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\def\N{{\mathbb N}}
\def\Z{{\mathbb Z}}
\def\C{{\mathbb C}}

\def\a{\alpha}
\def\b{\beta}
\def\l{\lambda}
\def\cB{{\mathcal B}}
\def\cA{{\mathcal A}}
\def\cR{{\mathcal R}}
\def\A{{\mathbb A}}
\def\B{{\mathbb B}}
\def\moins{\raise 1pt\hbox{{$\scriptstyle -$}}}
\def\plus{\raise 1pt\hbox{{$\scriptstyle +$}} }
\def\s{\scriptstyle}
\def\bmoins{\fbox{\raise 1pt\hbox{{$\scriptstyle -$}}1}}

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\begin{document}

\title{\Huge\sc
 Addition of $\pm 1$
\ :\  Application to Arithmetic}

\maketitle

\begin{center}
Alain Lascoux\footnote{Partially supported by the EC's IHRP
Program, within the Research Training Network ``Algebraic Combinatorics
in Europe'', grant HPRN-CT-2001-00272.} \\
CNRS, Institut Gaspard Monge, Universit\'e de Marne-la-Vall\'ee\\
77454 Marne-la-Vall\'ee Cedex, France\\[2pt]
 and \\[2pt]
Center for Combinatorics, LPMC\\
 Nankai University, Tianjin 300071, P.R. China\\
Alain.Lascoux@univ-mlv.fr
\end{center}

\begin{abstract}

We show that some classical identities about multiplicative functions,
and the Riemann zeta-function, may conveniently be interpreted 
in terms of the addition of $\pm 1$ to some alphabets.

\medskip
\centerline{\bf R\'esum\'e}

Nous montrons que des identit\'es classiques concernant 
des fonctions multiplicatives de la th\'eorie des nombres,
et la fonction zeta de Riemann, s'interpr\`etent commod\'ement 
en terme de l'op\'eration~: ``Ajouter $\pm 1$ \`a un alphabet''.

\end{abstract}

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\markright{\its S\'eminaire Lotharingien de
Combinatoire \bfs 52 \rms (2004), Article~B52a\hfill}
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\section{Introduction}

In good courses of classical analysis, like the one of Jean-Yves Thibon
for the Marne-La-Vall\'ee students,
one finds exercises of the following type:

\medskip
\noindent{\bf Exercise 99.} Show the following two identities: 
\begin{eqnarray*}
\zeta(s)^3/\zeta(2s)   &= & \sum \tau(n^2)\,  n^{-s}   \\[3pt]
 \zeta(s)^4/\zeta(2s)  &=&  \sum \tau(n)^2\,  n^{-s} \ ,
\end{eqnarray*}
where $\zeta(s)$ is the { Riemann zeta function},
and $\tau(n)$ is the \emph{number of divisors} of $n$.

Such identities seem to be totally out of reach of
my own students\footnote{and also of me, because J.-Y.~Thibon
does not write the solution.},
who have  learnt from me only symmetric functions, division by $1$,
and divided differences.
 
I shall  however show that addition of $\pm 1$ is sufficient
to generate such identities,
and the link with symmetric function theory will be provided by
the notion of \emph{multiplicative function}:

$f : \N \to $ commutative ring is \emph{multiplicative} if and only if
 $f(mn)= f(m)f(n)$ whenever
$m,n$ are relatively prime.

Equivalently, the \emph{prime decomposition} of $n$ determines
the value of $f(n)$:
$$ n=\prod_{p\text{ prime}} p^{k_p}\quad   \Rightarrow \quad  f(n)
         = \prod_{p\text{ prime}}  f\left( p^{k_p}  \right)  \ . $$


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But instead of considering the sequence
$$  [f(p^0),\, f(p^1),\, f(p^2),\, f(p^3),\, \ldots]  \ ,$$
or the generating function $\sum_{k=0}^\infty f(p^k) z^{k}$,
as does Lagrange, 
\emph{Symmetric Function Theory} \cite{Lascoux}
 tells us that we better introduce
an \emph{alphabet} $\A_p$, such that, for $k\in\N$, 
$f(p^k) $ is the $k$-th 
\emph{complete function} in this alphabet,
denoted by $S^k(\A_p)$, i.e.,
such that
$$ \sum_{k=0}^\infty f(p^k) z^{k} =\sigma_z(\A_p) := \sum_{k=0}
^\infty z^k S^k(\A_p) =
\prod_{a\in\A_p} (1\moins za)^{-1} .$$

In other words,
each generating function is {\it formally} factorized and gives rise to
an alphabet.

In summary, \emph{a multiplicative function is determined by a 
collection of alphabets $\cA= \{\A_p, p\text{ prime}\}$}.
In this sutation, this multiplicative function will be
denoted by $\mu_{\cA}(.)$.   

\medskip
It may be appropriate, before going any further, to recall a few facts
about symmetric functions.
An alphabet $\A=\{a_1,a_2,\ldots\}$ is written as the sum
of its elements: $\A=a_1+a_2+\cdots$, and the generating function
of complete functions in $\A$  is 
$$ \sigma_z(\A)=
\sum _{k=0} ^{\infty}z^kS^k(\mathbb A)= \prod_i (1-za_i)^{-1} \ . $$

Adding a letter $x$ to $\A$ is denoted by $\A \to \A+x$.
Adding three copies of the letter $1$ is denoted by $\A \to \A+3$.

More generally, given alphabets $\A,\B$, 
and complex numbers $k,r$,
the generating function $\sigma_z(k\A\pm\B +rx)$ is equal to
\begin{equation}
 \sigma_z(k\A\pm \B+rx)  =
 \prod_{\a\in\A} (1\moins z\a)^{-k}
\, \prod_{\b\in \B} (1\moins z\b)^{\mp 1}
(1-zx)^{-r}  \ . 
\end{equation}

We may need to specialize a letter to $3$,
but this must not be confused with taking three copies of $1$.
To allow nevertheless specializing a letter to a complex number $r$
 inside a symmetric function, without introducing 
intermediate variables, we write \fbox{$r$} for this specialization.
\emph{Boxes have to be treated as single variables.}

In other words, 
given a letter $x$, and  complex numbers $r, \alpha, \beta$, then
$$  \sigma_z\left( \alpha\, \fbox{$r$} +\beta\, x \right)  =
  (1- z\, r)^{-\alpha}\, (1-x)^{-\beta}  \ .$$


When all alphabets $\A_p$ are equal to the same alphabet $\A$,
we write $\A$ for the collection of $\cA=\{ \A,\A,\ldots\}$.
We shall also need the collection where $\A_p=\fbox{$p$}$,
that we denote by $\cA= \fbox{$\mathbf p$}$. It corresponds to the
multiplicative function $f(n)=n$.   



We shall now formulate properties of multiplicative functions
in terms of another classical object in number theory.

Given a function $f$ from $\N$ in a commutative ring, 
one can associate to it a \emph{Dirichlet series}%
\footnote{We shall not address convergence issues,
but shall treat the Dirichlet series as formal series.
However, for all identities that we consider here, the formal treatment
is {\it sufficient\/} to also deduce their validity
as identities for analytic series. For, all our series converge in some domain,
which must in fact be a (right) half plane (see \cite[Theorem~11.1]{Apostol}),
and then it follows from uniqueness of coefficients (see 
\cite[Theorem~11.3]{Apostol}) that the formal identity is at the same time
an analytic identity.}
$$ F(f;s) =   \sum_{n\geq 1} f(n) n^{-s}  \ , $$
the most notable being the \emph{Riemann zeta function}
$\zeta(s) := \sum_{n\geq 1}  n^{-s}$.


The product of Dirichlet series corresponds to  the 
\emph{convolution} of functions on $\N$: 
$$ F(f;s) F(g;s) = F(f\star g;s)
  = \sum_n \biggl(\sum_{d|n} f(d)g(n/d)\biggr) n^{-s} \ .$$

When $f$ is a multiplicative function, say $f=\mu_{\cA}$,
we  write $\zeta(\cA;s)$ for the corresponding 
Dirichlet series.  In this notation, the  Riemann zeta function is 
$\zeta(1;s)$, that is, each alphabet  $\A_p$ is equal to $1$.
%More generally, when $r$ is a positive integer, and $\alpha\in\C$, 
%$\zeta(rs-\alpha) := \sum_{n\geq 1} n^\alpha n^{-sr}$.
%The corresponding multiplicative function is such that
%$\mu(p^k)=0$ if $k$ is not a multiple of $r$, and
% $\mu(p^{kr})= p^{k\alpha}$. The alphabet $\A_p$ is such that
% $\sigma_z(\A_p)= (1-z^r p^\alpha)^{-1}$.


The product of two series $\zeta(\cA;s)$, $\zeta(\cB;s)$
is the series $\zeta(\cA+\cB;s)$, because for every $p$,
$\sigma_z(\A_p) \sigma_z(\B_p)= \sigma_z(\A_p+\B_p)$. 
Thus addition of alphabets
corresponds to products of Dirichlet series.

Notice that the convolution of multiplicative functions
with the classical M\"obius function   $\mu_{-1}$ or its
inverse $\mu_{1}$ (cf. \cite{Apostol}),
or the product of a Dirichlet series and $\zeta(s)^{-1}$ or
$\zeta(s)$, 
 are now interpreted as the addition
to each alphabet $\A_p$
 of $\moins 1$
or $1$  respectively:
\begin{eqnarray}
  \mu_{\cA-1}(n)   =  \sum_{d:\, d|n} \mu(n/d)\, \mu_{\cA}(d)
  & \Leftrightarrow&   \zeta(\cA-1,s) = \zeta(\cA,s) /\zeta(s)
    \\ 
  \mu_{\cA+1}(n)   =  \sum_{d:\, d|n} \mu_{\cA}(d)
 & \Leftrightarrow&   \zeta(\cA+1,s) = \zeta(\cA,s)\, \zeta(s)
    \  .
\end{eqnarray}

\section{Adding 1}

We are now in the position to
solve the exercise stated at the beginning of this text.

For the alphabets $2-\fbox{$\moins 1$}$ and 
$3-\fbox{$\moins 1$}$ we have
$$\sigma_z\left(2-\fbox{$\moins 1$}  \right)  = 
(1-(\moins 1)z) (1-z)^{-2} = 1 + 3 z + 5 z^2  + 7 z^3  + \cdots   $$
$$ \sigma_z\left(3-\fbox{$\moins 1$}  \right)  =  (1-(\moins 1)z)
(1-z)^{-3} =   1 + 4 z + 9 z^2  + 16 z^3 +\cdots\ . $$

Since  $\tau(p^{2k})=  2k+1$, it follows that 
$$ \zeta(s)^3/\zeta(2s)= \zeta\left( 2-\fbox{$\moins 1$}\, ,s  \right) = 
\sum_{n\ge1} \tau(n^2)\,  n^{-s} \ .  $$
Similarly,   $\tau(p^k)^2 = (k\plus 1)^2$ implies  
$$ \zeta(s)^4/\zeta(2s)= \zeta\left( 3-\fbox{$\moins 1$}\, ,s  \right) = 
\sum_{n\ge1} \tau(n)^2\,  n^{-s} \ , $$
and this proves the two required formulas.

In fact, many of the 
 classical arithmetical multiplicative functions 
occur in products of zeta functions, as shows the following table.
We need only identify the generating functions 
$\sigma_z(\A_p)$ for all alphabets appearing in the left column,
to ensure the validity of the expansions given in the right
column.   

$$ \begin{array}{|c|c|c|c|}
\s alphabet & \s series &\s \sigma_z(\A_p) & \s expansion \\
 -1  &  1/\zeta(s)  & 1-z 
                       &  \sum  \mu_{-1}(n)\, n^{-s}   \\[3pt]
 2   &  \zeta(s)^2 & (1\moins z)^{-2} 
                            & \sum  \tau(n)\,  n^{-s}   \\[3pt]
 r   &  \zeta(s)^r  &  (1\moins z)^{-r}
                            & \sum \tau_r(n)\,  n^{-s}   \\[3pt]
 -\bmoins & \zeta(s)/\zeta(2s)   &  1-(\moins z) 
   & \sum  \left|\mu_{-1}(n)\right| n^{-s}   \\[3pt]
 1- \bmoins & \zeta(s)^2/\zeta(2s)  & (1\plus z)/(1\moins z) 
   & \sum 2^{\omega(n)}\, n^{-s}  \\[3pt]
 2- \bmoins & \zeta(s)^3/\zeta(2s)   & (1\plus z)/(1\moins z)^2
                    & \sum \tau(n^2)\,  n^{-s}   \\[3pt]
 3- \bmoins & \zeta(s)^4/\zeta(2s)  & (1\plus z)/(1\moins z)^3 
                      & \sum \tau(n)^2\,  n^{-s}   \\ [3pt]
 1 +\fbox{$p^\alpha$}  & \zeta(s)\, \zeta(s-\alpha)  &
      (1\moins z)^{-1} (1\moins zp^\alpha)^{-1}
    & \sum \left(\sum_{d|n} d^\alpha\right)  n^{-s}   \\ [3pt]
 \fbox{$p$} -1 &  \zeta(s-1)/\zeta(s)  &
      (1\moins z)/ (1\moins zp)  
                    &  \sum \phi(n)\,  n^{-s}
\end{array}
$$

In the table, we use the function  $\tau(n)$,
which, as before, is the number of divisors of $n$, and,
more generally, $\tau_r(n)$, $r>2$, is the number of 
integral vectors $[n_1,\ldots, n_r]$ such that 
$n=n_1\cdots n_r$. In addition,
we denote by $\omega(n)$ the number of prime divisors of $n$
and by $\phi(n)$ the number of integers less than $n$ and prime to $n$. 

Let us elaborate in some more detail about the relation between
the third column (generating function for the alphabet)
and the fourth column (Dirichlet series), in the order the alphabets are
listed in the table:
\begin{itemize}
\item  $1-z$, i.e. $\mu(p)=-1$, $\mu(p^k)=0$, $k\geq 2$. 
            This is this the classical M\"obius function; 
\item $(1-z)^{-2}$, $\mu(p^k)= k+1=\tau(p^k)$ is the number of divisors
of $p^k$;
\item $(1-z)^{-r}$, $\mu(p^k)= \binom{k+r-1}{k-1}$ is the number of 
$r$-tuple $[n_1,\ldots, n_r]$ such that $n_1\cdots n_r = p^k$;
\item $ 1-(\moins z)$, $\mu(p)=1$, $\mu(p^k)=0$, $k\geq 2$;
\item $ (1+z)/(1-z)$, $\mu(p^k)=2$, $k\geq 1$;
\item $ (1+z)/(1-z)^2$, $\mu(p^k)=2k+1$ is the number of divisors of
   $(p^k)^2$;
\item $ (1+z)/(1-z)^3$, $\mu(p^k)=(k+1)^2$ is the square of the 
number $\tau(p^k)$ of divisors of $p^k$;
\item $(1-z)^{-1} (1-zp^\alpha)^{-1}$, $\mu(p^k)= 1+p^\alpha+\cdots 
+ p^{k\alpha}$
 is the sum of the $\alpha$-th powers of the divisors $\{1,p,\ldots, p^k\}$
 of $p^k$;
\item  $(1-z)/(1-zp)$, $\mu(p^k)= p(p^{k-1}-1)$ is the number of integers
less than $p^k$ and prime to $p$.
\end{itemize}

A natural generalization of the classical 
M\"obius function proceeds by taking
some complex number $\a$, instead of $\moins 1$,
 and setting  all $\A_p=\a$. Recall that
$$ S^k(\a) = \a (\a+1)\cdots (\a +k-1) / \, k! = \binom{\a+k-1}{k} \ . $$
Hence,
\begin{equation}
 \mu_{\a}\biggl(\prod_p  p^k\biggr) = \prod_p \binom{\a+k-1}{k} \ .
\end{equation}
These M\"obius functions of higher order have been considered
by Fleck \cite{Fleck, Dickson}\footnote{and still reappear 
in recent literature \cite{Brown}.}, 
who proved the convolution formula
 $ \mu_{\a} \star \mu_{-1} = \mu_{\a -1}  $.

Taking Dirichlet series instead of alphabets, Fleck's results
 translate into the seemingly more elaborate formula
\begin{equation}
\sum_{n=1}^\infty \mu_{\a}(n) n^{-s} = \zeta(s)
    \sum_{n=1}^\infty \mu_{\a-1}(n) n^{-s}   \ ,
\end{equation}
which is nothing else but
\begin{equation}
   \alpha = 1 +  (\alpha -1)   \ . 
\end{equation}

\def\boiteH{\mathbin{\hbox{$\bigcirc\kern-.78em$\scriptsize H\hskip3pt}} }

\section{Hadamard Products}

One can use other operations on alphabets than just adding
$\pm 1$.  For example, 
the ``pointwise'' product of two multiplicative functions is still
a multiplicative function.  It corresponds to the Hadamard product
of generating series: 
$$ \left(\sum_{i\ge0} z^i S^i(\A)\, ,\, \sum_{i\ge0} z^i S^i(\B) \right)
  \longrightarrow   \sum_{i\ge0} z^i S^i(\A) S^i(\B) \ ,  $$
or the Hadamard product of Dirichlet series, 
and induces an ``Hadamard product''
%on pair of alphabets : $ (\A,\B) \to \A\, \Disk{${\scriptsize H}$}\, \B$.
on pair of alphabets : $ (\A,\B) \to \A
\boiteH \B$.

Indeed, given two finite alphabets $\A$ and $\B$, the Hadamard product of
$\sigma_z(\A)$ and $\sigma_z(\B)$ is a rational function,
with denominator 
$$ \sigma_z(-\A\B):=\prod_{a\in\A}\prod_{b\in\B}(1-zab)^{-1} \ . $$ 
However, in general the numerator
will be too complicated to be applied to the theory
of the Riemann zeta function. I shall nevertheless give three 
explicit cases that I
found in the literature.

The first one is due to Ramanujan:
\begin{equation}
 \frac{ \zeta(s)\zeta(s-\alpha)\zeta(s-\beta)\zeta(s-\alpha-\beta)}{
     \zeta(2s-\alpha-\beta)} = \sum_{n\ge1} \frac{\sum_{d|n}d^\alpha 
               \sum_{d|n}d^\beta }{n^s} \ .
\end{equation}   
Indeed, 
$$ \frac{1-z^2xy}{(1-z)(1-zx)(1-zy)(1-zxy)} =
 \sum_{i\ge0} (1+x+\cdots+x^i)(1+y+\cdots+y^i) z^i    \ ,$$
which means that the Hadamard product of 
\begin{equation*} 
  (1-z)^{-1}(1-zx)^{-1}  \quad \text{and} \quad 
  (1-z)^{-1}(1-zy)^{-1}
\end{equation*}
 is the left-hand side. This left-hand side can be written
$\sigma_z(1+x+y+xy-\Omega)$, with $\Omega$ an alphabet of
cardinality $2$ such that $S^1(\moins\Omega)=0$ and 
  $S^2(\moins\Omega)=\moins xy$. 


When $x=p^\alpha$, $y=p^\beta$,  the generating function 
$\sigma_z(1+x+y+xy-\Omega)$ specializes
to the component depending on $p$ of 
$ \frac{ \zeta(s)\zeta(s-\alpha)\zeta(s-\beta)\zeta(s-\alpha-\beta)}{
     \zeta(2s-\alpha-\beta)}$, and, thus eventually, 
Ramanujan's identity is indeed just
\begin{equation}
  (1\plus x) \boiteH (1\plus y)  
=  (1\plus x \plus y\plus xy) - \Omega \ . 
\end{equation}


\bigskip
The second example uses the Hadamard powers of the simplest\footnote{
The Hadamard product of $f(z)$ with $(1-z)^{-1}$ being $f(z)$, 
one has to argue between $(1-z)^{-2}$ and $(1-2z)^{-1}$
for the title of being the simplest non-trivial series. Notice that 
not only  $1/(1-z)$ is stable under Hadamard powers,
but so is $1/(1-z^\alpha)$ for any positive integer $\alpha$.} 
series, $(1-z)^{-2}$.


By definition the $k$-th Hadamard power of 
$1+2z+3z^2+\cdots$ is 
$$ 1 +2^kz +3^k z^2 +4^k z^3 +\cdots =  {\mathcal E}_k(z)\, (1-z)^{k+1} \ ,$$
where ${\mathcal E}_k(z)$ is the $k$-th {\it Eulerian polynomial}, 
dear to combinatorialists \cite[Ch.~7]{Riordan}.

Define now, for any positive integer $k$, the function 
\begin{equation}
  \eta_k(s)  =
   \frac{1}{1-2^{-s}} 
\underset {p\equiv 1\ (\text{mod } 4) }{\prod_{p\text{ prime}} }
\frac{{\mathcal E}_k(p^{-s} ) }{
     1-p^{-s}}
\underset {r\equiv 3\ (\text{mod } 4) }{\prod_{r\text{ prime}} }
 \frac{1}{1-r^{-2s}}  \ .
\end{equation}

Therefore, thanks to our computations of the Hadamard powers of\break
$(1-z)^{-2}$ and $(1-z^2)^{-1}$, we see that 
$ \eta_k(s)$ is the $k$-th Hadamard power of $\eta_1(s)$.

The alphabets corresponding to $\eta_1(s)$ are
$$  \A_2= 1\ ,\quad \A_p = 2\ ,\quad \A_r = 1+\bmoins \ , $$
since $\sigma_z(1) =1/(1-z)$, $\sigma_z(2) =1/(1-z)^2$,
$\sigma_z\left(1+\bmoins  \right) = 1/(1-z^2)$.

Subtracting $1$ from all the alphabets, one gets a simpler function:
\begin{equation}  L(s) := \eta_1(s)/ \zeta(s) = 
  \prod_{p\equiv 1} \frac{1}{1-p^{-s}}\, 
      \prod_{r\equiv 3} \frac{1}{1+r^{-s}}  
    =  \sum_{n=0}^\infty   \frac{(\moins 1)^n}{(2n+1)^s} \ ,  
\end{equation}
corresponding to the alphabets
$$  \A_2= 0\ ,\quad \A_p = 1\ ,\quad \A_r = \bmoins \ . $$


Since the Hadamard square of $(1-z)^{-2}$ is equal to
$$ (1+z)\, (1-z)^{-3}  = \sigma_z\left( 3- \bmoins \right)    \ ,$$
the function $\eta_2(s)$  is associated to the alphabets
$$  \A_2= 1\ ,\quad \A_p =  3- \bmoins \ ,\quad \A_r = 1+\bmoins \ . $$

The Hadamard cube of $(1-z)^{-2}$ being
$$  (1+4z+z^2)\, (1-z)^{-4}  = 
            \sigma_z\left(4-\alpha -\beta\right)   \ ,$$
with $\alpha=\fbox{$\moins 2+\sqrt{3}$}$ , 
$\beta=\fbox{$\moins 2-\sqrt{3}$}$ ,
the function $\eta_3(s)$  is associated to the alphabets
$$  \A_2= 1\ ,\quad \A_p =  4- \alpha-\beta\ ,\quad \A_r = 1+\bmoins \ . $$

Finally, the fourth Hadamard power being
\begin{align*}
 (1+11z+11z^2+z^3)\, (1-z)^{-5}  &= 
    (1+z)(1+10z+z^2)\, (1-z)^{-5} \\
&=
  \sigma_z\left(5- \bmoins \moins\alpha \moins\beta\right)   \ ,
\end{align*}
with $\alpha=\fbox{$\moins 5+ 2\sqrt{6}$}$ ,
$\beta=\fbox{$\moins 5- 2\sqrt{6}$}$ ,
the function $\eta_4(s)$  is associated to the alphabets
$$  \A_2= 1\ ,\quad \A_p =  5- \bmoins 
-\alpha-\beta\ ,\quad \A_r = 1+\bmoins \ . $$


\bigskip
Let us close this text by considering higher powers $(1-z)^{-r}$, with 
$r>2$.  I shall restrict myself to their Hadamard squares.

Since 
$$  \frac{1}{(1-z)^r} \boiteH \frac{1}{(1-z)^r}  =
    P_{r-1}\left(\frac{1+z}{1-z} \right)\,  \frac{1}{(1-z)^r}  \, , $$
where $P_j$ is the {\it Legendre polynomial}
\cite{Riordan} of degree $j$, one has,
using the function $\tau_r(n)$ defined above,
\begin{equation}
   \sum_{n=1}^\infty \bigl(\tau_r(n)\bigr)^2\,  n^{-s}  =
        \zeta(s)^r\,   \prod_p  \left(P_{r-1} \left( \frac{1+p^s}{1-p^s}
 \right)  \right)  \ . 
\end{equation}

\begin{thebibliography}{1}

\bibitem{Apostol}
T. Apostol, \emph{Introduction to Analytic Number Theory}, Springer (1976).

\bibitem{Brown}
T. Brown, L.C. Hsu, J. Wang, P.J.-S. Shihue,
\emph{ On a certain kind of generalized number-theoretical M\"obius function}, Math.\ Scientist {\bf 25} (2000) 72-77.

\bibitem{Dickson}
L.E. Dickson, \emph{ History of the theory of numbers}, vol.~{\bf 1},
 Chelsea reprint (1952).

\bibitem{Fleck}
A. Fleck, \emph{\"Uber gewisse allgemeine zahlentheoretische
Funktionen, insbesondere eine der Funktion $\mu(m)$ verwandte Funktion
$\mu_k(m)$}, Sitzungsber.\ Berl.\ Math.\ Ges.\ {\bf 15} (1916) 3--8.

\bibitem{Lascoux} A. Lascoux,
\emph{Symmetric functions and combinatorial operators on polynomials},
CBMS/AMS Lectures Notes {\bf 99}, (2003).

\bibitem{Riordan}  J. Riordan,
\emph{Combinatorial identities}, Wiley (1968).

\bibitem{Titchmarsh1}
E.C. Titchmarsh, \emph{The zeta-function of Riemann},
   Cambridge University Press (1930).  

\bibitem{Titchmarsh2}
E.C. Titchmarsh, \emph{The theory of functions},
  Oxford University Press (1932).

\end{thebibliography}

\end{document} 






