# A note on generalized mobius (s-functions

A note on generalized mobius (s-functions
- Download this document for free, or read online. Document in PDF available to download.

In 1 the concept of a conjugate pair of sets of positive integers is introduced. Briefly, if Z denotés the set of positive integers and P and Q denote non-empty subsets of Z such that: if n1 pertenece a Z, n2 pertenece a Z, n1,n2 = 1, then 1 n = n1n2 pertenece a Presp. Q n1 pertenece a P,n2 pertenece a P resp. Q, and, if in addition, for each integer n pertenece a Z there is a unique factorization of the form 2 n = ab , a pertenece a P, b pertenece a Q, we say that each of the sets P and Q is a direct factor set of Z, and that P,Q is a conjugate pair. It is clear that P intersección Q = {11}. Among the generalized functions studied in 1 ,

Tipo de documento: Artículo - Article

Palabras clave: Integers, subsets, factorization, conjugate pair, functions

Source: http://www.bdigital.unal.edu.co

## Teaser

Revista Colombiana de Matematioas
Volumen II,1968,pags.6-11
A NOTE ON GENERALIZED MOBIUS f-FUNCTIONS
by
V.S.

ALBIS
In [1]
the ooncept of a conjugate pair of sets of
posi tive integers is int~oduoed •
Briefly, if
tes the set of positive integers and
non-empty subsets of
(n ,n )
l 2
(1)
1,
n
Z
P
such that.

if
and
Z
Q
deno-
nl E Z, n
denote
2
E Z,
then
n n E P(resp.Q) = n EP,n EP (resp.

Q),
l 2
l
2
=
and, if in addition, for each integer
n E Z
there is
a unique factorization of the form
(2)
n = ab , a
P,
E
b
E Q,
we say that each of the sets
factor set of
Z, and that
It is clear that
tions studied in
(3)
pp(n)
=
pnQ
[lJ,
E
=
P
(p,Q)
and
Q
is a direct
is a conjugate pair.
{11.

Among the generalized func-
we find
~(n-d)
din
dEP
a generalization of MOBIUS Jl-function.

The following
results are also proved in [1]:
(i) pP
E
dIn
dEQ
6
is a multiplicative function.
if
n = 1
if
n
fJp(n-d)
1
Here we shall show that
is the unique arithmeti-
flp
cal function satisfying (ii) above.
Let
r-
be such
that
(4 )
,.11 (n-d) = p (n)
I:
dIn
dEQ
~*
If
=
O.
~~(pk)
=
for every prime
~*
So let
(4)
llows from
~(p)
n = 1
if
n 1
is multiplicative, it suffices to prove that
~p(p )
k
-f:
if
p*(p)
that
relation holds for
(5)
(ii)
and
k
~(pU)
and every integer
be a multiplicative function; it fothat
for every prime
induction on
p
~(pk)
Jlp(l) = Jloll (1), thus
p.

We will now show by
=
r~(pk).

Suppose this
u k O.

From
(ii) we obtain
= _ ~
pp(pU-pi)
l l.

u
piE-Q
because
1
=
P
o
E
Q.

On the other hand, from
tain
_ I:
( 6)
J1*
(pu-pi)
l i u
rE-Q
0
because 1 = p E Q.

But by the induction hypothesis,
f1p(pU-i) = Jlt(pU-i) ( i = 1,•••, u) • Thus the rigth members in
(5) and (6)
are equal, so that
Jip(pU) = fl~(PlJ.)
In view of the above result, it suffices to show
that any function
p~
(4) is multiplicati-
satisfying
ve, thus proving the following
rg:§QB!2~Ll.
...