International Journal of Mathematical Archive-4(6), 2013, 148-150
Available online through www.ijma.info ISSN 2229 – 5046
A Remark on Arithmetic Function χ ∗
Bhabesh Das*
Department of Mathematics, Gauhati University, Guwahati-781014, Assam, India
(Received on: 19-05-13; Revised & Accepted on: 10-06-13)
ABSTRACT
In
this
paper
we
introduce
a
new
function χ ,
∗
arithmetic
which
is
χ (n) = n + (−1) d1 + (−1) d 2 + (−1) d3 + ...... + (−1) d k + 1 , where d1 , d 2 , d3 ,......d k
positive integer n . Using this arithmetic function we get some relations on σ , ϕ and χ -function.
∗
d1
d3
d2
dk
defined
as
are divisors of the
Mathematics Subject Classification: 11A25, 11A41
Keywords: Arithmetic function, divisor function, Euler’s totient function,
χ -function.
INTRODUCTION
An arithmetic function or number theoretic function f is one whose domain is the set of positive integers and whose
range is a subset of the complex numbers. Arithmetic function f is said to be multiplicative, if f ( mn) = f ( m) f ( n)
for g.c.d ( m, n) = 1 . Euler’s totient function is defined as=
ϕ ( n) n
1
∏ (1 − p ) , where
p runs through all prime
divisors of n .Let n be a natural number and d1 , d 2 , d 3 ,......d k be divisors of n such that d1 > d 2 > d 3 > ...... > d k .
σ is defined as σ (n) =n + d1 + d 2 + d3 + ...... + d k + 1 . K. Atanassov in
k
k +1
[1] introduced a new arithmetic function χ , which is defined as χ ( n) = n − d1 + d 2 − d 3 + ...... + (−1) d k + (−1) .
For any prime p , χ ( p )= p − 1 . In [1] author gave some results using this arithmetic function.
Well known Divisor arithmetic function
MAIN RESULTS
Here we shall introduce a new arithmetic function that is some how dual of χ . It will have the form
χ ∗ (n) = n + (−1) d d1 + (−1) d d 2 + (−1) d d3 + ...... + (−1) d d k + 1 .It
1
3
2
k
is
clear
that
for
any
prime p ,
χ ( p )= p + 1 and χ (1) = 0 .
∗
∗
In following table the values of functions χ , ϕ and
∗
σ
for the first 50 natural numbers are given.
Corresponding author: Bhabesh Das*
Department of Mathematics, Gauhati University, Guwahati-781014, Assam, India
International Journal of Mathematical Archive- 4(6), June – 2013
148
Bhabesh Das*/A Remark on Arithmetic Function χ / IJMA- 4(6), June-2013.
∗
n
χ ∗ ( n)
ϕ ( n)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
0
3
4
7
6
6
8
15
7
8
12
22
14
10
8
31
18
1
1
2
2
4
2
6
4
6
4
10
4
12
6
8
8
16
σ ( n)
1
3
4
7
6
12
8
15
13
18
12
28
14
24
24
31
18
n
χ ∗ ( n)
ϕ ( n)
σ ( n)
n
χ ∗ ( n)
ϕ ( n)
σ ( n)
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
15
20
32
12
14
24
54
21
16
16
42
30
26
32
63
20
20
6
18
8
12
10
22
8
20
12
18
12
28
8
30
16
20
16
39
20
42
32
36
24
60
31
42
40
56
30
72
32
63
48
54
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
24
67
38
22
24
80
42
34
44
62
14
26
48
118
43
33
24
12
36
18
24
16
40
12
42
20
24
22
46
16
42
20
48
91
38
60
56
90
42
96
44
84
78
72
48
124
57
93
It is easy to prove the followings
Proposition 1: For any natural number n ,
Proposition 2: =
If n
χ ∗ (n) ≤ σ (n).
2a , a ∈  , then χ ∗ (n) = σ (n) .
Proposition 3: If n = 2 p , where p is an odd prime, then
∗
χ=
( n)
σ ( n)
+2
3
Proposition 4: If n = pq , where p and q are distinct odd primes, then ϕ ( n) =
Proof: If n = pq , then
χ ∗ ( n) .
χ ∗ ( pq)= pq + (−1) p p + (−1) q q + 1
= pq − p − q + 1
=( p − 1)(q − 1)
= ϕ ( n)
Proposition 5: For any odd positive integer n ,
χ ∗ ( n) + σ ( n) =
2n + 2 .
Proposition 6: If n = 2 p , where p is an odd prime and a ∈  , then
a
∗
χ=
( n) σ ( p a ) + 2
Proof: If n = 2 p , then
a
χ ∗ ( n=
) (2 p a + 2 p a −1 + 2 p a − 2 + .... + 2 p ) − ( p a + p a −1 + p a − 2 + .... + p ) + 2 + 1
= 2( p a + p a −1 + p a − 2 + .... + p + 1) − ( p a + p a −1 + p a − 2 + .... + p ) + 1
= 2σ ( p a ) − (σ ( p a ) − 1) + 1
= σ ( pa ) + 2
Proposition 7: Any positive integer n = 2m, where m = p1 1 p2 2 ... pr r and
a
a
a
pi are distinct odd primes, then
∗
χ=
( n) σ ( m) + 2 .
Proposition 8: If n = pq , where p and q are two distinct odd primes such that p > q , then χ ∗ ( n) =
© 2013, IJMA. All Rights Reserved
p −1
χ ( n) .
p +1
149
Bhabesh Das*/A Remark on Arithmetic Function χ / IJMA- 4(6), June-2013.
∗
Proof: If n = pq , then
χ (n) = pq − p + q − 1= ( p + 1)(q − 1)
χ ∗ ( pq ) = ( p − 1)(q − 1) =
p −1
p −1
( p + 1)(q − 1) =
χ ( n)
p +1
p +1
REFERENCES
[1] Krassimir Atanassov, A remark on an arithmetic function. Part 2, NNTDM 15 (2009) 3, 21-22, CLBME - Bulg.
Academy of Sci., P.O. Box 12, Sofia-1113, Bulgaria.
[2] W.SIERPINSKI, Elementary theory of numbers, Warsawa, 1964.
[3] Nagell T., Introduction to number theory, John Wiley & Sons, New York, 1950.
Source of support: Nil, Conflict of interest: None Declared
© 2013, IJMA. All Rights Reserved
150