\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \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} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.5in} \setlength{\textheight}{8.9in} \newcommand{\seqnum}[1]{\href{http://www.research.att.com/cgi-bin/access.cgi/as/~njas/sequences/eisA.cgi?Anum=#1}{\underline{#1}}} \usepackage{epsfig} \usepackage{latexsym} \usepackage{amsfonts} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \begin{center} \vskip 1cm {\LARGE\bf How to Differentiate a Number} \vskip .3in \large Victor Ufnarovski \\ Centre for Mathematical Sciences \\ Lund Institute of Technology \\ P.O. Box 118 \\ SE-221 00 Lund \\ Sweden\\ \href{mailto:ufn@maths.lth.se}{ufn@maths.lth.se}\\ \vskip .1in Bo \AA hlander \\ KTH/2IT\\ Electrum 213 \\ 164 40 Kista \\ Sweden \\ \href{mailto:ahlboa@isk.kth.se}{ahlboa@isk.kth.se}\\ \end{center} \vskip .25in \newcommand{\N}{\bf N } \newcommand{\Pf}{\noindent{\it Proof. }} \newcommand{\fP}{{$\rule{2mm}{2mm} $ \medskip }} \newcommand{\ld}{\mbox{ld}} \newtheorem{Thm}{Theorem} \newtheorem{Lm}{Lemma} \newtheorem{Crl}{Corollary} \newtheorem{Cnj}{Conjecture} \noindent{\bf Abstract.} We define the derivative of an integer to be the map sending every prime to 1 and satisfying the Leibnitz rule. The aim of the article is to consider the basic properties of this map and to show how to generalize the notion to the case of rational and arbitrary real numbers. We make some conjectures and find some connections with Goldbach's Conjecture and the Twin Prime Conjecture. Finally, we solve the easiest associated differential equations and calculate the generating function. \vskip .25in \section{A derivative of a natural number} \label{sec:natural} Let $n$ be a positive integer. We would like to define a derivative $n'$ such that $(n,n')=1$ if and only if $n$ is square-free (as is the case for polynomials). It would be nice to preserve some natural properties, for example $(n^k)'=kn^{k-1}n'.$ Because $1^2=1$ we should have $1'=0$ and $n'=(1+1\cdots+1)'=0$, if we want to preserve linearity. But if we ignore linearity and use the Leibnitz rule only, we will find that it is sufficient to define $p'$ for primes $p.$ Let us try to define $n'$ by using two natural rules: \begin{itemize} \item $p'=1$ for any prime $p,$ \item $(ab)'=a'b+ab'$ for any $a,b\in \N$ (Leibnitz rule). \end{itemize} For instance, $$6'=(2\cdot 3)'=2'\cdot 3+ 2\cdot 3'=1\cdot 3 +2\cdot 1 = 5.$$ Here is a list of the first 18 positive integers and their first, second and third derivatives: $$\begin{array}{|c|ccccc ccccc ccccc ccc| } \hline n & 1 & 2& 3& 4& 5& 6& 7& 8& 9& 10& 11& 12 & 13& 14& 15& 16& 17 & 18\\ \hline n' &0 & 1& 1& 4& 1& 5& 1&12& 6& 7& 1& 16 & 1 & 9& 8& 32 & 1 & 21\\ n''& 0 & 0& 0& 4& 0& 1& 0&16& 5& 1& 0& 32 & 0 & 6& 12& 80 & 0 & 10\\ n'''&0 & 0& 0& 4& 0& 0& 0&32& 1& 0& 0& 80 & 0 & 5& 16& 176 & 0 & 7\\ \hline \end{array} $$ It looks quite unusual but first of all we need to check that our definition makes sense and is well-defined. \begin{Thm} \label{Thm1} The derivative $n'$ can be well-defined as follows: if $n=\prod_{i=1}^{k}{p_i^{n_i}}$ is a factorization in prime powers, then \begin{equation} \label{eq1}n'=n\sum_{i=1}^{k}{\frac{n_i}{p_i}}.\end{equation} It is the only way to define $n'$ that satisfies desired properties. \end{Thm} \Pf Because $1'=(1\cdot 1)'=1'\cdot 1+ 1\cdot 1'= 2\cdot 1',$ we have only one choice for $1'$ : it should be zero. Induction and Leibnitz rule show that if the derivative is well-defined, it is uniquely determined. It remains to check that the equation (\ref{eq1}) is consistent with our conditions. It is evident for primes and clear that (\ref{eq1}) can be used even when some $n_i$ are equal to zero. Let $a=\prod_{i=1}^{k}{p_i^{a_i}}$ and $b=\prod_{i=1}^{k}{p_i^{b_i}}.$ Then according to (\ref{eq1}) the Leibnitz rule looks as $$ab\sum_{i=1}^{k}{\frac{a_i+b_i}{p_i}} =\left( a\sum_{i=1}^{k}{\frac{a_i}{p_i}}\right) b +a\left( b\sum_{i=1}^{k}{\frac{b_i}{p_i}}\right)$$ and the consistency is clear. \fP For example $$(60)'=(2^2\cdot 3 \cdot 5)'= 60\cdot\left(\frac{2}{2}+ \frac{1}{3}+\frac{1}{5}\right) =60+20+12=92.$$ We can extend our definition to $0'=0$, and it is easy to check that this does not contradict the Leibnitz rule. Note that linearity does not hold in general; for many $a,b$ we have $(a+b)' \neq a'+b'.$ Furthermore $(ab)''\neq a''+2a'b'+b''$ because we need linearity to prove this. It would be interesting to describe all the pairs $(a,b)$ that solve the differential equation $(a+b)' = a'+b'.$ We can find one of the solutions, $(4,8)$ in our table above. This solution can be obtained from the solution $(1,2)$ by using the following result. \begin{Thm} If $(a+b)' = a'+b',$ then for any natural $k,$ we have $$(ka+kb)' = (ka)'+(kb)'.$$ The same holds for the inequalities $$(a+b)' \ge a'+b' \Rightarrow (ka+kb)' \ge (ka)'+(kb)',$$ $$(a+b)' \le a'+b' \Rightarrow (ka+kb)' \le (ka)'+(kb)'.$$ Moreover, all these can be extended for linear combinations, for example: $$(\sum \gamma_ia_i)'=\sum \gamma_i(a_i)' \Rightarrow (k\sum \gamma_ia_i)'=\sum \gamma_i(ka_i)'.$$ \end{Thm} \Pf The proof is the same for all the cases, so it is sufficient to consider only one of them, for example the case $\ge$ with two summands: $$(ka+kb)' =(k(a+b))'=k'(a+b)+k(a+b)'=$$ $$k'a+k'b+k(a+b)'\ge k'a+k'b+ka'+kb'= (ka)'+(kb)'.$$ \fP \begin{Crl} $$(3k)'=k'+ (2k)'; (2k)'\ge 2k'; (5k)'\le (2k)'+(3k)';(5k)'= (2k)'+3(k)'.$$ \end{Crl} \Pf $$ 3'=1'+ 2'; 2'\ge 1'+1'; 5'\le 2'+3';5'=2'+3\cdot 1'.$$ \fP Here is the list of all $(a,b)$ with $a1,$ $$n'\ge n\Rightarrow (kn)'>kn.$$ \end{Thm} \Pf $$(kn)' =k'n+kn'>kn'\ge kn.$$ \fP The following theorem shows that every $n>4$ that is divisible by $4$ satisfies the condition $n'>n.$ \begin{Thm}\label{thm:infp} If $n=p^p\cdot m$ for some prime $p$ and natural $m>1,$ then $n'=p^p(m+m')$ and $\lim_{k\rightarrow \infty} n^{(k)}= \infty.$ \end{Thm} \Pf According to the Leibnitz rule and (\ref{eq1}), $n'=(p^p)'\cdot m+ p^p\cdot m'=p^p(m+m')>n$ and by induction $ n^{(k)} \ge n+k.$ \fP The situation changes when the exponent of $p$ is less than $p.$ \begin{Thm}\label{thm:periods} Let $p^k$ be the highest power of prime $p$ that divides the natural number $n.$ If $01.$ On the other hand, if $p|n$ and $p|n'$ then $p^2|n.$ \fP \section{The equation $n'=n$} Let us solve some differential equations (using our definition of derivative) in positive integers. \begin{Thm} The equation $n'=n$ holds if and only $n=p^p,$ where $p$ is any prime number. In particular, it has infinitely many solutions in natural numbers. \end{Thm} \Pf If prime $p$ divides $n,$ then according to Theorem \ref{thm:periods}, at least $p^p$ should divide $n$ or else $n'\neq n.$ But Theorem \ref{thm:infp} implies that in this case $n=p^p,$ which according to (\ref{eq1}) is evidently equal to $n'.$ \fP Thus, considering the map $n\longrightarrow n'$ as a dynamical system, we have a quite interesting object. Namely, we have infinitely many fixed points, $0$ is a natural attractor, because all the primes after two differentiations become zero. Now it is time to formulate the first open problem. \begin{Cnj} \label{Cnj-main} There exist infinitely many composite numbers $n$ such that $n^{(k)}=0$ for sufficiently large natural $k.$ \end{Cnj} As we will see later, the Twin Prime Conjecture would fail if this conjecture is false. Preliminary numerical experiments show that for non-fixed points either the derivatives $n^{(k)}$ tend to infinity or become zero; however, we do not know how to prove this. \begin{Cnj} \label{Cnj-alt}Exactly one of the following could happen: either $n^{(k)}= 0$ for sufficiently large $k,$ or $\lim_{k\rightarrow \infty} n^{(k)}= \infty,$ or $n=p^p$ for some prime $p.$ \end{Cnj} According to Theorem \ref{thm:infp}, it is sufficient to prove that, for some $k$, the derivative $n^{(k)}$ is divisible by $p^p$ (for example by $4).$ In particularly we do not expect periodic point except fixed points $p^p$. \begin{Cnj}\label{cnj:periods} The differential equation $n^{(k)}=n$ has only trivial solutions $p^p$ for primes $p.$ \end{Cnj} Theorem \ref{thm:periods} gives some restrictions for possible nontrivial periods: if $p^k$ divides $n$ the period must be at least $k+1.$ Conjecture \ref{cnj:periods} is not trivial even in special cases. Suppose, for example, that $n$ has period $2,$ i.e. $m=n'\neq n$ and $ m'=n.$ According to Theorem \ref{thm:infp} and Theorem \ref{thm:periods}, both $n$ and $m$ should be the product of distinct primes: $n=\prod_{i=1}^{k}p_i,$ $m=\prod_{j=1}^{l}q_j,$ where all primes $p_i$ are distinct from all $q_j.$ Therefore, our conjecture in this case is equivalent to the following: \begin{Cnj}\label{cnj:period2}For any positive integers $k,l,$ the equation $$\left( \sum_{i=1}^{k}\frac{1}{p_i}\right)\left( \sum_{j=1}^{l}\frac{1}{q_j}\right)=1$$ has no solutions in distinct primes. \end{Cnj} \section{The equation $n'=a$} We start with two easy equations. \begin{Thm} The differential equation $n'=0$ has only one positive integer solution $n=1.$ \end{Thm} \Pf Follows immediately from (\ref{eq1}). \fP \begin{Thm}\label{Thmn1} The differential equation $n'=1$ in natural numbers has only primes as solutions. \end{Thm} \Pf If the number is composite then according to Leibnitz rule and the previous theorem, the derivative can be written as the sum of two positive integers and is greater than 1. \fP All other equations $n'=a$ have only finitely many solutions, if any. \begin{Thm}(\cite{Bar}) For any positive integer $n$ \begin{equation}\label{eqle} n' \le \frac{n\log_2 n}{2}.\end{equation} If $n$ is not a prime, then \begin{equation}\label{eqge} n' \ge 2 \sqrt{ n}.\end{equation} More generally, if $n$ is a product of $k$ factors larger than $1$, then \begin{equation}\label{eqgge}n'\ge kn^{\frac{k-1}{k}}.\end{equation} \end{Thm} \Pf If $n=\prod_{i=1}^{k}{p_i^{n_i}}$, then $$n\ge\prod_{i=1}^{k}{2^{n_i}}\Rightarrow \log_2 n \ge\sum_{i=1}^{k}{n_i}.$$ According to (\ref{eq1}) we now have $$ n'=n\sum_{i=1}^{k}{\frac{n_i}{p_i}}\le \frac{n\sum_{i=1}^{k}{n_i}}{2} \le \frac{ n\log_2 n}{2}.$$ If $n=n_1n_2n_3\cdots n_k$ then, according to the Leibnitz rule, $$n'=n_1'n_2n_3\cdots n_k+n_1n_2'n_3\cdots n_k+n_1n_2n_3'\cdots n_k+\ldots +n_1n_2n_3\cdots n_k'\ge$$ $$n_2n_3n_4\cdots n_k+n_1n_3n_4\cdots n_k+n_1n_2n_4\cdots n_k+\ldots+n_1n_2\cdots n_{k-1}=$$ $$n\left(\frac{1}{n_1}+\frac{1}{n_2}+\ldots+\frac{1}{n_k}\right) \ge n\cdot k\left(\frac{1}{n_1}\cdot\frac{1}{n_2}\cdots\frac{1}{n_k}\right)^{\frac{1}{k}}=k\cdot n\cdot n^{\frac{-1}{k}}=k\cdot n^{\frac{k-1}{k}}.$$ Here we have replaced the arithmetic mean by the geometric mean. \fP Note that bounds (\ref{eqle}) and (\ref{eqgge}) are exact for $n=2^k.$ \begin{Crl} \label{crl:finit} If the differential equation $n'=a$ has any solution in natural numbers, then it has only finitely many solutions if $a>1$. \end{Crl} \Pf The number $n$ cannot be a prime. According to (\ref{eqge}) the solutions must be no greater than $\frac{a^2}{4}.$ \fP What about the existence of solutions? We start with the even numbers. \begin{Cnj}\label{cnj:even} The differential equation $n'=2b$ has a positive integer solution for any natural number $b>1.$ \end{Cnj} A motivation for this is the famous \begin{Cnj}{\bf (Goldbach Conjecture)}\label{cnj:Goldbach} Every even number larger than $3$ is a sum of two primes. \end{Cnj} So, if $2b=p+q,$ then $n=pq$ is a solution that we need. Inequality (\ref{eqge}) helps us easy to prove that the equation $n'=2$ has no solutions. What about odd numbers larger than 1? It is easy to check with the help of (\ref{eqge}) that the equation $n'=3$ has no solutions. For $a=5$ we have one solution and more general have a theorem: \begin{Thm}\label{Thmtwins} Let $p$ be a prime and $a=p+2$. Then $2p$ is a solution for the equation $n'=a.$ \end{Thm} \Pf $(2p)'=2'p+2p'=p+2.$ \fP Some other primes also can be obtained as a derivative of a natural number (e.g. $7$), but it is more interesting which of numbers cannot. Here is a list of all $a\le 1000$ for which the equation $n'=a$ has no solutions (obtained using Maple and (\ref{eqge})): $$2,3,11,17,23,29,35,37,47,53,57,65,67,79,83,89,93,97,107,117,125,127,$$ $$137,145,149,157,163,173,177,179,189,197,205,207,209,217,219,223,233,$$ $$237,245,257,261,277,289,303,305,307,317,323,325,337,345,353,367,373,$$ $$377,379,387,389,393,397,409,413,415,427,429,443,449,453,457,473,477,$$ $$485,497,499,509,513,515,517,529,531,533,537,547,553,561,569,577,593,$$ $$597,605,613,625,629,639,657,659,665,673,677,681,683,697,699,709,713,$$ $$715,733,747,749,757,765,769,777,781,783,785,787,793,797,805,809,817,$$ $$819,827,833,835,845,847,849,853,857,869,873,877,881,891,895,897,907,$$ $$917,925,933,937,947,953,963,965,967,981,989,997.$$ Note that a large portion of them (69 from 153) are primes, one of them ($529=23^2$) is a square, and some of them (e.g. $765=3^2\cdot5\cdot 17$) have at least 4 prime factors. In general it is interesting to investigate the behavior of the ``integrating'' function $I(a)$ which calculates for every $a$ the set of solutions of the equation $n'=a$ and its weaker variant $i(a)$ that calculates the number of such solutions. As we have seen above $I(0)=\{ 0, 1\},$ $I(1)$ consist of all primes and $i(2)=i(3)=i(11)=\cdots =i(997)=0.$ Here is a list of the those numbers $a\le 100$ that have more than one ``integral'' (i.e. $i(a)\ge 2).$ For example $10$ has two ``integrals'' (namely $I(10)=\{ 21,25\}$) and $100$ has six $(I(100)=\{291, 979, 1411, 2059, 2419, 2491 \}$). \begin{verbatim} [10, 2], [12, 2], [14, 2], [16, 3], [18, 2], [20, 2], [21, 2], [22, 3], [24, 4], [26, 3], [28, 2], [30, 3], [31, 2], [32, 4], [34, 4], [36, 4], [38, 2], [39, 2], [40, 3], [42, 4], [44, 4], [45, 2], [46, 4], [48, 6], [50, 4], [52, 3], [54, 5], [55, 2], [56, 4], [58, 4], [60, 7], [61, 2], [62, 3], [64, 5], [66, 6], [68, 3], [70, 5], [71, 2], [72, 7], [74, 5], [75, 3], [76, 5], [78, 7], [80, 6], [81, 2], [82, 5], [84, 8], [86, 5], [87, 2], [88, 4], [90, 9], [91, 3], [92, 6], [94, 5], [96, 8], [98, 3], [100, 6]. \end{verbatim} Note that only three of them are primes. To complete the picture it remains to list the set of those $a<=100$ for which $i(a)=1.$ $$ 4, 5, 6, 7, 8, 9, 13, 15, 19, 25, 27, 33, 41, $$ $$ 43, 49, 51, 59, 63, 69, 73, 77, 85, 95, 99. $$ \begin{Thm}\label{thm:unboundi} The function $i(n)$ is unbounded for $n>1.$ \end{Thm} \Pf Suppose that $i(n)1$ for some constant $C.$ Then $$\sum_{k=2}^{2n}i(k)< 2Cn$$ for any $n.$ But for any two primes $p,q$ the product $pq$ belongs to $I(p+q)$ thus $$\sum_{k=2}^{2n} i(k)> {\sum_{ p\le q\le n}}' 1= \frac {\pi(n)(\pi(n)+1)}{2}>\frac{\pi(n)^2}{2},$$ where ${\sum}'$ means that the sum runs over the primes, and $\pi(n)$ is the number of primes not exceeding $n.$ This leads to the inequality $$2Cn > \frac{\pi(n)^2}{2} \Rightarrow \pi(n) < 2\sqrt{Cn},$$ which contradicts the known asymptotic behavior $\pi(n)\approx \frac{n}{\ln n}.$ \fP It would be interesting to prove a stronger result. \begin{Cnj}\label{cnj:everyi} For any nonnegative $m$ there exists infinitely many $a$ such that $i(a)=m.$ \end{Cnj} Another related conjecture is the following: \begin{Cnj}\label{cnj:infseq} There exists an infinite sequence $a_n$ of different natural numbers such that $a_1=1, (a_n)'=a_{n-1}$ for $n=2,3\ldots$ \end{Cnj} Here is an example of possible beginning of such a sequence: $$1 \leftarrow 7 \leftarrow 10 \leftarrow 25 \leftarrow 46 \leftarrow 129 \leftarrow 170 \leftarrow 501 \leftarrow 414 \leftarrow 2045. $$ The following table shows the maximum of $i(n)$ depending of the number $m$ of (not necessary different) prime factors in the factorization of $n$ for $n\le 1000.$ \[ {\begin{array}{|c|ccccccccc|} \hline m & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 &9 \\ \hline i(n)& 8 & 22 & 35 & 46& 52 & 52 & 40 & 47 & 32\\ \hline \end{array}} \] The next more detailed picture shows the distribution of $i(n)$ depending of the number $m$ for $ i(n)<33.$ Note that maximum possible $i(n)$ is equal $52,$ so we have only part of a possible table. We leave to the reader the pleasure of making some natural conjectures. \[ {\begin{array}{|c|ccccccccc|} \hline i(n)\backslash m&1 &2 &3 &4 &5 &6 &7 &8 &9 \\ \hline 0 &69& 49& 28& 6& 1& 0& 0& 0& 0\\ 1 &46& 89& 35& 8& 3& 1& 0& 0& 0\\ 2& 25& 44&18& 7& 1& 0& 0& 0& 0\\ 3&13& 16& 17& 7& 0& 0& 0& 0& 0\\ 4&9& 12& 8& 5& 2& 0&1& 0& 0\\ 5 &2& 6& 3& 4& 0& 1& 0& 0& 0\\ 6 &1& 7& 8& 1& 2& 0& 0& 0& 0\\ 7 &1& 10&4& 3& 2& 1& 0& 0& 0\\ 8 &2& 3& 8& 3& 2& 2& 0& 1& 0\\ 9 &0& 8& 6& 7& 4& 0& 0&0& 0\\ 10 & 0& 3& 7& 5& 1& 1& 0& 0& 0\\ 11 & 0& 8& 13& 2& 1& 2& 0& 0& 0\\ 12 &0& 4& 4& 5& 2& 0& 1& 0& 1\\ 13 &0& 3& 10& 5& 2& 2& 1& 0& 0\\ 14 &0& 7& 7& 5& 4& 1& 1& 0&0\\ 15 &0& 8& 8& 3& 3& 1& 0& 0& 0\\ 16 &0& 1& 15& 6& 5& 1& 0& 0& 0\\ 17 &0& 10& 4& 8&2& 0& 0& 0& 0\\ 18 &0& 3& 4& 5& 2& 1& 1& 0& 0\\ 19 &0& 4& 5& 9& 4& 2& 1& 1&0\\ 20 &0& 3& 7& 1& 0& 1& 0& 1& 0\\ 21 &0& 0& 5& 2& 4& 3& 0& 1& 0\\ 22 &0& 1& 2& 5& 1& 0& 1& 0& 0\\ 23 &0& 0& 4& 1& 1& 1& 2& 0& 0\\ 24 &0& 0& 1& 6& 3& 1& 0& 0& 0\\ 25& 0& 0& 3& 2& 1& 1& 0& 0& 0\\ 26 &0& 0& 1& 2& 4& 1& 0& 1& 0\\ 27 & 0& 0& 2& 1& 2& 1& 0& 0& 0\\ 28 & 0& 0& 1& 1& 0& 1& 1& 0& 0\\ 29 & 0& 0& 2& 2& 1& 0& 1& 0& 0\\ 30 & 0& 0& 1& 1& 1& 0& 0& 0& 0\\ 31 & 0& 0& 1& 4& 3& 0& 0& 0& 0\\ 32& 0& 0& 0& 6& 1& 1& 1& 0& 1\\ \hline \end{array}} \] \section{The equation $n''=1$} The main conjecture for the second-order equations is the following: \begin{Cnj}\label{cnj:diff2} The differential equation $n''=1$ has infinitely many solutions in natural numbers. \end{Cnj} Theorem \ref{Thmtwins} shows that $2p$ is a solution if $p,p+2$ are primes. So the following famous conjecture would be sufficient to prove. \begin{Cnj}({\bf prime twins})\label{cnj:twins} There exists infinitely many pairs $p,p+2$ of prime numbers. \end{Cnj} The following problem is another alternative which would be sufficient: \begin{Cnj}({\bf prime triples})\label{cnj:triples} There exists infinitely many triples $p,q,r$ of prime numbers such that $P=pq+pr+qr$ is a prime. \end{Cnj} Such a triple gives a solution $n=pqr$ to our equation, because $n'=P.$ In reality all the solutions can be described as follows. \begin{Thm} A number $n$ is a solution of the differential equation $n''=1$ if and only if the three following conditions are valid: \begin{enumerate} \item The number $n$ is a product of different primes: $n=\prod_{i=1}^{k}{p_i}.$ \item $\sum_{i=1}^{k}{{1}/{p_i}}=\frac{p}{n},$ where $p$ is a prime. \item If $k$ is even, then the smallest prime of $p_i$ should be equal to $2.$ \end{enumerate} \end{Thm} \Pf If $n=p^2m$ for some prime $p$ then $n'=p(2m+pm')$ is not prime and according to Theorem \ref{Thmn1} the number $n$ cannot be a solution. So, it is a product of different primes. Then the second condition means that $n'$ is a prime and by Theorem \ref{Thmn1} it is necessary and sufficient to be a solution. As to the number $k$ of factors it cannot be even if all primes $p_i$ are odd, because $n'$ in this case is (as the sum of $k$ odd numbers) even and larger than two. \fP \section{Derivative for integers} \label{sec:negative} It is time to extend our definition to integers. \begin{Thm} A derivative is uniquely defined over the integers by the rule $$(-x)'=-x'.$$ \end{Thm} \Pf Because $(-1)^2=1$ we should have (according to the Leibnitz rule) $2(-1)'=0$ and $(-1)'=0$ is the only choice. After that $(-x)'=((-1)\cdot x)'=0\cdot x' +(-1)\cdot x'=-x'$ is the only choice for negative $-x$ and as a result is true for positive integers also. It remains to check that the Leibnitz rule is still valid. It is sufficient to check that it is valid for $-a$ and $b$ if it was valid for $a$ and $b.$ It follows directly: $$((-a)\cdot b)'=-(a\cdot b)'=-(a'\cdot b+ a\cdot b')=-a'\cdot b+ (- a)\cdot b'= (-a)'\cdot b+ (- a)\cdot b'.$$ \fP \section{Derivative for rational numbers} The next step is to differentiate a rational number. We start from the positive rationals. The shortest way is to use (\ref{eq1}). Namely, if $x=\prod_{i=1}^{k}{p_i^{x_i}}$ is a a factorization of a rational number $x$ in prime powers, (where some $x_i$ may be negative) then we put \begin{equation} \label{eqr}x'=x\sum_{i=1}^{k}{\frac{x_i}{p_i}}\end{equation} and the same proof as in Theorem \ref{Thm1} shows that this definition is still consistent with the Leibnitz rule. Here is a table of derivatives of $i/j$ for small $i,j.$ $${\begin{array}{|r|rccccccccc|} \hline i\slash j & 1 & 2 & 3 & 4 & 5 & 6& 7 & 8 & 9 & 10 \\ \hline & & & & & & & & & & \\ 1&0 & {\displaystyle \frac {-1}{4}} & {\displaystyle \frac {-1}{9} } & {\displaystyle \frac {-1}{4}} & {\displaystyle \frac {-1}{ 25}} & {\displaystyle \frac {-5}{36}} & {\displaystyle \frac { -1}{49}} & {\displaystyle \frac {-3}{16}} & {\displaystyle \frac {-2}{27}} & {\displaystyle \frac {-7}{100}} \\ [2ex] 2 & 1 & 0 & {\displaystyle \frac {1}{9}} & {\displaystyle \frac {-1 }{4}} & {\displaystyle \frac {3}{25}} & {\displaystyle \frac { -1}{9}} & {\displaystyle \frac {5}{49}} & {\displaystyle \frac {-1}{4}} & {\displaystyle \frac {-1}{27}} & {\displaystyle \frac {-1}{25}} \\ [2ex] 3 & 1 & {\displaystyle \frac {-1}{4}} & 0 & {\displaystyle \frac {-1 }{2}} & {\displaystyle \frac {2}{25}} & {\displaystyle \frac { -1}{4}} & {\displaystyle \frac {4}{49}} & {\displaystyle \frac {-7}{16}} & {\displaystyle \frac {-1}{9}} & {\displaystyle \frac {-11}{100}} \\ [2ex] 4& 4 & 1 & {\displaystyle \frac {8}{9}} & 0 & {\displaystyle \frac {16}{25}} & {\displaystyle \frac {1}{9}} & {\displaystyle \frac {24}{49}} & {\displaystyle \frac {-1}{4}} & {\displaystyle \frac {4}{27}} & {\displaystyle \frac {3}{25} } \\ [2ex] 5& 1 & {\displaystyle \frac {-3}{4}} & {\displaystyle \frac {-2}{9} } & -1 & 0 & {\displaystyle \frac {-19}{36}} & {\displaystyle \frac {2}{49}} & {\displaystyle \frac {-13}{16}} & {\displaystyle \frac {-7}{27}} & {\displaystyle \frac {-1}{4}} \\ [2ex]6 & 5 & 1 & 1 & {\displaystyle \frac {-1}{4}} & {\displaystyle \frac {19}{25}} & 0 & {\displaystyle \frac {29}{49}} & {\displaystyle \frac {-1}{2}} & {\displaystyle \frac {1}{9}} & {\displaystyle \frac {2}{25}} \\ [2ex] 7 & 1 & {\displaystyle \frac {-5}{4}} & {\displaystyle \frac {-4}{9} } & {\displaystyle \frac {-3}{2}} & {\displaystyle \frac {-2}{ 25}} & {\displaystyle \frac {-29}{36}} & 0 & {\displaystyle \frac {-19}{16}} & {\displaystyle \frac {-11}{27}} & {\displaystyle \frac {-39}{100}} \\ [2ex] 8& 12 & 4 & {\displaystyle \frac {28}{9}} & 1 & {\displaystyle \frac {52}{25}} & {\displaystyle \frac {8}{9}} & {\displaystyle \frac {76}{49}} & 0 & {\displaystyle \frac {20}{ 27}} & {\displaystyle \frac {16}{25}} \\ [2ex]9 & 6 & {\displaystyle \frac {3}{4}} & 1 & {\displaystyle \frac {-3 }{4}} & {\displaystyle \frac {21}{25}} & {\displaystyle \frac { -1}{4}} & {\displaystyle \frac {33}{49}} & {\displaystyle \frac {-15}{16}} & 0 & {\displaystyle \frac {-3}{100}} \\ [2ex] 10 & 7 & 1 & {\displaystyle \frac {11}{9}} & {\displaystyle \frac {-3 }{4}} & 1 & {\displaystyle \frac {-2}{9}} & {\displaystyle \frac {39}{49}} & -1 & {\displaystyle \frac {1}{27}} & 0\\ & & & & & & & & & & \\ \hline \end{array}}$$ A natural property is the following: \begin{Thm}\label{thm:frac} For any two rationals $a,b$ we have $$\left(\frac{a}{b}\right)'= \frac{a'b-ab'}{b^2}. $$ A derivative can be well defined for rational numbers using this formula and this is the only way to define a derivative over rationals that preserves the Leibnitz rule. \end{Thm} \Pf If $a=\prod_{i=1}^{k}{p_i^{a_i}},b=\prod_{i=1}^{k}{p_i^{a_i}}$ then we have $$ \left(\frac{a}{b}\right)'=(\prod_{i=1}^{k}{p_i^{a_i-b_i}})'= (\prod_{i=1}^{k}{p_i^{a_i-b_i}})\sum_{i=1}^{k}{\frac{a_i-b_i}{p_i}}=$$ $$ \left(\frac{a}{b}\right)\sum_{i=1}^{k}{\frac{a_i}{p_i}}- \left(\frac{ab}{b^2}\right)\sum_{i=1}^{k}{\frac{b_i}{p_i}}= \frac{a'}{b}-\frac{ab'}{b^2}= \frac{a'b-ab'}{b^2}. $$ Let us check uniqueness. If $n$ is an integer then $n\cdot \frac{1}{n} =1$ and the Leibnitz rule demands $$n'\cdot \frac{1}{n} +n\left(\frac{1}{n}\right)'=0\Rightarrow \left(\frac{1}{n}\right)' =- \frac{n'}{n^2}.$$ After that $$\left(\frac{a}{b}\right)'=\left(a\cdot \frac{1}{b}\right)'=a'\cdot \frac{1}{b} +a\cdot \left(\frac{1}{b}\right)'= \frac{a'}{b}-a\cdot \left(\frac{b'}{b^2}\right) = \frac{a'b-ab'}{b^2} $$ is the only choice that satisfies the Leibnitz rule. This proves uniqueness. To prove that such a definition is well-defined, it is sufficient to see that $$\left(\frac{ac}{bc}\right)' = \frac{(ac)'(bc)-(ac)(bc)'}{(bc)^2}= \frac{(a'c+ac')(bc)-(ac)(b'c+bc')}{b^2c^2}=$$ $$ \frac{(a'bc^2+abc'c)- (ab'c^2+abcc')}{b^2c^2}=\frac{a'b-ab'}{b^2} $$ has the same value. \fP For negative rationals we can proceed as above and put $(-x)'=-x'.$ \section{Rational solutions of the equation $x'=a.$} Unexpectedly the equation $x'=0$ has nontrivial rational solutions, for instance $x=4/27.$ We can describe all of them. \begin{Thm} Let $k$ be some natural number, $\left\{ p_i, i=1,\ldots k\right\} $ be a set of different prime numbers and $\left\{ \alpha_i, i=1,\ldots k\right\} $ be a set of integers such that $\sum_{i=1}^{k}{\alpha_i}=0.$ Then $$x=\pm\prod_{i=1}^{k}{p_i^{\alpha_ip_i}}$$ are solutions of the differential equation $x'=0$ and any other nonzero solution can be obtained in this manner. \end{Thm} \Pf Because $(-x)'=-x'$ it is sufficient to consider positive solutions only. Let $x=\prod_{i=1}^{k}{p_i^{a_i}}$ Then from (\ref{eqr}) $$\sum_{i=1}^{k}{\frac{a_i}{p_i}}=0 \Rightarrow \sum_{i=1}^{k}{{a_i}\cdot{Q_i}}=0,$$ where $Q_i=\left({\prod_{j=1}^{k}{p_j} }\right)/{p_i}$ is not divisible by ${p_i}.$ Thus $a_i$ should be divisible by ${p_i}$ and $\alpha_i=\frac{a_i}{p_i}.$ \fP Other equations are more difficult. \begin{Cnj}\label{cnj:rat1} The equation $x'=1$ has only primes as positive rational solutions. \end{Cnj} Note that there exists a negative solution, namely $x=-\frac{5}{4}.$ One possible solution of this equation would be $x=\frac{n}{p^p}$ for some natural $n$ and prime $p.$ Because $x'=\frac{n'-n}{p^p}$ in this case we can reformulate the conjecture as \begin{Cnj}\label{cnj:natpp} Let $p$ be a prime. The equation $n'=n+p^p$ has no natural solutions except $n=qp^p,$ where $q$ is a prime. \end{Cnj} Note, that according to Theorem \ref{thm:periods} if a solution $n$ is divisible by $p$ it should be divisible by $p^p.$ Therefore $n=mp^p$ and $p^p(m'+m)=p^p(m+1)$ by Theorem \ref{thm:infp} and $m$ should be a prime. Thus it is sufficient to prove that any solution is divisible by $p.$ We do not expect that it is possible to integrate every rational number, though we do not know a counterexample. \begin{Cnj}\label{cnj:norat} There exists $a$ such that the equation $x'=a$ has no rational solutions. \end{Cnj} The first natural candidates do not verify the conjecture: $$(-\frac{21}{16})'=2;(-\frac{13}{4})'=3;(-\frac{22}{27})'=\frac{1}{3}. $$ \section{Logarithmic derivative} \label{sec:Log} One thing that is still absent in our picture is the analogue of the logarithm -- the primitive of $\frac{1}{n}.$ Because our derivative is not linear we cannot expect that the logarithm of the product is equal to the sum of logarithms. Instead this is true for its derivative. So let us define a logarithmic derivative $\ld(x)$ as follows. If $x=\prod_{i=1}^{k}{p_i^{x_i}}$ for different primes $p_i$ and some integers $x_i,$ then $$\ld(x)=\sum_{i=1}^{k}{\frac{x_i}{p_i}},\ \ld (-x)=\ld(x),\ \ld (0)=\infty.$$ In other words $$\ld(x)=\frac{x'}{x}.$$ \begin{Thm} For any rational numbers $$\ld(xy)=\ld(x)+\ld(y).$$ \end{Thm} \Pf $$\ld(xy)=\frac{(xy)'}{xy}=\frac{x'y+xy'}{xy}=\frac{x'}{x} +\frac{y'}{y}= \ld(x)+\ld(y).$$ \fP It is useful to divide every integer number into large and small parts. Let $\mbox{\rm sign}(x)x =|x|=\prod_{i=1}^{k}{p_i^{x_i}}$ and $x_i=a_ip_i+r_i,$ where $0\le r_i0.$ Then \begin{itemize} \item The equation \begin{equation} \label{eq:genexp} x'=\alpha x\end{equation} has nonzero rational solutions if and only if $b$ is a product of different primes or $b=1.$ \item If $x_0$ is a nonzero particular solution (\ref{eq:genexp}) and $y$ is any rational solution of the equation $y'=0$ then $x=x_0y$ is also a solution of (\ref{eq:genexp}) and any solution of (\ref{eq:genexp}) can be obtained in this manner. \item To obtain a particular solution of the equation (\ref{eq:genexp}) it is sufficient to decompose $\alpha$ into the elementary fractions: $$\alpha =\frac{a}{b} =\lfloor \alpha \rfloor +\sum_{i=1}^{k}\frac{c_i}{p_i},$$ where $b=\prod_{i=1}^{k}{p_i}, 1\le |c_i|n.$ So it is sufficient to put $t=1$ to get the desired $\lfloor\frac{n}{{p^j}}\rfloor$ in every summand. \fP If we use the same $L(x)$ that we used in the proof of the Theorem \ref{thm:genfunc}, we get the classical Legendre theorem that calculates the maximal power of a prime $p$ in $n!.$ On the other hand if we use $L(x)=\ld (x)$ we will be able, following Barbeau \cite{Bar}, to estimate $\sum_{i=1}^{n}\ld (i).$ Let $m=\lfloor\log_2n\rfloor.$ Then we can change infinity in our sums to $m.$ Using standard estimates $${\sum_{p\le n}}'\frac{1}{p} = O(\ln m), $$ $$ {\sum_{p> n}}'\frac{n}{p(p-1)}<\sum_{k> n}\frac{n}{k(k-1)}= \sum_{k> n}n\left(\frac{1}{k}-\frac{1}{k-1}\right)\le 1,$$ $${\sum_{p\le n}}'\frac{n}{p^{m+1}(p-1)}<{\sum_{p\le n}}'\frac{2n}{2^{m+1}p(p-1)}< \sum_{k\le n}\frac{2n}{nk(k-1)}\le 2,$$ we get $$\sum_{i=1}^{n}\ld (i)={\sum_{p\le n}}'\frac{1}{p}\sum_{j=1}^{m}\left\lfloor\frac{n}{{p^j}}\right\rfloor= {\sum_{p\le n}}'\frac{1}{p}\left(\sum_{j=1}^{m}\frac{n}{{p^j}}+O(m)\right)=$$ $${\sum_{p\le n}}'\frac{n}{p^{m+1}}\left(\frac{p^{m}-1}{{p-1}}\right)+O(\ln m)O(m)= {\sum_{p }}'\frac{n}{p(p-1)} -$$ $$-{\sum_{p>n}}'\frac{n}{p(p-1)}-{\sum_{p\le m}}'\frac{n}{p^{m+1}(p-1)}+O(\ln m)O(m)=$$ $${\sum_{p}}'\frac{n}{p(p-1)}+O(m\ln m).$$ \begin{Thm}\cite{Bar} Let $$ C={\sum_{p}}'\frac{1}{p(p-1)}=0.749\ldots$$ Then $$\ld (n!)=\sum_{i=1}^{n}\ld (i)=Cn+ O\left((\ln n)(\ln\ln n)\right)$$ $$\sum_{k=1}^{n}k'=\frac{C}{2}n^2+ O(n^{1+\delta})$$ for any $\delta >0.$ \end{Thm} \Pf The first formula is already proved. As to the second we have $$\sum_{k=1}^{n}k'=\sum_{k=1}^{n}k\cdot\ld(k)=\sum_{k=1}^{n}\sum_{i=k}^n \ld(i)=$$ $$ \sum_{k=1}^{n}(\ld({n!})-\ld({(k-1)!}))=n\ld(n!)- \sum_{k=1}^{n-1}\ld({k!})=$$ $$n(Cn+O(n^{\delta}))- \sum_{k=1}^{n-1}(Ck+O(n^{\delta}))=$$ $$Cn^2-C\frac{n(n-1)}{2}+O(n^{1+\delta})=\frac{C}{2}n^2+ O(n^{1+\delta}).$$ \fP We leave to the reader the pleasure to play with $\zeta_D(s)=\sum\frac{n'}{n^s}.$ \section{Logical dependence of the conjectures} \label{sec:conjd} Here we would like to exhibit some of the logical dependence between the different conjectures we have mentioned above. As we see the Conjectures \ref{cnj:infseq} and \ref{cnj:diff2} seem to be the key problems. \begin{Thm} The following picture describes the logical dependence between the different conjectures. $$(\ref{Cnj-alt}) \Rightarrow (\ref{cnj:periods}) \Rightarrow (\ref{cnj:period2}),$$ $$(\ref{cnj:rat1}) \Rightarrow (\ref{cnj:natpp}), $$ $$ (\ref{cnj:even}) \Leftarrow (\ref{cnj:Goldbach},\mbox{\it Goldbach}),$$ $$ (\ref{cnj:nobiject}) \Leftarrow (\ref{cnj:noinject}), $$ $$ \begin{array}{ccccc} & & (\ref{cnj:triples}, {\mbox Triples})& &(\ref{cnj:infseq})\\ & &\Downarrow & &\Downarrow\\ (\ref{cnj:twins},\mbox{\it Twins})& \Rightarrow& (\ref{cnj:diff2})& \Rightarrow &(\ref{Cnj-main})\\ \end{array}.$$ Additionally if Conjecture \ref{Cnj-main} is valid then either Conjecture \ref{cnj:infseq} or Conjecture \ref{cnj:diff2} is valid (or both). \end{Thm} \Pf The only nontrivial dependence is the last one. Suppose that Conjecture \ref{cnj:diff2} is wrong, but Conjecture \ref{Cnj-main} is true. We need to show that Conjecture \ref{cnj:infseq} is valid. Let $\Gamma$ be the tree having vertices $1$ (the root), the primes $p$ with $i(p)>0$ and all composite $n$ with $n^{(k)}=0$ for some $k\ge 1.$ Further, let $\Gamma$ have edges from $n$ to $n'.$ By Conjecture \ref{Cnj-main}, $\Gamma$ is infinite. By Theorem \ref{Thmn1} and Corollary \ref{crl:finit} the degree at each vertex different from $1$ is finite. Also the vertex $1$ has finite degree since \ref{cnj:diff2} is false. By K\"{o}ning infinity lemma $\Gamma$ contains an infinite chain, ending in $1,$ which is Conjecture \ref{cnj:infseq}. \fP \section{Concluding remarks} \label{sec:History} This article is our expression of the pleasure being a mathematician. We have written it because we found the subject to be very attractive and wanted to share our joy with others. To our surprise we did not find many references. In the article of A. Buium \cite{Buium} and other articles of this author (which are highly recommended) we at least have found that there exists authors who can imagine a derivative without the linearity property. But the article of E. J. Barbeau \cite{Bar} was the only article that has direct connection to our topic. Most of the material from this article we have repeated here (not always citing). We omitted only the description of the numbers with derivatives that are divisible by $4$ and his conjecture that for every $n$ there exists a prime $p$ such that all derivatives $n^{(k)}$ are divisible by $p$ for sufficiently large $k.$ In fact according to Theorems \ref{thm:infp}, \ref{thm:periods} it is equivalent to be divisible by $p^p$ for sufficiently large $k.$ Thus this conjecture is a bit stronger then Conjecture \ref{Cnj-alt}. The definition of the arithmetic derivative itself and its elementary properties was already in the Putnam Prize competition (it was Problem 5 of the morning session in March 25, 1950, \cite{Gleason} ) and probably was known in folklore even earlier. What we have done is mainly to generalize this definition in different directions, to solve some differential equations, to calculate the generating function and to invite the reader to continue work in this area. We are grateful to our colleagues for useful discussion, especially to G. Almkvist, A. Chapovalov, S. Dunbar, G. Galperin, S. Shimorin and the referee, who helped to improve the text. We are especially grateful to J. Backelin, who helped us to reduce the number of conjectures by suggesting ideas that translated them into theorems. \begin{thebibliography}{99} \bibitem[1]{Bar}{E. J. Barbeau,} Remark on an arithmetic derivative, {\em Canad. Math. Bull.} {\bf 4} (1961), 117--122. \bibitem[2]{Buium}{A. Buium,} Arithmetic analogues of derivations, {\em J. Algebra } {\bf 198} (1997), 290--299. \bibitem[3]{Con}{J. H. Conway and R. K. Guy}, {\em The Book of Numbers,} Springer, 1996. \bibitem[4]{Gleason}{A. M. Gleason, R. E. Greenwood, and L. M. Kelly}, {\em The William Lowell Putnam Mathematical Competition: Problems and Solutions 1938--1964}, Mathematical Association of America, 1980. \bibitem[5]{ref:Zoo} {J. Renze, S. Wagon, and B. Wick,} The Gaussian zoo, {\em Experiment. Math.} {\bf 10:2} (2001), 161--173. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2000 {\it Mathematics Subject Classification}: Primary 11A25; Secondary 11A41, 11N05, 11N56, 11Y55. \noindent \emph{Keywords: Arithmetic derivative, Goldbach's Conjecture, the Twin Prime Conjecture, prime numbers, Leibnitz rule, integer sequence, generating function.} \bigskip \hrule \bigskip \noindent (Concerned with sequence \seqnum{A003415}.) \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received April 4, 2003; revised version received July 27, 2003. Published in {\it Journal of Integer Sequences}, September 17, 2003. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.math.uwaterloo.ca/JIS/}. \vskip .1in \end{document}