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\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:
 $$\begin{array}{|c|ccccc ccccc ccccc ccc| }
 \hline
 n &amp; 1  &amp; 2&amp; 3&amp; 4&amp; 5&amp; 6&amp; 7&amp; 8&amp; 9&amp; 10&amp; 11&amp; 12 &amp; 13&amp; 14&amp; 15&amp; 16&amp; 17 &amp; 18\\
 \hline
 n' &amp;0  &amp; 1&amp; 1&amp; 4&amp; 1&amp; 5&amp; 1&amp;12&amp; 6&amp;  7&amp;  1&amp; 16 &amp; 1 &amp; 9&amp;  8&amp;  32 &amp; 1 &amp; 21\\
 n''&amp; 0 &amp; 0&amp; 0&amp; 4&amp; 0&amp; 1&amp; 0&amp;16&amp; 5&amp;  1&amp;  0&amp; 32 &amp; 0 &amp; 6&amp; 12&amp;  80 &amp; 0 &amp; 10\\
 n'''&amp;0 &amp; 0&amp; 0&amp; 4&amp; 0&amp; 0&amp; 0&amp;32&amp; 1&amp;  0&amp;  0&amp; 80 &amp; 0 &amp; 5&amp; 16&amp;  176 &amp; 0 &amp; 7\\
 \hline
 \end{array}  $$

 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 $a&lt;b\le 100,\gcd(a,b)=1,$ for which $(a+b)'=a'+b':$
$$
                               (1, 2),
                               (4, 35),
                               (4, 91),
                               (8, 85),
                               (11, 14),
                               (18, 67), (26, 29),
$$
$$                              
                               (27, 55),
                               (35, 81),
                               (38, 47),
                               (38, 83),
                               (50, 79),
                               (62, 83),
                               (95, 99).
$$

A similar result is
\begin{Thm} For any natural $k&gt;1,$
  $$n'\ge n\Rightarrow (kn)'&gt;kn.$$
\end{Thm}
\Pf 
$$(kn)' =k'n+kn'&gt;kn'\ge kn.$$
\fP

The following theorem shows that every $n&gt;4$ that is divisible by $4$ 
satisfies the condition $n'&gt;n.$
  \begin{Thm}\label{thm:infp} If $n=p^p\cdot m$ for some prime $p$ and natural $m&gt;1,$ 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}), $n'=(p^p)'\cdot m+ p^p\cdot m'=p^p(m+m')&gt;n$ 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 $0&lt;k&lt;p,$ then  $p^{k-1}$ is the  highest power of  $p$ that
  divides $n'.$
In particular, all the numbers $n, n',n'',\ldots,n^{(k)}$ are distinct.
 \end{Thm}

\Pf Let $n=p^km$. Then $n'= kp^{k-1}m+p^km'=p^{k-1}(km+pm')$,
and the expression inside parentheses is not
 divisible by $p.$ \fP

 \begin{Crl}
  A positive integer $n$ is square-free if and only if $(n,n')=1.$
\end{Crl}

\Pf If $p^2|n,$ then $p|n'$ and $(n,n')&gt;1.$ 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  $a&gt;1$.
   \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 $b&gt;1.$ \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 $a&lt;=100$ 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 $n&gt;1.$
\end{Thm}
\Pf Suppose that $i(n)&lt;C$ for all $n&gt;1$ for some constant $C.$ Then
$$\sum_{k=2}^{2n}i(k)&lt; 2Cn$$
for any $n.$  But for any two primes $p,q$ the product $pq$ belongs to
$I(p+q)$ thus
$$\sum_{k=2}^{2n} i(k)&gt; {\sum_{ p\le q\le n}}' 1=
\frac {\pi(n)(\pi(n)+1)}{2}&gt;\frac{\pi(n)^2}{2},$$
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
$$2Cn &gt; \frac{\pi(n)^2}{2}  \Rightarrow \pi(n) &lt; 2\sqrt{Cn},$$
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 $n\le 1000.$
\[ 
{\begin{array}{|c|ccccccccc|}
\hline
m &amp; 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 &amp; 6 &amp; 7 &amp; 8 &amp;9  \\
\hline
i(n)&amp; 8 &amp; 22 &amp; 35 &amp; 46&amp; 52 &amp; 52 &amp; 40 &amp; 47 &amp; 32\\
\hline
\end{array}} 
\]

The next more detailed picture shows the distribution of $i(n)$ depending of
the number $m$ for $ i(n)&lt;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&amp;1 &amp;2  &amp;3 &amp;4 &amp;5 &amp;6 &amp;7 &amp;8 &amp;9  \\
\hline
0 &amp;69&amp; 49&amp; 28&amp; 6&amp; 1&amp; 0&amp; 0&amp; 0&amp; 0\\
1 &amp;46&amp; 89&amp; 35&amp; 8&amp; 3&amp; 1&amp; 0&amp; 0&amp; 0\\
2&amp; 25&amp; 44&amp;18&amp; 7&amp; 1&amp; 0&amp; 0&amp; 0&amp; 0\\
3&amp;13&amp; 16&amp; 17&amp; 7&amp; 0&amp; 0&amp; 0&amp; 0&amp; 0\\
4&amp;9&amp; 12&amp; 8&amp; 5&amp; 2&amp; 0&amp;1&amp; 0&amp; 0\\
5 &amp;2&amp; 6&amp; 3&amp; 4&amp; 0&amp; 1&amp; 0&amp; 0&amp; 0\\
6 &amp;1&amp; 7&amp; 8&amp; 1&amp; 2&amp; 0&amp; 0&amp; 0&amp; 0\\
7 &amp;1&amp; 10&amp;4&amp; 3&amp; 2&amp; 1&amp; 0&amp; 0&amp; 0\\
8 &amp;2&amp; 3&amp; 8&amp; 3&amp; 2&amp; 2&amp; 0&amp; 1&amp; 0\\
9 &amp;0&amp; 8&amp; 6&amp; 7&amp; 4&amp; 0&amp; 0&amp;0&amp; 0\\
10 &amp; 0&amp; 3&amp; 7&amp; 5&amp; 1&amp; 1&amp; 0&amp; 0&amp; 0\\
11 &amp; 0&amp; 8&amp; 13&amp; 2&amp; 1&amp; 2&amp; 0&amp; 0&amp; 0\\
12 &amp;0&amp; 4&amp; 4&amp; 5&amp; 2&amp; 0&amp; 1&amp; 0&amp; 1\\
13 &amp;0&amp; 3&amp; 10&amp; 5&amp; 2&amp; 2&amp; 1&amp; 0&amp; 0\\
14 &amp;0&amp; 7&amp; 7&amp; 5&amp; 4&amp; 1&amp; 1&amp; 0&amp;0\\
15 &amp;0&amp; 8&amp; 8&amp; 3&amp; 3&amp; 1&amp; 0&amp; 0&amp; 0\\
16 &amp;0&amp; 1&amp; 15&amp; 6&amp; 5&amp; 1&amp; 0&amp; 0&amp; 0\\
17 &amp;0&amp; 10&amp; 4&amp; 8&amp;2&amp; 0&amp; 0&amp; 0&amp; 0\\
18 &amp;0&amp; 3&amp; 4&amp; 5&amp; 2&amp; 1&amp; 1&amp; 0&amp; 0\\
19 &amp;0&amp; 4&amp; 5&amp; 9&amp; 4&amp; 2&amp; 1&amp; 1&amp;0\\
20 &amp;0&amp; 3&amp; 7&amp; 1&amp; 0&amp; 1&amp; 0&amp; 1&amp; 0\\
21 &amp;0&amp; 0&amp; 5&amp; 2&amp; 4&amp; 3&amp; 0&amp; 1&amp; 0\\
22 &amp;0&amp; 1&amp; 2&amp; 5&amp; 1&amp; 0&amp; 1&amp; 0&amp; 0\\
23 &amp;0&amp; 0&amp; 4&amp; 1&amp; 1&amp; 1&amp; 2&amp; 0&amp; 0\\
24 &amp;0&amp; 0&amp; 1&amp; 6&amp; 3&amp; 1&amp; 0&amp; 0&amp; 0\\
25&amp; 0&amp; 0&amp; 3&amp; 2&amp; 1&amp; 1&amp; 0&amp; 0&amp; 0\\
26 &amp;0&amp; 0&amp; 1&amp; 2&amp; 4&amp; 1&amp; 0&amp; 1&amp; 0\\
27 &amp; 0&amp; 0&amp; 2&amp; 1&amp; 2&amp; 1&amp; 0&amp; 0&amp; 0\\
28 &amp;                    0&amp; 0&amp; 1&amp; 1&amp; 0&amp; 1&amp; 1&amp; 0&amp; 0\\
29 &amp;                     0&amp; 0&amp; 2&amp; 2&amp; 1&amp; 0&amp; 1&amp; 0&amp; 0\\
30 &amp;                     0&amp; 0&amp; 1&amp; 1&amp; 1&amp; 0&amp; 0&amp; 0&amp; 0\\
31 &amp;                    0&amp; 0&amp; 1&amp; 4&amp; 3&amp; 0&amp; 0&amp; 0&amp; 0\\
32&amp;                    0&amp; 0&amp; 0&amp; 6&amp; 1&amp; 1&amp; 1&amp; 0&amp; 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 &amp; 1 &amp; 2 &amp; 3 &amp; 4 &amp; 5 &amp; 6&amp; 7 &amp; 8 &amp; 9 &amp; 10 \\
\hline
                 &amp;   &amp;   &amp;   &amp;   &amp;   &amp;  &amp;   &amp;   &amp;   &amp; \\
1&amp;0 &amp; {\displaystyle \frac {-1}{4}}  &amp; {\displaystyle \frac {-1}{9}
}  &amp; {\displaystyle \frac {-1}{4}}  &amp; {\displaystyle \frac {-1}{
25}}  &amp; {\displaystyle \frac {-5}{36}}  &amp; {\displaystyle \frac {
-1}{49}}  &amp; {\displaystyle \frac {-3}{16}}  &amp; {\displaystyle 
\frac {-2}{27}}  &amp; {\displaystyle \frac {-7}{100}}  \\ [2ex] 2 &amp;
1 &amp; 0 &amp; {\displaystyle \frac {1}{9}}  &amp; {\displaystyle \frac {-1
}{4}}  &amp; {\displaystyle \frac {3}{25}}  &amp; {\displaystyle \frac {
-1}{9}}  &amp; {\displaystyle \frac {5}{49}}  &amp; {\displaystyle 
\frac {-1}{4}}  &amp; {\displaystyle \frac {-1}{27}}  &amp; 
{\displaystyle \frac {-1}{25}}  \\ [2ex] 3 &amp;
1 &amp; {\displaystyle \frac {-1}{4}}  &amp; 0 &amp; {\displaystyle \frac {-1
}{2}}  &amp; {\displaystyle \frac {2}{25}}  &amp; {\displaystyle \frac {
-1}{4}}  &amp; {\displaystyle \frac {4}{49}}  &amp; {\displaystyle 
\frac {-7}{16}}  &amp; {\displaystyle \frac {-1}{9}}  &amp; 
{\displaystyle \frac {-11}{100}}  \\ [2ex] 4&amp;
4 &amp; 1 &amp; {\displaystyle \frac {8}{9}}  &amp; 0 &amp; {\displaystyle 
\frac {16}{25}}  &amp; {\displaystyle \frac {1}{9}}  &amp; 
{\displaystyle \frac {24}{49}}  &amp; {\displaystyle \frac {-1}{4}} 
 &amp; {\displaystyle \frac {4}{27}}  &amp; {\displaystyle \frac {3}{25}
}  \\ [2ex] 5&amp;
1 &amp; {\displaystyle \frac {-3}{4}}  &amp; {\displaystyle \frac {-2}{9}
}  &amp; -1 &amp; 0 &amp; {\displaystyle \frac {-19}{36}}  &amp; {\displaystyle 
\frac {2}{49}}  &amp; {\displaystyle \frac {-13}{16}}  &amp; 
{\displaystyle \frac {-7}{27}}  &amp; {\displaystyle \frac {-1}{4}} 
 \\ [2ex]6 &amp;
5 &amp; 1 &amp; 1 &amp; {\displaystyle \frac {-1}{4}}  &amp; {\displaystyle 
\frac {19}{25}}  &amp; 0 &amp; {\displaystyle \frac {29}{49}}  &amp; 
{\displaystyle \frac {-1}{2}}  &amp; {\displaystyle \frac {1}{9}}  &amp; 
{\displaystyle \frac {2}{25}}  \\ [2ex] 7 &amp;
1 &amp; {\displaystyle \frac {-5}{4}}  &amp; {\displaystyle \frac {-4}{9}
}  &amp; {\displaystyle \frac {-3}{2}}  &amp; {\displaystyle \frac {-2}{
25}}  &amp; {\displaystyle \frac {-29}{36}}  &amp; 0 &amp; {\displaystyle 
\frac {-19}{16}}  &amp; {\displaystyle \frac {-11}{27}}  &amp; 
{\displaystyle \frac {-39}{100}}  \\ [2ex] 8&amp;
12 &amp; 4 &amp; {\displaystyle \frac {28}{9}}  &amp; 1 &amp; {\displaystyle 
\frac {52}{25}}  &amp; {\displaystyle \frac {8}{9}}  &amp; 
{\displaystyle \frac {76}{49}}  &amp; 0 &amp; {\displaystyle \frac {20}{
27}}  &amp; {\displaystyle \frac {16}{25}}  \\ [2ex]9 &amp;
6 &amp; {\displaystyle \frac {3}{4}}  &amp; 1 &amp; {\displaystyle \frac {-3
}{4}}  &amp; {\displaystyle \frac {21}{25}}  &amp; {\displaystyle \frac {
-1}{4}}  &amp; {\displaystyle \frac {33}{49}}  &amp; {\displaystyle 
\frac {-15}{16}}  &amp; 0 &amp; {\displaystyle \frac {-3}{100}}  \\ [2ex] 10 &amp;
7 &amp; 1 &amp; {\displaystyle \frac {11}{9}}  &amp; {\displaystyle \frac {-3
}{4}}  &amp; 1 &amp; {\displaystyle \frac {-2}{9}}  &amp; {\displaystyle 
\frac {39}{49}}  &amp; -1 &amp; {\displaystyle \frac {1}{27}}  &amp; 0\\
              &amp;   &amp;   &amp;   &amp;   &amp;   &amp;  &amp;   &amp;   &amp;   &amp; \\
\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 $0\le r_i&lt;p_i.$ We define
$$P(x)=\mbox{\rm sign}(x)\prod_{i=1}^{k}{p_i^{a_ip_i}},R(x)=\prod_{i=1}^{k}{p_i^{r_i}},
A(x)=\sum_{i=1}^{k}{{a_i}}.$$
\begin{Thm} The following properties hold
\begin{itemize}
 \item  $\ld(x)=A(x)+\ld (R(x)).$
 \item $x'=A(x)x+P(x)(R(x))'=x(A(x)+\ld (R(x))).$
\item If $x$ is a nonzero integer, then
 $$x| x'\Leftrightarrow \ld(x) \in{\bf Z} \Leftrightarrow R(x)=1.$$
\item if $\left(\frac{a}{b}\right)'$ is an integer, and $\gcd(a,b)=1$
  then $R(b)=1.$
\end{itemize}
\end{Thm}
\Pf  First we have
$$\ld(x)=\ld(P(x)R(x))=\ld(P(x))+\ld(R(x))=A(x)+\ld(R(x)).$$
Using this we get
$$x'=x\ld(x)=x(A(x)+\ld (R(x)))=xA(x)+x\ld(R(x)) =$$
$$xA(x)+P(x)R(x)\ld(R(x)) =A(x)x+P(x)(R(x))'.$$
  If $R(x)\neq 1$ then the sum 
$$\ld(R(x))=\sum_{i=1}^{k}{\frac{r_i}{p_i}}$$ cannot be an integer.
Otherwise 
$$\ld(R(x))\prod_{i=1}^kp_i=\sum_{i=1}^{k}{{r_i}{Q_i}},$$
and if $0&lt;r_j&lt;p_j$ then an integer on the left hand side is divisible
by $p_j,$ but on the right hand side is not because
$Q_j=\frac{\prod_{i=1}^kp_i}{p_j}$
and all primes $p_i$ are different. The last statement follows from
Theorem \ref{thm:frac}. \fP 

Now we are able to  solve the equation $x'=\alpha x$ with rational $\alpha$ in
 the rationals. We have already solved  this equation in the case $\alpha=0,$
so let $\alpha\neq 0.$
\begin{Thm}
  Let $\alpha=\frac{a}{b}$ be a rational number with
  $\gcd(a,b)=1,b&gt;0.$
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 $b=\prod_{i=1}^{k}{p_i}, 1\le |c_i|&lt;p_i.$ Then 
$$x_0 =4^{\lfloor\alpha\rfloor}\prod_{i=1}^{k}{p_i^{c_i}}$$ is a particular
solution.  (Of course the number $4$ can be replaced by $p^p$ for any
prime $p$).

\end{itemize}
\end{Thm}
\Pf The equation (\ref{eq:genexp}) is equivalent to the equation 
$$\ld(x)=\alpha \Leftrightarrow A(x)+\ld(R(x))=
\alpha =\frac{a}{b}.$$ Because $A(x)$ is an integer and 
$\ld(R(x))=\sum_{i=1}^{k}\frac{r_i}{p_i},$ 
the natural number $b$  should be equal to the product of the
different primes or should be  equal to $1.$ Suppose that $b$ is of this type. Then 
$$\ld\left(4^{\lfloor\alpha\rfloor}\prod_{i=1}^{k}{p_i^{c_i}}\right)=\lfloor\alpha\rfloor
+\sum_{i=1}^{k}\frac{c_i}{p_i}=\alpha$$ and we obtain a desired particular solution.
 If $y'=0$ and $x_0$ any particular solution then
$$(x_0y)'=x_0'y+x_0y'=\alpha x_0y,$$ also satisfies  (\ref{eq:genexp}). Finally, if $x'=\alpha x$ and $y=\frac{x}{x_0}$ then 
$$\ld(y)=\ld (x)-\ld(x_0)=0$$  means that $y$ is a solution of the
equation $y'=0.$
\fP

For instance the equation $x'=\frac{x}{4}$ has no solutions,
$x_0=\frac{2}{3}$ is a partial solution of the equation
$x'=\frac{x}{6}$
and to obtain all nonzero solutions we need to multiply $x_0$ with any $y$ such
that
$R(y)=1, A(y)=0.$

                 \section{How to differentiate irrational numbers}
\label{sec:nonrat}
The next step is to try to generalize our definition to irrational
numbers. The equation (\ref{eq1})  can still be used in the more general
situation. But first we need to think about the correctness of the
definition.
\begin{Lm}
 Let $\{ p_1,\ldots, p_k \} $ be a set of different primes and
 $\{ x_1,\ldots,x_k\} $ a set of rationals. Then
$$P=\prod_{i=1}^{k}{p_i^{x_i}}=1 \Leftrightarrow x_1=x_2=\cdots = x_k=0.$$
 \end{Lm}
\Pf It is evident if all $x_i$ are integers, because the primes are
different. If they are rational, let us choose a natural $m$ such that
all $y_i=mx_i$ are integer. Then $P^m=1$ too and we get
$y_i=0\Rightarrow x_i=0.$ \fP

Now we can extend our definition to any real number $x$ that can be
written
as a product $x=\prod_{i=1}^{k}{p_i^{x_i}}$ for different primes
$p_i$
and some nonzero rationals $x_i$.  The previous lemma shows that this form is
unique and as above we can define
$$x'=x\sum_{i=1}^{k}{\frac{x_i}{p_i}}.$$ The proof for the Leibnitz rule is
still valid too and we skip it. For example we have
$$(\sqrt{3})'=(3^{1/2})'=3^{1/2}\frac{1/2}{3}=\frac{\sqrt{3}}{6}.$$
More generally we have the following convenient formula:
\begin{Thm}
  Let $x,y$ be rationals and $x$ be positive. Then
\begin{equation} \label{eq:potens}
 (x^y)'=yx^{y-1}x'=\frac{yx'}{x}x^{y}=yx^y\ld(x).\end{equation}
\end{Thm}
\Pf If  $x=\prod_{i=1}^{k}{p_i^{x_i}},$ then



$$(x^y)'=(\prod_{i=1}^{k}{p_i^{yx_i}})'=
x^y\sum_{i=1}^{k}{\frac{yx_i}{p_i}}=
{y}x^{y-1}x\sum_{i=1}^{k}{\frac{x_i}{p_i}}={y}x^{y-1}x'.$$
\fP

An interesting corollary is



\begin{Crl}
  Let $a,b,c,d$ be rationals such that $a^b=c^d$ ($a,c$ being
  positive).
Then $$b\cdot\ld(a)=d\cdot\ld(c)$$ and
$$a'bc=c'ad.$$
In particular, for the case $a=b,c=d$, we have 
$$a^a=c^c\Rightarrow a'=c'.$$
\end{Crl}\fP

 As an example we can check directly that
$x^y=y^x$ has the solutions
$$x=\left(\frac{m+1}{m}\right)^m;y=\left(\frac{m+1}{m}\right)^{m+1},$$
thus 
$$ \frac{x'}{x^2}=\frac{y'}{y^2},$$
 so the  equation $x'=\frac{x^2}{4}$ has at least two
solutions obtained from $m=1$.

Another example is
$$(1/2)^{1/2}=(1/4)^{1/4} \Rightarrow \left(\frac{1}{2}\right)'
=\left(\frac{1}{4}\right)'=
-\left(\frac{1}{4}\right).$$ 
In general it is not  difficult to prove that all  rational
solutions
of the equation $x^x=y^y$ have the form 
$$x=\left(\frac{m}{m+1}\right)^m, y=\left(\frac{m}{m+1}\right)^{m+1}$$
for some natural $m.$ Direct calculations give the same result as above:
$$x'=m\left(\frac{m}{m+1}\right)^{m-1}\left(\frac{m}{m+1}\right)'=
\left(
  m+1\right)\left(\frac{m}{m+1}\right)^m\left(\frac{m}{m+1}\right)'=y'$$
and shows that this works even for rational $m.$

It would be natural to extend our definition to  infinite products:
if $x=\prod_{i=1}^{\infty}p_i^{x_i}$ is convergent then it is easy to
show
that the sum $x\sum_{i=1}^{\infty}\frac{x_i}{p_i}$ is also convergent.
However, the problem is that the sum is not necessary convergent to zero, when
$x=1.$
  This is a reason why such a natural generalization of the derivative
 is not well-defined. Maybe a more natural approach is to restrict
 possible products, but we still do not know a nice solution of the
 problems that arise. But there is another way, which we consider in Section \ref{sec:Gen}. 

\section{Arithmetic Derivative for UFD}
\label{sec:UFD}
The definition of the derivative and most of the proofs are based only
on the fact that that every natural number has a unique factorization
into primes. So it is not difficult to transfer it to an arbitrary UFD
(unique factorization domain) $R$ using the same definition:
$p'=1$ for every ``canonical'' prime (irreducible) element,
the Leibnitz rule and additionally
$u'=0$ for all units (invertible elements) in $R.$ For example we can
do it for a polynomial ring $K[x]$ or for the Gaussian numbers $a+bi.$
In the first case the canonical irreducible polynomials are monic, in
the second the canonical primes are ``positive'' primes  \cite{Con}.
This leads to a
well-defined
derivative for the field of fractions. Note also, that even the condition
UFD is not necessary -- we only  need to have a well-defined derivative,
i.e. independent of factorization.
We do not plan to develop the theory in this more abstract direction 
and restrict ourselves by the following trivial (but interesting) result.
\begin{Thm} Let $K$ be a field of characteristic zero 
  and with the derivative  $f'$ in $K[x]$ is defined
  as above. Let $\frac{d}{dx}$ be a usual derivative.
  Then $f'(x) =\frac{df(x)}{dx}$ if and only if the polynomial
  $f(x)$ is a product of linear factors. 
\end{Thm}

\Pf Because both derivatives are equal to zero on constants they coincide
on linear polynomials. If $f(x)$ has no linear irreducible factors
then $f'(x)$ has smaller degree then $\frac{df(x)}{dx}.$ Otherwise
$f(x)=l(x)g(x)$ for some linear polynomial $l(x)$ and
$$f'(x)-\frac{df(x)}{dx} =l'(x)g(x)+l(x)g'(x)-\frac{dl(x)}{dx}g(x)-
l(x)\frac{dg(x)}{dx}= $$
$$l(x)\left(g'(x)-\frac{dg(x)}{dx}\right)$$
and we can use induction.  
\fP

So, for the complex polynomials both definitions coincide. On the
other hand $(x^2+x+1)' = \frac{d}{dx}(x^2+x+1)=1$ in ${\bf{Z_2}}[x],$
though  $(x^2+x+1)$ is irreducible, thus characteristic restrictions
are essential.

Let us now  look at the Gaussian numbers. We leave to the
reader the pleasure of creating similar conjectures as for integers, for example
the analogs of Goldbach and prime twins conjectures (twins seem to be
pairs with distance $\sqrt{2}$ between two elements; more history and variants can be found in  ``The Gaussian zoo'' \cite{ref:Zoo}). We go
into another direction.

Note, that because $2+i$ and
$2-i$ are ``positive'' primes and $5=(2+i)(2-i),$
 we should have $5'=(2+i)+(2-i)=4,$ but this does not
coincide with the earlier definition. So it may be is time to change our
point of view radically.


                 \section{Generalized derivatives}
                 \label{sec:Gen}
Our definition is based on two key points -- the Leibnitz rule and 
$p'=1$ for primes. If we skip the second one and use the Leibnitz rule
only
we get a more general definition of  $D(x).$ Now, if 
 $x=\prod_{i=1}^{k}{p_i^{x_i}},$ then 
$$D(x)=x\sum_{i=1}^{k}{\frac{x_iD(p_i)}{p_i}},$$
and we can again repeat most of the proofs above. But it is much more
natural to use another approach.
\begin{Thm}
 Let $R$ be a commutative ring without zero divisors
 and let $L:R^*\longrightarrow R^+$ be a homomorphism of
  its multiplicative semigroup to the additive group. Then a map
$$D:R\longrightarrow R, D(x)=xL(x),D(0)=0$$ satisfies the Leibnitz
rule.
Conversely,
if $D(xy)=D(x)y+xD(y)$ then $L(x)=\frac{D(x)}{x}$ is a homomorphism.
If $R$ is a field  then $L$ is a
group homomorphism and $$D\left(\frac{x}{y}\right)=
\frac{D(x)y-xD(y)}{y^2}.$$ 
\end{Thm}
\Pf $$D(xy)-D(x)y-xD(y)=xyL(xy)-xL(x)y-xyL(y)=xy(L(xy)-L(x)-L(y))$$
and we see that the Leibnitz rule is equivalent to the homomorphism
condition. If $R$ is a field then the semigroup homomorphism is
automatically the
group homomorphism and $L(1/x)=-L(x)$ which is sufficient 
to get $$D\left(\frac{1}{y}\right)=\left(\frac{1}{y}\right)\left(-L(y)\right)=
\frac{-D(y)}{y^2}.$$ Then it remains to repeat the proof of  Theorem
\ref{thm:frac}. \fP

\begin{Crl}
  There exist infinitely many possibilities to extend the derivative
  $x',$ constructed in  Section \ref {sec:nonrat} on $\bf{Q}$ to all real
  numbers  preserving  the Leibnitz rule.
\end{Crl}
\Pf We start from the positive numbers. It is sufficient to extend
$ld(x).$
 Note that the  multiplicative group of positive real numbers is isomorphic
 to the additive group and both of them are vector spaces over
 rationals.  In  Section \ref {sec:nonrat} a map $ld(x)$ is defined
 over a subspace and there are infinitely many possibilities to extend a
 linear
map from a subspace to the whole space. Obviously it would be a group
homomorphism and this gives a derivative for positive numbers. For the negative
numbers we proceed as in  Section \ref{sec:negative}. \fP

Note that the  Axiom of Choice is being used here.
It would be nice to find some ``natural'' extension, which preserves
 condition (\ref{eq:potens}), but note that no such
extension can be continuous. To show this let us consider a sequence
$$x_n=\frac{2^{a_n}}{3^n},a_n=\lfloor n\log_2 3\rfloor.$$ It is bounded and
has a convergent subsequence (even convergent to $1.)$ But
  $$\lim_{n\rightarrow
  \infty}(x_n)'=\lim_{n\rightarrow
  \infty}x_n\left( \frac{a_n}{2}- \frac{n}{3}\right)=\infty.$$

An example of continuous generalized derivative gives us $D(x)=x\ln
x.$
It is easy to construct a surjective generalized derivative in the set
of integers, and is impossible to make it injective (because
$D(1)=D(-1)=D(0)=0).$  But probably  even the following
conjecture is true.
\begin{Cnj}\label{cnj:nobiject}
  There is no generalized derivative $D(x)$ which is bijection between
  the set of natural numbers and the set of nonnegative integers.
\end{Cnj}

We can even hope for a stronger variant:
\begin{Cnj}\label{cnj:noinject}
  For any generalized derivative $D(x)$ on the set of integers there
  exist
two different positive integers which have the same derivative.
\end{Cnj}

Returning to the generalized derivatives in ${\bf Q}$ or ${\bf R}$ let us
investigate their structure as a set.
\begin{Thm}
  If $D_1, D_2$ are two generalized derivatives and $a,b$ are 
  some real numbers then 
$aD_1+bD_2$ and $[D_1,D_2]=D_1D_2-D_2D_1$ are generalized derivatives
too.
Nevertheless the set of all generalized derivatives is {\em not} a
Lie algebra.
\end{Thm}
\Pf We have
$$(aD_1+bD_2)(xy)=aD_1(xy)+bD_2(xy)=aD_1(x)y+axD_1(y)+bD_2(x)y+bxD_2(y)=$$
$$aD_1(x)y+bD_2(x)y+xaD_1(y)+xbD_2(y)=(aD_1+bD_2)(x)y+x(aD_1+bD_2)(y).$$
In the same way:
$$[D_1,D_2](xy)=(D_1D2-D_2D_1)(xy)=D_1D_2(xy)-D_2D_1(xy)=$$
$$D_1(D_2(x)y+xD_2(y))-
D_2(D_1(x)y+xD_1(y))=$$
$$D_1(D_2(x))y+D_2(x)D_1(y)+D_1(x)D_2(y)+xD_1(D_2(y))-$$
$$(D_2(D_1(x))y+D_1(x)D_2(y)+D_2(x)D_1(y)+xD_2(D_1(y)))=$$
$$D_1(D_2(x))y+
xD_1(D_2(y))
-D_2(D_1(x))y-xD_2(D_1(y))=$$
$$[D_1,D_2](x)y+x[D_1,D_2](y).$$
But the commutator is not bilinear: in general
$$[aD_1+bD_2,D_3]\neq a[D_1,D_3]+b[D_2,D_3]$$
so we have no Lie algebra structure. 
\fP

 Let us define $D_{(p_i)}$ as a
derivative which maps a prime $p_i$ to $1$ and other primes $p_j$ to zero.
Then $[D_{(p_i)},D_{(p_j)}]=0,$ but already $[3D_{(2)},D_{(3)}]=-D_{(2)}.$
Nevertheless every generalized derivative $D$ can be uniquely written as
$$D=\sum_{i=1}^ {\infty}D(p_i)D_{(p_i)}.$$

\section{The generating function}
\label{sec:genfunc}

Let $D(x)$ be a generalized derivative over the reals and
$L(x)=\frac{D(x)}{x}$
 be corresponding logarithmic derivative. Let 
$$H_D(t)=\sum_{n=0}^{\infty}D(n)t^n,H_L(t)=\sum_{n=1}^{\infty}L(n)t^n$$
be their generating functions.
\begin{Thm} \label{thm:genfunc}
The generating functions $H_D(t),H_L(t)$ can be  be
  calculated as follows:
 
$$H_D(t)=t\frac{d}{dt}\left( H_L(t)\right).$$
$$H_L(t)={\sum_{p }}'L(p)\sum_{j=1}^{\infty}\frac{t^{p}}{1-t^{p^j}},$$ where the
first sum runs over all primes.
  \end{Thm}
\Pf The first formula is equivalent to the condition
$D(n)=n\cdot L(n).$ As to the second formula it is sufficient to prove
it for the  special case when $L(p)=1$ for some prime $p$
 and $L(q)=0$ for all other primes. Then we need to prove that
$$\sum_{n=0}^{\infty} L(n)t^n
=\sum_{j=1}^{\infty}\frac{t^{p}}{1-t^{p^j}}.$$
If $n=p^km$ and $\gcd (p,m)=1$ then $t^n$ appears  exactly in $k$ sums
$$\frac{t^{p}}{1-t^{p^j}}=\sum_{i=1}^{\infty}t^{ip^j}$$ for $j=1,2,\ldots,k.$ It only remains to
note that $L(n)=k.$ \fP

\begin{Crl}
  $$L(n!)=\sum_{i=1}^{n}L(i)={\sum_{p\le n}}'L(p)\sum_{j=1}^{\infty}\left\lfloor\frac{n}{{p^j}}\right\rfloor.$$
\end{Crl}
\Pf If we replace every $\frac{t^p}{1-t^{p^j}}$  by $\sum_{i=1}^{ \lfloor\frac{n}{{p^j}}\rfloor}t^{ip^j}$
we do not change the coefficients in $t^k$ for $k\le n$ and make them
equal to zero for $k&gt;n.$ 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
$${\sum_{p\le n}}'\frac{1}{p} = O(\ln m), $$
$$
{\sum_{p&gt; n}}'\frac{n}{p(p-1)}&lt;\sum_{k&gt; n}\frac{n}{k(k-1)}=
\sum_{k&gt; n}n\left(\frac{1}{k}-\frac{1}{k-1}\right)\le 1,$$

$${\sum_{p\le n}}'\frac{n}{p^{m+1}(p-1)}&lt;{\sum_{p\le n}}'\frac{2n}{2^{m+1}p(p-1)}&lt;
\sum_{k\le n}\frac{2n}{nk(k-1)}\le 2,$$
we get
$$\sum_{i=1}^{n}\ld (i)={\sum_{p\le
  n}}'\frac{1}{p}\sum_{j=1}^{m}\left\lfloor\frac{n}{{p^j}}\right\rfloor=
{\sum_{p\le
  n}}'\frac{1}{p}\left(\sum_{j=1}^{m}\frac{n}{{p^j}}+O(m)\right)=$$
$${\sum_{p\le
  n}}'\frac{n}{p^{m+1}}\left(\frac{p^{m}-1}{{p-1}}\right)+O(\ln m)O(m)=
{\sum_{p }}'\frac{n}{p(p-1)} -$$
$$-{\sum_{p&gt;n}}'\frac{n}{p(p-1)}-{\sum_{p\le m}}'\frac{n}{p^{m+1}(p-1)}+O(\ln m)O(m)=$$
$${\sum_{p}}'\frac{n}{p(p-1)}+O(m\ln m).$$
\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 $\delta &gt;0.$
\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.
$$(\ref{Cnj-alt}) \Rightarrow (\ref{cnj:periods})
 \Rightarrow (\ref{cnj:period2}),$$
$$(\ref{cnj:rat1})
 \Rightarrow (\ref{cnj:natpp}), $$
$$ (\ref{cnj:even}) \Leftarrow (\ref{cnj:Goldbach},\mbox{\it
  Goldbach}),$$ 
$$ (\ref{cnj:nobiject}) \Leftarrow (\ref{cnj:noinject}),  $$
$$ 
\begin{array}{ccccc}
&amp;  &amp; (\ref{cnj:triples}, {\mbox Triples})&amp;
&amp;(\ref{cnj:infseq})\\
&amp; &amp;\Downarrow &amp; &amp;\Downarrow\\
  (\ref{cnj:twins},\mbox{\it Twins})&amp;
 \Rightarrow&amp; (\ref{cnj:diff2})&amp; \Rightarrow &amp;(\ref{Cnj-main})\\
\end{array}.$$ 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 $i(p)&gt;0$ 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


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