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\begin{center}
\vskip 1cm{\LARGE\bf Arithmetic Subderivatives: \\
\vskip .1in
Discontinuity and Continuity}
\vskip 1cm
\large
Pentti Haukkanen and Jorma K. Merikoski \\
Faculty of Information Technology and Communication Sciences \\
FI-33014 Tampere University\\ 
Finland\\
\href{mailto:pentti.haukkanen@tuni.fi}{\tt pentti.haukkanen@tuni.fi} \\
\href{mailto:jorma.merikoski@tuni.fi}{\tt jorma.merikoski@tuni.fi} \\
\ \\
Timo Tossavainen\\
Department of Arts, Communication and Education\\ 
Lulea University of Technology\\
SE-97187 Lulea \\
Sweden \\
\href{mailto:timo.tossavainen@ltu.se}{\tt timo.tossavainen@ltu.se}
\end{center}

\vskip .2 in

\begin{abstract}
We first prove that any arithmetic subderivative of a rational number defines
a function that is everywhere discontinuous in a very strong sense. Second,
we show that although the restriction of this function to the set of integers is
continuous (in the relative topology), it is not Lipschitz continuous. Third,
we see that its restriction to a suitable infinite set is Lipschitz continuous.
This follows from the solutions of certain arithmetic differential equations.
\end{abstract}

\section{Introduction}

Let $0\ne x\in\Q$. There exists a unique sequence
$(\nu_p(x))_{p\in\PP}$ of integers (with only finitely many nonzero terms)
such that
\begin{equation}
\label{a}
x=(\sgn{x})\prod_{p\in\mathbb P}p^{\nu_p(x)}.
\end{equation}
Here $\mathbb P$ stands for the set of primes, and $\sgn{x}=x/|x|$.
Define $\sgn{0}=0$ and $\nu_p(0)=0$ for all $p\in\PP$; then Eq.~(\ref{a}) 
also holds for $x=0$.

Let $\emptyset\ne S\subseteq\Q$. The {\em arithmetic subderivative} \cite{MHT}
of $x\in\Q$ with respect to~$S$ is
$$
x'_S=D_S(x)=x\sum_{p\in S}\frac{\nu_p(x)}{p}.
$$
The {\em arithmetic partial derivative} \cite{Ko, HMT} of~$x$ with
respect to $p\in\mathbb P$ is $x'_p=D_p(x)=D_{\{p\}}(x)$. The {\em arithmetic
derivative} \cite{Sh,Ba,UA} of~$x$ is $x'=D(x)=D_{\PP}(x)$.
Arithmetic differential equations and arithmetic partial differential equations
have been studied \cite{UA,Ko,HMT,PS}.

We have $D_S(p)=1$ for all $p\in S$. If $x$ is near to~$p$ and suitably
chosen, then $D_S(x)$ can be far from one. So it is natural to expect that
$D_S$ is discontinuous at any $a\in\PP$.  However, it is not
equally apparent that $D_S$ is discontinuous at an arbitrary $a\in\Q$.

We say that a real-valued function~$f$, defined on an open real (or
rational) interval~$I$,
is {\em superdiscontinuous} at $a\in I$ if it attains values of an arbitrarily large absolute
value arbitrarily near to~$a$ and on its both sides. More precisely, given any
$\delta,M>0$, there are $x,y\in I$ such that
$$
a-\delta<x<a<y<a+\delta\quad{\rm and}\quad |f(x)|,|f(y)|>M.
$$
If $x$ (respectively, $y$) exists, then $f$ is {\em superdiscontinuous from the left} ({\em right}) at~$a$.
We prove in Section~\ref{main} that $D_S$ is superdiscontinuous at any~$a\in\Q$.
We need two lemmas.

\begin{lemma}
\label{dense}
Let $p,q\in\PP$, $p\ne q$. The set $\{ \pm p^mq^n\mid m,n\in\Z \}$ is dense in~$\Q$.
\end{lemma}
\begin{proof}
See the proof of \cite[Lemma~30]{HMT}.
\end{proof}

\begin{lemma}
\label{super}
Let $f$ be a real-valued function, defined on an open real {\rm (}or
rational{\rm )} interval~$I$, and let $a\in I$. If $f$ is superdiscontinuous at each
point of $I\setminus\{a\}$, then it is superdiscontinuous also at~$a$.
\end{lemma}
\begin{proof}
Let $\delta,M>0$.
It is no restriction to assume that $a\pm\delta\in I$. Since $f$ is 
superdiscontinuous (from the right) at $a-\delta$, there is $x\in I$ satisfying
$a-\delta<x<a$ and $|f(x)|>M$. Similarly, since $f$ is 
superdiscontinuous (from the left) at $a+\delta$, there is $y\in I$ satisfying
$a<y<a+\delta$ and $|f(y)|>M$.
\end{proof}

Let $f$ and $I$ be as above. The function~$f$ is {\em Lipschitz continuous}
(in~$I$) if there exists $L>0$ satisfying
\begin{equation}
\label{lipcont}
|f(x)-f(y)|\le L|x-y|
\end{equation}
for all $x,y\in I$. Since a Lipschitz continuous function is continuous in the ordinary
sense, the results in Section~\ref{main} imply that $D_S$ is Lipschitz discontinuous.
But the restriction of~$D_S$ to~$\Z$, denoted by~$D_S|_\Z$,
is continuous (in the relative topology); so, is it Lipschitz continuous? We show
in Section~\ref{lipschitz} that it is not.
This raises a further question: Is there an infinite set $A\subset\Q$ such that
$D_S|_A$ is Lipschitz continuous? Applying the previously known solutions of
the differential equations $x'=0$ and $x'_p=ax$, we give a positive answer.
This, in turn, motivates us to
consider the more general differential equation $x'_S=ax$. We solve it in Section~\ref{diffeq}.
Finally, we complete our paper with concluding remarks in Section~\ref{remarks}.

\section{Superdiscontinuity of $D_S$}
\label{main}

We state in Theorem~\ref{twoprimesout} an additional assumption, in order
to make the proof easier.

\begin{theorem}
\label{twoprimesout}
Let $\emptyset\ne S\subseteq\PP$. If there are distinct $p,q\in\PP\setminus S$,
then $D_S$ is superdiscontinuous at any $a\in\Q$. 
\end{theorem}
\begin{proof}
\leavevmode
\ \\
\noindent{\em Case 1:}
$a>0$. Let $\delta,M>0$, $s\in S$, and $k\in\mathbb{Z}_+$.
By Lemma~\ref{dense}, there are $m,n\in\mathbb{Z}$ such that
$$
\frac{a}{s^k}<p^mq^n<\frac{a}{s^k}+\frac{\delta}{s^k}.
$$
Then $x=s^kp^mq^n$ satisfies
$$
a<x<a+\delta.
$$
The subderivative
$$
D_S(x)=ks^{k-1}p^mq^n>ks^{k-1}\frac{a}{s^k}=\frac{ka}{s}>M
$$
if
$$
k>\frac{sM}{a}.
$$
Consequently, superdiscontinuity from the right follows.

To prove superdiscontinuity from the left, we use the above notation
but let $\delta<a$ (which is no restriction). Now, by Lemma~\ref{dense}, there are
$m,n\in\mathbb{Z}$ such that
$$
\frac{a}{s^k}-\frac{\delta}{s^k}<p^mq^n<\frac{a}{s^k}.
$$
Hence,
$$
a-\delta<x<a,
$$
and
$$
D_S(x)=ks^{k-1}p^mq^n>ks^{k-1}\frac{a-\delta}{s^k}=\frac{k(a-\delta)}{s}>M
$$
if
$$
k>\frac{sM}{a-\delta}.
$$
This verifies superdiscontinuity from the left.

\bigskip

\noindent{\em Case 2:} $a<0$. Apply the fact that $D_S(a)=-D_S(-a)$.

\bigskip

\noindent{\em Case 3:} $a=0$. Apply Lemma~\ref{super}.
\end{proof}

What about trying the same idea for $S=\PP\setminus\{p\}$, $p\in\PP$?
It is enough to assume that $a>0$.
Let $q\in S$ and $T=S\setminus\{q\}$; then $p,q\notin T$. Let
$\delta,M,s$, and $k$ be as above.
Again, there are $m$ and $n$ such that $x=s^kp^mq^n$ satisfies
$$
a<x<a+\delta.
$$
If
$$
k>\frac{sM}{a},
$$
then
$$
D_S(x)=D_T(x)+\frac{n}{q}s^kp^mq^n>M+ns^kp^mq^{n-1}.
$$
Consequently, if $n\ge 0$ (i.e., $\nu_q(x)\ge 0$ for some $q\in S$), then
$D_S(x)>M$, and superdiscontinuity from the right follows.
Unfortunately, a problem arises if $n<0$ (i.e., $\nu_q(x)<0$ for all $q\in S$).
Then $D_S(x)<0$, and we get nothing reasonable. 
It seems that our method cannot be revised to work in this case.
Instead of a single~$x$, we apparently should consider a suitable sequence~$(x_i)$.
We do this in the proof of the next theorem, where $S$ is arbitrary.
Theorem~\ref{twoprimesout} could now be omitted, but we find its proof
interesting on its own and therefore keep it.

\begin{theorem}
\label{twoprimesin}
Let $\emptyset\ne S\subseteq\PP$. Then $D_S$ is superdiscontinuous at any $a\in\Q$.
\end{theorem}

\begin{proof} 

We can assume that $a\ne 0$; then superdiscontinuity at $a=0$ follows from
Lemma~\ref{super}. Let $p,q\in\PP$, $p\ne q$, and denote $m=\nu_p(a)$, $n=\nu_q(a)$. Then
$$
a=p^mq^nr,
$$  
where $r\ne 0$ has $\nu_p(r)=\nu_q(r)=0$.
We show that there is a sequence~$(x_i)$ of rational
numbers with
\begin{equation}
\label{xi}
x_i\rightarrow a,\quad x_1,x_2,\ldots>a,
\end{equation}
and
\begin{equation}
\label{liminfty}
|D_S(x_i)|\to\infty.
\end{equation}
By a simple modification of this proof, it can be seen that there also
exists a sequence~$(x_i)$ satisfying
$$
x_i\rightarrow a,\quad x_1,x_2,\ldots<a
$$
and (\ref{liminfty}).

By Lemma~\ref{dense}, there are integer sequences $(m_i)$ and~$(n_i)$
such that the sequence
$$
(x_i),\, x_i=p^{m_i}q^{n_i}
$$
satisfies~(\ref{xi}). To prove~(\ref{liminfty}), we must know their limiting behaviour.

\bigskip

\noindent{\em Case 1:} Both $(m_i)$ and $(n_i)$ are bounded from above. Then they are bounded
also from below,
since otherwise $x_i\to 0$ or $(x_i)$ diverges, contradicting~(\ref{xi}).
Therefore, $(x_i)$ has only finitely many different terms, implying that its limit~$a$ is one
of them. This again contradicts~(\ref{xi}). Consequently, this case is impossible.

\bigskip

\noindent{\em Case 2:} The sequence $(m_i)$ is not bounded from above. If, instead,
$(n_i)$ is not bounded from above, we can proceed similarly.
Now, $(m_i)$ has a subsequence~$(m_{i_k})$ satisfying $m_{i_k}\to\infty$.
Since $p^{m_{i_k}}\to\infty$ but $x_{i_k}\to a\ne 0$,
necessarily $q^{n_{i_k}}\to 0$, i.e., $n_{i_k}\to-\infty$.
Including also the case when $(n_i)$ is not bounded from above, we therefore have
\begin{equation}
\label{mknklimits}
m_{i_k}\to\pm\infty\quad{\rm and}\quad n_{i_k}\to\mp\infty.
\end{equation}
In order to keep the notation simple, we let $(x_i)$ denote the subsequence~$(x_{i_k})$. Then
Eq.~(\ref{mknklimits}) reads
\begin{equation}
\label{mnlimits}
m_i\rightarrow\pm\infty\quad{\rm and}\quad n_i\rightarrow\mp\infty.
\end{equation}

\smallskip

By the convergence, the sequence $(x_i)$ is bounded also from above. Let $u\in\Q$ satisfy
$$
x_1,x_2,\ldots<u.
$$
We can assume in~Eq.~(\ref{mnlimits}) that
\begin{equation}
\label{mnlims}
m_i\rightarrow -\infty\quad{\rm and}\quad n_i\rightarrow\infty.
\end{equation}
If the signs are opposite, then a simple
modification applies. There is $i_0\in\Z_+$ such that $n_i>0$ for all $i\ge i_0$.
Denote
$$
c_i=\frac{m_i-m}{n_i},\quad{\rm where}\quad i\ge i_0.
$$
Then
\begin{eqnarray}
\label{cibnds}
a<x_i<u\iff p^mq^nr<p^{m_i}q^{n_i}<u\iff\qquad\qquad\qquad\qquad\nonumber
\\
q^{n-n_i}r< p^{m_i-m}<p^{-m}q^{-n_i}u\iff\nonumber
\\
(n-n_i)\log_pq+\log_pr<m_i-m<-m-n_i\log_pq+\log_pu\iff\nonumber
\\
(n-n_i+ \log_qr)\log_pq< m_i-m<-m+(-n_i+\log_qu)\log_pq\quad\nonumber
\\
\iff\Big(\frac{n+\log_qr}{n_i}-1\Big)\log_pq<c_i<
-\frac{m}{n_i}+\Big(\frac{\log_qu}{n_i}-1\Big)\log_pq.
\end{eqnarray}
By Eq.~(\ref{mnlims}),
$$
\Big(\frac{n+\log_qr}{n_i}-1\Big)\log_pq\to-\log_pq,\quad
-\frac{m}{n_i}+\Big(\frac{\log_qu}{n_i}-1\Big)\log_pq\to-\log_pq.
$$
Hence, by Eq.~(\ref{cibnds}),
\begin{equation}
\label{ci}
c_i\rightarrow -\log_pq.
\end{equation}
The function $D_S$ has not yet had any role. Let us now focus on it.

If $p,q\in S$ (i.e., there are at least two elements in~$S$), then
\begin{eqnarray}
\label{ds}
D_S(x_i)=x_i\Big(\frac{m_i}{p}+\frac{n_i}{q}\Big)=
x_i\Big(\frac{m+c_in_i}{p}+\frac{n_i}{q} \Big)=\nonumber
\\
x_i\frac{m}{p}+
x_i\Big(\frac{c_i}{p}+\frac{1}{q}\Big)n_i=:A_i+B_i.
\end{eqnarray}
By (\ref{xi}),
\begin{equation}
\label{alphai}
A_i\to a\frac{m}{p}.
\end{equation}
Further, (\ref{xi}) and (\ref{ci}) imply that
$$
x_i\Big(\frac{c_i}{p}+\frac{1}{q}\Big)\to a\Big(\frac{1}{q}-\frac{\log_pq}{p}\Big)=
\frac{a}{p}\Big(\frac{p}{q}-\log_pq\Big)\ne 0,
$$
because $a\ne 0$ and $p^p\ne q^q$. Consequently, by~Eq.~(\ref{mnlims}),
\begin{equation}
\label{betai}
|B_i|\to\infty.
\end{equation}
Finally, Eqs.~(\ref{betai}), (\ref{alphai}), and (\ref{ds}) imply (\ref{liminfty}).

The case of $S=\{p\}$ (i.e., there is only one element in~$S$) remains.
Then
$$
D_S(x_i)=x_i\frac{m_i}{p}.
$$
Since $x_i\to a(\ne 0)$ and $m_i\to\pm\infty$, the result
(\ref{liminfty}) again follows.
\end{proof}

\section{Lipschitz discontinuity of $D_S|_{\Z}$}
\label{lipschitz}

We recall Dirichlet's theorem on arithmetic progressions.

\begin{theorem}
\label{arprog}
Let $a\in\Z$, $b\in\Z_+$. If $\gcd{(a,b)}=1$, then there exist infinitely many primes of the form
$a+kb$, where $k\in\Z_+$.
\end{theorem}
\begin{proof}
See the proof of \cite[Theorem~7.9]{Ap}.
\end{proof}

Since all integer points are isolated (i.e., for all $a\in\Z$, there is $r>0$ such that
$N(a,r)\cap\Z=\{a\}$, where $N(a,r)=\{x\in\Q\mid |x-a|<r\}$), the restriction $D_S|_\Z$ is continuous (in the relative
topology). However, we prove that even $D_S|_{\Z_+}$ is Lipschitz discontinuous.

\begin{theorem} Let $\emptyset\ne S\subseteq\PP$. The restriction $D_S|_{\Z_+}$ is Lipschitz discontinuous. 
\end{theorem}
\begin{proof}
Let $L>0$. We show that there exist $x,y\in\Z_+$ such that
\begin{equation}
\label{claim}
x-y=1\quad{\rm and}\quad D_S(x)-D_S(y)>L.
\end{equation}
Let $p\in S$. Choose $n\in\Z_+$ with
\begin{equation}
\label{n}
np^{n-1}>L+1.
\end{equation}
By Theorem~\ref{arprog}, there is $k\in\Z_+$ such that
$$
y=kp^n-1\in\PP.
$$
Let
$
x=kp^n.
$
Then
\begin{equation}
\label{xminusy}
x-y=1,\quad D_S(y)\in\{0,1\},
\end{equation}
and
\begin{equation}
\label{dsx}
D_S(x)=kD_S(p^n)+p^nD_S(k)\ge kD_S(p^n)=knp^{n-1}\ge np^{n-1}.
\end{equation}
Now, Eqs.~(\ref{xminusy}), (\ref{dsx}), and (\ref{n}) imply Eq.~(\ref{claim}).
\end{proof}

On the other hand, we give two examples showing that if $S=\PP$ or
$S=\{p\}$, $p\in\PP$, then there is an infinite set~$A\subset\Q$ such
that $D_S|_A$ is Lipschitz continuous.

\begin{example}
\label{uaex}
Let $x\in\Q$. Then \cite[Theorem~15]{UA}
\begin{equation}
\label{de}
x'=0
\end{equation}
if (not ``only if'', contrary to this theorem)
\begin{equation}
\label{sol}
x=0\quad{\rm or}\quad x=\pm\prod_{p\in\PP}p^{\xi_p(x)p},
\end{equation}
where $(\xi_p(x))_{p\in\PP}$ is an integer sequence with a finite number of
nonzero terms such that
$$
\sum_{p\in\PP}\xi_p(x)=0.
$$
We present a correct ``only if'' part in Theorem~\ref{diffeqS}. Anyway, the set
$$
A=\big\{x\in\Q\mid x\,{\rm satisfies\,(\ref{sol})}\}
$$
is clearly infinite. Since the restriction $D|_A$ is identically zero by~Eq.~(\ref{de}),
it is Lipschitz continuous.
\end{example}

\begin{example}
\label{hmtex}
Let $p\in\PP$ and $a\in\Q$ satisfy
\begin{equation}
\label{ap}
ap\in\Z,
\end{equation}
and let $x\in\Q$. Then \cite[Theorem~3]{HMT}
\begin{equation}
\label{dep}
x'_p=ax
\end{equation}
if and only if
\begin{equation}
\label{solp}
x=cp^{ap},\quad{\rm where}\quad\nu_p(c)=0.
\end{equation}
The set
$$
A=\{x\in\Q\mid x\,{\rm satisfies\,(\ref{solp})}\}
$$
is clearly infinite. By~Eq.~(\ref{dep}), we have $D_p|_A(x)=ax$. It is
Lipschitz continuous with $L=|a|$.
\end{example}

\section{The differential equation $x'_S=ax$}
\label{diffeq}

We extend Examples \ref{uaex} and~\ref{hmtex}.

\begin{theorem}
\label{diffeqS}
Let $\emptyset\ne S\subseteq\PP$ and $a,x\in\Q$.
Then
\begin{equation}
\label{deS}
x'_S=ax
\end{equation}
if and only if
\begin{equation}
\label{solS}
x=c\prod_{p\in S}p^{\xi_p(x)p}.
\end{equation}
Here
\begin{equation}
\label{c}
\nu_p(c)=0\quad{\it for\,all}\quad p\in S,
\end{equation}
and {\rm (}$c=0$ or{\rm )}
$(\xi_p(x))_{p\in S}$ is a sequence of rational numbers with a finite number of nonzero terms {\rm (}this is needed if $S$ is infinite{\rm )} satisfying
\begin{equation}
\label{expcnd}
\xi_p(x)p\in\Z\quad{\it for\,all}\quad p\in S
\end{equation}
and
\begin{equation}
\label{sumcnd}
\sum_{p\in S}\xi_p(x)=a.
\end{equation}
\end{theorem}
\begin{proof}
If $x$ is as in Eq.~(\ref{solS}), then, by Eq.~(\ref{sumcnd}), we get
$$
x'_S=x\sum_{p\in S}\frac{\xi_p(x)p}{p}=x\sum_{p\in S}\xi_p(x)=xa,
$$
verifying Eq.~(\ref{deS}).

Conversely, suppose that $x$ satisfies Eq.~(\ref{deS}). We show that it
satisfies Eq.~(\ref{solS}). The case of $x=0$ is trivial. If the claim holds
for~$x$, then it also holds for~$-x$. Therefore, it is enough to consider
$$
x=\prod_{p\in\mathbb P}p^{\nu_p(x)}.
$$
Denoting $\nu_p=\nu_p(x)$ for short, we can write
$$
x=c\prod_{p\in S}p^{\nu_p}=c\prod_{p\in S}p^{\xi_pp},
$$
where $c$ satisfies Eq.~(\ref{c}) and
$$
\xi_p=\frac{\nu_p}{p},\quad p\in S.
$$
Now,
$$
\sum_{p\in S}\xi_p=\sum_{p\in S}\frac{\nu_p}{p}=
\frac{x'_S}{x}=\frac{ax}{x}=a
$$
by Eq.~(\ref{deS}), i.e., Eq.~(\ref{sumcnd}) holds. Also Eq.~(\ref{expcnd})
is obviously satisfied. Thus Eq.~(\ref{solS}) follows.
\end{proof}

The set
$$
A=\{x\in\Q\mid x\,{\rm satisfies\,(\ref{solS})}\}
$$
is clearly infinite. By Eq.~(\ref{deS}), the restriction
$D_S|_A(x)=ax$. It is Lipschitz continuous with $L=|a|$.

If $S=\PP$ and $a=0$, then Eq.~(\ref{solS}) reads
$$
x=c\prod_{p\in\PP}p^{\xi_p(x)p}.
$$
Here
$$
\sum_{p\in\PP}\xi_p(x)=0
$$
and $\nu_p(c)=0$ for all $p\in\PP$, i.e., $c\in\{0,\pm 1\}$.
Thus we encounter Eq.~(\ref{sol}), but it is enough that $\xi_p(x)p\in\Z$ for
all $p\in\PP$, not necessarily $\xi_p(x)\in\Z$.

If $S=\{p\}$, then
$$
a=\sum_{q\in\{p\}}\xi_q(x)=\xi_p(x).
$$
Hence, by Eq.~(\ref{solS}),
$$
x=c\prod_{q\in\{p\}}p^{\xi_q(x)q}=cp^{\xi_p(x)p}=cp^{ap},
\quad{\rm where}\quad\nu_p(c)=0,
$$
repeating Eq.~(\ref{solp}). The condition~(\ref{expcnd}) reduces to~(\ref{ap}).

\section{Concluding remarks}
\label{remarks}

We proved that $D_S$ is superdiscontinuous at any $a\in\Q$. We also proved that its restriction
to a suitable infinite set~$A$ is Lipschitz continuous.
This happens if, for example, the set~$A$ consists of the solutions of $x'=0$
or those of
\begin{equation}
\label{xprimep}
x'_p=ax,
\end{equation}
where $ap\in\Z$. These equations have been discussed
in the literature, while the more general equation,
\begin{equation}
\label{xprimeS}
x'_S=ax,
\end{equation}
apparently
has not. Anyway, it also provides us a suitable~$A$ if there is a sequence
$(\xi_p(x))_{p\in S}$ satisfying Eqs.~(\ref{expcnd}) and~(\ref{sumcnd}).

According to
Example~\ref{hmtex}, $ap\in\Z$ is a sufficient condition for Eq.~(\ref{xprimep})
to have nontrivial solutions. In fact, it is also necessary
\cite[Theorem~3]{HMT}. But what about Eq.~(\ref{xprimeS})?
It has nontrivial solutions if and only if
there are $p_1,\dots,p_k\in S$ such that $ap_1\cdots p_k\in\Z$.
We will present the proof in a forthcoming paper.

\begin{thebibliography}{9}

\bibitem{Ap} T.~M.~Apostol, {\em Introduction to Analytic Number Theory},
Springer, 1976.

\bibitem{Ba} E.~J.~Barbeau, Remarks on an arithmetic derivative,
\emph{Canad. Math. Bull.}~\textbf{4} (1961), 117--122.

\bibitem{HMT} P.~Haukkanen, J.~K.~Merikoski, and T.~Tossavainen, On
arithmetic partial differential equations, {\em J. Integer Sequences}~\textbf{19} (2016),
\href{https://cs.uwaterloo.ca/journals/JIS/VOL19/Tossavainen/tossa6.html}
{Article 16.8.6}.

\bibitem{Ko} J.~Kovi\v c, The arithmetic derivative and antiderivative,
\emph{J. Integer Sequences}~\textbf{15} (2012),
\href{https://cs.uwaterloo.ca/journals/JIS/VOL15/Kovic/kovic4.html}
{Article 12.3.8}.

\bibitem{MHT} J.~K.~Merikoski, P.~Haukkanen, and T.~Tossavainen,
Arithmetic subderivatives and Leibniz-additive functions,
\emph{Ann. Math. Informat.} {\bf 50} (2019).  Available at
\url{http://ami.ektf.hu}.

\bibitem{Sh} J.~Mingot Shelly, Una cuesti\'on de la teor\'ia de los n\'umeros,
\emph{Asociaci\'on Espa\~nola, Granada} (1911), 1--12.

\bibitem{PS} R.~K.~Pandey and R.~Saxena, On some conjectures about
arithmetic partial differential equations, {\em J. Integer Sequences}~\textbf{20}
(2017),
\href{https://cs.uwaterloo.ca/journals/JIS/VOL20/Pandey/pandey10.html}
{Article 17.5.2}.

\bibitem{UA} V.~Ufnarovski and B.~\AA hlander, How to differentiate
a number, \emph{J. Integer Sequences}~\textbf{6} (2003),
\href{http://www.cs.uwaterloo.ca/journals/JIS/VOL6/Ufnarovski/ufnarovski.html}
{Article 03.3.4}.

\end{thebibliography}

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\noindent 2010 {\it Mathematics Subject Classification:}
Primary 11A25; Secondary 26A15.

\noindent {\it Keywords:} 
arithmetic subderivative, arithmetic partial derivative, arithmetic derivative,
continuity, Lipschitz continuity.

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\noindent (Concerned with sequences
\seqnum{A000040} and
\seqnum{A003415}.)

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\vspace*{+.1in}
\noindent
Received  June 18 2019; 
revised version received October 9 2019.
Published in {\it Journal of Integer Sequences}, October 15 2019.

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\noindent
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