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# Certain combinatoric Bernoulli polynomials and convolution sums of divisor functions

https://doi.org/10.1186/1687-1847-2013-310

• Accepted: 23 September 2013
• Published:

## Abstract

It is known that certain convolution sums can be expressed as a combination of divisor functions and Bernoulli formula. One of the main goals in this paper is to establish combinatoric convolution sums for the divisor sums ${\stackrel{ˆ}{\sigma }}_{s}\left(n\right)={\sum }_{d|n}{\left(-1\right)}^{\frac{n}{d}-1}{d}^{s}$. Finally, we find a formula of certain combinatoric convolution sums and Bernoulli polynomials.

MSC:11A05, 33E99.

## Keywords

• Bernoulli numbers
• convolution sums

## 1 Introduction

The symbols and denote the set of natural numbers and the ring of integers, respectively. The Bernoulli polynomials ${B}_{k}\left(x\right)$, which are usually defined by the exponential generating function
$\frac{t{e}^{xt}}{{e}^{t}-1}=\sum _{k=0}^{\mathrm{\infty }}{B}_{k}\left(x\right)\frac{{t}^{k}}{k!},$
play an important and quite mysterious role in mathematics and various fields like analysis, number theory and differential topology. The Bernoulli polynomials satisfy the following well-known identities:
$\begin{array}{rcl}\sum _{j=0}^{N}{j}^{k}& =& \frac{{B}_{k+1}\left(N+1\right)-{B}_{k+1}\left(0\right)}{k+1}\phantom{\rule{1em}{0ex}}\left(k\ge 1\right)\\ =& \frac{1}{k+1}\sum _{j=0}^{k}{\left(-1\right)}^{j}\left(\genfrac{}{}{0}{}{k+1}{j}\right){B}_{j}{N}^{k+1-j}.\end{array}$
(1.1)
The Bernoulli numbers ${B}_{k}$ are defined to be ${B}_{k}:={B}_{k}\left(0\right)$. For $n\in \mathbb{N}$, $k\in \mathbb{Z}$, we define some divisor functions
It is well known that ${\sigma }_{k}^{\ast }\left(n\right)={\sigma }_{k}\left(n\right)-{\sigma }_{k}\left(\frac{n}{2}\right)$ and ${\stackrel{ˆ}{\sigma }}_{k}\left(n\right)={\sigma }_{k}\left(n\right)-2{\sigma }_{k}\left(\frac{n}{2}\right)$ [[1], (1.13)]. The identity
$\sum _{k=1}^{n-1}\sigma \left(k\right)\sigma \left(n-k\right)=\frac{5}{12}{\sigma }_{3}\left(n\right)+\left(\frac{1}{12}-\frac{1}{2}n\right)\sigma \left(n\right)$
for the basic convolution sum first appeared in a letter from Besge to Liouville in 1862 [2]. Hahn [[1], (4.8)] considered
(1.2)
For some of the history of the subject, and for a selection of these articles, we mention [3, 4] and [5], and especially [6, 7] and [8]. The study of convolution sums and their applications is classical, and they play an important role in number theory. In this paper, we investigate the combinatorial Bernoulli numbers and convolution sums. For k and n being positive integers, we show that the sum
$\sum _{j=2}^{k}\left(\genfrac{}{}{0}{}{2k+1}{2j}\right){B}_{2j}{\stackrel{ˆ}{\sigma }}_{2k+1-2j}\left(n\right)$

can be evaluated explicitly in terms of divisor functions and a combinatorial convolution sum. We prove the following.

Theorem 1 Let k, n be positive integers. Then
$\begin{array}{c}\sum _{j=2}^{k}\left(\genfrac{}{}{0}{}{2k+1}{2j}\right){B}_{2j}{\stackrel{ˆ}{\sigma }}_{2k+1-2j}\left(n\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\left(2k+1\right){\sigma }_{2k+1}^{\ast }\left(n\right)-\left(\frac{2k+3}{2}\right){\stackrel{ˆ}{\sigma }}_{2k+1}\left(n\right)-\left(\frac{k\left(2k+1\right)}{6}\right){\stackrel{ˆ}{\sigma }}_{2k-1}\left(n\right)\hfill \\ \phantom{\rule{2em}{0ex}}-\left(2k+1\right)\sum _{s=0}^{k-1}\left(\genfrac{}{}{0}{}{2k}{2s+1}\right)\sum _{m=1}^{n-1}{\stackrel{ˆ}{\sigma }}_{2k-2s-1}\left(m\right){\stackrel{ˆ}{\sigma }}_{2s+1}\left(n-m\right).\hfill \end{array}$
Remark 2 Let n be positive integers. In Theorem 1, replace k by 1, we find easily that
$\sum _{m=1}^{n-1}\stackrel{ˆ}{\sigma }\left(m\right)\stackrel{ˆ}{\sigma }\left(n-m\right)=\frac{1}{2}{\sigma }_{3}^{\ast }\left(n\right)-\frac{5}{12}{\stackrel{ˆ}{\sigma }}_{3}\left(n\right)-\frac{1}{12}\stackrel{ˆ}{\sigma }\left(n\right),$
(1.3)
and in particular, if $q\in N$, $p=2q+1$, an odd prime integer, then
$\sum _{m=1}^{q}\stackrel{ˆ}{\sigma }\left(m\right)\stackrel{ˆ}{\sigma }\left(p-m\right)=\frac{1}{6}q\left(q+1\right)\left(2q+1\right)=\sum _{m=1}^{q}{k}^{2}=\frac{1}{3}{B}_{3}\left(q+1\right).$
(1.4)

Equations (1.3) and (1.4) are in (1.2) and [[9], Corollary 2.4]. Using these combinatoric convolution sums, we obtain the following.

Theorem 3 If k is a positive integer, then
$\sum _{u+v+w=2k+1}{2}^{v-2k-1}\left(\genfrac{}{}{0}{}{2k+1}{u,v,w}\right){B}_{v}\cdot {\left(2l+1\right)}^{w}={B}_{2k+1}\left(l+1\right),$

where $\left(\genfrac{}{}{0}{}{2k+1}{u,v,w}\right)=\frac{\left(2k+1\right)!}{u!v!w!}$ and $l=1,2,3,4,5,6,7,8,9,10,11,12$.

Thus, we can pose a general question regarding Bernoulli polynomials.

Question For all $k,l\in \mathbb{N}$, does the identity
$\sum _{u+v+w=2k+1}{2}^{v-2k-1}\left(\genfrac{}{}{0}{}{2k+1}{u,v,w}\right){B}_{v}\cdot {\left(2l+1\right)}^{w}={B}_{2k+1}\left(l+1\right)\phantom{\rule{1em}{0ex}}\text{hold}?$

The problem of convolution sums of the divisor function ${\sigma }_{1}\left(n\right)$ and the theory of Eisenstein series has recently attracted considerable interest with the emergence of quasimodular tools. In connection with the classical Jacobi theta and Euler functions, other aspects of the function ${\sigma }_{1}\left(n\right)$ are explored by Simsek in [10]. Finally, we prove the following.

Theorem 4 If a (≥2) and k are positive integers, then
1. (i)
$\begin{array}{c}\sum _{s=0}^{k-1}\left(\genfrac{}{}{0}{}{2k}{2s+1}\right)\sum _{m=1}^{{2}^{a}-1}{\sigma }_{2k-2s-1,1}\left(\frac{m}{2};2\right){\sigma }_{2s+1}\left({2}^{a}-m\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\left(\frac{1}{2\left(2k+1\right)}\right){B}_{2k+1}\left({2}^{a}\right)+\left(\frac{1-{2}^{2k}}{2\left(2k+1\right)}\right)\sum _{i=1}^{a-1}{B}_{2k+1}\left({2}^{i}\right)+\frac{1}{4}\left(\frac{{2}^{\left(2k+1\right)a}-1}{{2}^{2k+1}-1}-{2}^{a+1}+1\right),\hfill \end{array}$

2. (ii)
$\begin{array}{c}\sum _{s=0}^{k-1}\left(\genfrac{}{}{0}{}{2k}{2s+1}\right)\sum _{m=1}^{{2}^{a}-1}{\sigma }_{2k-2s-1,0}\left(\frac{m}{2};2\right){\sigma }_{2s+1}\left({2}^{a}-m\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\left(\frac{1+{2}^{2k}}{2\left(2k+1\right)}\right)\sum _{i=1}^{a-1}{B}_{2k+1}\left({2}^{i}\right)+\frac{1}{4}\left(\frac{{2}^{\left(2k+1\right)a}-1}{{2}^{2k+1}-1}\right)\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{1}{2}\left(\frac{{2}^{2k\left(a+1\right)}-{2}^{2k\left(a+1\right)-1}+{2}^{2ka-1}-1}{{2}^{2k}-1}\right)\hfill \\ \phantom{\rule{2em}{0ex}}+{2}^{a-2}\left(\frac{3-{2}^{\left(2k-1\right)\left(a+1\right)}-{2}^{\left(2k-1\right)a+1}}{{2}^{2k-1}-1}\right)+{2}^{a-1}-\frac{1}{4},\hfill \end{array}$

3. (iii)
$\begin{array}{c}\sum _{s=0}^{k-1}\left(\genfrac{}{}{0}{}{2k}{2s+1}\right)\sum _{m=1}^{{2}^{a}-1}{\sigma }_{2k-2s-1}\left(\frac{m}{2}\right){\sigma }_{2s+1}\left({2}^{a}-m\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\left(\frac{1}{2\left(2k+1\right)}\right){B}_{2k+1}\left({2}^{a}\right)+\left(\frac{1}{\left(2k+1\right)}\right)\sum _{i=1}^{a-1}{B}_{2k+1}\left({2}^{i}\right)\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{1}{2}\left(\frac{{2}^{\left(2k+1\right)a}-1}{{2}^{2k+1}-1}\right)+\frac{1}{2}\left(\frac{{2}^{2k\left(a+1\right)}-{2}^{2k\left(a+1\right)-1}+{2}^{2ka-1}-1}{{2}^{2k}-1}\right)\hfill \\ \phantom{\rule{2em}{0ex}}+{2}^{a-2}\left(\frac{3-{2}^{\left(2k-1\right)\left(a+1\right)}-{2}^{\left(2k-1\right)a+1}}{{2}^{2k-1}-1}\right).\hfill \end{array}$

## 2 Properties of convolution sums derived from divisor functions

Proposition 5 ([8])

Let k, n be positive integers. Then
$\begin{array}{rcl}\sum _{j=2}^{k}\left(\genfrac{}{}{0}{}{2k+1}{2j}\right){B}_{2j}{\sigma }_{2k+1-2j}\left(n\right)& =& -\left(\frac{2k+3}{2}\right){\sigma }_{2k+1}\left(n\right)-\left(2k+1\right)\left(\frac{k}{6}-n\right){\sigma }_{2k-1}\left(n\right)\\ +\left(2k+1\right)\sum _{s=0}^{k-1}\left(\genfrac{}{}{0}{}{2k}{2s+1}\right)\sum _{m=1}^{n-1}{\sigma }_{2k-2s-1}\left(m\right){\sigma }_{2s+1}\left(n-m\right).\end{array}$

Proposition 6 ([7, 11])

Let k, n be positive integers. Then
$\begin{array}{c}\left(\text{i}\right)\phantom{\rule{1em}{0ex}}\sum _{s=0}^{k-1}\left(\genfrac{}{}{0}{}{2k}{2s+1}\right)\sum _{m=1}^{n-1}{\sigma }_{2k-2s-1}^{\ast }\left(m\right){\sigma }_{2s+1}^{\ast }\left(n-m\right)=\frac{1}{2}\left({\sigma }_{2k+1}^{\ast }\left(n\right)-n{\sigma }_{2k-1}^{\ast }\left(n\right)\right),\hfill \\ \left(\text{ii}\right)\phantom{\rule{1em}{0ex}}\sum _{s=0}^{k-1}\left(\genfrac{}{}{0}{}{2k}{2s+1}\right)\sum _{m=1}^{n}{\sigma }_{2k-2s-1}\left(2m-1\right){\sigma }_{2s+1}\left(2n-2m+1\right)=\frac{1}{4}{\sigma }_{2k+1}^{\ast }\left(2n\right).\hfill \end{array}$
Proof of Theorem 1 Let $k,n\in \mathbb{N}$. By Proposition 5 and Proposition 6, we obtain
$\begin{array}{rcl}T& :=& \sum _{s=0}^{k-1}\left(\genfrac{}{}{0}{}{2k}{2s+1}\right)\sum _{m=1}^{n-1}{\stackrel{ˆ}{\sigma }}_{2k-2s-1}\left(m\right){\stackrel{ˆ}{\sigma }}_{2s+1}\left(n-m\right)\\ =& \sum _{s=0}^{k-1}\left(\genfrac{}{}{0}{}{2k}{2s+1}\right)\sum _{m=1}^{n-1}\left({\sigma }_{2k-2s-1}\left(m\right)-2{\sigma }_{2k-2s-1}\left(\frac{m}{2}\right)\right)\\ ×\left({\sigma }_{2s+1}\left(n-m\right)-2{\sigma }_{2s+1}\left(\frac{n-m}{2}\right)\right).\end{array}$
It is easily checked that
$\begin{array}{c}{\sigma }_{2k-2s-1}\left(\frac{m}{2}\right){\sigma }_{2s+1}\left(n-m\right)+{\sigma }_{2k-2s-1}\left(m\right){\sigma }_{2s+1}\left(\frac{n-m}{2}\right)\hfill \\ \phantom{\rule{1em}{0ex}}={\sigma }_{2k-2s-1}\left(\frac{m}{2}\right){\sigma }_{2s+1}\left(\frac{n-m}{2}\right)+{\sigma }_{2k-2s-1}\left(m\right){\sigma }_{2s+1}\left(n-m\right)\hfill \\ \phantom{\rule{2em}{0ex}}+\left({\sigma }_{2k-2s-1}\left(m\right)-{\sigma }_{2k-2s-1}\left(\frac{m}{2}\right)\right)\left({\sigma }_{2s+1}\left(n-m\right)-{\sigma }_{2s+1}\left(\frac{n-m}{2}\right)\right).\hfill \end{array}$
Thus,
$\begin{array}{rcl}T& =& \sum _{s=0}^{k-1}\left(\genfrac{}{}{0}{}{2k}{2s+1}\right)\sum _{m=1}^{n-1}\left({\sigma }_{2k-2s-1}\left(m\right){\sigma }_{2s+1}\left(n-m\right)+4{\sigma }_{2k-2s-1}\left(\frac{m}{2}\right){\sigma }_{2s+1}\left(\frac{n-m}{2}\right)\right)\\ -2\sum _{s=0}^{k-1}\left(\genfrac{}{}{0}{}{2k}{2s+1}\right)\sum _{m=1}^{n-1}\left({\sigma }_{2k-2s-1}\left(\frac{m}{2}\right){\sigma }_{2s+1}\left(\frac{n-m}{2}\right)+{\sigma }_{2k-2s+1}\left(m\right){\sigma }_{2s+1}\left(n-m\right)\right)\\ -2\sum _{s=0}^{k-1}\left(\genfrac{}{}{0}{}{2k}{2s+1}\right)\sum _{m=1}^{n-1}\left({\sigma }_{2k-2s-1}\left(m\right)-{\sigma }_{2k-2s-1}\left(\frac{m}{2}\right)\right)\\ ×\left({\sigma }_{2s+1}\left(n-m\right)-{\sigma }_{2s+1}\left(\frac{n-m}{2}\right)\right)\\ =& 2\sum _{s=0}^{k-1}\left(\genfrac{}{}{0}{}{2k}{2s+1}\right)\sum _{m=1}^{n-1}\left({\sigma }_{2k-2s-1}\left(m\right)-{\sigma }_{2k-2s-1}\left(\frac{m}{2}\right)\right)\left({\sigma }_{2s+1}\left(n-m\right)-{\sigma }_{2s+1}\left(\frac{n-m}{2}\right)\right)\\ -\sum _{s=0}^{k-1}\left(\genfrac{}{}{0}{}{2k}{2s+1}\right)\sum _{m=1}^{n-1}\left({\sigma }_{2k-2s-1}\left(m\right){\sigma }_{2s+1}\left(n-m\right)-2{\sigma }_{2k-2s-1}\left(\frac{m}{2}\right){\sigma }_{2s+1}\left(\frac{n-m}{2}\right)\right)\\ =& 2\sum _{s=0}^{k-1}\left(\genfrac{}{}{0}{}{2k}{2s+1}\right)\sum _{m=1}^{n-1}{\sigma }_{2k-2s-1}^{\ast }\left(m\right){\sigma }_{2s+1}^{\ast }\left(n-m\right)\\ -\left\{\left(\frac{2k+3}{4k+2}\right){\sigma }_{2k+1}\left(n\right)+\left(\frac{k}{6}-n\right){\sigma }_{2k-1}\left(n\right)+\frac{1}{2k+1}\sum _{j=2}^{k}\left(\genfrac{}{}{0}{}{2k+1}{2j}\right){B}_{2j}{\sigma }_{2k+1-2j}\left(n\right)\right\}\\ +2\left\{\left(\frac{2k+3}{4k+2}\right){\sigma }_{2k+1}\left(\frac{n}{2}\right)+\left(\frac{k}{6}-\frac{n}{2}\right){\sigma }_{2k-1}\left(\frac{n}{2}\right)\\ +\frac{1}{2k+1}\sum _{j=2}^{k}\left(\genfrac{}{}{0}{}{2k+1}{2j}\right){B}_{2j}{\sigma }_{2k+1-2j}\left(\frac{n}{2}\right)\right\}\\ =& \left({\sigma }_{2k+1}^{\ast }\left(n\right)-n{\sigma }_{2k-1}^{\ast }\left(n\right)\right)-\left(\frac{2k+3}{4k+2}\right)\left\{{\sigma }_{2k+1}\left(n\right)-2{\sigma }_{2k+1}\left(\frac{n}{2}\right)\right\}\\ -\frac{k}{6}\left\{{\sigma }_{2k-1}\left(n\right)-2{\sigma }_{2k-1}\left(\frac{n}{2}\right)\right\}+n\left\{{\sigma }_{2k-1}\left(n\right)-{\sigma }_{2k-1}\left(\frac{n}{2}\right)\right\}\\ -\frac{1}{2k+1}\sum _{j=2}^{k}\left(\genfrac{}{}{0}{}{2k+1}{2j}\right){B}_{2j}\left\{{\sigma }_{2k+1-2j}\left(n\right)-2{\sigma }_{2k+1-2j}\left(\frac{n}{2}\right)\right\}\\ =& {\sigma }_{2k+1}^{\ast }\left(n\right)-\left(\frac{2k+3}{4k+2}\right){\stackrel{ˆ}{\sigma }}_{2k+1}\left(n\right)-\frac{k}{6}{\stackrel{ˆ}{\sigma }}_{2k-1}\left(n\right)-\frac{1}{2k+1}\sum _{j=2}^{k}\left(\genfrac{}{}{0}{}{2k+1}{2j}\right){B}_{2j}{\stackrel{ˆ}{\sigma }}_{2k+1-2j}\left(n\right).\end{array}$

This proves the theorem. □

Example 7 Let n be a positive integer. In Theorem 1, put $k=2$, we get
$\sum _{m=1}^{n-1}\stackrel{ˆ}{\sigma }\left(m\right){\stackrel{ˆ}{\sigma }}_{3}\left(n-m\right)=\frac{1}{8}{\sigma }_{5}^{\ast }\left(n\right)-\frac{7}{80}{\stackrel{ˆ}{\sigma }}_{5}\left(n\right)-\frac{1}{24}{\stackrel{ˆ}{\sigma }}_{3}\left(n\right)+\frac{1}{240}\stackrel{ˆ}{\sigma }\left(n\right).$
Corollary 8 Let k, n be positive integers. Then, we obtain
1. (i)
$\begin{array}{c}\sum _{j=2}^{k}\left(\genfrac{}{}{0}{}{2k+1}{2j}\right){B}_{2j}{\stackrel{ˆ}{\sigma }}_{2k+1-2j}\left(2n\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\left(\frac{2k+1}{2}\right){\sigma }_{2k+1}^{\ast }\left(2n\right)-\left(\frac{2k+3}{2}\right){\stackrel{ˆ}{\sigma }}_{2k+1}\left(2n\right)-\left(\frac{k\left(2k+1\right)}{6}\right){\stackrel{ˆ}{\sigma }}_{2k-1}\left(2n\right)\hfill \\ \phantom{\rule{2em}{0ex}}-\left(2k+1\right)\sum _{s=0}^{k-1}\left(\genfrac{}{}{0}{}{2k}{2s+1}\right)\sum _{m=1}^{n-1}{\stackrel{ˆ}{\sigma }}_{2k-2s-1}\left(2m\right){\stackrel{ˆ}{\sigma }}_{2s+1}\left(2n-2m\right),\hfill \end{array}$

2. (ii)
$\begin{array}{c}\sum _{j=2}^{k}\left(\genfrac{}{}{0}{}{2k+1}{2j}\right){B}_{2j}\left\{{\sigma }_{2k+1-2j}\left(n\right)+{\sigma }_{2k+1-2j}\left(\frac{n}{2}\right)\right\}\hfill \\ \phantom{\rule{1em}{0ex}}=-{\sigma }_{2k+1}\left(n\right)-2\left(k+1\right){\sigma }_{2k+1}\left(\frac{n}{2}\right)-\frac{\left(2k+1\right)\left(k-3n\right)}{6}{\sigma }_{2k-1}\left(n\right)\hfill \\ \phantom{\rule{2em}{0ex}}-\frac{\left(2k+1\right)\left(k-6n\right)}{6}{\sigma }_{2k-1}\left(\frac{n}{2}\right)\hfill \\ \phantom{\rule{2em}{0ex}}+2\left(2k+1\right)\sum _{s=0}^{k-1}\left(\genfrac{}{}{0}{}{2k}{2s+1}\right)\sum _{m=1}^{n-1}{\sigma }_{2k-2s-1}\left(\frac{m}{2}\right){\sigma }_{2s+1}\left(n-m\right),\hfill \end{array}$

3. (iii)
$\begin{array}{c}\sum _{j=2}^{k}\left(\genfrac{}{}{0}{}{2k+1}{2j}\right){B}_{2j}{\sigma }_{2k+1-2j}^{\ast }\left(n\right)\hfill \\ \phantom{\rule{1em}{0ex}}=2k{\sigma }_{2k+1}^{\ast }\left(n\right)-\frac{k\left(2k+1\right)}{6}{\sigma }_{2k-1}^{\ast }\left(n\right)-\frac{n\left(2k+1\right)}{2}{\stackrel{ˆ}{\sigma }}_{2k-1}\left(n\right)\hfill \\ \phantom{\rule{2em}{0ex}}-2\left(2k+1\right)\sum _{s=0}^{k-1}\left(\genfrac{}{}{0}{}{2k}{2s+1}\right)\sum _{m=1}^{n-1}{\sigma }_{2k-2s-1}^{\ast }\left(m\right){\stackrel{ˆ}{\sigma }}_{2s+1}\left(n-m\right)\hfill \\ \phantom{\rule{1em}{0ex}}=-2\left(k+1\right){\sigma }_{2k+1}^{\ast }\left(n\right)-\frac{\left(k-9n\right)\left(2k+1\right)}{6}{\sigma }_{2k-1}^{\ast }\left(n\right)+\frac{n\left(2k+1\right)}{2}{\sigma }_{2k-1}\left(\frac{n}{2}\right)\hfill \\ \phantom{\rule{2em}{0ex}}+2\left(2k+1\right)\sum _{s=0}^{k-1}\left(\genfrac{}{}{0}{}{2k}{2s+1}\right)\sum _{m=1}^{n-1}{\sigma }_{2k-2s-1}^{\ast }\left(m\right){\sigma }_{2s+1}\left(n-m\right).\hfill \end{array}$

Proof (i) We note that
$\begin{array}{c}\sum _{s=0}^{k-1}\left(\genfrac{}{}{0}{}{2k}{2s+1}\right)\sum _{m=1}^{n-1}{\sigma }_{2k-2s-1}^{\ast }\left(2m\right){\sigma }_{2s+1}^{\ast }\left(n-2m\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{s=0}^{k-1}\left(\genfrac{}{}{0}{}{2k}{2s+1}\right)\sum _{m=1}^{n-1}{\sigma }_{2k-2s-1}^{\ast }\left(m\right){\sigma }_{2s+1}^{\ast }\left(n-m\right)\hfill \\ \phantom{\rule{2em}{0ex}}-\sum _{s=0}^{k-1}\left(\genfrac{}{}{0}{}{2k}{2s+1}\right)\sum _{m=1}^{n-1}{\sigma }_{2k-2s-1}^{\ast }\left(2m-1\right){\sigma }_{2s+1}^{\ast }\left(n-\left(2m-1\right)\right).\hfill \end{array}$

(ii) and (iii) are applied in a similar way. □

## 3 Bernoulli polynomials and convolution sums

Proposition 9 ([12])

Let k, n be positive integers. Then
$\begin{array}{c}\sum _{s=0}^{k-1}\left(\genfrac{}{}{0}{}{2k}{2s+1}\right)\sum _{m=1}^{n-1}{2}^{2k-2s-1}{\sigma }_{2k-2s-1}\left(m/4\right){\sigma }_{2s+1}\left(n-m\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{4}{\sigma }_{2k+1}\left(n/2\right)-\frac{1}{4}\left({\sigma }_{2k}\left(n\right)-{2}^{2k+1}{\sigma }_{2k}\left(n/2\right)-{2}^{2k}{\sigma }_{2k}\left(n/4\right)\right)\hfill \\ \phantom{\rule{2em}{0ex}}-\frac{n}{4}\left({\sigma }_{2k-1}\left(n\right)+{2}^{2k}{\sigma }_{2k-1}\left(n/4\right)\right)\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{1}{2}{\sigma }_{2k,1}\left(n;2\right)+\frac{{2}^{2k-1}}{2k+1}\sum _{j=0}^{2k}\left(\genfrac{}{}{0}{}{2k+1}{j}\right){B}_{j}{\sigma }_{2k+1-j}\left(n/2\right)\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{1}{2\left(2k+1\right)}\sum _{j=0}^{2k}\left(\genfrac{}{}{0}{}{2k+1}{j}\right){B}_{j}{2}^{2k+1-j}{\sigma }_{2k+1-j}\left(n/4\right)\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{1}{2\left(2k+1\right)}\sum _{j=0}^{2k}\left(\genfrac{}{}{0}{}{2k+1}{j}\right){B}_{j}{\sigma }_{2k+1-j,1}\left(n;2\right)\hfill \\ \phantom{\rule{2em}{0ex}}-\frac{1}{2\left(2k+1\right)}\sum _{u+v+w=2k+1}{2}^{v-1}\left(\genfrac{}{}{0}{}{2k+1}{u,v,w}\right){B}_{v}{\sigma }_{w,1}\left(n;2\right).\hfill \end{array}$

It is well known that ${\sigma }_{2k-2s-1,0}\left(\frac{m}{2};2\right)={2}^{2k-2s-1}{\sigma }_{2k-2s-1}\left(\frac{m}{4}\right)$. Using Proposition 9, we get this lemma.

Lemma 10 Let k, n be positive integers. Then
1. (i)
$\begin{array}{c}\sum _{s=0}^{k-1}\left(\genfrac{}{}{0}{}{2k}{2s+1}\right)\sum _{m=1}^{n-1}{\sigma }_{2k-2s-1,0}\left(\frac{m}{2};2\right){\sigma }_{2s+1}\left(n-m\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{4}{\sigma }_{2k+1}\left(\frac{n}{2}\right)+\frac{1}{4\left(2k+1\right)}{\sigma }_{2k+1,0}\left(n;2\right)+\frac{1}{2\left(2k+1\right)}{\sigma }_{2k+1,0}\left(\frac{n}{2};2\right)\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{1}{2\left(2k+1\right)}{\sigma }_{2k,1}\left(n;2\right)-\frac{1}{4}{\sigma }_{2k}\left(n\right)+\frac{1}{2}{\sigma }_{2k,0}\left(n;2\right)-\frac{1}{4}{\sigma }_{2k,0}\left(n;2\right)+\frac{1}{4}{\sigma }_{2k,1}\left(n;2\right)\hfill \\ \phantom{\rule{2em}{0ex}}-\frac{n}{4}{\sigma }_{2k-1}\left(n\right)+\frac{k}{6}{\sigma }_{2k-1,0}\left(n;2\right)+\left(\frac{k}{12}-\frac{n}{8}\right){\sigma }_{2k-1,0}\left(\frac{n}{2};2\right)\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{k}{12}{\sigma }_{2k-1,1}\left(n;2\right)+\frac{{2}^{2k}}{2\left(2k+1\right)}\sum _{j=2}^{k}\left(\genfrac{}{}{0}{}{2k+1}{2j}\right){B}_{2j}{\sigma }_{2k+1-2j}\left(\frac{n}{2}\right)\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{1}{2\left(2k+1\right)}\sum _{j=2}^{k}\left(\genfrac{}{}{0}{}{2k+1}{2j}\right){B}_{2j}{\sigma }_{2k+1-2j,0}\left(\frac{n}{2};2\right)\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{1}{2\left(2k+1\right)}\sum _{j=2}^{k}\left(\genfrac{}{}{0}{}{2k+1}{2j}\right){B}_{2j}{\sigma }_{2k+1-2j,1}\left(n;2\right)\hfill \\ \phantom{\rule{2em}{0ex}}-\frac{1}{2\left(2k+1\right)}\sum _{u+v+w=2k+1}{2}^{v-1}\left(\genfrac{}{}{0}{}{2k+1}{u,v,w}\right){B}_{v}{\sigma }_{w,1}\left(n;2\right),\hfill \end{array}$

2. (ii)
$\begin{array}{c}\sum _{s=0}^{k-1}\left(\genfrac{}{}{0}{}{2k}{2s+1}\right)\sum _{m=1}^{n-1}{\sigma }_{2k-2s-1,1}\left(\frac{m}{2};2\right){\sigma }_{2s+1}\left(n-m\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\left(\frac{1}{2\left(2k+1\right)}\right){\sigma }_{2k+1}\left(n\right)+\left(\frac{k+1}{2k+1}\right){\sigma }_{2k+1}\left(\frac{n}{2}\right)+\left(\frac{k-3n}{12}\right){\sigma }_{2k-1}\left(n\right)\hfill \\ \phantom{\rule{2em}{0ex}}+\left(\frac{k-6n}{12}\right){\sigma }_{2k-1}\left(\frac{n}{2}\right)-\frac{1}{4}{\sigma }_{2k+1}\left(\frac{n}{2}\right)\hfill \\ \phantom{\rule{2em}{0ex}}-\left(\frac{1}{4\left(2k+1\right)}\right){\sigma }_{2k+1,0}\left(n;2\right)-\left(\frac{1}{2\left(2k+1\right)}\right){\sigma }_{2k+1,0}\left(\frac{n}{2};2\right)\hfill \\ \phantom{\rule{2em}{0ex}}-\left(\frac{1}{2\left(2k+1\right)}\right){\sigma }_{2k,1}\left(n;2\right)+\frac{1}{4}{\sigma }_{2k}\left(n\right)\hfill \\ \phantom{\rule{2em}{0ex}}-\frac{1}{2}{\sigma }_{2k,0}\left(n;2\right)+\frac{1}{4}{\sigma }_{2k,0}\left(n;2\right)-\frac{1}{4}{\sigma }_{2k,1}\left(n;2\right)\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{n}{4}{\sigma }_{2k-1}\left(n\right)-\frac{k}{6}{\sigma }_{2k-1,0}\left(n;2\right)-\left(\frac{k}{12}-\frac{n}{8}\right){\sigma }_{2k-1,0}\left(\frac{n}{2};2\right)-\frac{k}{12}{\sigma }_{2k-1,1}\left(n;2\right)\hfill \\ \phantom{\rule{2em}{0ex}}-\frac{{2}^{2k}}{2\left(2k+1\right)}\sum _{j=2}^{k}\left(\genfrac{}{}{0}{}{2k+1}{2j}\right){B}_{2j}{\sigma }_{2k+1-2j}\left(\frac{n}{2}\right)\hfill \\ \phantom{\rule{2em}{0ex}}-\frac{1}{2\left(2k+1\right)}\sum _{j=2}^{k}\left(\genfrac{}{}{0}{}{2k+1}{2j}\right){B}_{2j}{\sigma }_{2k+1-2j,0}\left(\frac{n}{2};2\right)\hfill \\ \phantom{\rule{2em}{0ex}}-\frac{1}{2\left(2k+1\right)}\sum _{j=2}^{k}\left(\genfrac{}{}{0}{}{2k+1}{2j}\right){B}_{2j}{\sigma }_{2k+1-2j,1}\left(n;2\right)\hfill \\ \phantom{\rule{2em}{0ex}}+\frac{1}{2\left(2k+1\right)}\sum _{u+v+w=2k+1}{2}^{v-1}\left(\genfrac{}{}{0}{}{2k+1}{u,v,w}\right){B}_{v}{\sigma }_{w,1}\left(n;2\right).\hfill \end{array}$

Remark 11 (i) Using Lemma 10, we obtain
$\begin{array}{c}\sum _{u+v+w=2k+1}{2}^{v-1}\left(\genfrac{}{}{0}{}{2k+1}{u,v,w}\right){B}_{v}{\sigma }_{w,1}\left(n;2\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\left(\frac{2k+1}{2}\right)\left\{{\sigma }_{2k+1}\left(\frac{n}{2}\right)-{\sigma }_{2k}\left(n\right)+2{\sigma }_{2k,0}\left(n;2\right)+{\sigma }_{2k,0}\left(\frac{n}{2};2\right)\right\}\hfill \\ \phantom{\rule{2em}{0ex}}-\left(\frac{n\left(2k+1\right)}{2}\right)\left\{{\sigma }_{2k-1}\left(n\right)+2{\sigma }_{2k-1,0}\left(\frac{n}{2};2\right)\right\}+\left(2k+1\right){\sigma }_{2k,1}\left(n;2\right)\hfill \\ \phantom{\rule{2em}{0ex}}+{2}^{2k}\sum _{j=0}^{2k}\left(\genfrac{}{}{0}{}{2k+1}{j}\right){B}_{j}{\sigma }_{2k+1-j}\left(\frac{n}{2}\right)+\sum _{j=0}^{2k}\left(\genfrac{}{}{0}{}{2k+1}{j}\right){B}_{j}{\sigma }_{2k+1-j,0}\left(\frac{n}{2};2\right)\hfill \\ \phantom{\rule{2em}{0ex}}+\sum _{j=0}^{2k}\left(\genfrac{}{}{0}{}{2k+1}{j}\right){B}_{j}{\sigma }_{2k+1-j,1}\left(n;2\right)\hfill \\ \phantom{\rule{2em}{0ex}}-2\left(2k+1\right)\sum _{s=0}^{k-1}\left(\genfrac{}{}{0}{}{2k}{2s+1}\right)\sum _{m=1}^{n-1}{\sigma }_{2k-2s-1,0}\left(\frac{m}{2};2\right){\sigma }_{2s+1}\left(n-m\right)\hfill \\ \phantom{\rule{1em}{0ex}}=-\left(\frac{2k+1}{2}\right)\left\{{\sigma }_{2k+1}\left(\frac{n}{2}\right)-2n{\sigma }_{2k-1,1}\left(\frac{n}{2};2\right)+{\sigma }_{2k,1}\left(\frac{n}{2};2\right)\right\}\hfill \\ \phantom{\rule{2em}{0ex}}-\sum _{j=0}^{2k}\left(\genfrac{}{}{0}{}{2k+1}{j}\right){B}_{j}{\sigma }_{2k+1-j}\left(n\right)+\left({2}^{2k}-1\right)\sum _{j=0}^{2k}\left(\genfrac{}{}{0}{}{2k+1}{j}\right){B}_{j}{\sigma }_{2k+1-j}\left(\frac{n}{2}\right)\hfill \\ \phantom{\rule{2em}{0ex}}+\sum _{j=0}^{2k}\left(\genfrac{}{}{0}{}{2k+1}{j}\right){B}_{j}{\sigma }_{2k+1-j,0}\left(\frac{n}{2};2\right)+\sum _{j=0}^{2k}\left(\genfrac{}{}{0}{}{2k+1}{j}\right){B}_{j}{\sigma }_{2k+1-j,1}\left(n;2\right)\hfill \\ \phantom{\rule{2em}{0ex}}+2\left(2k+1\right)\sum _{s=0}^{k-1}\left(\genfrac{}{}{0}{}{2k}{2s+1}\right)\sum _{m=1}^{n-1}{\sigma }_{2k-2s-1,1}\left(\frac{m}{2};2\right){\sigma }_{2s+1}\left(n-m\right).\hfill \end{array}$
1. (ii)
If n is an odd integer, then
$\begin{array}{c}\sum _{u+v+w=2k+1}{2}^{v-1}\left(\genfrac{}{}{0}{}{2k+1}{u,v,w}\right){B}_{v}{\sigma }_{w,1}\left(n;2\right)\hfill \\ \phantom{\rule{1em}{0ex}}=2\left(2k+1\right)\sum _{s=0}^{k-1}\left(\genfrac{}{}{0}{}{2k}{2s+1}\right)\sum _{m=1}^{n-1}{\sigma }_{2k-2s-1,1}\left(\frac{m}{2};2\right){\sigma }_{2s+1}\left(n-m\right).\hfill \end{array}$
(3.1)

2. (iii)
In (3.1), put $k=1$, we get
$\sum _{u+v+w=3}{2}^{v-1}\left(\genfrac{}{}{0}{}{3}{u,v,w}\right){B}_{v}{\sigma }_{w,1}\left(n;2\right)=12\sum _{m=1}^{n-1}{\sigma }_{1,1}\left(\frac{m}{2};2\right){\sigma }_{1}\left(n-m\right),$

and thus,
$\sum _{m=1}^{n-1}{\sigma }_{1,1}\left(\frac{m}{2};2\right){\sigma }_{1}\left(n-m\right)=\frac{1}{24}\left({\sigma }_{3}\left(n\right)-{\sigma }_{1}\left(n\right)\right).$
In (3.1), replace k by 2, we find that
$\begin{array}{c}\sum _{u+v+w=5}{2}^{v-1}\left(\genfrac{}{}{0}{}{5}{u,v,w}\right){B}_{v}{\sigma }_{w,1}\left(n;2\right)\hfill \\ \phantom{\rule{1em}{0ex}}=40\left(\sum _{m=1}^{n-1}{\sigma }_{3,1}\left(\frac{m}{2};2\right){\sigma }_{1}\left(n-m\right)+\sum _{m=1}^{n-1}{\sigma }_{1,1}\left(\frac{m}{2};2\right){\sigma }_{3}\left(n-m\right)\right),\hfill \end{array}$
and thus,
$\begin{array}{c}\sum _{m=1}^{n-1}{\sigma }_{3,1}\left(\frac{m}{2};2\right){\sigma }_{1}\left(n-m\right)+\sum _{m=1}^{n-1}{\sigma }_{1,1}\left(\frac{m}{2};2\right){\sigma }_{3}\left(n-m\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{240}\left(3{\sigma }_{5}\left(n\right)-10{\sigma }_{3}\left(n\right)+7{\sigma }_{1}\left(n\right)\right).\hfill \end{array}$
Proof of Theorem 3 If $n=1$, compare both sides of (3.1), we obtain
$\sum _{u+v+w=2k+1}{2}^{v-1}\left(\genfrac{}{}{0}{}{2k+1}{u,v,w}\right){B}_{v}=0.$
(3.2)
If we put $n=3$ in (3.1), we obtain
$\begin{array}{c}\sum _{u+v+w=2k+1}{2}^{v-1}\left(\genfrac{}{}{0}{}{2k+1}{u,v,w}\right){B}_{v}\cdot \left(1+{3}^{w}\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{u+v+w=2k+1}{2}^{v-1}\left(\genfrac{}{}{0}{}{2k+1}{u,v,w}\right){B}_{v}\cdot {3}^{w}\hfill \\ \phantom{\rule{1em}{0ex}}=2\left(2k+1\right)\sum _{s=0}^{k-1}\left(\genfrac{}{}{0}{}{2k}{2s+1}\right).\hfill \end{array}$
(3.3)
From (3.1) and (3.3), we get
$\sum _{u+v+w=2k+1}{2}^{v-1}\left(\genfrac{}{}{0}{}{2k+1}{u,v,w}\right){B}_{v}\cdot {3}^{w}=\left(2k+1\right){2}^{2k}.$
(3.4)
By combining (1.1) and (3.4), we obtain
$\sum _{u+v+w=2k+1}{2}^{v-1}\left(\genfrac{}{}{0}{}{2k+1}{u,v,w}\right){B}_{v}\cdot {3}^{w}={2}^{2k}{B}_{2k+1}\left(2\right).$

Others cases follow in a similar way. This completes the proof. □

Proof of Theorem 4 (i) If a is a positive integer, then
$\sum _{u+v+w=2k+1}{2}^{v-1}\left(\genfrac{}{}{0}{}{2k+1}{u,v,w}\right){B}_{v}\cdot {\sigma }_{w,1}\left({2}^{a};2\right)=\sum _{u+v+w=2k+1}{2}^{v-1}\left(\genfrac{}{}{0}{}{2k+1}{u,v,w}\right){B}_{v}=0$
by (3.2). According to Remark 11(i), we deduce that
$\begin{array}{rcl}0& =& 2\left(2k+1\right)\sum _{s=0}^{k-1}\left(\genfrac{}{}{0}{}{2k}{2s+1}\right)\sum _{m=1}^{{2}^{a}-1}{\sigma }_{2k-2s-1,1}\left(\frac{m}{2};2\right){\sigma }_{2s+1}\left({2}^{a}-m\right)\\ -\sum _{j=0}^{2k}\left(\genfrac{}{}{0}{}{2k+1}{j}\right){B}_{j}{\sigma }_{2k+1-j}\left({2}^{a}\right)+\left({2}^{2k}-1\right)\sum _{j=0}^{2k}\left(\genfrac{}{}{0}{}{2k+1}{j}\right){B}_{j}{\sigma }_{2k+1-j}\left(\frac{{2}^{a}}{2}\right)\\ +\sum _{j=0}^{2k}\left(\genfrac{}{}{0}{}{2k+1}{j}\right){B}_{j}{\sigma }_{2k+1-j,0}\left(\frac{{2}^{a}}{2};2\right)-\left(\frac{2k+1}{2}\right)\left\{{\sigma }_{2k+1}\left(\frac{{2}^{a}}{2}\right)-2\left({2}^{a}\right)+1\right\}.\end{array}$
If $a=1$, it is clearly evident. We suppose that $a>1$. We check that
$\begin{array}{c}\sum _{j=0}^{2k}\left(\genfrac{}{}{0}{}{2k+1}{j}\right){B}_{j}\cdot {\left({2}^{a}\right)}^{2k+1-j}\hfill \\ \phantom{\rule{1em}{0ex}}=\sum _{j=0}^{2k}\left(\genfrac{}{}{0}{}{2k+1}{j}\right){\left(-1\right)}^{j}{B}_{j}\cdot {\left({2}^{a}\right)}^{2k+1-j}+2\left(\genfrac{}{}{0}{}{2k+1}{1}\right){B}_{1}\cdot {\left({2}^{a}\right)}^{2k}\hfill \\ \phantom{\rule{1em}{0ex}}=\left(2k+1\right)\sum _{j=0}^{{2}^{a}}{j}^{2k}+2\left(2k+1\right)\left(\frac{-1}{2}\right){\left({2}^{a}\right)}^{2k}\hfill \\ \phantom{\rule{1em}{0ex}}={B}_{2k+1}\left({2}^{a}\right)\hfill \end{array}$
(3.5)
by (1.1).
1. (ii)

and (iii) are applied in a similar way. □

Remark 12 If p is a prime integer, then
$\sum _{u+v+w=2k+1}{2}^{v-2}\left(\genfrac{}{}{0}{}{2k+1}{u,v,w}\right){B}_{v}\cdot {p}^{w}=\left(2k+1\right)\sum _{s=0}^{k-1}\left(\genfrac{}{}{0}{}{2k}{2s+1}\right)\sum _{m=1}^{p-1}{\sigma }_{2k-2s-1,1}\left(\frac{m}{2};2\right){\sigma }_{2s+1}\left(p-m\right)$

by (3.1) and (3.2).

## Authors’ Affiliations

(1)
National Institute for Mathematical Sciences, Yuseong-daero 1689-gil, Yuseong-gu, Daejeon, 305-811, South Korea
(2)
Department of Elementary Mathematics Education, Necatibey Faculty of Education, Balikesir University, Balikesir, 10100, Turkey

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