Ying SHI Quanhui YANG

1 Introduction

LetA3be the set of all positive integerspqr,wherep,q,rare primes and are not all equal.For any integern=pqr∈A3,we define a functionwbywhereP(m)is the largest prime factor ofm.Definew0(n)=nandfor all integersi≥1.

In 2006,Goldring[4]proved that any elementn∈A3isw-periodic,i.e.,there exists an integeri≥0,such thatwi(n)=20.Denote the smallest such integeriby ind3(n).Goldring[4]proved thatwhereπ(x)denotes the number of primes not exceedingx.Later,Chen and Shi[1]improved Goldring’s result and proved thatfor alln∈A3.

LetPbe the set of all positive primes.An integermis called a parent ofnifw(m)=n.Write

Chen and Shi[2]proved that for any positive integerk,there are in finitely many elements ofB3which have at leastkparents inB3,and that there exist in finitely many elements ofB3which have no parents inB3.

Later,Jia[5]studied parents ofand obtained some interesting results.

Recently,Chen,Shi and Wu[3]proved that there exist in finitely manyn∈B3which have at leastn1.1886parents inB3.

For an integern=p1p2···pk,wherepi(1≤i≤k)are primes in the descending order and are not all equal,define

Clearly,Ω3(n)=w(n).In[6],Shi generalized Goldring’swfunction to the function Ωk.Definefor all integersi≥1.We callna simple integer for Ωkif there exists an integeri≥0,such thatis a prime power.For integersk≥3,let

where Ω(n)is the total number of prime factors ofn.If 2?k,Ω(n)=kandnis not a prime power,then Ωk(n)is also not a prime power.Otherwise,we havewherep1p2···pk=nandp,pi(1≤i≤k)are primes.It follows thatand thenp=2 orp=p1.Ifp=2,then,byit follows that 2|p1,and thenp1=p.Hence,n=pk,a contradiction.Therefore,the definition ofAkis consistent with that of the previous setA3.

An elementnofAkis Ωk-periodic if there exists a nonnegative integersand a positive integert,such thatThe smallest such integersis called the index of periodicity ofn,denoted by indk(n).The arrayis called a circular array ofAkiftelementssatisfyIn general,we regard all arrays such asas an equal array,denoted bywherebiis an element in this circular array.An elementnofAkis said to lie in the circular arrayultimately,if there exists an integerj≥0,such thatThe whole circular array inAkis denoted by

In[6],Shi proved the following theorem.

Theorem AEvery element of Akis periodic and each lies in some circular array ultimately.When k≥5,In addition,

In this paper,based on the method in[1],we prove the following result.

Theorem 1.1Let k be an odd integer with k≥4.For any integer n with k prime factors not all equal,we have

Remark 1.1Ifkis even,then the iteration of the arithmetic function Ωkmay stop.For example,k=4,Ω4(3×7×13×17)=54.Therefore,we only consider the odd case.

2 Preliminary Lemmas

Lemma 2.1Let X≥3be an integer and α be a real number with0<α<1.For any integer n with k(k≥4)prime factors not all equal,let n=p1p2···pk,where pi(1≤i≤k)are primes and p1≥p2≥···≥pk.If p1≤X and p2≤αX,then there exists an integer i with1≤i≤3,such thatthenIfpk>2,thenThus we may assume thatp1≥5,p1>αXandpk=2.

Now we consider the following two cases.

Case 1p1+2 is composite.

By the definition of Ωk(n)andpk=2,we have

Ifp2=2,thenthenWe also have

ProofIfp1≤αX,thenIfp1=3>αX,i.e.,n=3·2k−1,

Fori=2,3,···,k−1,ifpi+1>2,thenthenαX+2.

Hence,we obtain

Case 2p1+2 is prime.

Subcase 2.1n=p1·2k−1.It follows that

Sincep1>3,andp1,p1+2 are both primes,we have 3|p1+4.Hence,P(p1+4)

Clearly,we have

and

Therefore,

Subcase 2.2where 3≤i≤kandpi−1≥3.Then

Letwhereqi(1≤i≤k)are primes andq1≥q2≥···≥qk.Clearly,q1=p1+2≤X+2 and forj=2,3,···,k,

Sincep1>3,andp1,p1+2 are both primes,we have 3Thus,forj=2,k,ifthenthenWe also havefor

Byit follows that

Therefore,by all the cases above,there exists an integeriwith 1≤i≤3,such that

This completes the proof of Lemma 2.1.

Lemma 2.2Let X≥3,k≥4be integers and α<1be a positive real number.Let n=p1p2···pk,where pi(1≤i≤k)are primes in the descending order and are not all equal.If p1≤X and pj≤αX for some integer j with2≤j≤k,then there exists a positive integer i with1≤i≤4j−3,such that

ProofIfp1=3,thenn=3s·2k−sfor some integerswith 1≤s≤k−1.ByX≥3 andj≥2,we have

Thus we may assume thatX≥p1≥5.We shall prove it by induction onj.

By Lemma 2.1,the result is true forj=2.Now we suppose that it is true forj=l−1,where 2≤l−1

Now we assume thatp1≤Xandpl≤αX.We consider the following cases.

Case 1pk≥3.

For 1≤s≤l−2,we haveForl−1≤s≤k−1,we haveWe also have

Letwhereare primes andThenand

By the induction hypothesis,there exists an integeriwith 1≤i≤4l−7,such that

Hence,there exists an integeriwith 1≤i≤4l−6,such that

Case 2pk=2 andpl≥3.

If 1≤s≤l−2,thenIfl−1≤s≤k−1,thenP(ps+ps+1)≤We also haveHenceq1≤X+2.

Letwhereqi(1≤i≤k)are primes andSuppose thatBy the induction hypothesis,there exists an integeriwith 1≤i≤4l−7,such that

Hence,there exists an integeriwith 1≤i≤4l−6,such that

Subcase 2.1p1+2 is composite.

It follows thatHenceand we are done with the proof.

Subcase 2.2p1+2 is prime.

Ifthen we are done with the proof.

Now we assume that

Sincepk=2 andpl≥3,there exists an integertwithl≤t≤k−1,such thatpt≥3 andNoting thatand,we haveByp1>3,and sincep1andp1+2 are both primes,we have thatq1+2=p1+4 is composite.Now we go back to Case 1 ifqk≥3 and Subcase 2.1 ifqk=2.The maximal upper bound in these two cases appears in Subcase 2.1.Hence,there exists an integeriwithsuch that

Case 3pk=2 andpl=2.

In this case,we have

Subcase 3.1At least one ofpl−1+2 andp1+2 is composite.

Letp+2 be composite,wherep=p1orThenP(p+2)≥3 andLet,whereare primes and.Ifthen we use the induction hypothesis.Now suppose thatIt follows thatHence 3Ifl=k,thenqk≥3 and we go back to Case 1.Ifl

Subcase 3.2Bothpl−1+2 andp1+2 are primes.

It follows thatpl−1≥3,and then 2fori=1,2,···,l−2.

Subcase 3.2.1are not all primes.

We assume thatis composite,where 1≤j≤l−2.Then

kIfandl

Subcase 3.2.2are all primes.

(1)pl−1=3.

Ifthen,by 5≤p1≤X,we havepl−1=3≤αX.Thus,by the induction hypothesis,the result is true.Now we assume thatNoting that 2≤ αX,we haveBy the induction hypothesis andthere exists an integeriwith 1≤i≤4l−7,such that

(2)pl−1>3.

Sincep1>3,andp1andp1+2 are primes,we have(mod 3).Noting thatis a prime greater than 3 andp2>3,we havep2≡2(mod 3).Otherwise,ifp2≡1(mod 3),then,a contradiction.Similarly,we have(mod 3).It follows that(mod 3),pl−1+2≡1(mod 3)and(mod 3)fori=1,2,···,l−2.

Now we consider

For alli,jwith 1≤i,j≤l−2,we have

Hence,primes

are all odd,and none of them is more than

Letwhereare primes and.Then(if it exists)and there exist integersr,swith 1≤r

By all the cases above,Lemma 2.2 is true forj=l.That is,ifpl≤αXandp1≤X,then there exists an integeriwith 1≤i≤4l−3,such that

This completes the proof of Lemma 2.2.

Lemma 2.3Let k be an odd integer with k≥4.Then for any integer n with k prime factors not all equal,there exists an integer i with1≤i≤2logP(n)+4k−2,such that

ProofSuppose that,whereare primes in the descending order and are not all equal.Then

Now,we discuss the following cases.Case 1p1≥p2≥···≥pk≥3.

Let

Subcase 1.1At least one element ofis composite.

Since the largest prime factor of this composite element ofis less thanby Lemma 2.2,there exists an integeriwith 1≤i≤4k−2,such that

Subcase 1.2All elements ofPare primes.

In this case,it is clear that every element ofis an odd prime.Now we arrange thesekodd primes in the descending order and denote them by.Then we consider

If all these numbers are odd primes,then we arrange them in the descending order and denote them byContinue this process until they are not all primes.Suppose that for the(t+1)th time,there exists an integerswith 1≤s≤k,such thatis composite.

Sinceare odd primes fori=1,2,···,k,we have

That is,

Byit follows that

Thus

Continuing this argument,for all integersjwith 1≤j≤t,we have

Hence

Ifthen by,we havep1=p2=···=pk,a contradiction.So 2t+1≤p1,and thent<2logp1.Sinceis composite andLemma 2.2,there exists an integeriwith 1≤i≤2logP(n)+4k−2,such that

Case 2pk=2.

Ifp1=3,thenfori=1 and the result is obviously true.Now we assume thatp1≥5.Then.By Lemma 2.2,there exists an integeriwith 1≤i≤4k−3,such that

By all the cases above,for any integernwithkprime factors not all equal,there exists an integeriwith 1≤i≤2logP(n)+4k−2,such that

This completes the proof of Lemma 2.3.

3 Proof of Theorem 1.1

For any integernwithkprime factors not all equal,let,wherepi(1≤i≤k)are primes in the descending order and are not all equal.TakeandBy Lemma 2.3,there exist positive integerssuch that,for all integerst≥1,

and

where

By(3.1),we have

If,then by Theorem A,indk(n)is bounded and the result is true.Now we suppose that.Take a positive integert0,such that

Then

By Theorem A,for every elementn∈Ak,there exists a positive integerinsuch thatlies in some circular arrayultimately.

Let

Then there exists an integerjwith 1≤j≤c0,such thatlies in some circular array(2a3b5c)Ωkultimately.

By(3.2)–(3.3),we have

Thus,by(3.4)–(3.5),we have

where the constantc1depends only onk.

Therefore,

This completes the proof of Theorem 1.1.

AcknowledgementThe authors would like to thank Professor Yonggao Chen for his valuable suggestions and useful discussions,especially on the key skills used in this paper.

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