\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \usepackage{ifthen} \usepackage{pgfpages} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.5in} \setlength{\textheight}{8.9in} \newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \begin{center} \vskip 1cm{\LARGE\bf The Arithmetic Derivative and Antiderivative } \vskip 1cm \large Jurij Kovi\v{c}\\ Institute of Mathematics, Physics, and Mechanics \\ University of Ljubljana, Slovenia\\ \href{mailto:amadea.kovic@siol.net}{\tt amadea.kovic@siol.net} \\ \end{center} \vskip .2 in \begin{abstract} The notion of the arithmetic derivative, a function sending each prime to 1 and satisfying the Leibnitz rule, is extended to the case of complex numbers with rational real and imaginary parts. Some constraints on the solutions to some arithmetic differential equations are found. The homogeneous arithmetic differential equation of the $k$-th order is studied. The factorization structure of the antiderivatives of natural numbers is presented. Arithmetic partial derivatives are defined and some arithmetic partial differential equations are solved. \end{abstract} \section{Introduction: Basic concepts} The goal of this paper is: (i) to review what is known about the arithmetic derivative \cite{Barb, Stay, Ufna}, (ii) to define some related new concepts, and (iii) to prove some new results, mostly related to the conjectures formulated in \cite{Barb, Ufna}. \subsection{Definition of the arithmetic derivative} Barbeau \cite{Barb} defined the arithmetic derivative as the function $D: \mathbb Z \rightarrow \mathbb Z$, defined by the rules: $D(1) = D(0) = 0$ $D(p) = 1$ for any prime $p \in \mathbb P := \{2,3,5,7,\ldots, p_{i},\ldots \}$. $D(ab) = D(a)b + aD(b)$ for any $a,b \in \mathbb N$ (the Leibnitz rule) $D(-n) = -D(n)$ for any $n \in \mathbb N$. Ufnarovski and \AA hlander \cite{Ufna} extended $D$ to rational numbers by the rule: $$ D(\frac{a}{b}) = (\frac{a}{b})' = \frac{a'b - ab'}{b^{2}} $$ They also defined $D$ for real numbers of the form $x = \prod_{i=1}^{k}p_{i}^{\, x_{i}} $, where $p_{i}$ are different primes and $ x_{i}\in \mathbb Q$, by the additional rule: $$D(x) = x' = x\sum_{i=1}^{k}\frac{x_{i}}{p_{i}}.$$ They showed that the definition of $D$ can be extended to all real numbers and to the arbitrary unique factorization domain. I extend the definition of the arithmetic derivative to the numbers from the unit circle in the complex plane $\mathbb E = \{e^{i\varphi} ,\varphi \in [0,2\pi] \} $ and to the Gaussian integers ${\mathbb Z}[i] = \{a + bi, a,b \in \mathbb Z\}$. It is not possible to extend the definition of the arithmetic derivative to the Gaussian integers by the demand $D(q) = q' = 0$ for all irreducible Gaussian integers $q$ (the analog of primes). If one wants the Leibnitz rule to still remain valid, one easily gets a contradiction: $2 = (1 + i)(1 - i)$, hence $D(2) = 1\cdot(1 - i) + (1 + i)\cdot 1 = 2$, but $D(2) = 1$ because $2$ is a prime. However, one can extend $D$ to ${\mathbb Q}[i] = \{a + bi, a,b \in \mathbb Q\}$ using the polar decomposition of complex numbers as follows: \begin{proposition} \label{prop1} i) For any $\varphi = (m/n)\pi$, where $m,n \in \mathbb Q$ and $n \neq 0$, the equation $D(e^{i\varphi }) = 0$ holds. ii) Let $\mathbb E = \{z \in \mathbb C; |z| = 1 \} $ be the unit circle in the complex plane. There are uncountably many functions $D: \mathbb E \rightarrow \mathbb E$ satisfying the condition $D(zw) = D(z)w + zD(w)$ for all $z,w \in \mathbb E$. iii) If one defines the arithmetic derivative on $\mathbb E$ as follows: $D(e^{i\varphi}) = 0 $ for all $e^{i\varphi} \in \mathbb E$, then the definition of the arithmetic derivative can be uniquely extended to ${\mathbb Q}[i]$ so that the Leibnitz rule and the quotient rule remain valid. \end{proposition} \begin{proof} i) Take any $\varphi = (m/n)\pi$, where $m,n \in \mathbb Z$ and $n \neq 0$. Then $e^{in\varphi} = \pm1$, hence $D(e^{in\varphi}) = 0$. The Leibnitz rule implies $D(e^{in\varphi}) = D(e^{i\varphi})(e^{i\varphi} )^{n-1}$ and because $(e^{i\varphi} )^{n-1}\neq 0 $ it must be that $D(e^{i\varphi}) = 0$. ii) Sketch of the proof: One can easily see that there is a 1-1 correspondence between the functions $D: \mathbb E \rightarrow \mathbb C$ satisfying the Leibnitz rule $D(zw) = D(z)W + zD(w)$ and the additive functions $L: \mathbb R \rightarrow \mathbb C$ with a period of $2\pi$ (the correspondence is given by the formulas $L(s) := \frac{D(e^{is})}{e^{is}}$ and $D(e^{is}) := L(s)e^{is}$). It is also easy to see that for any additive function (satisfying Cauchy's functional equation $L(s + t) = L(s) + L(t)$) it is: $L(qx) = q \cdot L(x) $ for all $q \in \mathbb Q$ and $x \in \mathbb R$. Because $\mathbb R$ and $\mathbb C$ are linear spaces over $\mathbb Q$, this means that every additive function $L: \mathbb R \rightarrow \mathbb R$ is a $\mathbb Q$-linear transformation. Additive functions from $\mathbb R$ into $\mathbb C$ with a period of $2\pi$ are therefore in a 1-1 correspondence with $\mathbb Q$-linear transformations $L: \mathbb R \rightarrow \mathbb C$ such that $L(2\pi) = 0$. A linear transformation is uniquely determined with its values on any of the bases of its domain. Let $B$ be a base of the space $\mathbb R$ such that $\pi \in B$ (its existence is guaranteed by Zorn's lemma). Thus the $\mathbb Q$-linear transformations $L: \mathbb R \rightarrow \mathbb C$ such that $L(2\pi) = L(\pi) = 0$ are in a 1-1 correspondence with the set of all functions $L: B \rightarrow \mathbb C$ such that $L(\pi) = 0$. The base $B$ is obviously not countable, hence the same also holds for the set of all functions $L: B\setminus \{\pi\} \rightarrow \mathbb C$. By the chain of proven 1-1 correspondences, this also holds for the set of all functions $D: \mathbb E \rightarrow \mathbb E$ satisfying the Leibnitz rule. iii) If one writes such numbers in their polar form $a + bi = re^{i\varphi}$, then $D(re^{i\varphi}) = D(r)e^{i\varphi} + rD(e^{i\varphi} )$ $=D(r)e^{i\varphi} = D((a^{2} + b^{2})^{1/2})e^{i\varphi}$ because $D(e^{i\varphi}) = 0$ and $r = (a^{2} + b^{2})^{1/2}$. The Leibnitz rule is still valid because for any $z = re^{i\varphi }$ and $w = se^{i\psi } $ it is: $D(zw) = D((re^{i\varphi })(se^{i\psi} )) = D(rs)e^{i\varphi + \psi} = (D(r)s + r(D(s))e^{i\varphi + \psi} = $ $D(r)e^{i\varphi} se^{i\psi} + re^{i\varphi}D(s)e^{i\psi } = D(z)w + zD(w). $ The quotient rule is also still valid because $\frac{D(z)w - zD(w)}{w^{2}} = \frac{D(r)e^{i\varphi}se^{i\psi } - re^{i\varphi}D(s)e^{i\psi }}{s^{2}e^{i2\varphi}} = \frac{(D(r)s - rD(s))e^{i\varphi}}{s^{2}e^{i\psi }} = D(\frac{r}{s})e^{i(\varphi - \psi)} = D(\frac{r}{s}e^{i(\varphi - \psi)})= D(\frac{re^{i\varphi}}{se^{i\psi}}) = D(\frac{z}{w})$. \end{proof} Stay \cite{Stay} generalized the concept of the arithmetic derivative still further, practically for any number, using advanced techniques such as exponential quantum calculus. In this paper I focus on the arithmetic derivative as a function defined for natural numbers and rational numbers. \subsection{Higher derivatives and the logarithmic derivative} \emph{Higher derivatives} $n^{(k)}$ are defined inductively: $n^{(2)} = D^{2}(n) = D(D(n)) = n'' = (n')'$, $n^{(k+1)} = D^{k+1}(n) = D(D^{k}(n))$. Many conjectures on the arithmetic derivative focus on the behavior of sequences $(n, n', n'',\ldots ) $. Ufnarovski and \AA hlander \cite{Ufna} conjectured that for each $n \in \mathbb N$ exactly one of the following can happen: either $n^{(k)} = 0$ for sufficiently large $k$, or $lim _{k\rightarrow \infty} = \infty$, or $n = p^{p}$ for some prime $p$ (in this case $n^{(k)} = n$ for each $k \in \mathbb N$). They introduced the function $L$, called the \emph{logarithmic derivative}, satisfying the condition $$L(x) = \frac{x'}{x} = \frac{D(x)}{x}.$$ For any $x = \prod_{i=1}^{k}p_{i}^{\, x_{i}} $, where $p_{i}$ are different primes and $ x_{i}\in \mathbb Q$, $L$ satisfies the condition $$L(x) = \sum_{i=1}^{k}\frac{x_{i}}{p_{i}}.$$ For every prime $p$ and every $m,n \in \mathbb N$ the following formulas hold: $L(p) = \frac{1}{p},$ $L(p^{\frac{m}{n}}) = \frac{m}{np}, $ $L(1) = 0.$ From the definition of the logarithmic derivative also follow the formulas $L(-x) = L(x), L(0) = \infty$ and $D(x) = L(x)\cdot x $ \cite{Ufna}, hence $D^{2}(x) = D(L(x))x + L(x)D(x)$. I also use the notation $L(x) = x^{*} $. The logarithmic derivative is an additive function: $L(xy) = L(x) + L(y)$ for any $x, y \in \mathbb Q$. Consequently, using a table of values $L(p) = \frac{1}{p}$ (computed to sufficient decimal places!) and the formula $D(x) = L(x)\cdot x$, it is easy to find $D(n)$ for $n \in \mathbb N $ having all its prime factors in the table. For example, $D(5\cdot11^{3} ) = L(5\cdot11^{3})\cdot5\cdot11^{3}$ and because $ L(5\cdot11^{3}) = L(5) + 3L(11) = 0.2000 + 3\cdot0.0909 = 0.4727$, one can calculate $D(5\cdot11^{3}) = \lceil0.4727\cdot6655 \rceil = \lceil3145.8185 \rceil = 3146.$ {%\small \begin{center} \begin{tabular}{|c||c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline % after \\: \hline or \cline{col1-col2} \cline{col3-col4} ... $P_{N}$ & $2$& $3$& $5$& $7$& $11$ &$13$& $17$ \\ \hline $\frac{1}{P_{N}}$ & $0.5000$& $0.3333$& $0.2000$& $0.1429$& $0.090$& $0.0769$ &$0.0588$ \\ \hline \end{tabular} \end{center} } \begin{center} Table 1: Values of the logarithmic derivative for the first seven primes \end{center} \section{Brief review of known results and conjectures}\label{chap2} Barbeau \cite{Barb} proved that if $n$ is not a prime or unity then $D(n) \geq 2\sqrt{n}$ with equality only if $n = p^{2}, p \in \mathbb P$. He showed that, for integers possessing a proper divisor of the form $p^{p}, p \in \mathbb P$, $\lim _{k\rightarrow \infty} D^{k}(n) = \infty.$ Ufnarovski and \AA hlander \cite{Ufna} translated some famous conjectures in number theory (e.g., the Goldbach conjecture, the Prime twins conjecture) into conjectures about the arithmetic derivative. They formulated many other conjectures, mostly related to (arithmetic) differential equations; for example that the equation $x' = 1$ has only primes as positive rational solutions (but it has a negative rational solution $x = -\frac {5}{4}$). They also conjectured that the equation $n'' = n$ has no other solutions than $n = p^{p}, p \in \mathbb P$ in natural numbers and that there are some rational numbers without antiderivatives (or "integrals"). I study these conjectures in Section~\ref{chap3}. \subsection{Arithmetic differential equations and integrals} Most of the known results by Ufnarovski and \AA hlander \cite{Ufna} focus on arithmetic differential equations of the first and second order. For example: The only solutions to the equation $n' = n$ in natural numbers are $n = p^{p}$, where $p$ is any prime. The nonzero solutions to $x' = 0$ are the rational numbers of the form: $x = \pm\prod p_{i}^{\alpha_{i}p_{i}} $, where $p_{1}, \ldots ,p_{k} $ are different primes and $\{\alpha_{1},\ldots, \alpha_{n}\}$ is a set of integers such that $\Sigma_{i=1}^{k} \alpha_{i} = 0$. The rational solutions to differential equations $x' = x\alpha$ for all rational numbers $\alpha = a/b$, where $\gcd(a,b) = 1$ and $b > 0$, are all of the form $x = x_{0}y$, where $x_{0}$ is a nonzero particular solution and $y$ is any rational solution of the equation $y' = 0 $. Let $I(a)$ denote all the solutions of differential equation $n' = a$ for $n \in \mathbb N$ and let $i(a)$ denote the number of such solutions, called the integrals of $n$. Because $n'\geq 2\surd n $ if $n$ is not a prime or unity \cite{Barb}, the solutions satisfy $n \leq \frac {a^{2}}{4} $, hence $i(a)< \infty$ for any $a \in \mathbb N$. Because $\gcd(n,n') = 1$ if and only if $n$ is square-free, all integrals of primes are products of different primes: $p_{1}\cdots p_{k} $. It will be seen (Corollary~\ref{coro1}) that there are primes without integrals (e.g., primes 2, 3, 17). If $D(n)$ were known for each $n \in \mathbb N$, one would know which natural numbers are primes (because the equation $n' = 1$ has only primes as solutions in natural numbers). Ufnarovski and \AA hlander \cite{Ufna} gave a list of all $a \leq 1000 $ having no integral, a list of those numbers $a \leq 100$ having more than one integral, and a list of those $a \leq 100$ for which $i(a) = 1$. It will be seen (Corollary~\ref{coro2}) that something can also be said about the possible factorization structure of the integrals of a given natural number. \section{New results}\label{chap3} Ufnarovski and \AA hlander \cite{Ufna} solved the equation $x' = \alpha x$ in rational numbers for every rational number $\alpha$. Nonetheless it is interesting to know which natural numbers solve this equation when $\alpha = m$ is a natural number. \begin{proposition} \label{prop2} Let $m \in \mathbb N$. The solutions to the equation $x' = mx$ in natural numbers $x$ are exactly the numbers of the form $x = p_{1}^{p_{1}n_{1}} \cdots p_{k}^{p_{k}n_{k}}$, where $n_{1} + \dots + n_{k} = m.$ \end{proposition} \begin{proof} Let $x = p_{1}^{e_{1}} \cdots p_{k}^{e_{k} }$ be the factorization of $x$. The condition $x' = L(x)x = mx$ implies $L(x)(p_{1}\cdots p_{k} ) = (\frac{e_{1}}{p_{1}}+ \cdots + \frac{e_{k}}{p_{k}})(p_{1}\cdots p_{k} ) = m(p_{1}\cdots p_{k} ).$ Because the right side of this equation is divisible by any of the primes $p_{1}, \ldots ,p_{k}$, the left side must also be. This is possible only if for every prime $p_{i}$ the corresponding exponent $e_{i}$ is the multiple of this prime $e_{i} = n_{i}p_{i}$. The derivative of such a number $x = p_{1}^{p_{1}n_{1}} \cdots p_{k}^{p_{k}n_{k}} $ is $D(x) = D(p_{1}^{p_{1}n_{1}} \cdots p_{k}^{p_{k}n_{k}} ) = (n_{1} + \cdots + n_{k})x$ (by the Leibnitz rule and because $D(p^{p_{i}n_{i}}) = n_{i}p^{p_{i}n_{i}})$ and this is equal to $mx$ if and only if $n_{1} + \cdots + n_{k} = m.$ \end{proof} \subsection{The homogeneous differential equation of the $k'$th order} What are the solutions $x$ of the differential equation $$a_{k}x^{(k)} + a_{k-1}x^{(k-1)} + \cdots + a_{2}x^{(2)} + a_{1}x' + a_{0}x = 0$$ with rational coefficients $a_{i}$? In order to answer this question I introduce the concept of a logarithmic class. \begin{definition}\emph{Let $A$ be any chosen subring of the ring of complex numbers for which the arithmetic derivative is defined. The logarithmic class $N_{r,A}$ of the number $r$ consists of all numbers $x \in A$ with the same logarithmic derivative: $N_{r,A} = \{x \in A; x^{*} = L(x) = r\} $}. \end{definition} \begin{remark} If it is made perfectly clear which $A$ is being worked with the shorter notation $N_{r}$ can be used. For the purpose of this article, let $A$ be the set of all complex numbers with rational real and imaginary parts: $A = {\mathbb Q}[i] = \{a + bi, a,b \in \mathbb Q\}$. \end{remark} \begin{example} Besides the number $0$ and the complex numbers $z = e^{i\varphi} \in A$ on the unit circle in the Gaussian plane the class $N_{0}$ also contains rational solutions $x = \pm\prod p_{i}^{\alpha_{i}p_{i}} $, where $p_{1}, \ldots ,p_{k} $ are different primes and $\{\alpha_{1},\ldots, \alpha_{n}\}$ is a set of integers such that $\Sigma_{i=1}^{k} \alpha_{i} = 0$; for instance, $x = \frac{4}{27}$ \cite{Ufna}. All $x \in N_{0}$ solve the equation $x' = 0$. \end{example} The class $N_{1,\mathbb Q}$ contains all numbers $p^{p}$, where $p$ is any prime because $D(p^{p}) = p^{p}$. The class $N_{-1,\mathbb Q}$ contains all numbers $p^{-p} $, where $p$ is any prime because $D(p^{-p} ) = -p^{-p}. $ Do $N_{1,\mathbb Q}$ and $N_{-1,\mathbb Q}$ also contain other rational solutions? Yes: \begin{proposition}\label{prop3} If $(\frac{a}{b})' = \pm\frac{a}{b} \in \mathbb Q$ and if $a$ and $b$ have no common factors greater than 1, then $a = p_{1}^{p_{1}n_{1}} \cdots p_{k}^{p_{k}n_{k}}$, $b = q_{1}^{q_{1}s_{1}} \cdots q_{k}^{q_{k}n_{s}}$, and $a^{* }- b^{*} = \pm 1$, where $n_{i}, m_{j}, k,l $ are nonnegative integers and $a^{*}, b^{*}$ logarithmic derivatives. All such numbers are solutions because: $(\frac{a}{b})'= (\frac{a}{b})^{*}(\frac{a}{b}) = \pm(\frac{a}{b}).$ \end{proposition} \begin{proof} If $(\frac{a}{b})' = \pm\frac{a'b - b'a}{b^{2}} = \pm\frac{a}{b} $ then $a'b - b'a = \pm ab$, hence $\gcd(a,b) = 1$ implies $a' = ma$ and $b' = nb$, where $m = a^{*}$ and $n = b^{*}$ are natural numbers. Hence (by Proposition~\ref{prop2}) $a = p_{1}^{p_{1}n_{1}} \cdots p_{k}^{p_{k}n_{k}}$, where $n_{1} + \cdots + n_{k} = m,$ and $b = q_{1}^{q_{1}s_{1}} \cdots q_{k}^{q_{k}n_{s}}$, where $s_{1} + \cdots + s_{k} = n$. Because $(\frac{a}{b})' = \pm\frac{a}{b} $ implies $ (\frac{a}{b})^{*} = \pm 1$, it must be $a^{*} - b^{*} = \pm 1.$ \end{proof} \begin{proposition} \label{prop4} i) The derivative $D$ sends logarithmic classes into logarithmic classes as follows: $D(N_{r}) \subseteq N_{r^{*} + r}$. Consequently $D(N_{r}) \subseteq N_{r}$ if and only if $r^{*} = 0$ and therefore $r' = 0$. ii) If $r^{*} + r = 0$ and $r \neq 0$ then $D(N_{r}) = N_{0} - \{0\} \neq N_{0}$. So in this case $D(N_{r}) \supseteq N_{r^{*} + r}$ is not true. iii) If $r^{*} + r \neq 0$ then $D(N_{r}) = N_{r^{*} + r}$. \end{proposition} \begin{proof} If $r = 0$ then $N_{r} = N_{0} = N_{r^{*} + r}$. So let us now assume that $r\neq 0$. i) $D(N_{r}) \subseteq N_{r^{*} + r}$ is true because $x \in N_{r}$ implies $x'= rx$, hence $(x')' = r'x + rx' = \frac{r'x'}{r} + rx' = (\frac{r'}{r} + r)x' = (r^{*} + r )x' $. ii) It is possible indeed that $r^{*} + r = 0$ while $r \neq 0$ (an example is $r = \frac {1}{5}, r' = -\frac {1}{5^{2}}, r^{*} = - \frac {1}{5})$. If $r^{*} + r = 0$ then $0 \in N_{r^{*} + r}$ because $0' = 0$. Now suppose there is an $x \in N_{r}$ such that $D(x) = 0$. Wowever, this implies $x \in N_{0} $, hence $r = 0$ and there is a contradiction with the assumption $r \neq 0$. This means that in this case it is not true $D(N_{r}) \supseteq N_{r^{*} + r} = N_{0}$ because there is no $x \in N_{r}$ such that $D(x) = 0$. Now let it be proved that for each nonzero $y \in N_{r^{*} + r}$ there is an $x \in N_{r}$ such that $D(x) = y$. If there is any such $x \in N_{r}$, then it must be $y = x' = rx$, hence there is at most one such $x$ and it is defined by the formula $x = \frac{y}{r}$. Now suppose the derivative of this $x$ is not equal to $y$. This assumption leads to a contradiction because $D(\frac{y}{r}) = \frac{D(y)r - yD(r)}{r^{2}} = \frac{(r^{*} + r)yr - yr^{*}r}{r^{2}} = \frac{yr^{2}}{r^{2}}\neq y$ implies $1 \neq 1$. Because $D(x) = y = rx$, this $x$ is indeed a member of $N_{r}$. Thus it has been proved that $D(N_{r}) \supseteq N_{0} - \{0\}$. Because it is already known that i) is true and because it has been shown that there is no $x \in N_{r}$ such that $D(x) = 0$, the equation $D(N_{r}) = N_{0} - \{0\}$ holds. Moreover, because $0 \in N_{0}$, it is also true that $N_{0}$ is a proper subset of $N_{0} - \{0\}$. iii) If $r^{*} + r \neq 0$ then $r \neq 0$, and $N_{r^{*} + r}$ contains only nonzero elements because $0 \in N_{0} $. Now it is possible to repeat the reasoning as in ii) and for any $y \in N_{r^{*} + r}$ one finds an $x \in N_{r}$ such that $y = D(x)$. Hence $D(N_{r}) \supseteq N_{r^{*} + r}$ and this together with i) implies $D(N_{r}) = N_{r^{*} + r}$.\end{proof} \begin{remark} It is already known that there are many rational solutions to the equation $r' = 0$; for example, $r = 1, -1, 0,$ hence: $D(N_{1}) \subseteq N_{1}, D(N_{-1}) \subseteq N_{-1}$, $D(N_{0}) \subseteq N_{0}.$ \end{remark} \begin{proposition} \label{prop5} All the derivatives $x^{(k)}$ of any number $x$ can be expressed as functions of the logarithmic derivative $L(x) = x^{*}$ as follows: $x^{(k)} = f_{k}(x^{*})x$ where $f_{1}(x^{*}) = x^{*} $ and $f_{k+1}(x^{*}) = ((f_{k}(x^{*}))' + f_{k}(x^{*}))x^{*}.$ Thus $x' = x^{*}x, x^{(2)} = ((x^{*})' + (x^{*})^{2})x$ etc. \end{proposition} \begin{proof} This is true for $k = 1$. Suppose $x^{(k)} = f_{k}(x^{*})x$. Then by the Leibnitz rule: $x^{(k+1)} = (f_{k}(x^{*}))'x + f_{k}(x^{*})x' = ((f_{k}(x^{*}))' + f_{k}(x^{*})x^{*})x.$ \end{proof} \begin{proposition} \label{prop6} Any homogeneous differential equation $f(x) \equiv a_{k}x^{(k)} + a_{k-1}x^{(k-1)} + \cdots + a_{2}x^{(2)} + a_{1}x' + a_{0}x = 0$ reduces to an equation $g(x^{*}) = 0$. Consequently, if the set of nontrivial solutions to any homogeneous differential equation $f(x) = 0$ is not empty, then it consists of some classes $N_{r}$. \end{proposition} \begin{proof} Let $r = x^{*} = \frac{x'}{x}.$ Because $x^{(k)} = f_{k}(x^{*})x$, one can divide the differential equation by $x$ and get an equation of the form $g(r) = 0$. Thus $f(x) = 0$ if and only if $g(x^{*}) = 0.$ Hence whether $f(x) = 0$ or not depends only on the logarithmic class $N_{r} = N_{x^{*}}$. Note also that the degree of the polynomial $g$ is $k - 1$, one less than the degree of the polynomial $f$. \end{proof} \begin{example} The nonzero solutions to the equation $x'' - x = 0$ implying $((x^{*})' + (x^{*})^{2})x - x = 0$ satisfy the equation $((x^{*})' + (x^{*})^{2} - 1 = 0$ or $(x^{*})' = 1 - (x^{*})^{2} $. Thus the nonzero solutions to $x'' - x = 0$ exist if and only if the nonhomogeneous equation $r' = 1 - r^{2}$ can be solved. \end{example} \begin{proposition} \label{prop7} For every $r \in \mathbb Q$ the class $N_{r}$ is not empty because it contains the numbers $p^{pr}$, where $p$ is any prime. \end{proposition} \begin{proof} $L(p^{pr}) = r$ for any prime $p$, hence $p^{pr} \in N_{r}$. \end{proof} Thus in solving the homogenous differential equation $f(x) = 0$ one can always search for the solutions of the form $x = p^{pr}$. Of course, other solutions are also possible. \begin{remark} Numbers $p^{pr}$ behave in the first derivative just like the exponential function $e^{xr}$ whose derivative is $re^{xr}$. However, for the higher derivatives the analogy is no longer valid because $(p^{p}r)'' = (r' + r^{2})p^{pr}$, while $(e^{xr})'' = r^{2}e^{xr}$. This is one of the reasons why solving an arithmetic differential equation is harder than solving the analogous problem for functions. \end{remark} \begin{proposition}\label{prop8} i) In addition to the trivial solutions $x \in N_{0}$ the equation $x'' = 0$ also has solutions $x \in N_{1/p}$, where $p$ is any prime. ii) All solutions $x \in N_{r}$ of $x'' = 0$ satisfy the condition: $r^{*} + r = 0$ (hence $r \in N_{-r}$ and $r' = r^{*}r = -r^{2} $) and all such $x$ are solutions to $x'' = 0$. If $r = \frac{a}{b} \in \mathbb Q$ then $\frac{a}{b} = b^{*} - a^{*} = (\frac{b}{a})^{*}$, hence $(\frac{b}{a})' = 1$. \end{proposition} \begin{proof} i) If $x' = \frac{1}{p}$ then $x'' = - \frac{1}{p^{2}}x + \frac{1}{p}\frac{1}{p}x = 0 = (x')^{*}x'$, hence $(x')^{*} = 0 $. ii) Any solutions $x \in N_{r}$ of $x'' = 0$ satisfy the condition $D^{2}(N_{r}) \subseteq D(N_{r^{*} + r}) = D_{0}$, hence $r^{*} + r = 0$. Conversely, $r^{*} + r = 0 $ implies $D^{2}(N_{r}) \subseteq D(N_{r^{*} + r}) = D({N_{0}}) = \{0\}$, hence $x'' = 0$. If $r = \frac{a}{b}$ then $r' = \frac{a'b - ab'}{b^{2}} = \frac{a'b - ab'}{ab}\frac{a}{b}$, hence $(\frac{a}{b})^{*} = \frac{a'b - ab'}{ab} = -r = -\frac{a}{b}$, therefore $\frac{a'}{a} - \frac{b'}{b} = -\frac{a}{b}$ and $\frac{a}{b} = b^{*} - a^{*} = (\frac{b}{a})^{*} $. Hence $(\frac{b}{a})' = \frac{b}{a}^{*}\frac{b}{a} = \frac{a}{b}\frac{b}{a} = 1$. \end{proof} \begin{example} Ufnarovski and \AA hlander \cite{Ufna} observed that if $x = -\frac{5}{4}$ then $x' = 1$ (hence $x'' = 0 $). Then $r = x^{*} = - \frac{4}{5}$, $r' = \frac{-4\cdot5 - (-4)\cdot1}{5^{2}} = -\frac{16}{25} = \frac{4}{5}\frac{-4}{5},$ hence $r^{*} = \frac{4}{5}$ and $r^{*} + r = \frac{4}{5} + (-\frac{4}{5} ) = 0$. Hence all $x \in N_{-\frac{4}{5}}$ solve $x'' = 0$. \end{example} \subsection{The graph of derivatives of natural numbers} Of interest is the structure of the infinite directed graph $G_{D}$, whose vertices correspond to natural numbers $n$ and whose arcs $n\rightarrow D(n)$ connect the number and its derivative. The corresponding dynamic system: $n \rightarrow D(n)$ has two obvious attractors: $0$ and $\infty$. There are numbers $n$ with an increasing sequence of derivatives $n< n'< n''< \cdots < n^{(k)} < n^{(k+1)} < \cdots $ (e.g., $n = p^{pk}, p \in \mathbb P)$, so there are paths of infinite length in the graph $G_{D}$. Ufnarovski and \AA hlander \cite{Ufna} conjectured that the equation $n^{(k)} = n$ has only trivial solutions $p^{p}$, where $p \in \mathbb P$, satisfying $n' = n $. If this is true then the only cycles in $G_{D}$ are the loops in these fixed points. They have shown that if $m' = n$ and $n' = m$ then $m$ and $n$ must be square-free numbers: $n = \prod_{i=1}^{k} p_{i} $ and $m = \prod_{j=1}^{l}q_{j} $, where all $p_{i}$ are distinct from all $q_{i}$. I present further constraints (Propositions~\ref{prop9},~\ref{prop10},~\ref{prop11},~\ref{prop11a}) on the structure of possible solutions of the equation $n'' = n$ in natural numbers. However, the equation $x'' = x$ has non-trivial rational solutions of the form $x = p^{-p}$ where $p \in \mathbb P$ and they also satisfy the equation $x' = -x.$ Let us first show that any eventual nontrivial solutions $m,n$ (different from $m = n = p^{p}$ where $p \in \mathbb P$) of the system $n' = m$, $m' = n$ in natural numbers cannot be just a product of two primes; at least one of the numbers $m,n \in \mathbb N$ solving such a system must have at least three different prime factors. \begin{proposition} \label{prop9} The system $n' = m$, $m' = n$ has no solutions in natural numbers of the form $m = p_{1}p_{2}$ and $n = q_{1}q_{2} $, where $p_{1},p_{2},q_{1},q_{2} \in \mathbb P $. \end{proposition} \begin{proof} Because $p_{1},p_{2},q_{1},q_{2}$ must be different primes, it is not possible for both $m$ and $n$ to have the factor 2. Therefore one can assume that $p_{1}$ and $p_{2}$ are both odd. Then $m' = p_{1} + p_{2} = n = 2q_{2}$ and $n' = 2 + q_{2} = m = p_{1}p_{2}$, hence $2q_{2} = 2p_{1}p_{2} - 4 = p_{1} + p_{2}$ and $p_{1} = \frac{p_{2} + 4}{2p_{2} - 1} = 1 + \frac{5 - p_{2}}{2p_{2} - 1}$, and this implies $p_{2} \leq 5$ because $p_{1}$ must be a natural number. Moreover, it was assumed that $p_{2}$ is odd. However, for $p_{2} = 3$ one would get $p_{1} = 1 + \frac{2}{5} = \frac{7}{5} $ and for $p_{2} = 5 $ one would get $p_{1} = 1$. \end{proof} A computer search showed that there are no natural solutions to $ x'' = x$ such that $x < 10000 $. This result can be improved, at least if the smaller of the numbers $m$ and $n$ is odd. \begin{proposition} \label{prop10} Let $n' = m$, $m' = n$ and let $n = 2j - 1 < m.$ Then there are at least nine primes in the factorization of $n$ and $n > 3\cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29$. \end{proposition} \begin{proof} It is known that $m$ and $n$ must be products of different primes $n = \prod_{i=1}^{k} p_{i} $ and $m = \prod_{j=1}^{l}q_{j} $, where all $p_{i}$ are distinct from all $q_{i}$. Because $n < m = n' = L(n)n$ we have $L(n) = \sum_{i=1}^{k}\frac{1}{p_{i}} > 1$. Because $n$ is odd, all $p_{i}$ are greater than 2. The sum of the reciprocals of the first eight odd primes is: $\frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{11} + \frac{1}{13} + \frac{1}{17} + \frac{1}{19} + \frac{1}{23} = 0.9987\cdots < 1$. Thus $k \geq 9 $ and $n > 3\cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29$. \end{proof} If $n = 2j 1, s > 1 $ we have $p_{r} > r$ and $q_{s} > s$, hence $p_{1}< r$ and $q_{1} < s.$ Let $r + s = N$. Then $rs \leq (\frac{r + s}{2})^{2} = (\frac{N}{2})^{2}$. Thus $2q_{s} \leq p_{1}q_{s} < (\frac{N}{2})^{2} $ and $2p_{r} \leq q_{1}p_{r} <(\frac{N}{2})^{2},$ hence $ 2\max\{p_{r},q_{s}\} <\frac{N^{2}}{4}$. Let $P_{i}$ denote the $i$-th prime (thus $P_{1} = 2$, $P_{2} = 3$, $P_{3} = 5$, etc.) Because all the primes $p_{1}, \ldots,p_{r},q_{1}, \ldots ,q_{s}$ are distinct, we have $\max\{p_{r},q_{s}\} \geq P_{r+s } = P_{N}.$ Thus $2P_{N} = 2P_{r+s} \leq 2\max\{p_{r},q_{s}\} < rs \leq \frac{N^{2}}{4}.$ It is possible to directly check that $P_{N} \geq \frac{N^{2}}{8}$ if $ N\leq 33$. {%\small \begin{center}\label{table1} \begin{tabular}{|c||c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline % after \\: \hline or \cline{col1-col2} \cline{col3-col4} ... N & $1$ & $2$ & $3$ & $4$ & $5$& $6$& $7$& $8$& $9$& $10$& $11$& $12$\\ \hline $P_{N}$ & $2$ & $3$ & $5$ & $7$ & $11$& $13$& $17$& $19$& $23$& $27$& $31$& $37$\\ $N^{2}$ & $1$ & $4$ & $9$ & $16$ & $25$& $36$& $49$& $64$& $81$& $100$& $121$& $144$ \\ \hline \end{tabular} \end{center} } {%\small \begin{center} \begin{tabular}{|c||c|c|c|c|c|c|c|c|c|c|c|} \hline % after \\: \hline or \cline{col1-col2} \cline{col3-col4} ... N &$13$ & $14$ & $15$ & $16$ & $17$ & $18$ & $19$ & $20$& $21$& $22$& $23$ \\ \hline $P_{N}$ & $41$ & $43$ & $47$ & $53$ & $59$& $61$& $67$& $71$& $73$& $79$& $83$\\ $N^{2}$ & $169$& $196$& $225$& $256$ & $289$ & $324$ & $361$ & $400$& $441$& $484$& $529$ \\ \hline \end{tabular} \end{center} } {%\small \begin{center} \begin{tabular}{|c||c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|} \hline % after \\: \hline or \cline{col1-col2} \cline{col3-col4} ... N & $24$& $25$& $26$& $27$& $28$ &$29$& $30$ & $31$ & $32$ & $33$ & $34$\\ \hline $P_{N}$ & $89$& $97$& $101$& $103$& $107$& $109$ &$113$ &$127$ &$131$ &$137$ &$139$\\ $N^{2}$ & $576$& $625$& $676$& $729$ &$784$& $841$& $900$ & $961$ & $1024$ & $1089$ & $1156$\\ \hline \end{tabular} \end{center} } \begin{center} Table 2: The first 34 primes $p_{N}$ and squares $N^{2}$ \end{center} Hence it must be that $N \geq 34.$ If $\min\{p_{1},q_{1}\} \geq 3$, one can get a much better estimate from the condition: $p_{N} < \frac{N^{2}}{12}$, which is first fulfilled when $N \geq57 $. If $\min\{p_{1},q_{1}\} \geq 5$, then $p_{N} < \frac{N^{2}}{20} $, which is first fulfilled when $N \geq 110$. \end{proof} \begin{proposition}\label{prop11a} Suppose $n' = m$ and $n' = m$ and $n,m$ are both odd. Let $n$ have $r_{1}$ primes $p_{i}\equiv 1$ (mod 4) and $s_{1} $ primes $q_{j}\equiv -1$ (mod 4) and let $m$ have $r_{2}$ primes $p_{i}\equiv 1$ (mod 4) and $s_{2} $ primes $q_{j}\equiv -1$ (mod 4). Then $(r_{1}-s_{1})(r_{2}-s_{2}) \equiv 1$ (mod 4). \end{proposition} \begin{proof} Barbeau \cite[pp.\ 121--122]{Barb} proved that if $n = p_{1}\cdot p_{2}\dots p_{r}\cdot q_{1}\cdot p_{2}\dots q_{s}$, where $p_{i}\equiv 1$ (mod 4), $q_{j}\equiv -1$ (mod 4) are primes, not necessarily distinct, then $D(n) \equiv (-1)^{s}(r - s)$ (mod 4). Hence $m = n' = (-1)^{s_{1}}(r_{1}-s_{1})$ and $n = m' = (-1)^{s_{2}}(r_{2}-s_{2})$. Because $m$ and $n$ are odd, $r_{1}-s_{1}$ and $r_{2}-s_{2}$ are not even. However, $n \equiv (-1)^{s_{1}}$ (mod 4) and $m \equiv (-1)^{s_{2}}$ (mod 4). Hence $mn \equiv (-1)^{s_{1}}(-1)^{s_{2}}\equiv (-1)^{s_{1}}(-1)^{s_{2}}(r_{1}-s_{1})(r_{2}-s_{2})$ (mod 4). Consequently $(r_{1}-s_{1})(r_{2}-s_{2}) \equiv 1$ (mod 4). \end{proof} \subsection{Integrals of natural and rational numbers} Any $a$ such that $a' = b$ is called an integral of $b$. The set of all such integrals is denoted as $I(b)$. The same number can have different integrals: $25' = (5^{2})' = 2\cdot5 = 10$ and $21' = 3\cdot7 = 3 + 7 = 10$. Because $p' = 1$ for any prime, $I(1) = \mathbb P$. It is shown that 1 is the only natural number with infinitely many integrals among the natural numbers. \begin{proposition}\label{prop12} i) Let $b < 2\cdot3\cdot5\cdots P_{n}$, where $P_{n}$ is the $n$-th consecutive prime and let $a' = b$, where $a \in \mathbb N$. Then $a = p_{1}^{n_{1}}\cdots p_{m}^{n_{m}} $, where $m\leq n$ (hence $a$ is divisible by at most $n$ primes $p_{i}$). ii) If $a' = b > 1$, where $a \in \mathbb N$, then $a \leq \max(2\cdot3\cdot5\cdots P_{n}, b^{bn})$. Consequently every $b > 1$ has at most a finite number of integrals $a \in \mathbb N$. \end{proposition} \begin{proof} For each natural number $b$ there is a $n \in \mathbb N$ such that $b < 2\cdot3\cdot5\cdots P_{n}$. If $a = \prod_{i=1}^{^{m}} p_{i}^{n_{i}}$ has more than $n$ different prime factors $p_{i}$, then each summand of $a' = (\sum_{i=1}^{m} \frac {n_{i}}{p_{i}})\cdot b $ has at least $n$ prime factors, and the smallest of them is not smaller than $2\cdot3\cdot5\cdots P_{n}$, therefore in that case $a' > 2\cdot3\cdot5\cdots P_{n}> b$, so it cannot be $a' = b$. Thus $a = p_{1}^{n_{1}}\cdots p_{m}^{n_{m}}$ and $m \leq n$. ii) If $a$ contains a factor $p^{p}$, then $b = a'\geq a$, hence $ a \leq b < 2\cdot3\cdot5\cdots P_{n}$. The other possibility is that all exponents of $a = p_{1}^{n_{1}}\cdots p_{m}^{n_{m}}$ are smaller than their primes: $n_{i}< p_{i}$. It is necessary to consider two cases: If $m = 1$ then $a = p_{1}^{n_{1}}$ and $a' = n_{1}p_{1}^{n_{1}-1} = \frac{n_{1}a}{q_{1}} = b$. Now $b > 1$ implies $n_{1} \geq 2$ thus $p_{1}$ divides $b$, hence $p_{1}\leq b$ and $a = \frac{bp_{1}}{n_{1}} \leq bq_{1} \leq b^{2}$. If $m \geq 2$ then $a = p_{1}^{n_{1}}\cdots p_{m}^{n_{m}} < p_{1}^{p_{1}}\cdots p_{m}^{p_{m}} < b^{p_{1} + \cdots + p_{m}} 1$. \end{proof} \begin{corollary}\label{coro1} If $a' = p$ and $p$ is a prime, then $a = p_{1}\cdots p_{m}$ and all $p_{i}< p$. There are some primes without integrals $a \in \mathbb N$; for example 2,3,17. \end{corollary} \begin{proof} If $a = p^{2}c$, then $a' = p\cdot(2 + pc')$ is not a prime. If $a = p_{1}\cdots p_{m}$, then $p = a' > p_{i}$. Because $2 < 2\cdot3 $ and $3 < 2\cdot 3$, the only candidates for integrals of 2 and 3 are numbers of the form $a = p\cdot q$, where $p,q \in \mathbb P$. However, then $(p\cdot q)' = p + q > 5$, hence 2 and 3 can have no integrals $a \in \mathbb N$. Ufnarovski and \AA hlander \cite{Ufna} found their integrals in rational numbers: $(-\frac{21}{16})' = 2 $, $(-\frac{13}{4})' = 3 $. Any integral of $17 $ has at most 3 different prime factors because $17 < 2 \cdot3 \cdot5$. It cannot have only two different prime factors because if $a = pq$ then $a' = p + q$ and the sum of any two primes is not 17. However, if $a = pqr$, then $(pqr)' = pq + pr + qr \geq 2 \cdot 3 + 2 \cdot 5 + 3 \cdot 5 = 31 > 17.$ Thus there is no integral of 17. \end{proof} The numbers $s \in \mathbb N$ not divisible by any square $c^{2}$ where $c > 1$ are called square-free numbers. They can be either products of different primes or equal to 1. The set of square-free numbers is denoted $\mathbb S$. \begin{definition} \emph{Let $a = p_{1}^{n_{1}} \cdots p_{m}^{n_{m}}$, where $p_{i} \in \mathbb P $. The number $n_{i}$ is called the exponent of a prime $p_{i}$ in $a$. Let us define the following functions of $a$:} $s(a)$ \emph{is the greatest square-free divisor of} $a$, \emph{such that} $\gcd(s(a), \frac {a}{s(a)}) = 1$, \emph{$p(a) = p_{1}\cdots p_{m} $ is the product of all prime factors of $a$}, \emph{$f(a) = \frac {a}{p(a)} = \frac {a}{p_{1} \cdots p_{m}}$, $r(a) = a\cdot p(a) = a\cdot( p_{1}\cdots p_{m})$,} \emph{$h_{\max}(a) = \max \{n_{1}, \ldots , n_{m}\}$, $h_{\min}(a) = min \{n_{1}, \ldots ,n_{m}\}$. } \noindent \emph{For $a = 1$ we define} \emph{$s(a) = p(a) = f(a) = r(a) = h_{\max}(a) = h_{\min}(a) = 1.$} \end{definition} From this definition it follows that if $a \neq s(a)$ then $ h_{min}(\frac {a}{p(a)}) \geq 2.$ \begin{proposition} \label{prop13} i) Let $a \in \mathbb N$. Then $a' = f(a)p(a)a^{*}$, $p(a)a^{*} \in \mathbb N$ and $a = r(f(a))s(a)$. ii) If $a \in \mathbb P$ then $p(a)a^{*} = 1 $ and $f(a) = a'.$ If $a \in \mathbb N$ is not a prime then $p(a)a^{*} > 1 $ and $f(a) < a'.$ iii) If $a = p_{1}^{n_{1}} \cdots p_{m}^{n_{m}} $ and $n_{i}$ is not divisible by $p_{i}$ for all $i \in \{1,2,\dots ,m\}$ then $\gcd(f(a),p(a)a^{*}) = 1$ and $\gcd(a,a') = f(a)$. iv) If $h_{\min}(a)\geq 2$, then $s(a) = 1$, hence $a = r(f(a))$. If $h_{\max}(a) = 1$ then $p(a) = a$ and $f(a) = 1$, hence $r(f(a)) = r(1) = 1$. \end{proposition} \begin{proof} i)\, If $a = 1$ then $a' = a^{*} = 0$ implies $a' = f(a)p(a)a^{*}$. If $a = p_{1}^{n_{1}} \cdots p_{m}^{n_{m}}$ then $a' = aa^{*} = (\frac {a}{p(a)})p(a)a^{*}$ and $\frac {a}{p(a)} = f(a) = p_{1}^{n_{1}-1} \cdots p_{m}^{n_{m}-1}$. Obviously $p(a)a^{*} = (p_{1}\cdots p_{m})(\frac {n_{1}}{p_{1}} +\cdots + \frac {n_{m}}{p_{m}}) \in \mathbb N$. Because $f(a)$ is divisible by exactly those primes $p_{i}$ for which $p_{i}^{2}$ divides $a$, it is $r(f(a)) = \frac{a}{s(a)}$. Hence $a = r(f(a))s(a)$. ii) If $a \in \mathbb P$ then $p(a) = a $, $a' = 1$, $a^{*} = \frac {1}{a}$, $p(a)a^{*} = a \frac {1}{a} = 1$ and $f(a) = 1.$ Now suppose $a \in \mathbb N$ is not a prime. Then either $a = p_{1}^{n_{1}}$, where $n_{1} > 1$, or $a = p_{1}^{n_{1}} \cdots p_{m}^{n_{m}}$, where $ m \geq 2$. In the first case $p(a) = p_{1}$, $f(a) = p_{1}^{n_{1} -1}$, $a' = n_{1}p_{1}^{n_{1} -1}$, $a^{*} = \frac {n_{1}}{p_{1}}$, hence $p(a)a^{*} = p_{1}\frac {n_{1}}{p_{1}} = n_{1} > 1$. In the second case, $p(a)a^{*} = (p_{1}\cdots p_{m})(\frac {n_{1}}{p_{1}} +\cdots + \frac {n_{m}}{p_{m}})$ is a sum of at least two natural numbers, hence $p(a)a^{*} > 1 $. Therefore in both cases $f(a) = \frac {a'}{p(a)a^{*}} < a'$. iii) Now $p(a)a^{*} = (p_{1}\cdots p_{m})(\frac {n_{1}}{p_{1}} +\cdots + \frac {n_{m}}{p_{m}})$ is not divisible by any of the primes $p_{i} $ dividing $a$ because $n_{i}$ is not divisible by $p_{i}.$ Hence $p(a)a^{*} = q_{1}^{u_{1}} \cdots q_{t}^{u_{t}} $ and $a' = p_{1}^{n_{1}-1} \cdots p_{m}^{n_{m}-1}q_{1}^{u_{1}} \cdots q_{t}^{u_{t}} $, where all the primes $p_{1},\cdots ,p_{m}, q_{1},\cdots ,q_{t}$ are distinct. Hence $\gcd(f(a),p(a)a^{*}) = 1 $ and $\gcd(a,a') = f(a)$. iv) If $h_{\min}(a)\geq 2$, then $p(a) = p(f(a))$, hence $r(f(a)) = a$. If $h_{\max}(a) = 1$ then $p(a) = a $ and $f(a) = 1$, hence $ r(f(a)) = 1$. \end{proof} \begin{definition}If $a = p_{1}^{n_{1}} \cdots p_{m}^{n_{m}} $ where all $n_{i}< p_{i} $ then $a$ is in the set $\mathbb L $ whose elements are called "low" numbers. \end{definition} Now it is possible to describe the factorization and consequently obtain some bounds of the integrals of "low" numbers. \begin{corollary} \label{coro2} i) If $b \in \mathbb L$ then $I(b) \bigcap \mathbb N \subset \mathbb L$. Moreover, every $a \in I(b) \bigcap \mathbb N$ is of the form $a = r(c)s(a)$, where $b = cd$, $c < b$, $\gcd(c,d) = 1$, and $b = a' = r'(c)s(a) + r(c)s'(a)$. Therefore: i.a) If $s(a) = a$ then $r(c) = 1$ hence $c = 1$. Conversely, if $c = 1$ then $s(a) = a$. i.b) If $s(a) = 1$ then $a = r(c)$. i.c) If $r(c) \neq 1 $ then $s(a) \leq \frac{b - r(c)}{r'(c)} \leq \frac{b}{c}$, hence $a \leq r(b) - r(r(c)) \leq r(b) - 1$. i.d) If $r(c) \neq 1 $ and $s(a) $ is a prime then $s(a) = \frac{b - r(c)}{r'(c)}$. i.e) If $r(c) \neq 1 $ and $s(a) = p_{1}p_{2}$ then $\frac{b - r(c)}{r'(c) + r(c)} \leq s(a) \leq \frac{b - 5r(c)}{r'(c)}$. ii) For every $b = p_{1}^{n_{1}} \cdots p_{k}^{n_{k}} \in \mathbb L$ there are at most $2^{k} - 1$ different divisors $c$ of $b$ such that there is an integral $a$ of $b$ of the form $a = r(c)s(a)$. iii) If $b = p_{1}^{n_{1}} \cdots p_{k}^{n_{k}}$ and $n_{i} = p_{i} - 1 $ for all $i$, then any integral $a \in I(b) \bigcap \mathbb N$ is a square-free number: $a = s(a)$. \end{corollary} \begin{proof} i) If $a' = b$ and $a$ contains a factor $p_{i}^{p_{i}}$ then $a' = b$ is also divisible by $p_{i}^{p_{i}}$, too, hence $b$ is not in $\mathbb L$. Thus, because $b \in \mathbb L$, it must be $a \in \mathbb L.$ Hence $f(a)$ divides $a' = b$, $f(a) < a'$, and $r(f(a)) = \frac {a}{s(a)}$. Thus one writes $c = f(a)$, one really can get all the integrals of $b \in \mathbb L$ in the described form. i.a) and i.b) are obvious. i.c) If $r(c) \neq 1 $ then $r'(c)\neq 0$ and $s(a) = b - \frac{r(c)s'(a)}{r'(c)} \leq \frac{b - r(c)}{r'(c)} = \frac{b}{(cp(c))'} = \frac{b}{c'p(c) + cp'(c)} \leq \frac{b}{c}$. Hence $a = r(c)s(a) \leq c\cdot p(c)\frac{b - r(c)}{c} \leq p(b)b - p(c)r(c) = r(b) - r(r(c)) \leq r(b) - 1$ because $p(c) = p(r(c))$. i.d) If $r(c) \neq 1 $ and $s(a)$ is a prime, then $s'(a) = 1$ and $s(a) = \frac{b - r(c)}{r'(c)}.$ i.e) If $r(c) \neq 1 $ and $s(a) = p_{1}p_{2}$ then $s'(a) = p_{1} + p_{2} \leq 2 + 3 = 5$ hence $ s(a) = \frac{b - 5r(c)}{r'(c)}$. The inequality $\frac{b - r(c)}{r'(c) + r(c)} \leq s(a)$ follows from the fact that $p_{1} + p_{2}\leq p_{1}p_{2} $ for any two primes $p_{1}$ and $p_{2}$ implying $a' = (r(c)p_{1}p_{2})' = r'(c)p_{1}p_{2} + r(c)(p_{1} + p_{2}) \leq (r'(c) + r(c))p_{1}p_{2}.$ ii) Because each factor $p_{i}^{n_{i}}$ of $b = cd = p_{1}^{n_{1}} \cdots p_{k}^{n_{k}}$ belongs either to $c = f(a)$ or $d = \frac{b}{f(a)}$, and because $c < a'$, there are at most $2^{k} - 1$ possible factors $c$ dividing $b = a'$ such that $a = r(c)s(a)$. However, different integrals of $b$ may correspond to the same $c$, as in the example $(3\cdot13)' = (5\cdot11)' = 16$, in which both integrals $3\cdot13 $ and $5\cdot11$ correspond to $c = 1$. iii) If $p_{i} = n_{i} - 1 $ for all $i$, then $r(c) = 1$ or $r(c)$ contains at least one factor $p_{i}^{p_{i}}$, but then this is a contradiction because in that case $a = b'$ would also be divisible by $p_{i}^{p_{i}}$. Hence $r(c) = 1$ and $a = s(a)$. \end{proof} \begin{example} Let $b = 5^{2}11$. There are three possible factorizations $b = cd$ such that $\gcd(c,d)= 1$ and $1 \leq c < b$, corresponding to numbers $c_{1} = 1$, $c_{2} = 5^{2}$ and $c_{3} = 11$. Thus the candidates for $a \in I(b) \bigcap \mathbb N$ are: square-free numbers $a = r(1)s(a) = s(a) 2\cdot5^{2}11$, not divisible by $5$ or $11$, numbers of the form: $a = r(5^{2})s(a) = 5^{3}s(a)$, where $s(a) \leq \frac{b}{r'(c)} = \frac{5^{2}11}{3\cdot 5^{2}} = \frac{11}{3} \leq 4$, hence $s(a) \in \{1,2,3\}$, and numbers of the form $a = r(11)s(a) = 11^{2}s(a)$, where $s(a) \leq \frac{b}{r'(c)} = \frac{5^{2}11}{2\cdot 11} = \frac {5^{2}}{2}$, hence $s(a) \in \{1,2,3,5,6,7,10,11\}$. It can easily be seen that $s(a) > 1 $ because $s(a) = 1$ implies $r'(c) = b$, but it is $1' = 0 \neq b$, $(5^{3})' = 3\cdot5^{2} \neq b$ and $(11^{2})' = 2\cdot11 \neq b$. Thus in the case $c_{2} = 5^{2}$ it must be $s(a) \in \{2,3\} $ and in the case of $c_{3} = 11$ it is seen that $s(a) \in \{2,3,5,6,7,10,11\}$ is either a prime or a product of two primes. Hence in the case $a = 5^{3}s(a)$ one can use the formula for the prime $s(a) = \frac{b - r(c)}{r'(c)} = \frac{5^{2}11 - 5^{3}}{3\cdot5^{2}} = \frac{11 - 5} {3} = 2$ and one can verify directly that $(2\cdot5^{3})' = 5^{3} + 2\cdot3\cdot5^{2} = 5^{2}(5 + 6) = b$. Now let $a = 11^{2}s(a)$. If $s(a)$ is a prime one gets $s(a) = \frac{b - r(c)}{r'(c)} = \frac{5^{2}11 - 11^{2}}{2\cdot 11)} = \frac {25 - 11}{2} = 7$ and one can check directly that $(11^{2}\cdot7)' = 2\cdot11\cdot7 + 11^{2} = 11(14 + 11 ) = b$. It can easily be seen that $s(a)$ cannot be a product of two primes because that would imply a contradiction: $0 \leq s(a) \leq \frac{b - 5r(c)}{r'(c)} = \frac{5^{2}11 -5\cdot11^{2}}{2\cdot11} < 0$. \end{example} Square-free integrals corresponding to $c_{1} = 1$ are more difficult to find. It is necessary to check the derivatives of all square-free numbers $a \leq \frac{b^{2}}{4}$ and compare them with $b$. A similar estimate as in Proposition~\ref{prop10} can be made about positive solutions to the equation $(\frac{a}{b})' = \frac{2}{p}$, where $p \in \mathbb P $. \begin{definition} Let $P_{i+1}$ denote the $i$-th odd prime. For any $d \in \mathbb R$ let $O_{d} = 1\cdot\prod_{i=1}^{m} P_{i+1}$ denote the product of the first $m$ odd primes such that the sum of their reciprocals, denoted $R(d)$, is not smaller than $d$. \end{definition} Thus $R(d) = \sum_{i=1}^{m} \frac {1}{P_{i+1}} \geq d$. Because the series $\sum_{i=1}^{\infty}\frac{1}{P_{i+1}}$ diverges, $O_{d}$ is well defined for any real $d$. \begin{example} If $d \leq \frac{1}{3}$ then $O_{d} = P_{2} = 3.$ If $\frac{1}{3} \leq d \leq \frac{1}{3} + \frac{1}{5}$ then $O_{d} = P_{2}P_{3} = 3\cdot5 = 15.$ If $\frac{1}{3} + \frac{1}{5}\leq d \leq \frac{1}{3} + \frac{1}{5} + \frac{1}{7}$ then $O_{d} = P_{2}P_{3}P_{4} = 3\cdot5\cdot7 = 105.$ \end{example} \begin{proposition}\label{prop14} Let $(\frac{a}{b})' = \frac{2}{p}$, where $p \in \mathbb P$, $p> 2$, $\gcd(a,b) = 1$, $\frac{a}{b} > 0$. Then $a = p_{1}\cdots p_{m}$ is an odd square-free number with $m \geq 9$ prime factors $p_{i} \in \mathbb P$ and $b$ is of the form $b = q_{1}^{q_{1}n_{1}} \dots q_{s}^{q_{s}n_{s}}$ where $q_{i} \in \mathbb P $ and $n_{i} \in \mathbb N $. Moreover, $L(b) \in \mathbb N$, $L(a) > L(b) \geq 1$ and $a' > a > O_{1} \geq 3\cdot5\cdot7\cdot11\cdot13\cdot17\cdot19\cdot23\cdot29 $. \end{proposition} \begin{proof} $(\frac{a}{b})' = \frac{2}{p}$ implies $\frac{a'b - ab'}{b^{2}} = \frac{2}{p}$, hence $b = pc$, where $c \in \mathbb N$. Therefore $a'pc - ab' = 2c^{2}p$ and because $a$ and $b$ have no common prime factors, we have $b' = kc$. Hence $a'pc - akc = 2c^{2}p$ and by dividing this equation by $c$ one gets $a'p - ak = 2cp,$ hence $k = pd$ because $p$ divides $b$ so it cannot divide $a$. This implies $a'p - apd = 2cp$, hence $a' - ad = 2c$, and $b' = pdc = bd$. Thus $d = L(b) = b^{*}$. By Proposition~\ref{prop2}, $b = q_{1}^{q_{1}n_{1}} \cdots q_{s}^{q_{s}n_{s}} $, where $n_{1} + \cdots + n_{s} = d.$ Because $\frac{a}{b} > 0$, it can be assumed $a > 0$ and $b > 0$. It must be $a > 1$ because $a = 1$ implies $a' = 0$ and $-d = 2c$, implying a contradiction: $0 = 2c + d > 0$. From the equation $a' - ad = 2c$ follows $\gcd(a,a') = 1$, because any such common prime factor different from 2 would also divide $c$ and $b$ and this would contradict $\gcd(a,b) = 1$, and if 2 divides $a$ and $a'$ then it must be $a = 4e$, hence $a' = 4(e + e') $ and $4(e' + e - 4 ed) = 2c$ would imply that 2 also divides $b$. Now it can be seen that $a$ cannot be divisible by $2$. In that case $2$ would also divide $a'$ or $p$. However, this is impossible because $\gcd(a',a) = 1$ and $p > 2$. Because $\gcd(a,a') = 1$, $a$ must be a square-free number. Because $a > 1$, it must be: $a = p_{1}\cdots p_{m}$, and because $ a' - ad = 2c > 0$ it is $a' > ad \geq a $, hence $L(a) = \frac{a'}{a} = \sum_{i=1}^{m}\frac{1}{p_{i}} > d \geq 1$. Therefore $a > O_{d} \geq O_{1} \geq 3\cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29$ and $ m \geq 9$, because the sum of the reciprocals of the first eight odd primes is less than 1: $\frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{11} + \frac{1}{13} + \frac{1}{17} + \frac{1}{19} + \frac{1}{23} = 0.9987\cdots < 1$. \end{proof} Ufnarovski and \AA hlander \cite{Ufna} conjectured that some rational numbers have no integrals. The above proposition shows that the integrals of $\frac{2}{3}$ are not easy to find, if they exist at all. The same holds for any $\frac{2}{p}, p \in \mathbb P.$ In Proposition~\ref{prop4}.iii) I showed that if $y \in N_{y^{*}} = N_{r+r^{*}}$ and if $r + r^{*} \neq 0$ then there is exactly one $x \in N_{r}$ such that $x' = y$ and this is $x = \frac{y}{r}$. Hence to find a rational integral $x$ of any nonzero rational number $y$ one just has to find a rational number $r = x^{*}$ such that $r + r^{*} = y^{*}$. In other words, one has to find a logarithmic class $N_{r}$ such that $D(N_{r}) = N_{y^{*}}$. The equation $r + r^{*} = y^{*}$ translates into $r^{2} - y^{*}r + r' = 0$. Perhaps for some rational numbers $y$ this equation cannot be satisfied by any rational number $r$, hence such $y$ cannot have a rational integral $x$. Expressing $r$ as a function of its derivative $r'$, one gets at most two different solutions $r_{1,2} = \frac{y^{*} \pm \sqrt{(y^{*})^{2} - 4r'}}{2}$. Now $ r \in \mathbb Q$ implies $(y^{*})^{2} - 4r' \geq 0$ and $r' \leq \frac{(y^{*})^{2}}{4}.$ Because $x \in \mathbb Q$ implies $r = x^{*} \in \mathbb Q$ the expression $(y^{*})^{2} - 4r' = q^{2}$ must be a square of a rational number $q$. \begin{example} If $y = \frac{2}{3}$ then $y^{*} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$. If there is a positive rational number $x = \frac{a}{b} \in N_{r}$ such that $ x' = y = \frac{2}{3}$ then $r = x^{*} = (\frac{a}{b})^{*} = \frac{y}{x} = \frac{2}{3x}> 0$ satisfies the equation $r + r^{*} = y^{*} = \frac{1}{6}$. This equation translates into $r^{2} - \frac{1}{6}r + r' = 0$. For each $r'$ one gets at most two different solutions $r_{1,2} = \frac{y^{*} \pm \sqrt{(y^{*})^{2} - 4r'}}{2} = \frac{\frac{1}{6} \pm \sqrt{\frac{1}{36} - 4r'}}{2} = \frac{1}{12} \pm \sqrt{\frac{1}{144} - r'}$. Hence $r' \leq \frac{(y^{*})^{2}}{4} = \frac{1}{144}$. It is known from Proposition~\ref{prop14} that $r = a^{*} - b^{*} = \sum_{i=1}^{m}\frac{1}{p_{i}} - d > 0$, where $ d \in \mathbb N$. Thus $x = \frac{y}{r} > 0$. Now $r' = (\frac{2}{3x})' = \frac{3x - 6}{9x^{2}} = \frac{x - 2}{3x^{2}} = \frac{x - 2}{3x^{2}}\frac{3x}{2}\frac{2}{3x}$, hence $r^{*} = \frac{x - 2}{2x}$. If $x \leq 2$ then $r^{*} \leq 0 $ and $r = y^{*} - r^{*} \leq y^{*} = \frac{1}{6} $ leads to a contradiction: $x = \frac{y}{r} \geq \frac{y}{y^{*}} = \frac{\frac{2}{3}}{\frac{1}{6}} = 4$. Thus it must be $x > 2$, hence $r^{*} > 0$. Now $0 < r = y^{*} - r^{*} < y^{*} = \frac{1}{6} $ implies $x = \frac{y}{r} > \frac{y}{y^{*}} = \frac{\frac{2}{3}}{\frac{1}{6}} = 4$, hence $a > 4b$. Thus there is no positive rational number $x$ such that $x \leq 4$ and $x' = \frac{2}{3}$.\end{example} \subsection{Partial derivatives and partial differential equations} \begin{definition} \emph{Let $a = \prod_{i=1}^{k}p_{i}^{\, x_{i}}$ be the factorization of $a \in \mathbb N$ into primes. The partial derivative $\frac{\partial a} {\partial pi} = D_{p_{i}}(a)$ is defined as $\frac{\partial a} {\partial p_{i}} = a\frac{x_{i}}{p_{i}}.$ If $p \in \mathbb P$ is not in the factorization of $a$, then $\frac{\partial a} {\partial p_{i}} = 0.$} \end{definition} From this definition it immediately follows that $D(a) = a\sum_{i=1}^{k}\frac{\partial a} {\partial p_{i}}. $ One can also define higher partial derivatives; for example: $\frac{\partial ^{2} a}{\partial p_{i}\partial p_{j}} = D^{2}_{p_{i}p_{j}} = $ $\frac{\partial (\partial a/\partial p_{j})}{\partial p_{i}} $, $\frac{\partial ^{2} a}{\partial p_{i}\partial p_{j}} = D^{2}_{p_{i}^{2}} = $ $\frac{\partial (\partial a/\partial p_{i})}{\partial p_{i}} $, etc., and study partial differential equations. \begin{example} $D_{2}(2^{3}\cdot5^{4}) = 3\cdot2^{2}\cdot5^{4} $, $D_{5}(3\cdot2^{2}5^{4} ) = 3\cdot2^{2}\cdot4\cdot5^{3} = 3\cdot2^{4}\cdot5^{3} $ \noindent $D_{5}(2^{3}\cdot5^{4} ) = 2^{3}\cdot4\cdot5^{3} = 2^{5}\cdot5^{3}$ and $D_{2}(2^{5}\cdot5^{3} ) = 5\cdot2^{4}\cdot5^{3}$. Thus the order of applying $D_{p} $ and $D_{q}$ is important: $D_{p}D_{q}$ is not always equal to $D_{q}D_{p} $. \end{example} \begin{proposition}\label{prop15} If $a = p^{e}q^{f} $ and $\gcd(e,q) = 1$ and $ \gcd(f,p) = 1$, then $D_{p}D_{q}(a) = D_{q}D_{p}(a). $ \end{proposition} \begin{proof} In that case it is $D_{p}D_{q}(p^{e}q^{f}) = D_{p}(p^{e}fq^{f-1}) = ep^{e-1}fq^{f-1}$ and $D_{q}D_{p}(p^{e}q^{f} ) = D_{q}(ep^{e-1}q^{f} ) = ep^{e-1}fq^{f-1}$. \end{proof} \begin{proposition}\label{prop16} Let $n = p^{e}q^{f}c$, where $p,q$ are primes not dividing $c$. Then $D_{p}(p^{e}q^{f}c) = D_{q}(p^{e}q^{f}c) $ if and only if $e = kp $ and $f = kq$, where $ k \in \mathbb N$. \end{proposition} \begin{proof} From $D_{p}(p^{e}q^{f}c)= ep^{e-1}q^{f}c = fp^{e}q^{f-1}c = D_{q}(p^{e}q^{f}c) $ follows $eq = bp$, hence $e = kp$ and $f = kq$. Then $D_{p}(p^{kp}q^{kq}c) = k(p^{kp}q^{kq}c) = D_{q}(p^{e}q^{f}c).$ \end{proof} \begin{proposition}\label{prop17} i) The only solutions to the partial differential equation $D^{2}_{p^{2}}(n) = n$ in natural numbers are $n = p^{pk}c$, where $\gcd(c,p) = 1$. ii) If $n = p^{e}d$ where $\gcd(p,d) = 1$ and $e \geq p$ then $D^{2}_{p^{2}}(n)\geq n.$ iii) The only solutions to the partial differential equation $(D^{2}_{p^{2}} + D^{2}_{q^{2}})(n) = n$ in natural numbers are $n = p^{p}c$ and $n = q^{q}c$, where $\gcd(pq,c) = 1$. \end{proposition} \begin{proof} i) Let $n = p^{e}c$, where $e \in \mathbb N \bigcup \{0\}$ and $c \in \mathbb N$ is not divisible by $p$. If $e = 0$ then $D_{p}(n) = 0$, hence $D^{2}_{p^{2}}(n) = 0$. If $e = kp$ and $k \in \mathbb N$ then $D_{p}(n) = D_{p}(p^{pk}c) = pkp^{pk-1}c = p^{pk}c = n$ hence $D^{2}_{p^{2}}(n) = n$. If $e = 1$ then $D_{p}(n) = c$, hence $D^{2}_{p^{2}}(n) = D_{p}(c) = 0$. ii) If $e = kp$ and $k \in \mathbb N$ then $D^{2}_{p^{2}}(n) = n$, as is already known from i). If $\gcd(e,p) = 1$ and $e > p$ then $D_{p}(n) = D_{p}(p^{e}c) = ep^{e-1}c$, hence $D^{2}_{p^{2}}(n) = D_{p}(ep^{e-1}c) = e(e - 1)p^{e-2}c \geq (p + 1)pp^{e-2}c > p^{e}c = n$. iii). One can write $n = p^{e}q^{f}c $, where $\{e,f\} \subseteq \mathbb N \bigcup \{0\}$ and $c \in \mathbb N$ is not divisible by $p$ or $q$. So it is necessary to study the equation $(D^{2}_{p^{2}} + D^{2}_{q^{2}})( p^{e}q^{f}c) = p^{e}q^{f}c. $ If $f = 1$ then $D^{2}_{q^{2}}(p^{e}c) = 0$ and one gets the equation $D^{2}_{p}(n) = n$ whose only solution is $n = p^{p}c $, as is known from i). Likewise if $e = 1$ one gets $n = q^{q}c. $ In the case of $e \geq 2$ and $f \geq 2$, one can use the following argument: Obviously $\gcd(qc,p) = 1$ implies that $D^{2}_{p^{2}}(n)$ is divisible by $q^{f}c$. Likewise $\gcd(pc,q) = 1$ implies that $D^{2}_{q^{2}}(n)$ is divisible by $p^{e}c$. So one can write $D^{2}_{p^{2}}(n) = aq^{f}c$, $D^{2}_{q^{2}}(n) = bp^{e}c$, where $a,b \in \mathbb N \bigcup \{0\}$. Suppose $a \neq 0$ and $b \neq 0$. Hence one gets the equation $aq^{f}c + bp^{e}c = p^{e}q^{f}c$. This equation can be solved only if $a$ is divisible by $p^{e}$ and if $b$ is divisible by $q^{f}$. Suppose $a \neq 0$ and $b \neq 0$. Then $D^{2}_{p^{2}}(n) \geq n$ and $D^{2}_{q^{2}}(n) \geq n$, hence $D^{2}_{p^{2}}(n) + D^{2}_{q^{2}}(n)\geq 2n$ and the equality is not possible. Hence it must be either $a = 0$ or $b = 0$. However, this is not possible if $e \geq 2$ and $f \geq 2$. \end{proof} \section{Concluding remarks} I have proved some new results about the arithmetic derivative and integral. I have defined arithmetic partial derivatives and solved some arithmetic partial differential equations. I have shown that, for any solution to the system $m' = n, n' = m$ in natural numbers, at least one of the numbers $m$ and $n$ is not smaller than the number $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 31 \cdot 37 \cdot41 \cdot 43 \cdot 47 \cdot 53 \cdot 59$. The arithmetic derivative can be defined on sequences of numbers as follows: $D(a_{1},a_{2}, \ldots , a_{n}, \ldots ) = (D(a_{1}),D(a_{2}), \ldots , D(a_{n}), \ldots ) ).$ Thus taking any integer sequence $(a) = (a_{1},a_{2}, \ldots , a_{n}, \ldots )$ one can get an infinite family of derived sequences $D(a), D^{2}(a), \ldots, D^{k}(a), \ldots .$ I believe many other useful applications of the arithmetic derivative will be discovered. \section{Acknowledgements} I thank the referee for many suggestions that helped me to improve this paper. I am also grateful to Marko Petkov\v{s}ek for his explanation of why there are uncountably many functions on the unit circle in the complex plane satisfying the Leibnitz rule (Proposition 1.ii). \begin{thebibliography}{9} \bibitem{Barb} E. J. Barbeau, Remarks on arithmetic derivative, \emph{Canad. Math. Bull}, \textbf{4} (1961), 117--122. \bibitem{Stay} M. Stay, Generalized number derivatives, \emph{J. Integer Seq.}, \textbf{8} (2005), \href{http://www.cs.uwaterloo.ca/journals/JIS/VOL8/Stay/stay44.html}{Article 05.1.4}. \bibitem{Ufna} V. Ufnarovski and B. \AA hlander, How to differentiate a number, \emph{J. Integer Seq.}, \textbf{6} (2003), \href{http://www.cs.uwaterloo.ca/journals/JIS/VOL6/Ufnarovski/ufnarovski.html}{Article 03.3.4}. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2010 {\it Mathematics Subject Classification}: Primary 11A25; Secondary 11A41, 11N05, 11N56, 11Y55. \noindent \emph{Keywords: } arithmetic derivative, arithmetic antiderivative, arithmetic partial derivative, Leibnitz rule. \bigskip \hrule \bigskip \noindent (Concerned with sequences \seqnum{A000040}, \seqnum{A000290}, \seqnum{A051674}.) \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received May 19 2011; revised version received March 18 2012. Published in {\it Journal of Integer Sequences}, March 25 2012. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}