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\begin{center}
\vskip 1cm{\LARGE\bf 
On the Dirichlet Convolution of Completely \\
\vskip .1in
Additive Functions
}
\vskip 1cm
\large
Isao Kiuchi and Makoto Minamide \\
Department of Mathematical Sciences\\
Yamaguchi University\\
Yamaguchi 753-8512 \\
Japan\\
\href{mailto:kiuchi@yamaguchi-u.ac.jp}{\tt kiuchi@yamaguchi-u.ac.jp}\\
\href{mailto:minamide@yamaguchi-u.ac.jp}{\tt minamide@yamaguchi-u.ac.jp} \\
\end{center}

\vskip .2 in

\begin{abstract}
Let $k$ and $l$ be non-negative integers.  For two completely
additive functions $f$ and $g$, we consider various identities for the
Dirichlet convolution of the $k$th powers of $f$ and the $l$th powers
of $g$. Furthermore, we derive some asymptotic formulas for sums of
convolutions on the natural logarithms.
\end{abstract}

\section{Statements of results}
Let $f$ and $g$ be two arithmetical functions that are completely additive.
That is, these functions satisfy $f(mn)=f(m)+f(n)$  and $g(mn)=g(m)+g(n)$ for
all positive integers $m$ and $n$. 
We shall consider the arithmetical function 
\begin{align}\label{def-1}
D_{k,l}(n;f,g):=\sum_{d|n}f^{k}(d)g^{l}\left(\frac{n}{d}\right),
\end{align}
which represents 
the Dirichlet convolution of the $k$th power of 
$f$ and the $l$th power of $g$                              
for non-negative integers $k$ and $l$.
The above function provides a certain generalization
of the classical number-of-divisors function $d(n)$. 
In fact, 
$$
D_{0,0}(n;f,g)=d(n).
$$

The first purpose of this study is to investigate some recurrence formulas for $D_{k,l}(n;f,g)$ with respect to $k$ and $l$.
Since 
\begin{align}                                                                                                                  \label{f-2}  
\sum_{d|n}f(d)=\frac{1}{2}d(n)f(n),                                                                                                          
\end{align}
where $f$ is a completely additive function, we have 
\begin{align}                                                                                                                 \label{def-3}  
D_{1,1}(n;f,g)=\frac{1}{2}d(n)f(n)g(n)-\sum_{d|n}f(d)g(d).                                                                     
\end{align}  
Similarly, as in \eqref{def-3},
we use \eqref{def-1} for $D_{k,l+1}(n;f,g)$ to obtain
\begin{align*}
D_{k,l+1}(n;f,g)&= \sum_{d|n}f^{k}(d)g^{l}\left(\frac{n}{d}\right)g\left(\frac{n}{d}\right)  \\ 
                &=g(n)\sum_{d|n}f^{k}(d)g^{l}\left(\frac{n}{d}\right)-\sum_{d|n}f^{k}(d)g(d)g^{l}\left(\frac{n}{d}\right).  
\end{align*}
Hence, we deduce the following two recurrence formulas.                    
\begin{theorem} 
Let $k$ and $l$ be non-negative integers, and let $f$ and $g$ be completely additive functions. 
Then we have
\begin{align}
& D_{k,l+1}(n;f,g)+\sum_{d|n}f^{k}(d)g^{l}\left(\frac{n}{d}\right)g(d)=g(n)D_{k,l}(n;f,g),                                  \label{f-4}  \\  
& D_{k+1, l}(n;f,g)+\sum_{d|n}f^{k}(d)f\left(\frac{n}{d}\right)g^{l}\left(\frac{n}{d}\right)=f(n)D_{k,l}(n;f,g).            \label{f-5}
\end{align} 
\end{theorem} 

Now, we put $f=g$ in \eqref{f-4} (or \eqref{f-5}),  
and set  $D_{k,l}(n;f):=D_{k,l}(n;f,f)$. Then, we deduce the following corollary. 

\begin{corollary}\label{coro-ichi}
Using the same notation given above, we have 
\begin{align}
 D_{k,l+1}(n;f)+D_{k+1,l}(n;f)=f(n)D_{k,l}(n;f).                                                                            \label{f-6}
\end{align} 
Particularly, if $k=l$, we have
\begin{align}
D_{k+1, k}(n;f)=D_{k,k+1}(n;f)=\frac{1}{2}f(n)D_{k,k}(n;f).                                                                 \label{f-7}
\end{align}
\end{corollary}



Because the symmetric property  $D_{k,l}(n;f)=D_{l,k}(n;f)$, we only consider the function $D_{k,k+j}(n,f)$ for $j=1,2,\ldots$. 

  

\begin{example} The formulas \eqref{f-6} and \eqref{f-7} imply that 
\begin{align}
D_{k,k+2}(n;f)&=\frac{1}{2}f^{2}(n)D_{k,k}(n;f) - D_{k+1,k+1}(n;f),                                                             \nonumber \\ 
D_{k,k+3}(n;f)&=\frac{1}{2}f^{3}(n)D_{k,k}(n;f) - \frac{3}{2}f(n)D_{k+1,k+1}(n;f),                                             \nonumber \\
D_{k,k+4}(n;f)&=\frac{1}{2}f^{4}(n)D_{k,k}(n;f) -2f^{2}(n)D_{k+1,k+1}(n;f) + D_{k+2,k+2}(n;f),                                 \nonumber \\
D_{k,k+5}(n;f)&=\frac{1}{2}f^{5}(n)D_{k,k}(n;f) -\frac{5}{2}f^{3}(n)D_{k+1,k+1}(n;f) + \frac{5}{2}f(n)D_{k+2,k+2}(n;f),        \nonumber \\
D_{k,k+6}(n;f)&=\frac{1}{2}f^{6}(n)D_{k,k}(n;f) -3f^{4}(n)D_{k+1,k+1}(n;f) +\frac{9}{2}f^{2}(n)D_{k+2,k+2}(n;f)               \nonumber \\ 
              &\quad -D_{k+3,k+3}(n;f),                                                                                        \nonumber \\
D_{k,k+7}(n;f)&=\frac{1}{2}f^{7}(n)D_{k,k}(n;f)-  \frac{7}{2}f^{5}(n)D_{k+1,k+1}(n;f) +7f^{3}(n)D_{k+2,k+2}(n;f)              \nonumber \\
              &\quad -\frac{7}{2}f(n)D_{k+3,k+3}(n;f),                                                                         \nonumber \\
D_{k,k+8}(n;f)&=\frac{1}{2}f^{8}(n)D_{k,k}(n;f)- 4f^{6}(n)D_{k+1,k+1}(n;f) +10f^{4}(n)D_{k+2,k+2}(n;f)                        \nonumber \\
              &\quad  -8f^{2}(n)D_{k+3,k+3}(n;f) +D_{k+4,k+4}(n;f).                                                            \nonumber
\end{align}
\end{example}


Next, we shall demonstrate that the explicit evaluation of the function $D_{k,k+m}(n;f)$ ($m=2,3,\ldots$) 
can be expressed as a combination of the functions 
$D_{k,k}(n;f)$, $D_{k+1,k+1}(n;f)$, $D_{k+2,k+2}(n;f)$, $\ldots$, 
$D_{k+\lfloor \frac{m}{2}\rfloor,k+\lfloor \frac{m}{2}\rfloor}(n;f)$.  
 Hence, we shall give a recurrence formula 
between  $D_{k,k}(n;f), \ldots ,$ $D_{k+\lfloor\frac{m}{2}\rfloor,k+\lfloor\frac{m}{2}\rfloor}(n;f)$ and $D_{k,k+m}(n;f)$.   

\begin{theorem}\label{teiri-3} Let $k$ and $m$ be positive integers, and let $D_{k, k+m}(n;f)$ be the function defined by the above formula. Then we have
\begin{align}   
 D_{k,k+m}(n;f)                                                                                                                     
=\sum_{j=0}^{\lfloor \frac{m}{2} \rfloor }c_{k,j}^{(m)}f^{m-2j}(n)D_{k+j,k+j}(n;f),                                          \label{f-16}
\end{align}
where
\begin{align*}
c_{k,j}^{(m)}=\begin{cases} 
\displaystyle  \frac{1}{2}, & \text{if $j=0$;}\\
\displaystyle -\frac{m}{2}, & \text{if $j=1$;}\\
(-1)^{j}\displaystyle \frac{m}{2\cdot j!}\displaystyle \prod_{i=1}^{j-1}\left(m-(j+i)\right), 
& \text{
if $2\leq j \leq \displaystyle \lfloor\frac{m}{2}\rfloor $.}  
\end{cases}
\end{align*}
\end{theorem}



\begin{proof} By \eqref{f-7} in Corollary \ref{coro-ichi}, the equality \eqref{f-16} holds for $m=1$ and all $k\in\mathbb{N}$. 
Now, we assume that \eqref{f-16} is true for $m=1,2,\ldots, l$ and $k\in\mathbb{N}$. Using this assumption and \eqref{f-6} 
in Corollary \ref{coro-ichi}, we observe that
\begin{align*}
D_{k, k+l+1}(n;f)&=\sum_{j=0}^{\lfloor\frac{l}{2}\rfloor}c_{k,j}^{(l)}f^{l+1-2j}(n)D_{k+j,k+j}(n;f)\\
                  &\quad -\sum_{j=0}^{\lfloor\frac{l-1}{2}\rfloor}c_{k+1,j}^{(l-1)}f^{l-1-2j}(n)D_{k+1+j,k+1+j}(n;f).
\end{align*}
For even $l=2q$, we have
\begin{align*}
D_{k, k+2q+1}(n;f)&=\frac{1}{2}f^{2q+1}(n)D_{k,k}(n;f)\\
                   & \quad +\sum_{j=1}^{q}\left(c_{k,j}^{(2q)}-c_{k+1,j-1}^{(2q-1)}\right)f^{2q+1-2j}(n) D_{k+j,k+j}(n;f)
\end{align*}
and
\begin{align*}
c_{k,j}^{(2q)}-c_{k+1,j-1}^{(2q-1)}=(-1)^{j}\frac{2q+1}{2\cdot j!}\prod_{i=1}^{j-2}\left(2q-(j+i)\right)(2q-j)=c_{k,j}^{(2q+1)}.
\end{align*}
For odd $l=2q-1$, we observe that
\begin{align*}
D_{k,k+2q}(n;f) &= \frac{1}{2}f^{2q}(n)D_{k,k}(n;f)\\
&\quad +\sum_{j=1}^{q-1} \left(c_{k,j}^{(2q-1)}-c_{k+1, j-1}^{(2q-2)}\right)f^{2q-2j}(n)D_{k+j}(n;f)\\
& \quad + (-1)^{\lfloor\frac{2q}{2}\rfloor}D_{k+\lfloor \frac{2q}{2}\rfloor, k+\lfloor \frac{2q}{2}\rfloor }(n;f).
\end{align*} 
By our assumption, since
\begin{align*}
c_{k,j}^{(2q-1)}-c_{k+1,j-1}^{(2q-2)}
=(-1)^{j}\frac{2q}{2\cdot j!}\prod_{i=1}^{j-2}\left(2q-1-(j+i)\right) (2q-1-j)=c_{k,j}^{(2q)},
\end{align*}
we obtain the assertion \eqref{f-16} for all $k$ and $m \in\mathbb{N}$.
\end{proof}                                                   

Now, we consider another expression for $D_{k,l}(n;f,g)$ using the arithmetical function
\begin{align}                                                                                                         \label{def-H}
H_{k,m}(n;f,g):=\sum_{d|n}f^{k}(d)g^{m}(d).                                                                                  
\end{align}
If $f=g$, we set $H_{k+m}(n;f)=H_{k,m}(n;f,f)$.
The right-hand side of (\ref{def-H}) implies the Dirichlet convolution of $1$ and $f^{k}g^{m}$.
Since $g$ is a completely additive function, we have 
\begin{align*}
D_{k,l}(n;f,g)&=\sum_{d|n}f^{k}(d)\left(g(n)-g(d)\right)^{l}\\
              &=\sum_{d|n}f^{k}(d)\sum_{m=0}^{l}(-1)^{m}\begin{pmatrix}l\\
                       m\end{pmatrix}g^{l-m}(n)g^{m}(d). 
\end{align*}
From \eqref{def-H} and the above, we obtain the following theorem. 

\begin{theorem}\label{teiri-1-4} Let $k$ and $l$ be non-negative integers, and let $f$ and $g$ be completely additive functions. 
Then we have 
\begin{align*}
D_{k,l}(n;f,g)=\sum_{m=0}^{l}(-1)^{m}{l \choose m}g^{l-m}(n)H_{k,m}(n;f,g),                                               
\end{align*} 
where the function $H_{k,m}(n;f,g)$ is defined by \eqref{def-H}.
\end{theorem}


We immediately obtain the following corollary.

\begin{corollary}Let $k$ and $l$ be non-negative integers, and let $f$ and $g$ be completely additive functions. 
Then we have 
\begin{align}
D_{k,l}(n;f)=\sum_{m=0}^{l}(-1)^{m}{l \choose m}f^{l-m}(n)H_{k+m}(n;f).                                                  \label{f-19}
\end{align}
\end{corollary}

Note that
\begin{align}
H_{k,m}(n;f,g)&=\sum_{d|n}f^{k}\left(\frac{n}{d}\right)g^{m}\left(\frac{n}{d}\right)    \nonumber  \\
&=\sum_{i=0}^{k}\sum_{j=0}^{m}(-1)^{i+j} {k \choose i}{m \choose j}f^{k-i}(n)g^{m-j}(n)\sum_{d|n}f^{i}(d)g^{j}(d).      \label{f-20}
\end{align}
Applying \eqref{f-20} to Theorem \ref{teiri-1-4}, we have the following theorem. 
                
\begin{theorem} Let $k$ and $l$ be non-negative integers, and let $f$ and $g$ be completely additive functions. 
Then we have 
\begin{align*}
&D_{k,l}(n;f,g)  \nonumber  \\
&=\sum_{m=0}^{l}\sum_{i=0}^{k}\sum_{j=0}^{m}(-1)^{m+i+j} {l \choose m}{k \choose i}{m \choose j}
f^{k-i}(n)g^{l-j}(n)\sum_{d|n}f^{i}(d)g^{j}(d).                                                                         
\end{align*}
\end{theorem}



In the case where $f=g$, note that
\begin{align*}
H_{k+m}(n;f)&=\sum_{d|n}\left(f(n)-f(d)\right)^{k+m}\\
            &=\sum_{j=0}^{k+m}(-1)^{j}{k+m \choose j}f^{k+m-j}(n)\sum_{d|n}f^{j}(d).
\end{align*}
From \eqref{f-19} and the above, we obtain the following corollary.

\begin{corollary}\label{coro-17} Let $k$ and $l$ be non-negative integers, and let $f$ be a completely additive function. 
Then we have 
\begin{align}
D_{k,l}(n;f)=\sum_{m=0}^{l}\sum_{j=0}^{k+m}(-1)^{m+j}{l \choose m}{k+m \choose j}f^{k+l-j}(n)\sum_{d|n}f^{j}(d).        \label{f-22}
\end{align}
\end{corollary}

\section{Recurrence formula connecting $D_{k, l}(n;f)$ with $\sum_{d|n}f^{j}(d)$}

The second purpose of this study is to derive another expression for $D_{k,l}(n;f)$ that involves the divisor function $d(n)$. 
Before stating Theorem \ref{th-2-2}, we prepare the following lemma.  

\begin{lemma}\label{hodai-21}
Let $f$ be a completely additive function. There exist the constants $e_{q,q}$, $e_{q,q-1}, \ldots, e_{q,1}$ ($q=1,2,\ldots$) that satisfy the equation 
\begin{align}                                                                                                              \label{f-23}
\sum_{d|n}f^{2q-1}(d)=e_{q,q}d(n)f^{2q-1}(n) +\sum_{j=1}^{q-1}e_{q,q-j}f^{2q-2j-1}(n)\sum_{d|n}f^{2j}(d).                     
\end{align}
Moreover, the relations among sequences $(e_{q,q-j})_{j=1}^{q}$ are as follows.
\begin{align}
& e_{q,q}=\frac{1}{2} \left( 1-\sum_{j=1}^{q-1}{2q-1 \choose 2j-1}e_{j,j}\right)
=\frac{\left(2^{2q}-1\right)B_{2q}}{q},                                                                                    \label{f-24}\\ 
& e_{q,q-j}=\frac{1}{2}\left( {2q-1 \choose 2j} - \sum_{i=2 \atop i-j\geq 1}^{q-1}{2q-1 \choose 2i-1}e_{i,i-j}\right ),  \nonumber  
\end{align}
where $B_{n}$ denotes the $n$th Bernoulli number, which is defined by the Taylor expansion 
\begin{align*}
\frac{z}{e^{z}-1}=\sum_{n=1}^{\infty}\frac{B_{n}}{n!}z^{n}, \quad (|z|<2\pi).
\end{align*} 
\end{lemma}

\begin{proof}
By \eqref{f-2}, the case $q=1$ in \eqref{f-23} is trivial. 
Assume that there exist $e_{p,p}$, $e_{p,p-1}, \ldots , e_{p,1}$ ($p\leq q$) such that 
\begin{align}                                                                                                                 \label{f-26}
\sum_{d|n}f^{2p-1}(d)=e_{p,p}d(n)f^{2p-1}(n) +\sum_{j=1}^{p-1}e_{p,p-j}f^{2p-2j-1}(n)\sum_{d|n}f^{2j}(d).                      
\end{align}
Since
\begin{align*}
\sum_{d|n}f^{2q+1}(d) = \sum_{j=0}^{2q+1} (-1)^{j} {2q+1 \choose j} f^{2q+1-j}(n)\sum_{d|n}f^{j}(d),                           
\end{align*}
we have 
\begin{align}
\sum_{d|n}f^{2q+1}(d)&=\frac{1}{2}d(n)f^{2q+1}(n)
+\frac{1}{2}\sum_{j=1}^{q}{2q+1 \choose 2j}f^{2q+1-2j}(n)
\sum_{d|n}f^{2j}(d) \nonumber \\
&\quad - \frac{1}{2}\sum_{j=1}^{q}{2q+1 \choose 2j-1}f^{2q+2-2j}(n) \sum_{d|n}f^{2j-1}(d).                                      \label{f-27}
\end{align}
Applying \eqref{f-26} to \eqref{f-27}, we obtain  
\begin{align*}
\sum_{d|n}f^{2q+1}(d) =&\frac{1}{2}
\left( 1-\sum_{j=1}^{q}{2q+1 \choose 2j-1}e_{j,j}\right ) d(n)f^{2q+1}(n)\\
&+\frac{1}{2}\sum_{j=1}^{q}\left( {2q+1 \choose 2j}  -\sum_{i=2 \atop i-j\geq 1}^{q}{2q+1 \choose 2i-1}e_{i,i-j}\right )
f^{2q-2j+1}(n)\sum_{d|n}f^{2j}(d).
\end{align*}
By induction, this completes the proof, except for the second term on the right-hand side of \eqref{f-24}.



The first term on the right-hand side of \eqref{f-24} implies
\begin{align*}
e_{q,q}=1-\sum_{k=1}^{q}{2q-1 \choose 2k-1} e_{k,k}
       =1-\sum_{k=1}^{q}{2q \choose 2k}\frac{k}{q}e_{k,k}.
\end{align*}
Here we put $a(k)=k e_{k,k}$.  Then we have
\begin{align}
a(q)=q-\sum_{k=1}^{q} {2q \choose 2k}a(k).                                                                                     \label{f-28}
\end{align}
Since $a(1)=e_{1,1}=1/2$ and $\left(2^{2}-1\right)B_{2}=1/2$, 
we only need to show that $\left(2^{2k}-1\right)B_{2k}$ $(k=1,\ldots , q)$ satisfies
the recurrence formula \eqref{f-28}.
Consider the $n$th Bernoulli polynomial $B_{n}(x)$, which is defined by the following Taylor expansion:
\begin{align*}
\frac{ze^{xz}}{e^{z}-1}=\sum_{n=0}^{\infty}\frac{B_{n}(x)}{n!}z^{n} \quad (|z|<2\pi).
\end{align*}
The following relations are known among $B_{n}(1)$, $B_{n}(1/2)$ and $B_{n}$,
\begin{align*}
B_{n}(1)=B_{n}, \quad  \quad B_{n}\left(\frac{1}{2}\right)=-\left(1-2^{1-n}\right)B_{n}.                                    
\end{align*}
By the formula \cite[Thm.\ 12.12, p.\ 264]{A}
\begin{align}
B_{n}(y)=\sum_{k=0}^{n}{n \choose k}B_{k}y^{n-k},                                                                          \label{f-30}
\end{align}
we observe that 
\begin{align*}
B_{2q}(y)=y^{2q}-qy^{2q-1}+\sum_{k=2}^{2q}{2q \choose k}B_{k}y^{2q-k}.
\end{align*}
In this equation, we consider $y=1$ and $y=1/2$; then
\begin{align}
B_{2q}&=1-q+\sum_{k=1}^{q} {2q \choose 2k}B_{2k}                                                                     \label{f-31}
\end{align}
and 
\begin{align}                                                             
2^{2q}B_{2q}\left(\frac{1}{2}\right)
&=\left(2-2^{2q}\right)B_{2q}  \nonumber  \\
&=1-2q+\sum_{k=1}^{q}{2q \choose 2k}B_{2k}2^{2k}.                                                                      \label{f-32}            
\end{align}
Subtracting \eqref{f-32} from \eqref{f-31}, we obtain
\begin{align*}
\left(2^{2q}-1\right)B_{2q}
=q-\sum_{k=1}^{q} {2q \choose 2k}\left(2^{2k}-1\right)B_{2k}.                                                              
\end{align*}
This recurrence formula for $\left(2^{2k}-1\right)B_{2k}$'s is equivalent to \eqref{f-28}. This completes the proof of \eqref{f-23}.
\end{proof}



Applying Lemma \ref{hodai-21}  to \eqref{f-22} in Corollary \ref{coro-17},  we have
\begin{align}
D_{k,l}(n;f)&=\sum_{m=0}^{l}(-1)^{m}{l\choose m}\sum_{j=0}^{\lfloor \frac{k+m}{2}\rfloor }{k+m\choose 2j}f^{l+k-2j}(n)\sum_{d|n}f^{2j}(d) \nonumber \\
&\quad -\sum_{m=0}^{l}(-1)^{m}{l\choose m}\sum_{j=1}^{\lfloor \frac{k+m+1}{2}\rfloor }{k+m\choose 2j-1}f^{l+k+1-2j}(n)\sum_{d|n}f^{2j-1}(d).    \label{f-33}
\end{align}
The second term on the right-hand side of \eqref{f-33} gives us
\begin{align}
&-\left( \sum_{m=0}^{l}(-1)^{m}{l\choose m}\sum_{j=1}^{\lfloor \frac{k+m}{2}\rfloor }e_{j,j}{k+m\choose 2j-1}\right ) f^{l+k}(n)d(n) \nonumber \\
&-\sum_{m=0}^{l}\sum_{j=1}^{\lfloor \frac{k+m+1}{2}\rfloor }\sum_{i=1}^{j-1}(-1)^{m}{l\choose m}{k+m\choose 2j-1}e_{j,j-i}f^{l+k-2i}(n)\sum_{d|n}f^{2i}(d)
                                                                                                                                   \label{f-34}
\end{align}
using \eqref{f-23}. From \eqref{f-24}, \eqref{f-33} and \eqref{f-34}, we have the following theorem.

\begin{theorem}\label{th-2-2} Let $k$ and $l$ be non-negative integers, and let $f$ be a completely additive function. 
There exist the constants $e_{j,j}$, $e_{j,j-i}$ ($j=1,2,\ldots, \lfloor \frac{k+m+1}{2}\rfloor $, $1\leq j-i <j$) and $A_{k,l}$ such that  
\begin{align}
D_{k,l}(n;f) 
=& A_{k,l}f^{l+k}(n)d(n)  \nonumber  \\
&+ \sum_{m=0}^{l}\sum_{j=0}^{\lfloor\frac{k+m}{2}\rfloor}(-1)^{m} {l \choose m}{k+m \choose 2j}f^{l+k-2j}(n)\sum_{d|n}f^{2j}(d)            \label{f-35}    \\
&-\sum_{m=0}^{l}\sum_{j=1}^{\lfloor \frac{k+m}{2}\rfloor }\sum_{i=1}^{j-1}(-1)^{m}{ l \choose m}{k+m \choose 2j-1}
e_{j, j-i}f^{l+k-2i}(n)\sum_{d|n}f^{2i}(d),                                                               \nonumber                         
\end{align}
where 
\begin{align}
A_{k,l}&=\sum_{m=0}^{l}(-1)^{m-1}{l \choose m} \sum_{j=1}^{\lfloor \frac{k+m+1}{2}\rfloor }\frac{(2^{2j}-1)B_{2j}}{j}{k+m \choose 2j-1} \nonumber   \\
       &= 2\sum_{m=0}^{l}(-1)^{m}{l \choose m} \frac{2^{k+m+1}-1}{k+m+1}B_{k+m+1}.                                               \label{f-36} 
\end{align} 
\end{theorem}
\begin{proof}
We only need to show \eqref{f-36} to complete the proof of Theorem \ref{th-2-2}. We set 
\begin{align*}
A_{k,l}
&=\sum_{m=0}^{l}(-1)^{m-1}{l \choose m} J_{k,m},  \qquad  
J_{k,m}=\sum_{j=1}^{\lfloor \frac{k+m+1}{2}\rfloor }\frac{(2^{2j}-1)B_{2j}}{j}{k+m \choose 2j-1}.    
\end{align*}
From the identity $\frac{1}{j}{k+m\choose 2j-1}=\frac{2}{k+m+1}{k+m+1 \choose 2j}$, we have 
$$
J_{k,m}=\frac{2}{k+m+1}\sum_{j=1}^{\lfloor \frac{k+m+1}{2}\rfloor}(2^{2j}-1)B_{2j} {k+m+1 \choose 2j}.
$$ 
Since $B_{2j+1}=0$\ if\  $j\geq 1$ and\  $B_{1}=-\frac12$, we have
\begin{align}
J_{k,m}
&=\frac{2}{k+m+1}\sum_{j=0}^{k+m+1}(2^{2j}-1)B_{j} {k+m+1 \choose j} +1.                                                      \label{f-37}
\end{align}
Taking $y=\frac12$  and $y=1$ in \eqref{f-30}, we get 
$$
2^{n}B_{n}\left(\frac12\right)=\sum_{j=0}^{n}2^{j}{n\choose j}B_{j} \qquad {\rm{and}} \qquad 
B_{n}=\sum_{j=0}^{n}{n \choose j}B_{j},
$$
respectively. Hence we have 
\begin{align}
 2^{n}B_{n}\left(\frac12\right) - B_{n} 
=\sum_{j=0}^{n}\left(2^{j}-1\right){n\choose j}B_{j}.                                                                         \label{f-38}
\end{align}
On the other hand, it is known that 
\begin{align}
B_{n}(mx)=m^{n-1}\sum_{j=0}^{m-1}B_{n}\left(x+\frac{j}{m}\right).                                                             \label{f-39}
\end{align} 
See \cite[p.\ 275]{A}. Taking $x=0$ and $m=2$ in \eqref{f-39}, we obtain 
$$
 2^{n}B_{n}\left(\frac12\right) - B_{n}
=-\left(2^{n}-1\right)B_{n}. 
$$
Then we have, from \eqref{f-38} and the above,
$$
-\left(2^{n}-1\right)B_{n}
=\sum_{j=0}^{n}\left(2^{j}-1\right)B_{j}{n\choose j}. 
$$ 
Substituting this relation in \eqref{f-37}, we have 
$$  
J_{k,m}
=-\frac{2}{k+m+1}(2^{k+m+1}-1)B_{k+m+1}+1.                                                                        
$$  
Hence we have 
\begin{align*}
A_{k,l} = 2\sum_{m=0}^{l}(-1)^{m}{l \choose m} \frac{2^{k+m+1}-1}{k+m+1}B_{k+m+1}.                                      
\end{align*}
\end{proof}
 
From \eqref{f-35}, we obtain the following corollary.

\begin{corollary}\label{coro-23} For every positive integer $k$, there are constants $c_{k}, c_{k-1}, \ldots, c_{0}$ such that  
\begin{align}
D_{k,k}(n;f) = c_{k}d(n)f^{2k}(n) + \sum_{j=1}^{k}c_{k-j}f^{2k-2j}(n)\sum_{d|n}f^{2j}(d).                                      \label{f-41}
\end{align}
\end{corollary}


From \eqref{f-16} and \eqref{f-41}, we obtain the following corollary.

\begin{corollary}\label{coro-24} For every positive integer $k$ and $m$ $(\geq 2)$, there are constants 
$c_{0}, c_{1}, \ldots , c_{k},$ $\ldots ,  c_{k+\lfloor \frac{m}{2}\rfloor }$  as in Corollary \ref{coro-23}
and $c_{k,1}^{(m)}=-\frac{m}{2}$, $c_{k,2}^{(m)}=\frac{m(m-3)}{4}$, $c_{k,3}^{(m)}$, $\ldots$, $c_{k, \lfloor \frac{m}{2}\rfloor }^{(m)}$  
as in Theorem \ref{teiri-3} such that  
\begin{align}
&D_{k,k+m}(n;f)=\left( \frac12 c_{k}+\sum_{p=1}^{\lfloor \frac{m}{2}\rfloor }c_{k,p}^{(m)}c_{k+p}\right) d(n)f^{2k+m}(n)    \nonumber \\
& \quad + \frac12 \sum_{j=1}^{k}c_{k-j}f^{2k+m-2j}(n)\sum_{d|n}f^{2j}(d)                                                          \nonumber\\
& \quad + \sum_{p=1}^{\lfloor \frac{m}{2}\rfloor }c_{k,p}^{(m)}\sum_{j=1}^{k+p}c_{k+p-j}f^{2k+m-2j}(n)\sum_{d|n}f^{2j}(d).  \nonumber               
\end{align}
\end{corollary}



                                          
\begin{example} Corollaries \ref{coro-23} and \ref{coro-24} give us 
\begin{align*}
& D_{1,1}(n;f)=\frac{1}{2}d(n)f^{2}(n) -\sum_{d|n}f^{2}(d),\\
& D_{2,1}(n;f)=\frac{1}{4}d(n)f^{3}(n)-\frac{1}{2}f(n)\sum_{d|n}f^{2}(d),\\
& D_{3,1}(n;f)=-\frac{1}{4}d(n)f^{4}(n)+\frac{3}{2}f^{2}(n)\sum_{d|n}f^{2}(d)-\sum_{d|n}f^{4}(d),\\
& D_{2,2}(n;f)=\frac{1}{2}d(n)f^{4}(n)-2f^{2}(n)\sum_{d|n}f^{2}(d)+\sum_{d|n}f^{4}(d),\\
& D_{4,1}(n;f)=-\frac{1}{2}d(n)f^{5}(n)+\frac{5}{2}f^{3}(n)\sum_{d|n}f^{2}(d)-\frac{3}{2}f(n)\sum_{d|n}f^{4}(d), \\
& D_{3,2}(n;f)=\frac{1}{4}d(n)f^{5}(d)-f^{2}(n)\sum_{d|n}f^{2}(d)+\frac{1}{2}f(n)\sum_{d|n}f^{4}(d).
\end{align*}
\end{example}

\section{Applications}

The second author \cite[p.\ 330]{M} showed an asymptotic formula for $\sum_{n\leq x}D_{1,1}(n;\log)$  
\begin{align}
\sum_{n\leq x}D_{1,1}(n;\log)   &
=\frac{1}{6}x\log^{3}x -\frac{1}{2}x\log^{2}x + (1-2A_{1})x\log x  \nonumber \\
& \quad  + (2A_{1}-4A_{2}-1)x + O_{\varepsilon}\left(x^{\frac13 +\varepsilon}\right)                                             \label{f-43}
\end{align}
for $x>2$ and all $\varepsilon >0$.   
Here the constants $A_{1}$ and $A_{2}$ are coefficients of the Laurent expansion of the Riemann zeta-function $\zeta(s)$ 
in the neighbourhood $s=1$:
\begin{align*}
\zeta(s)=\frac{1}{s-1}+A_{0} +A_{1}(s-1) + A_{2}(s-1)^{2} + A_{3}(s-1)^{3} + \cdots.                        
\end{align*}
We use \eqref{f-7}, \eqref{f-43} and Abel's identity \cite[Thm.~4.2, p.\ 77]{A} to obtain 
\begin{align*}
\sum_{n\leq x}D_{2,1}(n;\log)   
&=\frac{1}{12}x\log^{4}x-\frac{1}{3}x\log^{3}x + (1+A_{1})x\log^{2}x   \nonumber  \\
&\qquad    -2(1+A_{2})x\log x + 2(1+A_{2})x +  O_{\varepsilon}\left(x^{\frac13 +\varepsilon}\right).                           
\end{align*} 


 
Furthermore, a generalization of \eqref{f-43} for the partial sums of $D_{k,k}(n;\log)$ 
for positive integers $k$  was considered by the second author \cite[Thm.\ 1.2, \ p.\ 326]{M}, 
who demonstrated that there exists a polynomial $P_{2k+1}$ of degree $2k+1$  such that  
\begin{align}
\sum_{n\leq x}D_{k,k}(n;\log) 
= xP_{2k+1}(\log x) + O_{k, \varepsilon}\left(x^{\frac{1}{3}+\varepsilon}\right)                                                 \label{f-46}
\end{align}
for every  $\varepsilon>0$.
Applying Theorem \ref{teiri-3} and the above formula \eqref{f-46}, we have the following theorem. 

 

\begin{theorem} There exists a polynomial $U_{2k+m+1}$ of degree $2k+m+1$  such that   
\begin{align*}
\sum_{n\leq x}D_{k,k+m}(n;\log)
=xU_{2k+m+1}(\log x) + O_{k,m,\varepsilon}\left(x^{\frac{1}{3}+\varepsilon}\right)                                                   
\end{align*}
for $m=2,3,\ldots$ and every $\varepsilon>0$. 
\end{theorem}

\section{Acknowledgment} The authors deeply thank the referee for carefully reading this paper and indicating some mistakes.

\begin{thebibliography}{9}

\bibitem{A}
T. M. Apostol.
\newblock {\it Introduction to Analytic Number Theory}.
\newblock Springer-Verlag, 1976.


\bibitem{M}
M. Minamide.
\newblock The truncated Vorono{\" i} formula for the derivative of the Riemann zeta-function.
\newblock {\it Indian J. Math.} {\bf 55} (2013), 325--352.  
\end{thebibliography} 


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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11A25; Secondary 11P99.

\noindent \emph{Keywords: } completely additive function, Dirichlet convolution.

\bigskip
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\vspace*{+.1in}
\noindent
Received April 23 2014;
revised versions received  May 15 2014; July 9 2014; July 20 2014; July 31 2014.
Published in {\it Journal of Integer Sequences}, August 5 2014.

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