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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





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


Introducción



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. ...





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