\input amstex
\documentstyle{amsppt}
\NoBlackBoxes

\leftheadtext{P.J. Grabner,
P. Kirschenhofer, H. Prodinger and 
R.F. Tichy}
\define\({\left(}
\define\){\right)}
\define\ups{\upsilon}
\define\lf{\lfloor}
\define\rf{\rfloor}
\topmatter
\title On the Moments of the Sum-of-Digits Function \endtitle
\author P.J. Grabner$^\dag$, 
P. Kirschenhofer, H. Prodinger and 
R.F. Tichy$^\dag$\endauthor
\thanks
{$^\dag$ These authors were
supported by the Austrian Science Foundation (Project Nr. P8274PHY)}
\endthanks
\address 
\hbox{
\vbox{\hbox to 6 truecm {P. Grabner and R. Tichy\hfill}
\hbox to 6 truecm{Institut f\"ur Mathematik\hfill}
\hbox to 6 truecm{TU Graz\hfill}
\hbox to 6 truecm{Steyrergasse 30\hfill}
\hbox to 6 truecm{8010 Graz, Austria\hfill}}
\vbox{\hbox to 6 truecm{P. Kirschenhofer and H. Prodinger\hfill}
\hbox to 6 truecm{Institut f\"ur Algebra und\hfill}
\hbox to 6 truecm{Diskrete Mathematik, TU Wien\hfill}
\hbox to 6 truecm{Wiedner Hauptstra\ss e 8--10\hfill}
\hbox to 6 truecm{1040 Wien, Austria\hfill}}}
\endaddress
\abstract  We are concerned 
with the asymptotic description of the higher 
moments of the sum-of-digits function. 
For the mean value a completely satisfactory result 
involving a continuous and no-where differentiable oscillation 
function is due to Trollope (1968) and 
H. Delange (1975). The asymptotic behaviour of the second moment of the 
binary sum-of-digits 
function was investigated by J. Cocquet (1986) and by 
P. Kirschenhofer (1990). A somewhat weaker result in the decimal number 
system is due to R. E. Kennedy and C. N. Cooper (1991). 
In the present paper we prove similar (more 
or less explicit) formulas in the case of $s$-th moments by a generalization of Delange`s 
approach. Furthermore we discuss a second method 
for proving such results: the so-called Mellin-Perron 
summation formula. 
\endabstract
\endtopmatter



\heading 1. Introduction\endheading

In a recent paper in the {\sl Fibonacci Quarterly\/}
R. E. Kennedy and C. N. Cooper \cite{K-C91} dealt with the second moment
of the sum-of-digits function in the decimal number system, explicitely
stating the question for higher moments as an open problem. We consider
this question in the sequel.


Let $\nu(n)$ represent the total number of $1$-digits in the binary
 representation
of the integer $n$. It is not hard to see that
$$M_1(N)=\sum_{n<N}\nu(n)=\frac 12N\log_2N+o(N\log N),\tag 1.1$$
since asymptotically the binary representations contain roughly as many $0$'s
as $1$'s. The Trollope--Delange formula is more surprising: It expresses $S(n)$
in an exact formula (cf. \cite{Tr68}, \cite{De75})
$$M_1(n)=\frac 12N\log_2N+N\delta_1(\log_2N),\tag 1.2$$
where $\delta_1(u)$, a `fractal function', is a continuous, periodic, nowhere
differentiable function with an explicit Fourier expansion involving the
Riemann zeta function.
                   
The argument given by Delange is based on a combinatorial decomposition of
binary representations of integers, followed by a computation of the Fourier
coefficients of the fractal function. This exact formula was used to analyze
register allocation algorithms, or equivalently the order of random `channel networks'
(cf.\cite{F-R-V79}).
It was later extended to some non-standard digital representations of
 integers. Cantor representations
were considered in \cite{K-T84}, and
Gray code \cite{F-R80} was studied for the purpose of analyzing sorting networks.
In \cite{Ki83} occurrences of blocks of digits in standard $q$-ary representations,
in \cite{K-P84} subblock occurrences in Gray code representation and
in \cite{Co83} the 
subsequence $u(3n)$ of the classical Thue-Morse sequence $u(n)$
 were investigated. In a recent
paper \cite{G-T91} this result was extended to digit expansions with respect
to linear recurrences. As general surveys we refer to Stolarsky's article
\cite{Sto77} and to \cite{K-P-T85}. As an especially important paper in this
area we mention Brillhart, Erd\H os and Morton's paper \cite{B-E-M83}.
 
The second moment $M_2(N)$ was investigated by J. Cocquet \cite{Co86},
and independently
in \cite{Ki90}. The following exact formula
is proved by Delange's method:
$$M_2(N)=\sum_{n<N}\nu(n)^2=\frac N4\log^2_2N+N\log_2N\(\frac14+\delta_1(\log_2N)\)+N\delta_2(\log_2N),\tag1.3$$
where $\delta_1$, $\delta_2$ are continuous nowhere differentiable functions of period $1$ and $\delta_1$
is the function that occurred in \thetag{1.2}.
By the same approach such exact formul\ae{} can be proved for arbitrary $q$-ary number systems:
$$\sum_{n<N}\nu_q(n)^2=\(\frac{q-1}2\)^2N\log^2_qN+N\log_qN\eta_1(\log_qN)+N\eta_2(\log_qN),\tag1.4$$
where $\eta_1$ and $\eta_2$ are continuous nowhere differentiable functions of period $1$ and
$\log_q$ denotes the logarithm to base $q$, as usual.
In \cite{K-C91} R.E. Kennedy and C.N. Cooper rediscovered a weaker result for the
decimal system:
$$\sum_{n<N}\nu_{10}(n)^2=20.25N\log_{10}N+o(N\log N).$$

In \cite{Co86} a similar formula is established for higher moments but
without proving the continuity of the fluctuations. 
In Section 2 we use Delange's approach to compute the 
third moment in the binary number
system, including the proof of the continuity
of the fluctuations. The $q$-ary case can be settled by similar but more elaborate 
computations.
In Section 3 we sketch this approach in the case of higher moments.
\footnote{While finishing the present paper the authors got to know
a manuscript due to J.-M. Dumont and A. Thomas where the corresponding
result is proved for digital systems generated by substitution automata.
This more general result is proved by different and less elementary
methods.} 
In the final Section 4
we present an exposition of a purely analytic approach using Mellin transform. In \cite{F-G-K-P-T92}
this approach was applied to establish exact and asymptotical formul\ae{} in various cases
of digital sums. By this method we get automatically the mean and the Fourier-coefficients of
the periodic functions. In detail we apply this method to the expectation (first moment) of
the $q$-ary sum-of-digits function. In the case of higher moments some analytic problems
remain to be solved.
 
 
\heading 2. Delange's Approach: The Third Moment\endheading




In order to compute the third moment $M_3$ of the binary sum-of-digits 
function we introduce
the following notation:
$$f_j(t)=\left\lf\frac t{2^j}\right\rf
-2\left\lf\frac t{2^{j+1}}\right\rf,\quad g_j(t)=f_j(t)-\frac 12.$$
Thus we have
$$\align M_3(N)&=\int_0^N\sum_{(j_1,j_2,j_3)} 
f_{j_1}(t)f_{j_2}(t)f_{j_3}(t)dt\cr
&=\sum_{k=0}^3{3\choose k}\frac1{2^{3-k}}\int_0^N\sum_{(j_1,\dots,j_k)}
g_{j_1}(t)\dots g_{j_k}(t)dt,\endalign$$
where the summation is extended over all $k$-tuples $(j_1,\dots,j_k)$ 
of non negative integers
$\leq L=\lf\log_2 N\rf$. Denoting the integral for $k=3$ by $I_3$ we obtain
$$
\aligned M_3(N)&=I_3+\frac32(L+1)M_2(N)-\frac 34(L+1)^2M_1(N)+
\frac N8(L+1)^3\cr
&=I_3+N\Bigl(\frac 18K^3+K^2\big(\frac38+\frac34\delta_1\big)
+K\big(\frac32\delta_2+
\frac38y\big)\cr
&+ \big( \frac32y\delta_2-\frac34y^2\delta_1+\frac18y^3\big)\Bigr),
\endaligned\tag 2.1
$$
where $\delta_1=\delta_1(\log_2N)$ and $\delta_2=\delta_2(\log_2N)$ 
are the periodic fluctuations
of the first and second moments occurring in \thetag{1.2} and \thetag{1.3}; 
$K=\log_2N$ and
$y=1-\{\log_2N\}$. We note that $\delta_1(0)=\delta_1(1)=
\delta_2(0)=\delta_2(1)=0$.
 
Now we compute the remaining integral
$$\align I_3&=\(\sum_{j_1=j_2=j_3}+\quad
3\sum_{j_1=j_2\not=j_3}+\quad
6\sum_{j_1<j_2<j_3}\)\int_0^Ng_{j_1}(t)g_{j_2}(t)g_{j_3}(t)dt\cr
&=\frac14\sum_j\int_0^Ng_j(t)dt+\frac34L\sum_j\int_0^Ng_j(t)dt+6\sum_{j_1<j_2<j_3}\int_0^Ng_{j_1}(t)g_{j_2}(t)g_{j_3}(t)dt.\endalign$$
Notice that the first two parts of this equality follow from the fact that $g_j(t)$ is piecewise
constant and $g_j(t)^2=\frac14$.
For the last summation we obtain
$$\sum_{j_1<j_2<j_3}=\sum_{0\leq j_1<j_2<j_3\leq L}2^{j_3+1}\int_0^N g(2^{j_3-j_1}u)g(2^{j_3-j_2}u)g(u)du,$$
where $g(u)=\lf2u\rf-2\lf u\rf-\frac12$.
Now transform the index of summation as follows:
$j:=j_3$, $d_1:=j_3-j_1$, $d_2:=j_3-j_2$. Thus we obtain
$$\sum_{0\leq j_1<j_2<j_3\leq L}=\sum_{j=0}^L2^{j+1}\sum_{1\leq d_1<d_2\leq j}\int_0^{\frac N{2^{j+1}}}
g(2^{d_1}u)g(2^{d_2}u)g(u)du.$$
Notice that the inner summation can be extended to infinity without changing the value and set
$k=L-j$. Hence
$$\align \sum_{0\leq j_1<j_2<j_3\leq L}&=2^{L+1}\sum_{k\geq0}\sum_{1\leq d_1<d_2}\int_0^{2^{k+\{\log_2N\}-1}}
g(u)g(2^{d_1}u)g(2^{d_2}u)du\cr
&=2^{1-\{\log_2N\}}\sum_{k\geq0}2^{-k}h_3(2^{\{\log_2N\}-1}2^k),\endalign$$
where $h_3$ is the periodic function defined by
$$h_3(x)=\sum_{1\leq d_1<d_2}\int_0^xg(u)g(2^{d_1}u)g(2^{d_2}u)du.\tag2.2$$
We observe that the integrals occurring in \thetag{2.2} are integrals over distinct Rade\-macher functions;
hence $h_3(1)=h_3(\frac12)=0$.
Thus setting
$$\psi(x)=2^{1-\{x\}}\sum_{k\geq0}2^{-k}h_3(2^{k}2^{\{x\}-1})\tag2.3$$
we obtain a continuous and $1$-periodic function with $\psi(0)=\psi(1)=0$. 
Therefore we have
$$I_3=N\(K\big(\frac34\delta_1-\frac38y\big)+\big(-\frac12\delta_1+
\frac x4+\frac34x\delta_1-
\frac38x^2+6\psi(1-y)\big)\).\tag2.4$$
 
By combining \thetag{2.1} and \thetag{2.4} we derive
$$
M_3(N)=N\(\frac18K^3+K^2
\big(\frac38+\frac34\delta_1\big)+K\big(\frac34\delta_1+
\frac32\delta_2\big)+\delta_3\),\tag2.5
$$
where
$$\delta_3=\delta_3(\log_2N)=\frac32y\delta_2-\frac34y^2\delta_1+\frac18y^3-\frac12\delta_1+\frac y4+\frac34y\delta_1-\frac38y^2
+6\psi(1-y).\tag 2.6$$
Thus we have proved

\proclaim{Theorem 1} The third moment of the binary sum-of-digits function satisfies
$$\align \frac1N\sum_{n<N}\nu_2(n)^3&=\frac18\(\log_2N\)^3+\(\frac38+\frac34\delta_1(\log_2N)\)\(\log_2N\)^2\cr
&+\(\frac34\delta_1(\log_2N)+\frac32\delta_2(\log_2N)\)\log_2N+\delta_3(\log_2N),\endalign$$
where $\delta_3$ is a continuous and periodic function with period $1$ defined by \thetag{2.6}.
\endproclaim

{\bf Remark.} Obviously this result can be extended to 
$q$-ary representations. However, the
computations are much more involved.
 
 
 
\heading 3. Delange's Approach: The General Case\endheading
 
In the following we sketch the computation of the $s$-th moments $M_s$
of the binary sum-of-digits function:

Adopting analogous notations as in the previous section we start from

$$
\align 
M_s(N)&=\int_0^N\sum_{(j_1,\dots,j_s)} f_{j_1}(t)\dots f_{j_s}(t)dt\cr
&=\sum_{k=0}^s{s\choose k}\frac1{2^{s-k}}\int_0^N\sum_{(j_1,\dots,j_k)}
g_{j_1}(t)\dots g_{j_k}(t)dt.
\endalign
$$

In the following we evaluate the integral $I_k$ appearing in the last
expression. To this end we observe that $g_j^2(t)\equiv\frac 14$, so that
it is meaningful to group together multiple occurences of $g$'s with
the same index $j$ and to distinguish odd and even frequencies since
$g_j^{2n}(t)\equiv\frac{1}{2^{2n}}$, while 
$g_j^{2n+1}(t)\equiv\frac{1}{2^{2n}}g_j(t)$.

The above described grouping of $g$'s with the same index induces
a {\sl set partition\/} of the subindex set $\{1,\dots,k\}$, where
we have to count separately the numbers $p$ resp. $r$ of blocks of
odd resp. even cardinality. Let us denote by
$$S(k;p,r)$$
the number of the above described set partitions. Then we may compute
$I_k$ as follows: (We use the abbreviation $(x)_k=x(x-1)\dots(x-k+1)$.)
$$
I_k=\sum_{p,r}\frac{p!}{2^{k-p}}S(k;p,r)(L+1-p)_r
\int_0^N\sum_{0\le i_1<\dots<i_p\le L}
g_{i_1}(t)\dots g_{i_p}(t)dt\tag 3.1
$$
(Observe that for given index values $i_1,\dots,i_p$ belonging to the $p$ blocks of
odd cardinality, there are $(L+1-p)_r$ possibilities to choose the $r$
index values belonging to the blocks of even cardinality differently
from  $i_1,\dots,i_p$.)

The sums
$$
\Sigma_p(N)=\sum_{0\le i_1<\dots<i_p\le L} \int_0^N
g_{i_1}(t)\dots g_{i_p}(t)dt
$$
can be treated in exactly the same manner as in Section 2 for the
instance $p=3$ and yield
$$
\frac 1N\Sigma_p(N)=\tau_p(\log_2N),
$$
where
$$
\tau_p(x)=2^{1-\{x\}}\sum_{i\ge 0}
2^{-i}h_p\left(2^{i}2^{-1+\{x\}}\right)
$$
with
$$
h_p(x)=\sum_{1\le i_1<i_2<\dots<i_{p-1}}\int_0^xg(u)g(u2^{i_1})\dots
g(u2^{i_{p-1}})du.
$$
$h_p(x)$ is a continuous function with $h_p(m)=0$ for all $m\in\Bbb Z$,
so that $\tau_p(x)$ is a periodic function of period 1.

Collecting all contributions and rewriting $L=\left\lfloor \log_2N
\right\rfloor$ as $L=\log_2N-\{\log_2N\}$ we obtain

\proclaim{Theorem 2} 
The $s$-th moment of the binary sum-of-digits function satisfies
$$
\frac1N\sum_{n<N}\nu_2(n)^s=
\frac1{2^s}\(\log_2N\)^s
+\sum_{i=0}^{s-1}\left(\log_2N\right)^i\eta_i(\log_2N)
$$
where $\eta_i(x)$ are continuous periodic functions of period 1 that can 
be expressed explicitely as described before.
\endproclaim

{\bf Remark 1.}  The numbers $S(k;p,r)$ do not seem to have been studied
in the literature. Of course they are related to the classical 
{\sl Stirling numbers of the second kind\/} \cite{Ri58} via
$$
\sum_{p+r=j}
S(k;p,r)=S(k,j).
$$
The trivariate generating function of the $S(k;p,r)$ is easily obtained
via the {\sl operator method for combinatorial constructions\/} as described 
systematically in \cite{F-V90}:
$$
\sum_{p,r,k}S(k;p,r)x^py^r\frac{z^k}{k!}=
\exp\big(x\sinh z+y(\cosh z-1)\big).\tag 3.2
$$
We note that related recurrences already occur in \cite{Co86}.

{\bf Remark 2.} From Theorem 2 it is not difficult to compute expressions for the higher
centralized moments.





 
 
\heading 4. Mellin Perron Approach: An Exposition\endheading



As announced in the introduction we present an analytic method to compute digital sums.
The method makes use of the so called Mellin Perron summation formula:
$$\sum_{n<N}a_n\(1-\frac nN\)=\frac 1{2\pi i}\int\limits_{c-i\infty}^{c+i\infty}
\sum_{n=1}^\infty\frac{a_n}{n^s}N^s\frac{ds}{s(s+1)}.\tag4.1$$
This formula is well known in analytic number theory and can be proved by an application
of Mellin transform (cf. \cite{Do50}, \cite{Ap84}, \cite{FGKPT92}).
We demonstrate the use of this method in the case of the expectation of the $q$-ary sum-of-digits
function.

\proclaim{Theorem 3} {\rm[Delange]} 
The sum-of-digits function $\nu_q(n)$ satisfies
$$S_q(N)=\sum_{n<N}\nu_q(n)=\frac {q-1}2N\log_qN+N\delta(\log_qN),$$
where $\delta(u)$ is representable by the Fourier series $\delta(u)=\sum_{k\in\Bbb Z}\delta_ke^{2\pi ikx}$
and
$$\align &\delta_0=\frac{q-1}2\(\log_q2\pi-\frac1{\log q}\)-\frac{q+1}4\cr
&\delta_k=-\frac{q-1}{\log q}\zeta(\chi_k)\quad\text{for }k\not=0,\endalign$$
where $\chi_k=\frac{2k\pi i}{\log q}$.
\endproclaim


\demo{Proof} Let $\ups_q(k)$ be the largest exponent such that 
$q^{\ups_q(k)}$ divides $k$
and $\nu_q(k)$ be the sum-of-digits function in the $q$-ary representation. 
We have
$\nu_q(k)-\nu_q(k-1)=1-(q-1)\ups_q(k)$, so that $S_q(N)$ resembles a double
summation of $\ups_q(k)$. Furthermore it is well-known and easy to see that
$$
\sum_{k\geq1}\frac{\ups_q(k)}{k^s}=\frac{\zeta(s)}{q^s-1}.\tag4.1
$$
Thus, from \thetag{4.1}, with $a_k=1-(q-1)\ups_q(k)$, we get
the basic integral representation
$$
S_q(n)=\frac n{2\pi i}
\int\limits_{2-i\infty}^{2+i\infty}\frac{q^s-q}{q^s-1}
\zeta(s)n^s\frac{ds}{s(s+1)}.\tag4.2
$$
The integrand in \thetag{4.2} has a double pole at $s=0$ and simple poles at $s=\chi_k$
(note that the pole $s=1$ of $\zeta(s)$ is cancelled out). Shifting the line of integration
to $\Re(s)=-\frac14$ and taking residues into account we get
$$S(n)=\frac12 n\log_qn+n\tau_0(\log_qn)-nR(n),\tag4.3$$
where the remainder term is
$$R(n)=\frac1{2\pi i}\int\limits_{-\frac14-i\infty}^{-\frac14+i\infty}\frac{q^s-q}{q^s-1}n^s\frac{ds}{s(s+1)},\tag4.4$$
so that there only remains to prove that $R(n)\equiv 0$ when $n$ is an
integer. The integral converges since $|\zeta(-\frac14+it)|\ll|t|^{\frac34}$
(cf. \cite{W-W27}).
 
Using the expansion
$$\frac{q^s-q}{q^s-1}=q+(q-1)\(q^s+q^{2s}+q^{3s}\cdots\)$$
in \thetag{4.4} which is legitimate since now $\Re(s)<0$, we find that $R(n)$
is a sum of terms of the form
$$\frac 1{2\pi i}\int\limits_{-\frac14-i\infty}^{-\frac14+i\infty}\zeta(s)(q^kn)^s\frac{ds}{s(s+1)},$$
and each of these terms is $0$ by virtue of
$$\frac 1{2\pi i}\int\limits_{-\frac14-i\infty}^{-\frac14+i\infty}\zeta(s)n^s\frac{ds}{s(s+1)}=0,$$
which is an immediate consequence of the summation formula for arithmetic progressions and \thetag{4.1}.
\qed\enddemo

{\bf Remark.} It is clear from the discussion above that an exact formula for a 
sum-of-digits function is obtained each time the analytical behaviour of the corresponding Dirichlet
generating function is sufficently known. In the case of higher moments of the sum-of-digits function
this procedure can be worked out only formally: serious convergence problems occur. In order
to overcome these difficulties one would have to know precise bounds for the rate of growth of
the analytic continuation of the Dirichlet series
$$\sum_{n=1}^\infty\frac{\nu(n)^k}{n^s}$$
along vertical lines.
 
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\enddocument

