generality, we may assume $p\ge q&gt;2$. Then we have $2p+2q&gt;3pq&gt;6p&gt;2p+2p\ge2p+2q$, a contradiction. 
\end{proof}

\begin{proposition}\label{prop:mod4}
    Suppose $pq$ is compatible to another semiprime. Then $pq\equiv\modd{3} {4}$. It follows that either $p\equiv1$ and $q\equiv3$ or $p\equiv3$ and $q\equiv\modd{1} {4}$.
\end{proposition}
\begin{proof}
    Since $p,q$ are prime, they can only be congruent to 1 or 3 mod 4. Otherwise, they are divisible by 2, which is ruled out by Proposition~\ref{prop:odd}. If $p\equiv1\equiv q$, then $rs\equiv0$ or $2$. But then $rs$ is divisible by 2, which contradicts Lemma~\ref{lemma:rs_odd}. The case of $p\equiv3\equiv q$ produces a contradiction similarly.
\end{proof}

Using Proposition~\ref{prop:mod4} and quadratic reciprocity, we see that either $p$ and $q$ are both quadratic residues modulo each other or neither of them is a quadratic residue modulo the other. Before addressing the residues modulo 10 in an attempt to explain the consistent pattern in our data, recall the following method for solving Diophantine equations of the form $ax+by+cxy=d$. Observe that this equation is equivalent to $(cx+b)(cy+a)=ab+cd$. Thus, if $a,b,c,d$ are known, we can look for solutions by factoring $ab+cd$ in all possible ways and setting the factors equal to $cx+b$ and $cy+a$. Using this method, we obtain the following lemma.

\begin{lemma}\label{lemma:conjugate_bound}
    Let $pq$ and $rs$ be compatible semiprimes with $p&gt;q$. Then $r,s\ge241$ and we have $3q&lt;r,s&lt;3p$.
\end{lemma}
\begin{proof}
    From the definition of compatible semiprimes, we have
    \begin{equation*}
        \frac{p+q}{2}=r+s=\frac{3pq+1}{2s}+s,
    \end{equation*}
    so that we come to a Diophantine equation of the form
    \begin{equation*}
        sp+sq-3pq=2s^2+1.
    \end{equation*}
    Now applying the method mentioned above with $a=b=s$, $c=-3$, and $d=2s^2+1$, we have
    \begin{equation*}
        (s-3p)(s-3q)=-5s^2-3.
    \end{equation*}
    By factoring the right hand side for each choice of $s$, it can now be verified computationally that the first prime $s$ satisfying this equality so that $p,q,r$ are also prime is $s=241$. Moreover, since $-5s^2-3&lt;0$, we have that $3q&lt;s&lt;3p$. By symmetry, this argument holds also for $r$.
\end{proof}

Let us now determine the possible congruences modulo 5, so that the congruences modulo 10 can be determined, thereby explaining the pattern we observe in our data.
\begin{proposition}\label{prop:mod5}
    Let $pq$ and $rs$ be compatible semiprimes. Then $p\equiv1$ and $q\equiv4$ or $q\equiv1$ and $p\equiv\modd{4} {5}$. Similarly, $r\equiv1$ and $s\equiv4$ or $s\equiv1$ and $r\equiv\modd{4} {5}$.
\end{proposition}
\begin{proof}
    Let us begin by noting that if a product of two primes is congruent to 5, then one of the primes is necessarily equivalent to 5. Thus, eliminating all such cases can be done by showing that neither prime is 5. Suppose for the sake of a contradiction that $q=5$. Then we have
    \begin{equation*}
        2(r+s)=p+5=\frac{2rs-1}{15}+5,
    \end{equation*}
    and so it follows that $15(r+s)=rs+37&gt;rs$. Thus, $\frac{1}{r}+\frac{1}{s}&gt;\frac{1}{15}$. But by Lemma~\ref{lemma:conjugate_bound}, we have $\frac{1}{r}+\frac{1}{s}\le\frac{2}{241}$, a contradiction. Thus, by symmetry, $p,q\ne5$. We now proceed by eliminating the remaining cases one by one.

    If $pq\equiv1$, then $rs\equiv2$ and the possible congruences are therefore $(p,q)\equiv(1,1),(2,3),(4,4)$ and $(r,s)\equiv(1,2),(3,4)$, none of which satisfy $p+q=2(r+s)$. If $pq\equiv2$, then $rs\equiv1$ and the possible congruences are therefore $(p,q)\equiv(1,2),(3,4)$ and $(r,s)\equiv(1,1),(2,3),(4,4)$, none of which satisfy $p+q=2(r+s)$. If $pq\equiv3$, then $rs\equiv0$, which violates Lemma~\ref{lemma:conjugate_bound}. Finally, if $pq\equiv4$, then $rs\equiv4$ and the possible congruences are therefore $(p,q)\equiv(1,4),(2,2),(3,3)$ and $(r,s)\equiv(1,4),(2,2),(3,3)$. The only ones satisfying $p+q=2(r+s)$ are $(p,q)\equiv(1,4)$ and $(r,s)\equiv(1,4)$.
\end{proof}

We are now ready to give some justification for why all of the entries in Tables~\ref{tab:an} and \ref{tab:conjugate_pairs} are congruent to 9 modulo 10. The next proposition shows that this is always the case for compatible semiprimes, and so if all elements of $(a_n)_{n=1}^\infty$ are given by semiprimes compatible with another semiprime, then every entry is congruent to 9 modulo 10. Of course, not all elements of $a_n$ are compatible semiprimes, given the exceptional case $n=12$, but even if we restrict our attention to the case that $D(n)$ is even, it is not known that all remaining elements of $(a_n)_{n=1}^\infty$ are semiprimes compatible with another semiprime.
\begin{proposition}
    Let $pq$ and $rs$ be compatible semiprimes. Then $p\equiv1$ and $q\equiv9$ or $p\equiv 9$ and $q\equiv\modd{1} {10}$. Similarly, $r\equiv1$ and $s\equiv9$ or $r\equiv 9$ and $s\equiv\modd{1} {10}$.
\end{proposition}
\begin{proof}
    By Lemma~\ref{lemma:rs_odd}, $pq$ is odd, so there exists an $n$ such that $pq=1+2n$. On the other hand, $pq\equiv\modd{4} {5}$ by Proposition~\ref{prop:mod5}, so there exists an $m$ such that $pq=4+5m$. It follows that $2n=3+5m$ and $m$ is therefore odd. Thus, there is an integer $k$ such that $m=2k+1$, and we have $pq=4+5m=9+10k\equiv\modd{9} {10}$. Similarly, we have $rs\equiv\modd{9} {10}$. The possible congruences are therefore $(p,q),(r,s)\equiv(1,9),(3,3),(7,7)$. The only choice satisfying $p+q=2(r+s)$ is $(p,q)\equiv(1,9)\equiv(r,s)$.
\end{proof}

\section{Sum of reciprocals of compatible semiprimes}\label{sec:sum}
It is not clear if there are infinitely many compatible semiprimes and therefore infinitely many solutions to our commutation problem. One method of proving that a given set has infinite cardinality is to show that a sum over that set diverges. For example, the sum of the reciprocals of the prime numbers diverges, showing that there are infinitely many prime numbers. Brun showed that the sum of the reciprocals of the twin primes actually converges, so that a similar argument cannot be made and the twin prime conjecture still remains unsolved. Let us prove a similar result related to our compatible semiprimes.

\begin{theorem}
    Let $S$ denote the set of semiprimes compatible with another semiprime. Then the sum $\sum_{s\in S}\frac{1}{s}$ converges.
\end{theorem}
\begin{proof}
    We bound the number of semiprimes which are mapped to a semiprime by the Collatz map and use Abel's summation formula to show that the sum of the reciprocals of these numbers converges. It then follows that the sum over $S$ converges, as $S$ forms a subset of this set. Let $x$ be large and fix a $y\le x$. For every semiprime $pq\le y$, define the set
    \begin{equation*}
        S_{pq}:=\bigg\{n\le x : pq\mid n \text{ or } pq\mid\frac{3n+1}{2}\bigg\}.
    \end{equation*}
    Observe that if $n&gt;y$ belongs to $S_{pq}$ for some $pq\le y$, then at least one of $n$ and $\frac{3n+1}{2}$ is not a semiprime. Indeed, since $pq\le y$ and $n&gt;y$ by assumption, there must be a third prime factor in either case. It follows that the number of semiprimes $rs\le x$ such that $\frac{3rs+1}{2}$ is also semiprime is bounded above by the quantity
    \begin{equation*}
        y+x-\lvert\cup_{pq\le y}S_{pq}\rvert.
    \end{equation*}
    Here the subtraction of the union of the sets $S_{pq}$ removes those points $n&gt;y$ that belong to $S_{pq}$ since we have shown that they do not produce semiprimes under the Collatz mapping. However, it also removes points $n\le y$ which may satisfy this hypothesis, and so we make up for this with the addition of a factor of $y$. Now let $\ell$ be an even number. By the inclusion-exclusion principle, we have
    \begin{equation}\label{eq:inclusion-exclusion}
        \lvert\cup_{pq\le y}S_{pq}\rvert\ge\sum_{pq\le y}\lvert S_{pq}\rvert-\sum_{p_1q_1&lt;p_2q_2\le y}\lvert S_{p_1q_1}\cap S_{p_2q_2}\rvert+\cdots-\sum_{p_1q_1&lt;\cdots&lt;p_\ell q_\ell\le y}\lvert S_{p_1q_1}\cap\cdots\cap S_{p_\ell q_\ell}\rvert.
    \end{equation}
    We must therefore endeavor to estimate $\lvert S_{p_1q_1}\cap\cdots\cap S_{p_jq_j}\rvert$ for arbitrary $j=1,\ldots,\ell$. Observe that $n\in S_{p_1q_1}\cap\cdots\cap S_{p_jq_j}$ if and only if $n\equiv0$ or $\frac{3n+1}{2}\equiv\modd{0} {p_iq_i}$ for every $i=1,\ldots,j$. By Proposition~\ref{prop:odd}, we may assume that $p_i,q_i\ne2$ so that the latter case is equivalent to $3n+1\equiv\modd{0} {p_iq_i}$, and rearranging produces $3n\equiv\modd{-1} {p_iq_i}$. By Proposition~\ref{prop:mod3}, we may assume $p_i,q_i\ne3$ so that $\gcd(3,p_iq_i)=1$. It follows that this congruence has a unique solution which can be obtained from Euler's theorem. Thus, for each $i$ there are exactly two possible residues modulo $p_iq_i$. Now, if $n$ satisfies one of the two possible congruences for every $i$, then so does $n+mp_1q_1\cdots p_jq_j$ for every $m\in\mathbb{Z}$. There are at most $\lfloor\frac{x}{p_1q_1\cdots p_jq_j}\rfloor+1$ and at least $\lfloor\frac{x}{p_1q_1\cdots p_jq_j}\rfloor$ values of $m$ such that $n+mp_1q_1\cdots p_jq_j\le x$ ($\lfloor x\rfloor$ denotes the integer part of $x$). Moreover, since the two congruences are chosen independently for each $i$, there are $2^j$ possible choices. Putting these observations together, it follows that there is a $\lambda_j\in[-1,1]$ such that
    \begin{equation*}
        \lvert S_{p_1q_1}\cap\cdots\cap S_{p_jq_j}\rvert=\frac{2^jx}{p_1q_1\cdots p_jq_j}+2^j\lambda_j.
    \end{equation*}
    Thus, we have
    \begin{equation*}
        x-\lvert\cup_{pq\le y}S_{pq}\rvert\le x-\sum_{pq\le y}\bigg(\frac{2x}{pq}+2\lambda_1\bigg)+\cdots+\sum_{p_1q_1&lt;\cdots&lt;p_\ell q_\ell\le y}\bigg(\frac{2^\ell x}{p_1q_1\cdots p_\ell q_\ell}+2^\ell\lambda_\ell\bigg).
    \end{equation*}
    Letting $\sigma_k$ denote that $k$-th symmetric sum over the set $\{\frac{2}{pq}:pq\le y\}$, we have
    \begin{equation*}
        x-\lvert\cup_{pq\le y}S_{pq}\rvert\le x(1-\sigma_1+\sigma_2-\cdots+\sigma_\ell)+y^\ell,
    \end{equation*}
    where we have used the fact that there are at most $(y/2)^\ell$ terms in \eqref{eq:inclusion-exclusion}, as well as the estimate $\lvert2^j\lambda_j\rvert\le2^\ell$. We need the following estimate on the alternating sum of the symmetric sums:
    \begin{equation*}
        1-\sigma_1+\sigma_2-\cdots-\sigma_k\le\prod_{j=1}^n(1-a_j)\le1-\sigma_1+\sigma_2-\cdots+\sigma_\ell
    \end{equation*}
    for all $k$ odd and $\ell$ even (the $k$-th symmetric sum is over the set $\{a_1,\ldots,a_n\}$ here). To see this, observe that the inequalities are trivially true for $n=1$ and proceed by induction on $n$. Since $\ell$ is even, we have in our case that
    \begin{equation*}
        x-\lvert\cup_{pq\le y}S_{pq}\rvert\le x\bigg(\prod_{pq\le y}\bigg(1-\frac{2}{pq}\bigg)+\sigma_\ell\bigg)+y^\ell.
    \end{equation*}
    Now, the product is easily bounded above. We have
    \begin{align*}
        \prod_{pq\le y}\bigg(1-\frac{2}{pq}\bigg)&amp;=\prod_{pq\le y}\bigg(\bigg(1-\frac{1}{pq}\bigg)^2-\frac{1}{(pq)^2}\bigg)\\
        &amp;\le\bigg(\prod_{pq\le y}\bigg(1-\frac{1}{pq}\bigg)\bigg)^2\\
        &amp;=\bigg(\exp\bigg(\log\bigg(\prod_{pq\le y}\bigg(1-\frac{1}{pq}\bigg)\bigg)\bigg)\bigg)^2,
    \end{align*}
    and now we proceed by estimating the logarithm. Using the complete additivity property, we have
    \begin{equation*}
        \log\bigg(\prod_{pq\le y}\bigg(1-\frac{1}{pq}\bigg)\bigg)=\sum_{pq\le y}\log\bigg(1-\frac{1}{pq}\bigg)\le-\sum_{pq\le y}\frac{1}{pq}.
    \end{equation*}
    Let $\pi_2(x)$ denote the number of semiprimes no greater than $x$. Using Abel's summation formula, we can write the partial sum of the reciprocal semiprimes as
    \begin{equation*}
        -\sum_{pq\le y}\frac{1}{pq}=-\frac{\pi_2(x)}{x}\bigg\vert_2^y-\int_2^y\frac{\pi_2(x)}{x^2}dx.
    \end{equation*}
    Ishmukhametov and Sharifullina showed that $\pi_2(x) = \sum_{k=1}^{\sqrt{x}}\lfloor\pi\bigl(\frac{x}{p_k}\bigr)-k+1\rfloor$ \cite{ishmukhametov2014}, but all we really need is the result of Landau that $\pi_2(y)\sim\frac{x\log(\log(x))}{\log(x)}$ \cite{landau1909}, from which it follows that
    \begin{equation}\label{eq:semiprime-sum}
        \begin{aligned}
        -\sum_{pq\le y}\frac{1}{pq}&amp;\sim-\frac{\log(\log(y))}{\log(y)}+\frac{\log(\log(2))}{\log(2)}-\int_2^y\frac{\log(\log(x))}{x\log(x)}dx\\
        &amp;=-\frac{\log(\log(y))}{\log(y)}+\frac{\log(\log(2))}{\log(2)}-\frac{1}{2}(\log(\log(y)))^2+\frac{1}{2}(\log(\log(2)))^2\\
        &amp;=-\frac12(\log(\log(y)))^2+O(1).
        \end{aligned}
    \end{equation}
    Thus, we have
    \begin{equation*}
        \exp\bigg(\log\bigg(\prod_{pq\le y}\bigg(1-\frac{1}{pq}\bigg)\bigg)\bigg)\le A\exp\bigl(-\frac12(\log(\log(y)))^2\bigr)
    \end{equation*}
    for some constant $A$ and it follows that
    \begin{equation*}
        \prod_{pq\le y}\bigg(1-\frac{2}{pq}\bigg)\le Ce^{-(\log(\log(y)))^2}\le \frac{C}{\log^m(y)}
    \end{equation*}
    for some constant $C$ and for every integer $m&gt;0$. All that remains in our estimate is to bound the symmetric sum $\sigma_\ell$. Observe that
    \begin{equation*}
        \bigg(\sum_{pq\le y}\frac{2}{pq}\bigg)^\ell=\sum_{p_1q_1,p_2q_2,\ldots,p_\ell q_\ell\le y}\frac{2^\ell}{p_1q_1\cdots p_\ell q_\ell}=\ell!\sum_{p_1q_1&lt;p_2q_2&lt;\cdots&lt;p_\ell q_\ell\le y}\frac{2^\ell}{p_1q_1\cdots p_\ell q_\ell}+\sum_{pq\le y}\frac{2^\ell}{(pq)^\ell},
    \end{equation*}
    where the factor of $\ell!$ is the number of ways to order the $\ell$ semiprimes. It follows that
    \begin{equation*}
        \sigma_\ell \le \frac{1}{\ell!}\bigg(\sum_{pq\le y}\frac{2}{pq}\bigg)^\ell,
    \end{equation*}
    and from \eqref{eq:semiprime-sum}, we have
    \begin{equation*}
        \sigma_\ell\le \frac{1}{\ell!}\bigl(D(\log(\log(y))\bigr)^2+B)^\ell
    \end{equation*}
    for some constants $D$ and $B$. Thus, using $\ell^\ell\ge\ell!$ and $y\le x$, it follows that
    \begin{align*}
        \sigma_\ell&amp;\le\bigl(\frac{1}{\ell}\bigl(D(\log(\log(y)))^2+B\bigr)\bigr)^\ell\\
        &amp;\le\bigl(\frac{1}{\ell}\bigl(D(\log(\log(x)))^2+B\bigr)\bigr)^\ell
    \end{align*}
    Let $\pi_s$ denote the number of semiprimes which are mapped to a semiprime under the Collatz map. Putting these bounds together, we see that
    \begin{equation*}
        \pi_s(x)\le y+x-\lvert\cup_{pq\le y}S_{pq}\rvert&lt;y+x\bigg(\frac{C}{\log^m(y)}+\bigg(\frac{1}{\ell}\bigl(D(\log(\log(x)))^2+B\bigr)\bigg)^\ell\bigg)+y^\ell.
    \end{equation*}
    But $y$, $m$, and $\ell$ are free to be chosen except for the restriction $y\le x$. We tame the behavior of the term containing $B$ and $D$ by choosing $\ell\approx2\bigl(D(\log(\log(x)))^2+B\bigr)$. Now choosing $y=x^{\frac{1}{D(\log(\log(x)))^2+B}}$, we see that
    \begin{equation}\label{eq:bound}
        \pi_s(x)\le\frac{Mx\bigl(D(\log(\log(x))\bigr)^2+B)^m}{(\log(x))^m}.
    \end{equation}
    for a constant $M$. Let us now show that the sum $\sum_{s\in C_s}\frac{1}{s}$ over the set $C_s$ of all semiprimes which are mapped to a semiprime under the Collatz map converges. By the Abel summation formula, we have
    \begin{equation*}
        \sum_{s\in C_s}\frac{1}{s}=\frac{\pi_s(x)}{x}\bigg\vert_{2}^\infty+\int_2^\infty\frac{\pi_s(x)}{x^2}dx.
    \end{equation*}
    Using \eqref{eq:bound}, the boundary term vanishes at infinity and the integral converges for large enough $m$. Thus, the sum converges. Since the compatible semiprimes form a subset, we finally have that $\sum_{s\in S}\frac{1}{s}$ converges, and this completes the proof.
\end{proof}



\section{Conclusion}\label{sec:conclusion}

We have studied a curious class of pairs of semiprime numbers which we call compatible. These numbers form solutions to the arithmetic differential equation
\begin{equation}\label{eq:arithmetic_diff}
    D\bigg(\frac{3n+1}{2}\bigg)=\frac{D(n)}{2},
\end{equation}
which is a special case of the problem of searching for numbers for which the arithmetic derivative and Collatz map commute. In this latter problem, we have found an exhaustive list of the first 30 numbers providing a solution, and all but one of them is a solution to the special case \eqref{eq:arithmetic_diff}. The compatible semiprimes form additional solutions to \eqref{eq:arithmetic_diff} which therefore solve the commutation problem as well, although the list we provide in Table~\ref{tab:conjugate_pairs} is likely not exhaustive.

To close out this work, we provide the following conjectures worth studying in future work. Note that Conjecture~\ref{conj:compatible} implies Conjecture~\ref{conj:diff} which implies Conjecture~\ref{conj:commute}. The latter logical implication holds because there are no even $n$ for which \eqref{eq:arithmetic_diff} holds. Indeed, such an $n$ can be written $n=2m$ so that \eqref{eq:arithmetic_diff} reduces to $2D(6m+1)=4D(m)+8m+1$. Since the left hand side is even while the right hand side is odd, such an $n$ cannot exist. Note also that Conjecture~\ref{conj:semiprime} implies Conjecture~\ref{conj:squarefree}.

\begin{conjecture}\label{conj:compatible}
    There are infinitely many compatible semiprimes.
\end{conjecture}

\begin{conjecture}\label{conj:diff}
    There are infinitely many solutions to the arithmetic differential equation \eqref{eq:arithmetic_diff}.
\end{conjecture}

\begin{conjecture}\label{conj:commute}
    There are infinitely many solutions to the commutation problem $D(C(n))=C(D(n))$.
\end{conjecture}

\begin{conjecture}\label{conj:semiprime}
    With the exception of 606938385, the only solutions to the commutation problem $D(C(n))=C(D(n))$ are given by compatible semiprimes.
\end{conjecture}

\begin{conjecture}\label{conj:squarefree}
    Every solution to the commutation problem $D(C(n))=C(D(n))$ is squarefree.
\end{conjecture}

It should also be noted that similar problems can be formulated for the generalized Collatz functions
\begin{equation*}
    C_{a,b}(n):=\begin{cases}
        \frac{an+b}{2}, &amp; \text{ if $n$ odd;}\\
        \frac{n}{2}, &amp; \text{ if $n$ even,}
    \end{cases}
\end{equation*}
with $a\equiv \modd{b} {2}$. An exhaustive brute force search for numbers $n\le10^7$ solving the analogous commutation problem produced no solutions for $a=5,b=1$, while the slight variant $a=5,b=3$ has the eleven solutions $n=$ 12419, 20171, 37727, 134579, 199259, 301799, 574319, 866891, 1580291, 1625411, 8014031, each of which is a squarefree semiprime. On the other hand, the case $a=7, b=1$ has only the solution $n=429$ in this range, and 429 is a product of three distinct prime factors. The slight variant $a=7,b=3$ produced no solutions and the variant $a=7,b=5$ has the seven solutions $n=29831, 38051, 76331, 568031, 888971, 1855871, 3095711$, each of which is a squarefree semiprime consisting of the product of a prime congruent to 3 and a prime congruent to 7 modulo 10. In each of these examples, the solutions are squarefree almost-primes, and for this reason we make the following conjecture.

\begin{conjecture}
    The solutions to the commutation problem $D(C_{a,b}(n))=C_{a,b}(D(n))$ are squarefree for every $a\equiv \modd{b} {2}$.
\end{conjecture}




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\end{thebibliography}

\pagebreak

\section*{Appendix}
\begin{table}[H]
\begin{center}
\begin{tabular}{ccccc}
         $pq$ &amp; Prime Factorization &amp; $p+q$ &amp; $rs$ &amp; Prime Factorization\\
         \hline
         6371331359 &amp; $16561\times384719$ &amp; 401280 &amp; 9556997039 &amp; $77801\times122839$\\
         \hline
         8975589239 &amp; $9241\times971279$ &amp; 980520 &amp; 13463383859 &amp; $29201\times461059$\\
         \hline
         10657078799 &amp; $20071\times530969$ &amp; 551040 &amp; 15985618199 &amp; $83059\times192461$\\
         \hline
         12860887439 &amp; $6089\times2112151$ &amp; 2118240 &amp; 19291331159 &amp; $18539\times1040581$ \\
         \hline
         13170630899 &amp; $8681\times1517179$ &amp; 1525860 &amp; 19755946349 &amp; $26839\times736091$\\ 
         \hline
         15512319959 &amp; $4111\times3773369$ &amp; 3777480 &amp; 23268479939 &amp; $12401\times1876339$\\
         \hline
         20713686659 &amp; $27179\times762121$ &amp; 789300 &amp; 31070529989 &amp; $108631\times286019$\\
         \hline
         23664471179 &amp; $11369\times2081491$ &amp; 2092860 &amp; 35496706769 &amp; $35099\times1011331$\\
         \hline
         24180093239 &amp; $26959\times896921$ &amp; 923880 &amp; 36270139859 &amp; $100291\times361649$\\
         \hline
         27169948919 &amp; $29209\times930191$ &amp; 959400 &amp; 40754923379 &amp; $110339\times369361$\\
         \hline
         30617640719 &amp; $9721\times3149639$ &amp; 3159360 &amp; 45926461079 &amp; $29629\times1550051$\\
         \hline
         41203174319 &amp; $39929\times1031911$ &amp; 1071840 &amp; 61804761479 &amp; $167971\times367949$\\
         \hline
         44444955899 &amp; $37831\times1174829$ &amp; 1212660 &amp; 66667433849 &amp; $144289\times462041$\\
         \hline
         45265630259 &amp; $34949\times1295191$ &amp; 1330140 &amp; 67898445389 &amp; $125941\times539129$\\
         \hline
         59663684459 &amp; $50839\times1173581$ &amp; 1224420 &amp; 89495526689 &amp; $241261\times370949$\\
         \hline
         60235146119 &amp; $51329\times1173511$ &amp; 1224840 &amp; 90352719179 &amp; $364621\times247799$\\
         \hline
         71047305239 &amp; $38231\times1858369$ &amp; 1896600 &amp; 106570957859 &amp; $130279\times818021$\\
         \hline
         72269351939 &amp; $56681\times1275019$ &amp; 1331700 &amp; 108404027909 &amp; $382271\times283579$\\
         \hline
         74550954899 &amp; $43291\times1722089$ &amp; 1765380 &amp; 111826432349 &amp; $153319\times729371$\\
         \hline
         74797926239 &amp; $56239\times1330001$ &amp; 1386240 &amp; 112196889359 &amp; $435481\times257639$\\
         \hline
         90923314259 &amp; $59141\times1537399$ &amp; 1596540 &amp; 136384971389 &amp; $247729\times550541$\\
         \hline
         91551073739 &amp; $28909\times3166871$ &amp; 3195780 &amp; 137326610609 &amp; $91141\times1506749$\\
         \hline
         92153208659 &amp; $34949\times2636791$ &amp; 2671740 &amp; 138229812989 &amp; $113041\times1222829$\\
         \hline
         100918727099 &amp; $44119\times2287421$ &amp; 2331540 &amp; 151378090649 &amp; $148861\times1016909$\\
         \hline
         105138878699 &amp; $63079\times1666781$ &amp; 1729860 &amp; 157708318049 &amp; $603689\times261241$\\
         \hline
         105567827399 &amp; $22441\times4704239$ &amp; 4726680 &amp; 158351741099 &amp; $69019\times2294321$\\
         \hline
         108216566279 &amp; $49201\times2199479$ &amp; 2248680 &amp; 162324849419 &amp; $170111\times954229$\\
         \hline
         108868846199 &amp; $67511\times1612609$ &amp; 1680120 &amp; 163303269299 &amp; $534581\times305479$\\
         \hline
         129264961559 &amp; $60271\times2144729$ &amp; 2205000 &amp; 193897442339 &amp; $219619\times882881$\\
         \hline
         131559702959 &amp; $47881\times2747639$ &amp; 2795520 &amp; 197339554439 &amp; $159349\times1238411$\\
         \hline
         138217769399 &amp; $48991\times2821289$ &amp; 2870280 &amp; 207326654099 &amp; $162971\times1272169$\\
         \hline
         138868264319 &amp; $78079\times1778561$ &amp; 1856640 &amp; 208302396479 &amp; $379649\times548671$\\
         \hline
         150260888399 &amp; $75209\times1997911$ &amp; 2073120 &amp; 225391332599 &amp; $310379\times726181$\\
         \hline
         152376066599 &amp; $38561\times3951559$ &amp; 3990120 &amp; 228564099899 &amp; $122029\times1873031$\\
         \hline
         178661152559 &amp; $39929\times4474471$ &amp; 4514400 &amp; 267991728839 &amp; $125731\times2131469$\\
         \hline
         182554595819 &amp; $72739\times2509721$ &amp; 2582460 &amp; 273831893729 &amp; $1023751\times267479$\\
\end{tabular}
\end{center}
\end{table}

\begin{table}[H]
\begin{center}
\begin{tabular}{ccccc}
$pq$ &amp; Prime Factorization &amp; $p+q$ &amp; $rs$ &amp; Prime Factorization\\
         \hline
         185904249719 &amp; $85889\times2164471$ &amp; 2250360 &amp; 278856374579 &amp; $368551\times756629$\\
         \hline
         192709329959 &amp; $71471\times2696329$ &amp; 2767800 &amp; 289063994939 &amp; $256369\times1127531$\\
         \hline
         203788963199 &amp; $84551\times2410249$ &amp; 2494800 &amp; 305683444799 &amp; $912349\times335051$\\
         \hline
         218613696239 &amp; $51151\times4273889$ &amp; 4325040 &amp; 327920544359 &amp; $164089\times1998431$\\
         \hline
         233568190619 &amp; $67741\times3447959$ &amp; 3515700 &amp; 350352285929 &amp; $229189\times1528661$\\
         \hline
         240001727519 &amp; $102911\times2332129$ &amp; 2435040 &amp; 360002591279 &amp; $711649\times505871$\\
         \hline
         260131119179 &amp; $103699\times2508521$ &amp; 2612220 &amp; 390196678769 &amp; $462571\times843539$\\
         \hline
         262091228279 &amp; $85889\times3051511$ &amp; 3137400 &amp; 393136842419 &amp; $313109\times1255591$\\
         \hline
         262593906359 &amp; $37649\times6974791$ &amp; 7012440 &amp; 393890859539 &amp; $116191\times3390029$\\
         \hline
         288964258199 &amp; $36871\times7837169$ &amp; 7874040 &amp; 433446387299 &amp; $113359\times3823661$\\
         \hline
         296122234379 &amp; $41341\times7162919$ &amp; 7204260 &amp; 444183351569 &amp; $127849\times3474281$\\
         \hline
         296955821639 &amp; $104119\times2852081$ &amp; 2956200 &amp; 445433732459 &amp; $1056479\times421621$\\
         \hline
         306450940139 &amp; $79411\times3859049$ &amp; 3938460 &amp; 459676410209 &amp; $270619\times1698611$\\
         \hline
         322643081699 &amp; $88169\times3659371$ &amp; 3747540 &amp; 483964622549 &amp; $309359\times1564411$\\
         \hline
         341144024159 &amp; $60919\times5599961$ &amp; 5660880 &amp; 511716036239 &amp; $194101\times2636339$\\
         \hline
         343031424959 &amp; $124991\times2744449$ &amp; 2869440 &amp; 514547137439 &amp; $724991\times709729$\\
         \hline
         364522651739 &amp; $125219\times2911081$ &amp; 3036300 &amp; 546783977609 &amp; $587579\times930571$\\
         \hline
         368068728719 &amp; $116689\times3154271$ &amp; 3270960 &amp; 552103093079 &amp; $1159201\times476279$\\
         \hline
         395384049899 &amp; $115891\times3411689$ &amp; 3527580 &amp; 593076074849 &amp; $1311619\times452171$\\
         \hline
         400481245739 &amp; $85619\times4677481$ &amp; 4763100 &amp; 600721868609 &amp; $286771\times2094779$\\
         \hline
         405113920919 &amp; $135119\times2998201$ &amp; 3133320 &amp; 607670881379 &amp; $706291\times860369$\\
         \hline
         412288472879 &amp; $136351\times3023729$ &amp; 3160080 &amp; 618432709319 &amp; $714529\times865511$\\
         \hline
         436306401359 &amp; $55529\times7857271$ &amp; 7912800 &amp; 654459602039 &amp; $172981\times3783419$\\
         \hline
         473023514939 &amp; $143981\times3285319$ &amp; 3429300 &amp; 709535272409 &amp; $697729\times1016921$\\
         \hline
         481853046599 &amp; $132529\times3635831$ &amp; 3768360 &amp; 722779569899 &amp; $536191\times1347989$\\
         \hline
         484957912139 &amp; $123239\times3935101$ &amp; 4058340 &amp; 727436868209 &amp; $1564081\times465089$\\
         \hline
         523044914519 &amp; $131111\times3989329$ &amp; 4120440 &amp; 784567371779 &amp; $1555999\times504221$\\
         \hline
         526841686199 &amp; $138511\times3803609$ &amp; 3942120 &amp; 790262529299 &amp; $560081\times1410979$\\
         \hline
         560607218879 &amp; $136519\times4106441$ &amp; 4242960 &amp; 840910828319 &amp; $527581\times1593899$\\
         \hline
         561826681079 &amp; $154351\times3639929$ &amp; 3794280 &amp; 842740021619 &amp; $709729\times1187411$\\
         \hline
         579247503719 &amp; $157831\times3670049$ &amp; 3827880 &amp; 868871255579 &amp; $740359\times1173581$\\
         \hline
         651630487619 &amp; $102059\times6384841$ &amp; 6486900 &amp; 977445731429 &amp; $336211\times2907239$\\
         \hline
         656472916739 &amp; $143159\times4585621$ &amp; 4728780 &amp; 984709375109 &amp; $539641\times1824749$\\
         \hline
         736241831879 &amp; $129119\times5702041$ &amp; 5831160 &amp; 1104362747819 &amp; $2468131\times447449$\\
         \hline
         750468526739 &amp; $97381\times7706519$ &amp; 7803900 &amp; 1125702790109 &amp; $3588229\times313721$\\
         \hline
         844661031299 &amp; $196139\times4306441$ &amp; 4502580 &amp; 1266991546949 &amp; $1134871\times1116419$\\
         \hline
         852127113419 &amp; $142231\times5991149$ &amp; 6133380 &amp; 1278190670129 &amp; $497509\times2569181$\\
         \hline
         852545495159 &amp; $141601\times6020759$ &amp; 6162360 &amp; 1278818242739 &amp; $494359\times2586821$\\
         \hline
         883609516259 &amp; $171529\times5151371$ &amp; 5322900 &amp; 1325414274389 &amp; $663331\times1998119$\\
\end{tabular}
\end{center}
\end{table}

\begin{table}[H]
\begin{center}
\begin{tabular}{ccccc}
$pq$ &amp; Prime Factorization &amp; $p+q$ &amp; $rs$ &amp; Prime Factorization\\
         \hline
         1067201903519 &amp; $210359\times5073241$ &amp; 5283600 &amp; 1600802855279 &amp; $941461\times1700339$\\
         \hline
         1069842593219 &amp; $107509\times9951191$ &amp; 10058700 &amp; 1604763889829 &amp; $4686961\times342389$\\
         \hline
         1086684073439 &amp; $220009\times4939271$ &amp; 5159280 &amp; 1630026110159 &amp; $1106491\times1473149$\\
         \hline
         1125176937359 &amp; $175391\times6415249$ &amp; 6590640 &amp; 1687765406039 &amp; $634241\times2661079$\\
         \hline
         1179415384139 &amp; $226199\times5214061$ &amp; 5440260 &amp; 1769123076209 &amp; $1644061\times1076069$\\
         \hline
         1205649411119 &amp; $227569\times5297951$ &amp; 5525520 &amp; 1808474116679 &amp; $1697191\times1065569$\\
         \hline
         1248830314859 &amp; $221021\times5650279$ &amp; 5871300 &amp; 1873245472289 &amp; $937481\times1998169$\\
         \hline
         1255569121199 &amp; $135119\times9292321$ &amp; 9427440 &amp; 1883353681799 &amp; $4272959\times440761$\\
         \hline
         1255865342579 &amp; $233509\times5378231$ &amp; 5611740 &amp; 1883798013869 &amp; $1693501\times1112369$\\
         \hline
         1267060830239 &amp; $184351\times6873089$ &amp; 7057440 &amp; 1900591245359 &amp; $663281\times2865439$\\
         \hline
         1353561815339 &amp; $225821\times5993959$ &amp; 6219780 &amp; 2030342723009 &amp; $2177449\times932441$\\
         \hline
         1409845008239 &amp; $244561\times5764799$ &amp; 6009360 &amp; 2114767512359 &amp; $1125169\times1879511$\\
         \hline
         1424445760319 &amp; $248071\times5742089$ &amp; 5990160 &amp; 2136668640479 &amp; $1823051\times1172029$\\
         \hline
         1428658219199 &amp; $209249\times6827551$ &amp; 7036800 &amp; 2142987328799 &amp; $783599\times2734801$\\
         \hline
         1459820224439 &amp; $256489\times5691551$ &amp; 5948040 &amp; 2189730336659 &amp; $1633531\times1340489$\\
         \hline
         1477666902419 &amp; $161641\times9141659$ &amp; 9303300 &amp; 2216500353629 &amp; $538939\times4112711$\\
         \hline
         1539898047899 &amp; $166919\times9225421$ &amp; 9392340 &amp; 2309847071849 &amp; $558209\times4137961$\\
         \hline
         1618340838659 &amp; $179369\times9022411$ &amp; 9201780 &amp; 2427511257989 &amp; $607951\times3992939$\\
         \hline
         1815603367859 &amp; $204311\times8886469$ &amp; 9090780 &amp; 2723405051789 &amp; $710089\times3835301$\\
         \hline
         2180605168559 &amp; $275521\times7914479$ &amp; 8190000 &amp; 3270907752839 &amp; $1087631\times3007369$\\
         \hline
         2271244344419 &amp; $285161\times7964779$ &amp; 8249940 &amp; 3406866516629 &amp; $2982799\times1142171$\\
         \hline
         2283132057839 &amp; $319321\times7149959$ &amp; 7469280 &amp; 3424698086759 &amp; $1617949\times2116691$\\
         \hline
         2312261656139 &amp; $299539\times7719401$ &amp; 8018940 &amp; 3468392484209 &amp; $1262731\times2746739$\\
         \hline
         2515975240439 &amp; $253769\times9914431$ &amp; 10168200 &amp; 3773962860659 &amp; $902521\times4181579$\\
         \hline
         2525623096319 &amp; $339161\times7446679$ &amp; 7785840 &amp; 3788434644479 &amp; $1929971\times1962949$\\
         \hline
         2759705814599 &amp; $339289\times8133791$ &amp; 8473080 &amp; 4139558721899 &amp; $1528771\times2707769$\\
         \hline
         2798487028499 &amp; $354469\times7894871$ &amp; 8249340 &amp; 4197730542749 &amp; $1826761\times2297909$\\
         \hline
         2914249202819 &amp; $355339\times8201321$ &amp; 8556660 &amp; 4371373804229 &amp; $2591551\times1686779$\\
         \hline
         3015166657439 &amp; $355361\times8484799$ &amp; 8840160 &amp; 4522749986159 &amp; $1608769\times2811311$\\
         \hline
         3029547727979 &amp; $371299\times8159321$ &amp; 8530620 &amp; 4544321591969 &amp; $2195071\times2070239$\\
         \hline
         3142431670799 &amp; $376631\times8343529$ &amp; 8720160 &amp; 4713647506199 &amp; $1982741\times2377339$\\
         \hline
         3207163248719 &amp; $355609\times9018791$ &amp; 9374400 &amp; 4810744873079 &amp; $1517939\times3169261$\\
         \hline
         3342365758439 &amp; $338609\times9870871$ &amp; 10209480 &amp; 5013548637659 &amp; $3777541\times1327199$\\
         \hline
         3388989990119 &amp; $363119\times9333001$ &amp; 9696120 &amp; 5083484985179 &amp; $3314219\times1533841$\\
         \hline
         3434499230039 &amp; $376679\times9117841$ &amp; 9494520 &amp; 5151748845059 &amp; $3068159\times1679101$\\
         \hline
         3727388517539 &amp; $374729\times9946891$ &amp; 10321620 &amp; 5591082776309 &amp; $3613559\times1547251$\\
         \hline
         3894649427759 &amp; $421009\times9250751$ &amp; 9671760 &amp; 5841974141639 &amp; $2484721\times2351159$\\
         \hline
         4196759425979 &amp; $430739\times9743161$ &amp; 10173900 &amp; 6295139138969 &amp; $2126191\times2960759$\\
         
    \end{tabular}
\end{center}
         \caption{Additional solutions to the commutation problem $D(C(n))=C(D(n))$ given by compatible semiprimes.}
         \label{tab:conjugate_pairs} 
\end{table}

\bigskip
\hrule
\bigskip

\noindent 2020 {\it Mathematics Subject Classification}:
Primary 11A25; Secondary 11A07, 11N56.

\noindent \emph{Keywords:} arithmetic derivative, Collatz, semiprime, arithmetic function, arithmetic differential equation.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequences
\seqnum{A001248},
\seqnum{A001359},
\seqnum{A005384},
\seqnum{A046132}, and
\seqnum{A376275}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received November 15 2024;
revised versions received February 9 2025; February 13 2025.
Published in {\it Journal of Integer Sequences}, February 14 2025.

\bigskip
\hrule
\bigskip

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


\end{document}
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