On the Smarandache Function and Square Complements

2004·0.19 MB·

by  Zhang Wenpeng

In Scientia Magna, Vol.1, No.1, 3 p.
ONTHESMARANDACHEFUNCTIONANDSQUARE
COMPLEMENTS ∗
ZhangWenpeng
ResearchCenterforBasicScience,Xi’anJiaotongUniversity,Xi’an,Shaanxi,P.R.China
[email protected]
XuZhefeng
DepartmentofMathematics,NorthwestUniversity,Xi’an,P.R.China
[email protected]
Abstract Themainpurposeofthispaperisusingtheelementarymethodtostudythemean
valuepropertiesoftheSmarandachefunction,andgiveaninterestingasymptotic
formula.
Keywords: Smarandachefunction;Squarecomplements;Asymtoticformula.
§1. Introduction
Let n be an positive integer, if a(n) is the smallest integer such that na(n)
isaperfectsquarenumber,thenwecalla(n)asthesquarecomplementsofn.
ThefamousSmarandachefunctionS(n)isdefinedasfollowing:
S(n) = min{m : m ∈ N,n|m!}.
In problem 27 of [1], Professor F. Smarandache let us to study the properties
of the square complements. It seems no one know the relation between this
sequence and the Smaradache function before. In this paper, we shall study
the mean value properties of the Smarandache function acting on the square
complements, and give an interesting asymptotic formula for it. That is, we
shallprovethefollowingconclusion:
∗ThisworkissupportedbytheN.S.F.(10271093,60472068)andtheP.N.S.F.ofP.R.China.
2 SCIENTIAMAGNAVOL.1,NO.1
Theorem. Foranyrealnumberx ≥ 3,wehavetheasymptoticformula
(cid:195) (cid:33)
(cid:88) π2x2 x2
S(a(n)) = +O .
12lnx ln2x
n≤x
§2. Proof of the theorem
To complete the proof of the theorem, we need some simple Lemmas. For
convenience,wedenotethegreatestprimedivisorofnbyp(n).
√
Lemma1. Ifnisasquarefreenumberorp(n) > n,thenS(n) = p(n).
Proof. (i)nisasquarefreenumber. Letn = p p ···p p(n),then
1 2 r
p |p(n)!, i = 1,2,···,r.
i
Son|p(n)!,butp(n)†(p(n)−1)!,son†(p(n)−1)!,thatis,S(n) = p(n);
√
(ii)p(n) > n. Letn = pα1pα2···pαrp(n),sowehave
1 2 r
√
pα1pα2···pαr < n
1 2 r
then
pαi|p(n)!, i = 1,2,···,r.
i
Son|p(n)!,butp(n)†(p(n)−1)!,soS(n) = p(n).
ThisprovesLemma1.
Lemma2. Letpbeaprime,thenwehavetheasymptoticformula
(cid:195) (cid:33)
(cid:88) x2 x2
p = +O .
√ 2lnx ln2x
x≤p≤x
Proof. Letπ(x)denotesthenumberoftheprimesuptox. Notingthat
(cid:181) (cid:182)
x x
π(x) = +O ,
lnx ln2x
fromtheAbel’sidentity[2],wehave
(cid:90)
(cid:88) √ √ x
p = π(x)x−π( x) x− π(t)dt
√
√ x
x≤p≤x
(cid:195) (cid:33)
x2 1 x2 x2
= − +O
lnx 2lnx ln2x
(cid:195) (cid:33)
x2 x2
= +O .
2lnx ln2x
ThisprovesLemma2.
Nowweprovethetheorem. Firstwehave
(cid:88) (cid:88)
S(a(n)) = S(n)|µ(n)|
n≤x m2n≤x
(cid:88) (cid:88)
= S(n)|µ(n)|. (1)
√
m≤ xn≤ x
m2
OntheSmarandachefunctionandsquarecomplements1 3
Totheinnersum,usingtheabovelemmasweget
(cid:88)
S(n)|µ(n)|
n≤ x
(cid:88)m2 (cid:179) (cid:180)
3
= p|µ(n)|+O x2
np≤ x
√m2
p≥ np
(cid:88) (cid:179) (cid:180)
3
= p|µ(n)|+O x2
np≤ x
√m2
p≥ x
(cid:88)m2 (cid:88) (cid:179) (cid:180)
3
= |µ(n)| p+O x2
√ √
n≤ x x ≤p≤ x
m2 m2 nm2 (cid:195) (cid:33)
(cid:88) |µ(n)|x2 (cid:88) |µ(n)|x2 x2
= + +O
2n2m4ln x √ 2n2m4ln x m4ln2x
n≤ln2x nm2 ln2x<n≤ x nm2
(cid:195) (cid:33) m2
ζ(2)x2 x2
= +O . (2)
2ζ(4)m4lnx m4ln2x
Combining(1)and(2),wehave
 
(cid:88) ζ(2)x2 (cid:88) 1 x2 (cid:88) 1
S(a(n)) = +O 
2ζ(4)lnx √ m4 ln2x √ m4
n≤x m≤ x m≤ x
(cid:195) (cid:33)
ζ(2)x2 x2
= +O .
2lnx ln2x
Notingthatζ(2) = π2,sowehave
6
(cid:195) (cid:33)
(cid:88) π2x2 x2
S(a(n)) = +O .
12lnx ln2x
n≤x
ThiscompletestheproofofTheorem.
Reference
[1]F.Smaradache,OnlyProblems,NotSolutions,XiquanPublishingHouse,
Chicago,1993.
[2] Tom M. Apostol, Introduction to Analytic Number Theory, Springer-
Verlag,NewYork,1976,pp. 77.
[3] Pan Chengdong and Pan Chengbiao, Element of the Analytic Number
Theory,SciencePress,Beijing,1991.