INTEGERS 13 (2013)
#A68
CONVOLUTION SUMS INVOLVING LEGENDRE-JACOBI
SYMBOL AND DIVISOR FUNCTIONS
Mohamed El Bachraoui
Dept. of Mathematical Sciences, United Arab Emirates University, Al-Ain,
United Arab Emirates
melbachraoui@uaeu.ac.ae
Received: 5/10/12, Revised: 1/20/13, Accepted: 9/18/13, Published: 10/10/13
Abstract
We use elementary tools to evaluate a variety of convolution sums which involve
the Legendre-Jacobi symbol and the divisor functions.
1. Introduction
Let N = {1, 2, 3, . . .} and let N0 = {0, 1, 2, . . .}. We will use the following definitions.
Definition 1. Let the function be defined on Q as follows:
then (n) = 0 , and if n 2 N, then
X
(n) =
d.
(0) = 1, if n 2 Q\N0 ,
d|n
Definition 2. Let n, r 2 N0 , let m 2 N, and let
X
r,m (n) =
d.
d|n
d⌘r mod m
We will be using the basic fact that for all m, n 2 N
0,m (n)
⇤
Definition 3. Let the function
⇤
(n) = 0 and if n 2 N, then
⇤
= m (n/m).
(1)
be defined on Q as follows: if n 2 Q \ N, then
(n) =
X
d|n,
n
d
d.
odd
By [10, Theorem 3.4] we have for all n 2 N that
⇤
(n) = (n)
(n/2).
(2)
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For an integer n and an odd prime number p we shall use n p to denote the
Legendre-Jacobi symbol of n with respect to p, see for instance Apostol [2] about
properties of n p . In this note we shall evaluate a variety of convolution sums
involving the Legendre-Jacobi symbol and the divisor functions. Our main tool is
the following theorem.
Theorem 1. [6] Let ↵ 2 N. Let A1 , A2 , . . . A↵ ✓ N and A = (A1 , A2 , . . . , A↵ )
and let fi : Ai ! C for i = 1, 2, . . . , ↵ be arithmetic functions and let f =
(f1 , f2 , . . . , f↵ ). If both
FA (q) =
↵ Y
Y
qn )
(1
fi (n)
n
=
pA,f (n)q n
n=0
i=1 n2Ai
and
1
X
↵ X
X
fi (n) n
q
n
i=1
n2Ai
converge absolutely and represent analytic functions in the unit disk |q| < 1, then
!
n
↵
X
X
npA,f (n) =
pA,f (n k)
fi,Ai (k) ,
i=1
k=1
where pA,f (0) = 1 and
X
fi,Ai (k) =
fi (d).
d|k
d2Ai
We note that Theorem 1 for the special case ↵ = 1, has been given in Apostol
[2] and in Robbins [9] to give formulas relating arithmetic functions to sums of
divisors functions. The authors’ key argument is that generating functions for the
arithmetic functions they considered have the form of infinite products ranging over
a single set of natural numbers. In our previous work [6] we used Theorem 1 to deal
with arithmetic functions whose generating functions involve finitely many infinite
products ranging over di↵erent sets of natural numbers and to derive a variety of
inductive formulas for such functions. Among our results in [6] we were able by
means of Theorem 1 to reproduce the following result of Liouville [7].
( 1)n s(n)n =
where
(n)
1,2 (n)
2
+
X
( 1)k+1
k 1
⇣
(
1, if n = m2
s(n) =
0, otherwise.
(n
k2 ) +
1,2 (n
⌘
k2 ) ,
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We note further that the infinite products along with their power series expansions
which we will use in this paper are all found as theorems in the recent paper of
Alaca, Alaca, and Williams [1]. Besides several new results on infinite products of
Ramanujan type in [1], the authors reproduced a variety of other results including
some due to Carlitz [5] and Ramanujan [8].
2. The Results
From now on we will suppose that n is a positive integer. While the corollaries in
this section are given along with their proofs, proofs for the theorems are provided
in the next section.
Theorem 2. We have
(a)
n
X1✓X
d2
n
X1✓X
d2
j=1
(b)
j=1
3 d
d|j
3
d|j
◆✓
(n j)
◆✓
(n
j
d
j)
✓
n
j
3
✓
9
◆◆
n
j
3
=
1
1 ⇣n⌘ n X 2
(n)
d
9
9
3
9
◆◆
=
n
3
Corollary 1. If p is an odd prime, then
(a)
p 1✓
X
X
j=1
(b)
p 1✓
3 d
j
d
◆✓
d|j
X X
j=1
d2
◆✓
d2
3
d|j
(p
✓
j)
(p j) 9
✓
p
j
3
p
j
3
3 d .
d|n
◆◆
◆◆
1X
d2
3
d|n
n
.
d
8
>
<0, if3 p = 3
= 1 9p , if p ⌘ 1 (mod 3)
>
: 1+p3
, if p ⌘ 1 (mod 3).
9
=
8
>
<6, if p 2= 3
(p 1)(p +1)
,
3
1)(p2 1)
,
3
>
: (p
if p ⌘ 1
if p ⌘
(mod 3)
1
(mod 3).
Proof. The case p = 3 is easily checked in both parts. Suppose now that p 6= 3.
Then as
(
1, if p ⌘ 1 (mod 3)
(p) = p + 1, (p/3) = 0, and
3 p =
1, if p ⌘ 1 (mod 3),
Theorem 2(a) yields
9
p 1⇣X
X
j=1
d|j
d2
3 d
⌘⇣
(p
j)
✓
p
◆
j ⌘
=p+1
3
(
1 p3 , if p ⌘ 1 (mod 3)
=
1 + p3 , if p ⌘ 1 (mod 3),
⇣
p
3 1 +p2
3 p
⌘
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which after division by 9 gives part (a). Further by Theorem 2 (b) we have
◆✓
✓
◆◆
p 1✓X
⌘
X
j
p j
p 1⇣
2
d
3
(p j) 9
=
3 p +p2
3 1
d
3
3
j=1 d|j
(
2
(p 1)(1+p )
, if p ⌘ 1 (mod 3)
= (p 1)(3 1+p2 )
, if p ⌘ 1 (mod 3),
3
which proves part (b).
Theorem 3. The following identities are true:
(a)
n
X1✓X
j=1
(b)
(c)
d2
4 d
d|j
◆✓
⇤
◆✓
⇤
⇤
(n j)+4
◆✓
j
d
4
d|j
n
X1✓X
j=1
4 d
d|j
n
X1✓X
j=1
d2
✓
(n j) 3
(n j) 4
✓
⇤
✓
n
◆◆
j
2
n
j
2
n
◆
=
p
j
j
(n)
+
4
✓
4
◆◆
2
⇤
=
n
j
4
1
4
⇤
⇤
◆◆
⇣n⌘
2
n
=
4
⇤
(n)+
nX 2
d
4
1X
⇣n⌘
2
4 d .
d|n
d2
4
d|n
+
nX
4
n
.
d
4 d .
d|n
Corollary 2. If p is an odd prime, then ;
(a)
p 1✓
X
X
j=1
(b)
4 d
◆✓
⇤
d2
4
j
d
◆✓
(p
d|j
p 1✓
X
X
j=1
d2
d|j
✓
j) + 4
⇤
j)
✓
(p
3
2
p
j
2
◆
=
(c)
p 1✓
X
X
j=1
d|j
4 d
◆✓
⇤
(p
j)
4
⇤
✓
p
j
2
=
(
4
✓
◆◆
(
◆◆
1 p3
,
4
1+p3
,
4
p
j
4
=
(p 1)
,
4
p 1
,
4
(mod 4)
1
(mod 4).
◆◆
(p 1)(p2 +1)
,
4
(p 1)(p2 1)
,
4
(
if p ⌘ 1
if p ⌘
if p ⌘ 1
if p ⌘
(mod 4)
1
(mod 4).
if p ⌘ 1 (mod 4)
if p ⌘ 1 (mod 4).
Proof. Suppose that p is an odd prime. Then Theorem 3 (a) implies
◆✓
✓
◆◆
p 1✓X
X
p j
⇤
4
d2
4 d
(p j) + ⇤
2
j=1 d|j
✓
◆
2
=p
4 1 +p
4 p
p 1
(
p3 1, if p ⌘ 1 (mod 4)
=
p3 1, if p ⌘ 1 (mod 4),
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5
which upon division by
3 (b) we get
4 yields the required formula in part (a). Next by Theorem
p 1✓X
X
j=1
d2
j
d
4
d|j
◆✓
(p
p
=
=
j)
1
(
4
✓
3
✓
p
j
2
◆
✓
4
4 p +p2
4 1
(p 1)(1+p2 )
, if p ⌘ 1
4
(p 1)( 1+p2 )
, if p ⌘
4
p
j
4
◆
◆◆
(mod 4)
1 (mod 4),
giving part (b). Further by Theorem 3 (c) we have
p 1✓X
X
j=1
4 d
d|j
◆✓
⇤
(p
j)
4
⇤
⇣p
◆
j⌘
2
=
=
⇣
p 1+
(
1 p
4
⌘
p 1
4 , if p ⌘ 1
p 1
4 , if p ⌘
p
1
(mod 4)
1 (mod 4),
which proves part (c).
Theorem 4. We have
(a)
n
X1✓X
d 5 d
n
X1✓X
d 5
j=1
(b)
j=1
d|j
d|j
◆✓
(n
◆✓
(n
j
d
✓
j)
j)
25
n
j
5
✓
◆◆
n
j
5
1
(n)
5
=
◆◆
= (n
1 ⇣n⌘
5
5
1)
X
d|n
d 5
nX
d 5 d .
5
d|n
n
.
d
Corollary 3. If p is an odd prime, then
(a)
p 1✓X
X
d 5 d
◆✓
p 1✓X
X
j
d 5
d
◆✓
j=1
(b)
j=1
d|j
d|j
(p j)
(p
j)
✓
p
25
j
5
✓
◆◆
p
j
5
8
>
<0, if2 p = 5
= 1 5p , if p ⌘ ±1 (mod 5)
>
: 1+p2
5 , if p ⌘ ±2 (mod 5).
◆◆
8
>
<17, if p = 5
= (p 1)(p + 5), if p ⌘ ±1 (mod 5)
>
:
(p 1)(p 5), if p ⌘ ±2 (mod 5).
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Proof. The results are easily checked if p = 5. Now suppose that p 6= 5 is an odd
prime. Then by Theorem 4(a) we find
✓
◆
✓
◆✓
✓
◆◆
p 5 1 +p 5 p
+p+1
p 1 X
X
p j
d 5 d
(p j)
=
5
5
j=1 d|j
(
1 p2
5 , if p ⌘ ±1 (mod 5)
= 1+p
2
5 , if p ⌘ ±2 (mod 5),
which proves part (a). Part (b) follows similarly by Theorem 4(b).
Theorem 5. The following formulas hold:
(a)
n
X1✓X
j=1
d 8 d
d|j
◆✓
⇤
(n
j) + 2
1
2
=
(b)
n
X1✓X
j=1
d 8
d|j
j
d
◆✓
(n
j)
(c)
n
X1✓X
j=1
8 d
d|j
◆✓
⇤
(n
⇤
3
=
n
2
◆
j
2
⇤
✓
◆
n
j
2
2
⇤
2
⇤
=
j
2
◆
⇤
(n) +
✓
n
◆◆
j
4
⇣n⌘
⇤
nX
d 8 d .
2
4
n
j
4
d|n
◆
✓
8
n
j
8
◆◆
j
.
d
d 8
n
✓
⇤
+4
2
d|n
✓
+8
⇣n⌘
(n) +
1X
n
j)
✓
⇤
8
⇣n⌘
2
⇤
✓
n
⇤
+8
j
◆◆
4
⇣n⌘
4
+n
X
8 d .
d|n
Corollary 4. If p is an odd prime, then
(a)
p 1✓X
X
j=1
d 8 d
d|j
◆✓
⇤
(p
j) + 2
=
(b)
p 1✓X
X
j=1
d|j
j
d 8
d
◆✓
(p
j)
(
3
=
(
⇤
✓
p
2
1 p2
2 ,
1+p2
2 ,
✓
j
p
+8
⇤
✓
p
j
4
◆◆
if p ⌘ ±1 (mod 8)
if p ⌘ ±3 (mod 8).
j
2
◆
◆
2
✓
p2 1
2 , if p ⌘ ±1
(p 1)2
, if p ⌘ ±3
2
p
j
4
◆
8
(mod 8)
(mod 8).
✓
p
j
8
◆◆
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INTEGERS: 13 (2013)
p 1✓X
X
(c)
j=1
8 d
d|j
◆✓
⇤
(p
j)
2
(
=
✓
⇤
p
j
2
◆
⇤
8
p 1
2 , if p ⌘ 1, 3
p 1
2 , if p ⌘ 1,
✓
p
j
4
◆◆
(mod 8)
3 (mod 8).
Proof. By Theorem 5(a) we get
p 1✓X
X
j=1
◆✓
d 8 d
d|j
⇤
(p
⇤
j) + 2
⇣p
⌘
1⇣ ⇣
=
p 8 1 +p 8 p
2
2
j⌘
(
⌘
1 =
p
⇤
+8
⇣p
◆
j⌘
4
1 p2
2 ,
1+p2
2 ,
if p ⌘ ±1 (mod 8)
if p ⌘ ±3 (mod 8),
which proves part (a). Parts (b) and (c) follow similarly by Theorem 5, parts (b)
and (c).
Theorem 6. The following identities are valid:
n
X1✓X
(a)
j=1
◆✓
d 12 d
d|j
(n
+8
= (n) + 3
(b)
n
X1✓X
j=1
⇣n⌘
+8
d 12
j
d
3
d|j
⇣n⌘
4
◆✓
j) + 3
✓
j
4
(n
n
2
2
n
j
3
1X
d|n
j
3
6
j)
✓
n
◆
✓
◆
✓
+ 12
⇣n⌘
+ 12
3
=
n
◆
✓
n
j
6
⇣n⌘
24
12
n
2
j
2
✓
◆
n
j
4
n
d 12
.
d
◆
24
n
X
✓
n
j
12
◆◆
d 12 d .
d|n
◆
6
✓
n
j
12
◆◆
Corollary 5. If p is an odd prime, then
(a)
p 1✓X
X
j=1
d|j
d 12 d
◆✓
(p
j) + 3
✓
p
j
◆
✓
p
j
3
◆
✓
p
j
◆
+8
+ 12
24
4
6
8
>
<4, if p = 3
= 1 p2 , if p ⌘ ±1 (mod 12)
>
:
1 + p2 , if p ⌘ ±5 (mod 12).
✓
p
j
12
◆◆
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INTEGERS: 13 (2013)
p 1✓X
X
(b)
j=1
◆✓
j
d
d 12
d|j
(p
j)
✓
2
✓
◆
p
j
2
✓
◆
◆
✓
◆◆
p j
p j
p j
3
2
6
3
4
12
8
>
<2,2 if p = 3
= p 2 1 , if p ⌘ ±1 (mod 12)
>
: (p 1)2
, if p ⌘ ±5 (mod 12).
2
Proof. Part (a) follows by Theorem 6(a) and part (b) follows by Theorem 6(b) using
the same sort arguments as before.
Theorem 7. We have
◆✓
n
X1✓X
15 d
(n
j=1
j)
6
d|j
=
⇣n⌘
(n) + 6
3
✓
n
j
3
⇣n⌘
+ 10
◆
n
◆
j
5
⇣n⌘
X
+n
15
15
5
✓
10
+ 15
✓
n
j
15
◆◆
15 d .
d|n
An immediate consequence is the next corollary.
Corollary 6. If p is an odd prime, then we have
p 1✓X
X
j=1
15 d
d|j
◆✓
(p
j)
6
✓
p
j
3
◆
✓
10
p
j
5
◆
+ 15
✓
p
j
15
◆◆
8
5, if p = 3
>
>
>
<9, if p = 5
=
>
p 1, if p ⌘ 1, 17, 19, 23, 7, 11, 13, 29 (mod 60)
>
>
:
p 1, if p ⌘ 7, 11, 13, 29, 1, 17, 19, 23 (mod 60).
Theorem 8. We have
◆✓
n
X1✓X
20 d
(n
j=1
d|j
5
=
(n) + 2
⇣n⌘
2
+4
j)
✓
n
⇣n⌘
4
2
j
5
+5
◆
✓
p
j
2
10
⇣n⌘
5
◆
4
n
j
✓
+ 10
✓
◆
10
⇣n⌘
10
n
j
4
+ 20
◆
✓
n
j
◆◆
20
⇣n⌘
X
20
+n
20
d|n
20 d .
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INTEGERS: 13 (2013)
An immediate consequence is the next corollary.
Corollary 7. If p is an odd prime, then
p 1✓X
X
j=1
◆✓
20 d
d|j
Theorem 9. We have
◆✓
n
X1✓X
24 d
(n
j=1
d|j
8
=
(n) + 2
⇣n⌘
2
+3
(p
j)
✓
2
✓
◆
j
2
✓
4
✓
2
n
j
8
⇣n⌘
3
◆
+8
✓
n
j
2
12
⇣n⌘
8
✓
p
j
4
◆
3
n
j
✓
◆
◆
✓
◆◆
p j
p j
p j
5
10
+ 20
5
10
20
8
>
<4, if p = 5
= p 1, if p ⌘ 1, 3, 7, 9 (mod 20)
>
:
p 1, if p ⌘ 1, 3, 7, 9 (mod 20).
j)
◆
p
✓
◆
12
⇣n⌘
12
n
j
3
+ 24
◆
✓
n
j
◆◆
24
⇣n⌘
X
24
+n
24
12
d|n
We have the following consequence.
Corollary 8. If p is an odd prime, then we have
p 1✓X
X
j=1
d|j
24 d
◆✓
(p
j)
✓
p
2
j
◆
✓
p
j
2
◆
3
p
j
✓
✓
◆
p
j
3
◆
✓
p
j
◆◆
8
12
+ 24
8
12
24
8
>
<2, if p = 3
= p 1, if p ⌘ 1, 5, 7, 11 (mod 24)
>
:
p 1, if p ⌘ 1, 5, 7, 11 (mod 24).
3. Proofs of the Theorems
Throughout this section we shall suppose that q 2 C such that |q| < 1.
24 d .
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INTEGERS: 13 (2013)
Proof of Theorem 2. The following two identities are due to Carlitz [5] and new
proofs have been reproduced in Alaca et al. [1, Theorems 2.2 and 2.3].
1
Y
(1
q n )9 (1
q 3n )
3
=
n=1
1
X
a(n)q n = 1
1 ⇣X
X
9
n=0
n=1
where
a(0) = 1 and a(n) =
9
X
d2
d2
3 d
d|n
⌘
qn ,
(3)
1
(4)
3 d .
d|n
1
Y
(1
q 3n )9 (1
qn )
3
=
n=1
1
X
b(n)q n =
n=0
1 ⇣X
X
n=1
where
X
b(0) = 1 and b(n) =
d2
n ⌘ n
q
d
3
d|n
d2
3
d|(n+1)
,
n+1
.
d
To prove part (a) we apply Theorem 1 to the identity (3) with the sets A1 = N and
A2 = 3N and the functions f1 (k) = 9k on A1 and f2 (k) = 3k on A2 to get
na(n) =
9 (n) + 3
0,3 (n)
+
n
X1
a(j)
9 (n
j) + 3
0,3 (n
j)
j=1
=
9 (n) + 9 (n/3)
9
n
X1
a(j)
(n
j)
j=1
✓
n
j
3
◆
(5)
,
where the second identity follows by the relation (1). Now use the definition of a(j)
and divide by 9 to conclude the desired formula. As to part (b), apply Theorem 1
to the identity (4) for the sets A1 = 3N and A2 = N and the functions f1 (k) = 9k
on A1 and f2 (k) = 3k on A2 to get
nb(n) =
n
X1
b(j)
9
0,3 (n
j) + 3 (n
j) = 3
j=0
n
X1
b(j)
(n
j)
9
j=0
Equivalently,
◆✓
n ✓X
X
j
2
d
3
(n + 1
d
j=1
j)
9
d|j
✓
n+1
3
j
◆◆
=
✓
n X 2
d
3
n
j
3
3
d|n+1
as desired.
◆
. (6)
n+1
,
d
2
Proof of Theorem 3. The following two identities are due to Carlitz [5], see [1,
Theorems 2.4 and 2.5] for new proofs.
1
Y
(1 q n )4 (1 q 2n )6 (1 q 4n )
n=1
4
=
1
X
n=0
a(n)q n = 1 4
1 ⇣X
X
n=1
d|n
d2
4 d
⌘
q n , (7)
11
INTEGERS: 13 (2013)
where
a(0) = 1 and a(n) =
4
X
d2
4 d .
d|n
1
Y
q 2n )6 (1
(1
q 4n )4 (1
qn )
4
=
n=1
1
X
b(n)q n =
n=0
where
n=1
X
b(0) = 1 and b(n) =
1 ⇣X
X
d2
d2
n ⌘ n
q
d
4
d|n
1
, (8)
n+1
.
d
4
d|(n+1)
By virtue of Theorem 1 applied to the formula (7) and identity (2) we obtain
na(n) =
n
X1
4 (n)
a(j)
12 (n/2) + 16 (n/4)+
4 (n
j)
12
j=1
=
4
⇤
(n)
16
⇤
✓
n
j
2
(n/2)
4
◆
n
X1
j=1
✓
+ 16
⇣
a(j)
⇤
n
(n
j
4
◆
(9)
j) + 4
⇤
◆✓
⇤
✓
n
2
◆
j ⌘
,
which is equivalent to
n
X
2
d
4 d =
⇤
(n)+4
⇤
(n/2) 4
n
X1✓X
j=1
d|n
2
d
4 d
d|j
(n j)+4
⇤
✓
n
j
2
◆◆
,
giving part (a). As to part (b), applying Theorem 1 to the relation (8) and proceeding as before we find
◆✓
✓
◆
✓
◆◆
n ✓X
X
j
n+1 j
n+1 j
d2
4
(n + 1 j) 3
4
d
2
4
j=1
d|j
=
n X 2
d
4
4
d|n+1
n+1
,
d
as desired. As to part (c), we have by [1, Theorem 4.5]
1
Y
(1
q 2n )10 (1
qn )
4
(1
q 4n )
4
=1+4
n=1
1 ⇣X
X
n=1
4 d
d|n
which by Theorem 1 and the relation (2) translates into
16
n
X1✓X
j=1
d|j
4 d
◆✓
⇤
(n
j)
4
⇤
⇣n
2
◆
j⌘
⌘
qn ,
(10)
12
INTEGERS: 13 (2013)
=
4
⇤
⇤
(n) + 16
⇣n⌘
2
+ 4n
X
4 d ;
d|n
yielding the desired identity.
2
Proof of Theorem 4. The following two identities are due to Ramanujan, see
Alaca et al. [1, Theorems 3.2 and 3.3] for proofs based on work by Bailey [3, 4]:
1
Y
q n )5 (1
(1
q 5n )
1
=1
5
n=1
1
Y
1 ⇣X
X
n=1
(1
q 5n )5 (1
qn )
1
=
n=1
1 ⇣X
X
n=1
⌘
qn .
d 5 d
d|n
d 5
d|n
n ⌘ n
q
d
1
(11)
.
(12)
By Theorem 1 applied to (11) we have
5n
X
d 5 d =
5 (n)+5 (n/5)+
n
X1✓
5
j=1
d|n
X
d 5 d
d|j
◆✓
✓
5 (n j)+5
n
j
5
◆◆
,
or equivalently,
n
X1✓X
j=1
d 5 d
d|j
◆✓
(n
✓
j)
n
j
5
◆◆
=
1
(n)
5
1 ⇣n⌘
5
5
nX
d 5 d ,
5
d|n
which gives part (a). As to part (b), apply Theorem 1 to identity (12) to get
◆✓
✓
◆◆
n ✓X
X
X
n+1
j
n+1 j
n
d 5
=
d 5
(n + 1 j) 25
,
d
d
5
j=1
d|n+1
d|j
and proceed as in part (a).
2
Proof of Theorem 5. We have the following two formulas of Carlitz [5] which appear
as Theorem 3.4 and Theorem 3.5 respectively in Alaca et al. [1]:
1
Y
(1
q n )2 (1
q 2n )(1
q 4n )3 (1
q 8n )
2
=1
2
n=1
1 ⇣X
X
n=1
1
Y
(1
q 2n )3 (1
q 4n )(1
q 8n )2 (1
qn )
2
=
n=1
1 ⇣X
X
n=1
d 8 d
d|n
d 8
d|n
⌘
qn ,
n ⌘ n
q
d
1
.
(13)
(14)
Part (a) follows by Theorem 1 applied to (13) and part (b) follows by Theorem 1
applied to (14). To prove part (c) use the same sort of argument with an application
of Theorem 1 to the following identity which is Theorem 4.6 in [1]:
1
Y
(1
n=1
q 2n )3 (1
q 4n )3 (1
qn )
2
(1
q 8n )
2
=1+2
1 ⇣X
X
n=1
d|n
8 d
⌘
qn .
13
INTEGERS: 13 (2013)
2
Proof of Theorem 6. We proceed as in the previous proofs. We have the following
two identities by Alaca et al. [1, Theorems 3.6 and 3.7]:
1
Y
q n )(1
(1
q 3n )(1
q 4n )2 (1
q 6n )2 (1
q 12n )
2
=1
n=1
1
Y
1 ⇣X
X
n=1
q 2n )2 (1
(1
q 3n )2 (1
q 4n )(1
q 12n )(1
qn )
2
=
n=1
1 ⇣X
X
n=1
d 12 d
d|n
d 12
d|n
⌘
qn ,
n ⌘ n
q
d
1
.
Then application of Theorem 1 to the first identity yields part (a) and application
of Theorem 1 to the second identity yields part (b).
2
Proof of Theorem 7. Just as before, apply Theorem 1 to the following result of
Alaca et al. [1, Theorem 4.2].
1
Y
(1
q 3n )2 (1
q 5n )2 (1
qn )
1
(1
q 15n )
1
=1+
n=1
1 ⇣X
X
n=1
15 d
d|n
⌘
qn .
2
Proof of Theorem 8. By [1, Theorem 4.3] we have
1
Y
(1
q 2n )(1
q 4n )(1
q 5n )(1
q 10n )(1
qn )
1
(1
q 20n )
1
= 1+
n=1
1 ⇣X
X
n=1
20 d
d|n
⌘
which by virtue of Theorem 1 for an odd prime p yields the desired identity.
qn ,
2
Proof of Theorem 9. Combine Theorem 1 with the following formula [1, Theorem
4.4]
1
Y
(1
q 2n )(1
n=1
q 3n )(1
q 8n )(1
q 12n )(1
qn )
1
(1
q 24n )
1
= 1+
1 ⇣X
X
n=1
d|n
24 d
⌘
qn .
2
Acknowledgment The author is grateful to the referee for valuable comments and
interesting suggestions.
References
[1] A. Alaca, S. Alaca, and K. Williams, Some infinite products of Ramanujan type, Canad. Math.
Bull. Vol. 52 (4), (2009), 481-492.
[2] T. M. Apostol, Introduction to Analytic Number Theory, Undergraduate Texts in Mathematics, Springer, first edition, 1976.
INTEGERS: 13 (2013)
14
[3] W. N. Bailey, A note on two of Ramanujan’s formulae, Quart. J. Math. Oxford Ser. (2)
3(1952), 29-31.
[4] W. N. Bailey, A further note on two of Ramanujan’s formulae, Quart. J. Math. Oxford Ser.
(2) 3(1952), 158-160.
[5] L. Carlitz, Note on some partition formulae, Quart. J. Math. Oxford Ser. (2) 4(1953), 168-172.
[6] M. El Bachraoui, Inductive formulas for some arithmetic functions, International Journal of
Number Theory 7, Number 8, (2011), 2261-2268.
[7] J. Liouville, Sur quelques formules générales qui peuvent être utiles dans la théorie des nombres, (7th article), J. Math. Pures Appl. 4, (1859), 1-8.
[8] S. Ramanujan, The lost notebook and other unpublished papers, Narosa Publishing House,
New Delhi, 1988.
[9] N. Robbins, Some identities connecting partition functions to other number theoretic functions, Rocky Mountain Journal of Mathematics, Volume 29, Number 1, (1999), 335-345.
[10] K. S. Williams, Number Theory in the Spirit of Liouville, Cambridge University Press, New
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