Second, we show a formula of the fourth order of convolution sums of divisor functions expressed by their divisor functions and linear combination of Bernoulli polynomials

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Vol. 22 (2021), No. 2, pp. 731–748 DOI: 10.18514/MMN.2021.3369

RELATIONS OF COMBINATORIC CONVOLUTION SUMS FOR RESTRICTED DIVISOR FUNCTIONS AND BERNOULLI

POLYNOMIALS

DAEYEOUL KIM AND NAZLI YILDIZ IKIKARDES Received 03 June, 2020

Abstract. In this paper, we study combinatoric convolution sums involving divisor functions.

First, we establish some explicit formulas for certain combinatoric convolution sums of divisor functions derived from Bernoulli polynomials. Second, we show a formula of the fourth order of convolution sums of divisor functions expressed by their divisor functions and linear combination of Bernoulli polynomials.

2010 Mathematics Subject Classification: 11A05, 33E99

Keywords: Bernoulli Polynomials, Convolution sums, Divisor functions, Linear combination of polynomials

1. INTRODUCTION AND STATEMENT OF MAIN RESULTS

The symbols N, Z and R denote the set of natural numbers, the ring of integers and the field of real numbers, respectively. Over the years Bernoulli polynomials have been used to prove important mathematical theorems. Since the 17th century, many mathematicians in different fields have been interested in Bernoulli polynomials. The Bernoulli polynomials Bk(x), which are usually defined by the exponential generating function

te^xt e^t− 1=

∞

∑

k=0

Bk(x) k! t^k,

play an important and quite mysterious role in mathematics and various fields like analysis, number theory and differential topology. The Bernoulli numbers Bk are defined to be Bk:= Bk(0).

The Bernoulli polynomial [7] is expressed through the respective numbers Bn(x) =

n

∑

k=0

n k

Bkx^n−k,

∗This work was supported by The Research Fund of Balikesir University, Project No: 2014/55.

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Bn(x + y) =

n

∑

k=0

n k

Bk(x)y^n−k, B^′_n(x) = nB_n−1(x),

Bn(1 − x) = (−1)ⁿBn(x) and B_n(1 + x) − Bn(x) = nxⁿ⁻¹. For n ∈ N and k ∈ Z, we define two divisor functions

σk(n) :=

∑

d|n

d^k, σˆk(n) :=

∑

d|n

(−1)ⁿ^d⁻¹d^k, σ^∗_k(n) :=

∑

d|n

n dodd

d^k.

The identity

n−1

∑

k=1

σ(k)σ(n − k) = 5

12σ₃(n) + 1 12−1

2n

σ(n)

for the basic convolution sum first appeared in a letter from Besge to Liouville in 1862 ([3]). Hahn [6, (4.8)] considered

36

∑

m<n

σ(m) ˆˆ σ(n − m) =

(−3 ˆσ(n) + 3eσ3(n), if n is odd,

−3 ˆσ(n) − 5eσ₃(n) + 4eσ₃(ⁿ₂), if n is even. (1.1) It is well-known that ˆσk(n) = σk(n) − 2σk(ⁿ₂) ([6, (1.13)]).

In this article, we are trying to focus on the combinatorial convolution sums. For positive integers k, l, n, p and q the combinatorial convolution sums

k−1 s=0

∑

2k 2s + 1

_n−1

m=1

∑

σˆ2k−2s−1(m) ˆσ2s+1(n − m) and

∑

1≤m≤p−1 1≤m^′≤q−1 a,b,c,d odd a+b+c+d=2l

2l a, b, c, d

σˆa(m) ˆσb(p − m) ˆσc(m^′) ˆσd(q − m^′)

can be evaluated explicitly in terms of divisor functions and a sum involving Bernoulli polynomials.

We are motivated by Ramanujan’s recursion formula for sums of the product of two Eisenstein series [2] and its proof, and also the following identities ([1],[4],[9]):

k−1

∑

s=0

2k 2s + 1

_N−1

∑

m=1

σ_2k−2s−1(m)σ2s+1(N − m)

=2k + 3

4k + 2σ2k+1(N) + k 6− N

σ2k−1(N)

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+ 1 2k + 1

k

∑

j=2

2k + 1 2 j

B_{2 j}σ2k+1−2 j(N).

Using these new formulas and addition theorem of Bernoulli polynomials, we derive the explicit formulas for quadnomial convolution sums of divisor functions. We define ˆB_2k+1(n) that if n is even, then ∑d|nB_2k+1(d) − 2 ∑d|ⁿ₂B_2k+1(d) and if n is odd, then ∑d|nB_2k+1(d). Similarly, ˆˆB_2k+1(n) that if n is even, then ∑d|nBˆ_2k+1(d) − 2 ∑d|ⁿ₂Bˆ_2k+1(d) and if n is odd, then ∑d|nBˆ_2k+1(d). More precisely, we prove the following results.

Theorem 1. If k ≥ 1 and n ≥ 2, then

k−1

∑

s=0

2k 2s + 1

_n−1

∑

m=1

ˆ

σ_2k−2s−1(m) ˆσ_2s+1(n − m)

=1

2σ_2k+1(n) −1

2σˆ_2k(n) − 1

2k + 1Bˆ_2k+1(n).

Theorem 2. If k ≥ 2 and N ≥ 1, then

k−1

∑

s=1

2k 2s + 1

_2N−1

∑

m=1

(−1)^mσˆ2k−2s−1(m) ˆσ2s+1(2N − m)

=1

2σ2k+1(N) −1

2σˆ2k(2N) − 1

2k + 1Bˆ_2k+1(2N).

Corollary 1. For N ≥ 1, we have

2N−1

∑

mm=1odd leven

ˆ

σ_l−1(m) ˆσ(2N − m)

= 1

2l[σl+1(N) + σl+1(2N)] −1

lσˆl(2N) − 2

l(l + 1)Bˆ_l+1(2N).

Corollary 2. For k ≥ 2 and N ≥ 1, we have

k−1

∑

s=1

2k 2s + 1

_2N−1

∑

m=1 m odd

ˆ

σ_2k−2s−1(m) ˆσ_2s+1(2N − m) = 2^2k−1σ^∗_2k+1(N)

−1

4[σ_2k+1(N) + σ_2k+1(2N)] −1

2σˆ_2k(2N) + 1

2k + 1Bˆ_2k+1(2N).

Theorem 3. Let m ∈ N, m ≥ 1 and x, y ∈ R. Then we have

m−1

∑

k=1

2m 2k

B_2k+1(x)B_2m−2k+1(y)

(2k + 1)(2m − 2k + 1) = 1

4m + 4{B_2m+2(y − x)

−B_2m+2(x + y)} − 1

2m + 1{xB_2m+1(y) + yB_2m+1(x)}

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+ 1

4m + 2{B_2m+1(x) + B_2m+1(y) + (x + y − 1)B_2m+1(x + y) +(x − y)B_2m+1(y − x)} .

The following theorem gives a formula of the fourth order of convolution sums of divisor functions expressed by their divisor functions and linear combination of Bernoulli polynomials.

Theorem 4. Let l, p, q ∈ N with greater than 1. Then

∑

2l a, b, c, d

= 1

8(2l + 1)n ˆˆB_2l+1(q + p + 1) + ˆˆB_2l+1(q − p + 1) − ˆˆB_2l+1(q + p) − ˆˆB_2l+1(q − p) +

∑

d|p d^′|q

d^′dB_2l+1(d^′+ d + 1) + B2l+1(d^′− d + 1) − B2l+1(d^′+ d) − B2l+1(d^′− d)

−

∑

d|p

dBˆ_2l+1(q + d + 1) + ˆB_2l+1(q − d + 1) − ˆB_2l+1(q + d) − ˆB_2l+1(q − d)

−

∑

d^′|q

d^′Bˆ_2l+1(d^′+ p + 1) + ˆB_2l+1(d^′− p + 1) − ˆB_2l+1(d^′+ p) − ˆB_2l+1(d^′− p) )

+ 1

4(2l + 1) n

2 ˆˆB_2l+1(p + q) + ˆˆB_2l+1(p − q) + ˆˆB_2l+1(q − p)

−

∑

d|p

dBˆ_2l+1(q + d) + ˆB_2l+1(q − d) −

∑

d^′|q

d^′Bˆ_2l+1(p + d^′) + ˆB_2l+1(p − d^′) )

+ 1

2(2l + 1) n

(p + q − 1) ˆˆB_2l+1(p + q) + (p − q) ˆˆB_2l+1(q − p)

− (2p − 1) ˆˆB_2l+1(q) − (2q − 1) ˆˆB_2l+1(p) + ˆB_2l+1(p)h

σ₁(q) − σ₀(q) + 2σ₀q 2

i + ˆB_2l+1(q)

h

σ1(p) − σ0(p) + 2σ0

p 2

io + 1

4σ₁(p) − σ₁

p 2

h

σ_2l+1(q) − σ_2l(q) + 2σ_2l

q 2

i

+h

σ2l+1(p) − σ_2l(p) + 2σ_2lp 2

i 1

4σ1(q) − σ₁q 2

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− 1

4(l + 1)h ˆˆB_2l+2(p + q) − ˆˆB_2l+2(q − p)i .

2. PROPERTIES OF COMBINATORIC CONVOLUTION SUMS OF DIVISOR FUNCTIONS DERIVED FROMBERNOULLI POLYNOMIALS

The purpose of this section is to give proofs of the main results. Also, we would like to enrich this section with examples. To proof of the theorems, we need the auxiliary results and lemmas.

Lemma 1. ([5]) Let k, n be positive integers. Then

k

∑

j=2

2k + 1 2 j

B_{2 j}σˆ_{2k+1−2 j}(n) = (2k + 1)σ^∗_2k+1(n)

− 2k + 3

2 ˆσ_2k+1(n) − k(2k + 1)

6 ˆσ_2k−1(n)

− (2k + 1)

k−1

∑

s=0

2k 2s + 1

_n−1

∑

m=1

ˆ

σ_2k−2s−1(m) ˆσ_2s+1(n − m).

Lemma 2. ([8]) Let m∈ N and x, y ∈ R. Then we have (a)

m

∑

k=0

m k

B_k+1(x) k+ 1

B_m−k+1(y) m− k + 1

= (x + y − 1)B_m+1(x + y)

m+ 1 −B_m+2(x + y)

m+ 2 − B_m+2(x)

(m + 1)(m + 2)− B_m+2(y) (m + 1)(m + 2). (b)

m

∑

k=0

(−1)^km k

B_k+1(x) k+ 1

B_m−k+1(y) m− k + 1

= (x − y)B_m+1(y − x)

m+ 1 +B_m+2(y − x)

m+ 2 + B_m+2(1 − x)

(m + 1)(m + 2)+ B_m+2(y) (m + 1)(m + 2).

Now we are ready to prove the main results of the article.

Proof of the Theorem1. Consider T :=

k

∑

j=2

2k + 1 2 j

B_{2 j}σˆ2k+1−2 j(n)

=

2k+1

∑

j=0

2k + 1 j

Bjσˆ_{2k+1− j}(n) −

k

∑

j=0

2k + 1 2 j + 1

B_{2 j+1}σˆ2k+1−(2 j+1)(n)

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−2k + 1 0

B₀σˆ_2k+1(n) −2k + 1 2

B₂σˆ_2k+1−2(n).

It is well-known that B₀= 1, B₁= −¹₂, B₂=¹₆ and B_{2 j+1}= 0 ( j ≥ 1). Thus

T =

2k+1

∑

j=0

2k + 1 j

Bjσˆ_{2k+1− j}(n) −2k + 1 0

B₀σˆ_2k+1(n)

−2k + 1 1

B₁σˆ_2k(n) −2k + 1 2

B₂σˆ_2k−1(n)

=

2k+1

∑

j=0

2k + 1 j

Bjσˆ2k+1− j(n) − ˆσ2k+1(n) +

k+1

2

σˆ2k(n) −k(2k + 1)

6 σˆ2k−1(n).

(2.1) Using Bn(x) = ∑ⁿ_l=0 ⁿ_lB_lx^n−l, we obtain

2k+1

∑

j=0

2k + 1 j

B_j

σ_{2k+1− j}(n) − 2σ_{2k+1− j}

n 2

=

2k+1

∑

j=0

2k + 1 j

B_j





∑

d|n

d^{2k+1− j}− 2

∑

d|ⁿ₂

d^{2k+1− j}





=

∑

d|n

B_2k+1(d) − 2

∑

d|ⁿ₂

B_2k+1(d)

= ˆB_2k+1(n).

(2.2) Combining (2.1), Lemma1and (2.2) we deduce that

2k+1

∑

j=0

2k + 1 j

B_jσˆ2k+1− j(n) − ˆσ2k+1(n) + (k +1

2) ˆσ2k(n) −k(2k + 1)

6 σˆ2k−1(n)

= ˆB_2k+1(n) − ˆσ_2k+1(n) + (k +1

2) ˆσ_2k(n) −k(2k + 1)

6 σˆ_2k−1(n)

− (2k + 1)σ^∗_2k+1(n) + 2k + 3 2

σˆ_2k+1(n) + k(2k + 1) 6

σˆ_2k−1(n)

= −(2k + 1)

k−1

∑

s=0

2k 2s + 1

_n−1

∑

m=1

ˆ

σ_2k−2s−1(m) ˆσ_2s+1(n − m),

(7)

and we obtain

k−1

∑

s=0

2k 2s + 1

_n−1

∑

m=1

σˆ_2k−2s−1(m) ˆσ_2s+1(n − m)

= 1

2σ_2k+1(n) −1

2σˆ_2k(n) − 1

2k + 1Bˆ_2k+1(n).

□ Remark1. With k = 1 in Theorem1, we recover (1.1).

Example1. With k = 1, 2, 3 in Theorem1look as (a) Bˆ₃(n) =3

4σ3(n) −3

2σˆ2(n) − 3

n−1

∑

m=1

σˆ1(m) ˆσ1(n − m),

(b) Bˆ₅(n) =5

2σ5(n) −5

2σˆ4(n) − 40

n−1

∑

m=1

σˆ1(m) ˆσ3(n − m),

(c) Bˆ₇(n) =7

2σ7(n) −7

2σˆ6(n) − 84

n−1

∑

m=1

σˆ1(m) ˆσ5(n − m)

− 140

n−1

∑

m=1

σˆ3(m) ˆσ3(n − m).

Lemma 3. ([4]) If k≥ 1 and N ∈ N, then

k−1

∑

s=0

2k 2s + 1

N

∑

m=1

ˆ

σ_2k−2s−1(2m − 1) ˆσ_2s+1(2N − 2m + 1) =1

4σ^∗_2k+1(2N).

Proof of the Theorem2. Using Theorem1, we obtain

k−1

∑

s=0

2k 2s + 1

_2N−1

∑

m=1

ˆ

σ_2k−2s−1(m) ˆσ_2s+1(2N − m) (2.3)

=1

2[σ_2k+1(2N) − ˆσ_2k(2N)] − 1

2k + 1Bˆ_2k+1(2N).

Using (2.3) and Lemma3, we get

k−1

∑

s=1

2k 2s + 1

_N−1

∑

m=1

ˆ

σ_2k−2s−1(2m) ˆσ_2s+1(2N − 2m) (2.4)

= 1

4[σ_2k+1(2N) + σ_2k+1(N)] −1

2σˆ_2k(2N) − 1

2k + 1Bˆ_2k+1(2N).

With (2.4) and Lemma3, we get

k−1

∑

s=1

2k 2s + 1

_2N−1

∑

m=1

(−1)^mσˆ2k−2s−1(m) ˆσ2s+1(2N − m)

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=1

2σ_2k+1(N) −1

2σˆ_2k(2N) − 1

2k + 1Bˆ_2k+1(2N).

□ Example2. With k = 2, 3, 4 in Theorem2, we have

(a)

40B3(x)B3(y) − 6 [B3(x) + B3(y)] = −12 [xB5(y) + yB5(x)]

+ 6 [(x + y − 1)B₅(x + y) + (x − y)B₅(y − x)] + 5 [B₆(y − x) + B₆(x + y)] , (b)

B₃(x)B₅(y) + B₅(x)B₃(y) = 1

16[B₈(y − x) − B₈(x + y)] −1

7[xB₇(y) + yB₇(x)]

+ 1

14[B7(x) + B7(y) + (x + y − 1)B7(x + y) + (x − y)B7(y − x)] , (c)

B₃(x)B₇(y) + B₇(x)B₃(y) = 3

120[B₁0(y − x) − B₁0(x + y)]

− 1

12[xB₉(y) + yB₉(x)] + 1

24[B₉(x) + B₉(y) +(x + y − 1)B9(x + y) + (x − y)B9(y − x)] −21

10B₅(x)B5(y).

Proof of the Corollary1. Using Theorem1and Theorem2, we obtain (2k)

2N−1

∑

mm=1odd

σˆ_2k−1(m) ˆσ(2N − m)

=1

2σ2k+1(N) +1

2σ2k+1(2N) − ˆσ2k(2N) − 2

2k + 1Bˆ_2k+1(2N). (2.5) Replacing 2k by l in (2.5), we get proof of the corollary. □ Proof of the Corollary2. Let m is odd. Using Theorem 1 and Theorem 2, we obtain

(2k)

2N−1

∑

mm=1odd

ˆ

σ_2k−1(m) ˆσ(2N − m)

+ 2

k−1

∑

s=1

2k 2s + 1

_2N−1

∑

mm=1odd

ˆ

σ_2k−2s−1(m) ˆσ_2s+1(2N − m)

= 1

2[σ_2k+1(2N) − σ_2k+1(N)] . (2.6)

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It is clearly known that σ^∗_k(2N) = σk(2N) − σk(N) and σ^∗_k(2N) = 2^kσ^∗_k(N). After we

used it in (2.6), we get proof of the corollary. □

Proof of the Theorem3. Replacing 2m by m in Lemma2, we get

m

∑

k=0

2m 2k

B_2k+1(x)B_2m−2k+1(y) (2k + 1)(2m − 2k + 1)

= 1

2(2m + 1){(x + y − 1)B2m+1(x + y) + (x − y)B_2m+1(y − x)}

+ 1

2(2m + 2){B_2m+2(y − x) − B_2m+2(x + y)} . Using Bk(1 − x) = (−1)^kBk(x) with k ≥ 0, we derive

m−1

∑

k=1

2m 2k

B_2k+1(x)B_2m−2k+1(y)

(2k + 1)(2m − 2k + 1)= 1

4m + 4{B_2m+2(y − x) − B_2m+2(x + y)}

− 1

2m + 1{xB_2m+1(y) + yB_2m+1(x)} + 1

4m + 2{B_2m+1(x) + B_2m+1(y) + +(x + y − 1)B2m+1(x + y) + (x − y)B2m+1(y − x)} .

□ Example3. With m = 2, 3, 4 in Theorem3, we have

(a)

40B₃(x)B₃(y) − 6 [B₃(x) + B₃(y)] = −12 [xB₅(y) + yB₅(x)]

+ 6 [(x + y − 1)B₅(x + y) + (x − y)B₅(y − x)] + 5 [B₆(y − x) + B₆(x + y)] , (b)

B₃(x)B₅(y) + B₅(x)B₃(y) = 1

16[B₈(y − x) − B₈(x + y)] −1

7[xB₇(y) + yB₇(x)]

+ 1

14[B₇(x) + B₇(y) + (x + y − 1)B₇(x + y) + (x − y)B₇(y − x)] , (c)

B₃(x)B₇(y) + B₇(x)B₃(y) = 3

120[B₁₀(y − x) − B₁₀(x + y)]

− 1

12[xB₉(y) + yB₉(x)] + 1

24[B₉(x) + B₉(y) +(x + y − 1)B₉(x + y) + (x − y)B₉(y − x)] −21

10B₅(x)B₅(y).

(10)

3. Divisor functions and linear combination of Bernoulli polynomials We consider the polynomial

∑

2l a, b, c, d

ˆ

σa(m) ˆσb(p − m) ˆσc(m^′) ˆσd(q − m^′)

once more. Let’s call this polynomial C. Actually, the polynomials C are combina- tions of Bernoulli polynomials and divisor functions.

Proof of the Theorem4. We note that

2m 2k

2k 2s + 1

2m − 2k 2s^′+ 1

= (2m)!

(2s + 1)!(2k − 2s − 1)!(2s^′+ 1)!(2m − 2k − 2s^′− 1)!

2m

2s + 1, 2k − 2s − 1, 2s^′+ 1, 2m − 2k − 2s^′− 1

=

2m a, b, c, d

.

(3.1) By Theorem1and (3.1), we obtain

C=

∑

2l a, b, c, d

=

l−1

∑

w=1

2l 2w

"

w−1

∑

s=0

2w 2s + 1

p−1

∑

m=1

σˆ2w−2s−1(m) ˆσ2s+1(p − m)

#

×

"

l−w−1

∑

s^′=0

2l − 2w 2s^′+ 1

q−1

∑

m^′=1

σˆ_2l−2w−2s^′₋₁(m^′) ˆσ_2s^′₊₁(q − m^′)

#

=

l−1

∑

w=1

2l 2w

( 1

2σ_2w+1(p) −1

2σ_2w(p) + σ_2wp 2

−

∑

d|p

B_2w+1(d) 2w + 1

+2

∑

d|₂^p

B_2w+1(d) 2w + 1







× 1

2σ2l−2w+1(q) −1

2σ2l−2w(q) + σ2l−2w

q 2

−

∑

d^′|q

B_2l−2w+1(d^′) 2l − 2w + 1 + 2

∑

d^′|^q₂

B_2l−2w+1(d^′) 2l − 2w + 1





 . Consider the terms of C, we obtain

C₍₁₎:=1 4

l−1

∑

w=1

2l 2w

σ_2w+1(p)σ_2l−2w+1(q),

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C₍₂₎:= −1 4

l−1

∑

w=1

2l 2w

σ2w+1(p)σ_2l−2w(q), C₍₃₎:=1

2

l−1

∑

w=1

2l 2w

σ2w+1(p)σ_2l−2w

q 2

,

C₍₄₎:= −1 4

l−1

∑

w=1

2l 2w

σ_2w(p)σ_2l−2w+1(q), C₍₅₎:=1

4

l−1

∑

w=1

2l 2w

σ_2w(p)σ_2l−2w(q),

C₍₆₎:= −1 2

l−1

∑

w=1

2l 2w

σ2w(p)σ2l−2w

q 2

,

C₍₇₎:=1 2

l−1

∑

w=1

2l 2w

σ_2w

p 2

σ_2l−2w+1(q), C₍₈₎:= −1

2

l−1

∑

w=1

2l 2w

σ_2w

p 2

σ_2l−2w(q),

C₍₉₎:=

l−1 w=1

∑

2l 2w

σ2w

p 2

σ2l−2w

q 2

,

C₍₁₀₎:= −1 2

∑

d|p l−1

∑

w=1

2l 2w

B_2w+1(d)

2w + 1 σ_2l−2w+1(q), C₍₁₁₎:=1

2

∑

d|p l−1

∑

w=1

2l 2w

B_2w+1(d)

2w + 1 σ_2l−2w(q), C₍₁₂₎:= −

∑

d|p l−1

∑

w=1

2l 2w

B_2w+1(d) 2w + 1 σ_2l−2w

q 2

,

C₍₁₃₎:= −1 2

∑

d^′|q l−1

∑

w=1

2l 2w

B_2l−2w+1(d^′)

2l − 2w + 1 σ_2w+1(p), C₍₁₄₎:=1

2

∑

d^′|q l−1

∑

w=1

2l 2w

B_2l−2w+1(d^′)

2l − 2w + 1 σ_2w(p), C₍₁₅₎:= −

∑

d^′|q l−1

∑

w=1

2l 2w

B_2l−2w+1(d^′) 2l − 2w + 1 σ_2w

p 2

,

C₍₁₆₎:=

∑

d|^p₂ l−1

∑

w=1

2l 2w

B_2w+1(d)

2w + 1 σ_2l−2w+1(q),

(12)

C₍₁₇₎:= −

∑

d|^p₂ l−1

∑

w=1

2l 2w

B_2w+1(d)

2w + 1 σ2l−2w(q), C₍₁₈₎:= 2

∑

d|^p₂ l−1

∑

w=1

2l 2w

B_2w+1(d) 2w + 1 σ_2l−2w

q 2

,

C₍₁₉₎:=

∑

d^′|^q₂ l−1

∑

w=1

2l 2w

B_2l−2w+1(d^′)

2l − 2w + 1 σ2w+1(p), C₍₂₀₎:= −

∑

d^′|^q₂ l−1

∑

w=1

2l 2w

B_2l−2w+1(d^′)

2l − 2w + 1 σ2w(p), C₍₂₁₎:= 2

∑

d^′|^q₂ l−1

∑

w=1

2l 2w

B_2l−2w+1(d^′) 2l − 2w + 1 σ_2w

p 2

,

C₍₂₂₎:=

∑

d|p d^′|q

l−1

∑

w=1

2l 2w

B_2w+1(d) 2w + 1

B_2l−2w+1(d^′) 2l − 2w + 1 ,

C₍₂₃₎:= 4

∑

d|^p₂ d^′|^q₂

l−1 w=1

∑

2l 2w

B_2w+1(d) 2w + 1

B_2l−2w+1(d^′) 2l − 2w + 1 ,

C₍₂₄₎:= −2

∑

d|^p₂ d^′|q

l−1

∑

w=1

2l 2w

B_2w+1(d) 2w + 1

B_2l−2w+1(d^′) 2l − 2w + 1 , and

C₍₂₅₎:= −2

∑

d|p d^′|^q₂

l−1

∑

w=1

2l 2w

B_2w+1(d) 2w + 1

B_2l−2w+1(d^′) 2l − 2w + 1 . From the binomial theorem,

l−1 w=1

∑

2l 2w

d^′2l−2wd^2w= 1 2 h

(d^′+ d)^2l+ (d^′− d)^2li

− d^′2l− d^2l, we obtain

C₍₁₎=1 8

∑

d|p d^′|q

d^′d[(d^′+ d)^2l+ (d^′− d)^2l] −1 4

∑

d|p d^′|q

dd^′2l+1−1 4

∑

d|p d^′|q

d^2l+1d^′.

By the property of Bernoulli polynomial,

B_n(x + 1) − Bn(x) = nxⁿ⁻¹,

(13)

we get

C₍₁₎= 1 8(2l + 1)

∑

d|p d^′|q

d^′dB_2l+1(d^′+ d + 1) + B_2l+1(d^′− d + 1)

−B_2l+1(d^′+ d) − B_2l+1(d^′− d) −1

4[σ₁(p)σ_2l+1(q) + σ_2l+1(p)σ₁(q)] . Similarly, we derive that

C₍₂₎= − 1 8(2l + 1)

∑

d|p d^′|q

dB_2l+1(d^′+ d + 1) + B_2l+1(d^′− d + 1) − B_2l+1(d^′+ d)

−B_2l+1(d^′− d) +1

4[σ1(p)σ2l(q) + σ2l+1(p)σ0(q)] , C₍₃₎= 1

4(2l + 1)

∑

d|p d^′|^q₂

dB_2l+1(d^′+ d + 1) + B_2l+1(d^′− d + 1) − B_2l+1(d^′+ d)

−B_2l+1(d^′− d) −1 2 h

σ₁(p)σ_2l

q 2

+ σ_2l+1(p)σ₀

q 2

i , C₍₄₎= − 1

8(2l + 1)

∑

d|p d^′|q

d^′B_2l+1(d^′+ d + 1) + B2l+1(d^′− d + 1) − B2l+1(d^′+ d)

−B_2l+1(d^′− d) +1

4[σ₀(p)σ_2l+1(q) + σ_2l(p)σ₁(q)] , C₍₅₎= 1

8(2l + 1)

∑

d|p d^′|q

B_2l+1(d^′+ d + 1) + B2l+1(d^′− d + 1)

−B_2l+1(d^′+ d) − B_2l+1(d^′− d) −1

4[σ₀(p)σ_2l(q) + σ_2l(p)σ₀(q)] , C₍₆₎= − 1

4(2l + 1)

∑

d|p d^′|^q₂

B_2l+1(d^′+ d + 1) + B_2l+1(d^′− d + 1) − B2l+1(d^′+ d)

−B_2l+1(d^′− d) +1 2 h

σ₀(p)σ_2lq 2

+ σ_2l(p)σ₀q 2

i , C₍₇₎= 1

4(2l + 1)

∑

d|^p₂ d^′|q

d^′B_2l+1(d^′+ d + 1) + B_2l+1(d^′− d + 1) − B_2l+1(d^′+ d)

−B_2l+1(d^′− d) −1 2 h

σ0

p 2

σ2l+1(q) + σ2l

p 2

σ1(q)

i ,

(14)

C₍₈₎= − 1 4(2l + 1)

∑

d|^p₂ d^′|q

B_2l+1(d^′+ d + 1) + B_2l+1(d^′− d + 1) − B_2l+1(d^′+ d)

−B_2l+1(d^′− d) +1 2 h

σ₀

p 2

σ_2l(q) + σ_2lp 2

σ₀(q)i

, C₍₉₎= 1

2(2l + 1)

∑

d|₂^p d^′|^q₂

B_2l+1(d^′+ d + 1) + B2l+1(d^′− d + 1) − B2l+1(d^′+ d)

−B_2l+1(d^′− d) −h σ₀

p 2

σ_2l

q 2

+ σ_2l

p 2

σ₀

q 2

i . We obtain

C₍₁₀₎= −1 2

l−1 w=1

∑

2l 2w

∑

d|p

B_2w+1(d)

2w + 1 σ2l−2w+1(q)

= − 1

2(2l + 1)

l−1 w=1

∑

(2l + 1)(2l)!

(2w + 1)!(2l − 2w)!

∑

d|p d^′|q

B_2w+1(d)σ2l+1−2w(q)

= − 1

4(2l + 1)

∑

d|p d^′|q

d^′B_2l+1(d + d^′) + B_2l+1(d − d^′)

+1

2σ₁(p)σ_2l+1(q) −1

4σ₀(p)σ_2l+1(q) + 1

2(2l + 1)σ₁(q)

∑

d|p

B_2l+1(d).

Similarly, like C₍₁₀₎, we get C₍₁₁₎= 1

4(2l + 1)

∑

d|p d^′|q

B_2l+1(d + d^′) + B_2l+1(d − d^′)

−1

2σ₁(p)σ_2l(q) +1

4σ₀(p)σ_2l(q) − 1

2(2l + 1)σ₀(q)

∑

d|p

B_2l+1(d),

C₍₁₂₎= − 1

2(2l + 1)

∑

d|p d^′| f racq2

B_2l+1(d + d^′) + B_2l+1(d − d^′) + σ₁(p)σ_2lq 2

−1

2σ₀(p)σ_2lq 2

+ 1

2l + 1σ₀

q 2

∑

d|p

B_2l+1(d),

C₍₁₃₎= − 1 4(2l + 1)

∑

d|p d^′|q

dB_2l+1(d^′+ d) + B_2l+1(d^′− d) +1

2σ_2l+1(p)σ₁(q)

(15)

−1

4σ_2l+1(p)σ₀(q) + 1

2(2l + 1)σ₁(p)

∑

d^′|q

B_2l+1(d^′),

C₍₁₄₎= 1 4(2l + 1)

∑

d|p d^′|q

B_2l+1(d^′+ d) + B_2l+1(d^′− d) −1

2σ_2l(p)σ₁(q)

+1

4σ_2l(p)σ₀(q) − 1

2(2l + 1)σ₀(p)

∑

d^′|q

B_2l+1(d^′),

C₍₁₅₎= − 1 2(2l + 1)

∑

d|^p₂ d^′|q

B_2l+1(d^′+ d) + B_2l+1(d^′− d) + σ_2lp 2

σ₁(q)

−1 2σ_2l

p 2

σ₀(q) + 1 2l + 1σ₀

p 2

∑

d^′|q

B_2l+1(d^′),

C₍₁₆₎= 1 2(2l + 1)

∑

d|^p₂ d^′|q

d^′B_2l+1(d^′+ d) + B_2l+1(d − d^′) − σ1

p 2

σ2l+1(q)

+1 2σ0

p 2

σ2l+1(q) − 1

2l + 1σ1(q)

∑

d|^p₂

B_2l+1(d),

C₍₁₇₎= − 1 2(2l + 1)

∑

d|^p₂ d^′|q

B_2l+1(d + d^′) + B_2l+1(d − d^′) + σ₁p 2

σ_2l(q)

−1 2σ₀

p 2

σ_2l(q) + 1

2l + 1σ₀(q)

∑

d|^p₂

B_2l+1(d),

C₍₁₈₎= 1 2l + 1

∑

d|^p₂ d^′|^q₂

B_2l+1(d + d^′) + B_2l+1(d − d^′) − 2σ1

p 2

σ_2l

q 2

+ σ0

p 2

σ2l

q 2

− 2

2l + 1σ0

q 2

∑

d|₂^p

B_2l+1(d),

C₍₁₉₎= 1 2(2l + 1)

∑

d|p d^′|^q₂

dB_2l+1(d^′+ d) + B_2l+1(d^′− d) − σ_2l+1(p)σ₁

q 2

+1

2σ2l+1(p)σ0

q 2

− 1

2l + 1σ1(p)

∑

d^′|^q₂

B_2l+1(d^′),

(16)

C₍₂₀₎= − 1 2(2l + 1)

∑

d|p d^′|^q₂

B_2l+1(d^′+ d) + B_2l+1(d^′− d) + σ2l(p)σ₁q 2

−1

2σ_2l(p)σ₀

q 2

+ 1

2l + 1σ₀(p)

∑

d^′|^q₂

B_2l+1(d^′)

and

C₍₂₁₎= 1 2l + 1

∑

d|^p₂ d^′|^q₂

B_2l+1(d^′+ d) + B_2l+1(d^′− d) − 2σ_2lp 2

σ₁

q 2

+ σ_2lp 2

σ₀

q 2

− 2 2l + 1σ₀

p 2

∑

d^′|^q₂

B_2l+1(d^′).

From Lemma2, then we get

C₍₂₂₎=

∑

d|p d^′|q

l−1

∑

w=1

2l 2w

B_2w+1(d) 2w + 1

B_2l−2w+1(d^′) 2l − 2w + 1

= 1

2(2l + 1)

∑

d|p d^′|q

(d + d^′− 1)B2l+1(d + d^′) + (d − d^′)B_2l+1(d^′− d)

−(2d − 1)B2l+1(d^′) − (2d^′− 1)B2l+1(d)

− 1

4(l + 1)

∑

d|p d^′|q

B_2l+2(d + d^′) − B2l+2(d^′− d) ,

C₍₂₃₎= 4 2(2l + 1)

∑

d|^p₂ d^′|^q₂

(d + d^′− 1)B_2l+1(d + d^′) + (d − d^′)B_2l+1(d^′− d)

−(2d − 1)B_2l+1(d^′) − (2d^′− 1)B_2l+1(d)

− 4

4(l + 1)

∑

d|^p₂ d^′|^q₂

B_2l+2(d + d^′) − B_2l+2(d^′− d) ,

C₍₂₄₎= − 2 2(2l + 1)

∑

d|^p₂ d^′|q

(d + d^′− 1)B2l+1(d + d^′) + (d − d^′)B2l+1(d^′− d)

−(2d − 1)B_2l+1(d^′) − (2d^′− 1)B_2l+1(d)

(17)

+ 2 4(l + 1)

∑

d|^p₂ d^′|q

B_2l+2(d + d^′) − B_2l+2(d^′− d) ,

C₍₂₅₎= − 2 2(2l + 1)

∑

d|p d^′|^q₂

(d + d^′− 1)B_2l+1(d + d^′) + (d − d^′)B_2l+1(d^′− d)

−(2d − 1)B2l+1(d^′) − (2d^′− 1)B2l+1(d)

+ 2

4(l + 1)

∑

d|p d^′|^q₂

B_2l+2(d + d^′) − B2l+2(d^′− d) .

Summing C_(i)(i = 1, ..., 25), we drive the Theorem. □ Example4. In Theorem4, for l = a = b = c = d = 1 and p = q = 2, we get

∑

2l a, b, c, d

σˆa(m) ˆσb(p − m) ˆσc(m^′) ˆσd(q − m^′) = 4!.

REFERENCES

[1] H. P. B. Cho, D. Kim, “Evaluation of a certain combinatorial convolution sum in higher level cases.” J. Math. Anal. Appl., vol. 406, no. 1, pp. 203–210, 2013, doi: https://doi.org/10.1016/j.jmaa.2013.04.052.

[2] B. C. Berndt, Ramanujan’s Notebooks, Part II. New York: Springer-Verlag, 1989. doi:

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[4] A. B. D. Kim, “Convolution identities for twisted Eisenstein series and twisted divisor functions.”

Fixed Point Theory and Applications, vol. 81, 2013, doi:https://doi.org/10.1186/1687-1812-2013- 81.

[5] N. I. D. Kim, “Certain combinatoric Bernoulli polynomials and convolution sums of divisor functions.” Advance Difference Equations, vol. 2013, no. 310, 2013.

[6] H. Hahn, “Convolution sums of some functions on divisors.” Rocky Mountain Journal of Mathem- atics, vol. 37, no. 5, pp. 1593–1622, 2007.

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[8] R. Z. W. Chu, “Convolutions of Bernoulli and Euler polynomials.” Sarajevo Journal of Mathemat- ics, vol. 6, no. 6(18), pp. 147–163, 2010.

[9] K. Williams, Number Theory in the Spirit of Liouville. Cambridge University Press, New York:

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Authors’ addresses Daeyeoul Kim

Jeonbuk National University, Institute of Pure and Applied Mathematics, Department of Mathemat- ics, 567 Baekje-daero, Deokjin-gu, Jeonju-si, Jeollabuk-do 54896, Republic of Korea

E-mail address: kdaeyeoul@jbnu.ac.kr Nazli Yildiz Ikikardes

(Corresponding author) Balikesir University, Necatibey Faculty of Education, Department of Math- ematics and Science Education, 10100 Balikesir, Turkey

E-mail address: nyildiz@balikesir.edu.tr, nyildizikikardes@gmail.com