The Explicit Formula for the Smarandache Functionand Solutions of Related Equations

LIAO Qunying, LUO Wenli

(Institute of Mathematics and Software Science, Sichuan Normal University, Chengdu 610066, Sichuan)

The Explicit Formula for the Smarandache Functionand Solutions of Related Equations

LIAO Qunying, LUO Wenli

(InstituteofMathematicsandSoftwareScience,SichuanNormalUniversity,Chengdu610066,Sichuan)

Letφ(n) andS(n) be the Euler function and Smarandache function for a positive integern, respectively. By using elementary methods and techniques, the explicit formula forS(pα) is obtained, wherepis a prime andαis a positive integer. As a corollary, some properties for positive integer solutions of the equationsφ(n)=S(nk) orσ(2αq)/S(2αq) are given, whereqis an odd prime andσ(n) is the sum of different positive factors forn.

Smarandache function; Euler function; Gauss function; perfect number

1 Introduction and Main Results

In 1918, Kempner[1]studied the formula of the value min{m:m∈N,n|m!}forafixedpositiveintegern.In1993,Smarandacheraisedsomeinterestingnumbertheoryproblems,andputforwordthedefinitionoftheSmarandachefunctionS(n)=min{m:m∈N,n|m!} for a positive integern. From the definition,S(1)=1,S(2)=2,S(3)=3, and so on. So far, there are some good related results[1-9]. For example, in [2], the distribution ofS(n) was discussed, and the asymptotic formula ofS(n) was given as follows

whereP(n) is the maximum prime factor ofn, andζ(s) is the Riemann-zeta function. In [3], Farris studied the bound ofS(n) and got the following upper and lower bounds

On the other hand, a lot of number theory equations related toS(n) have been studied in recent years. Especially, for a given positive integerk, many properties for positive integer solutions of the equationφ(n)=S(nk) were studied, whereφis the Euler function. Easy to see that this is equivalent to solve the equation

(*)

wherepis a prime, gcd(p,m)=1 andS(pαk)≥S(mk).

Theorem1.1Letpbeaprimeandαbeapositiveinteger.

1)Foranypositiveintegerrandα=pr,wehave

2)Foranypositiveintegerr, t∈[1,r]andα=pr-t,wehave

3)Foranypositiveintegerr, t∈[r+1,pr-pr-1]andα=pr-t.

(I)If

with

then we have

(1)

(II) If

witht∈[1,kn] and

then

(2)

Corollary 1.2 Letαbe a positive integer. If

then we haveS(2α)=α+n.

Fork=2,3,4, the solutions of the equation (*) have been discussed in [7]. In the present paper, we complement their results and obtain some necessary conditions for solutions of the equation (*).

Theorem 1.3 1) For any positive integerk, there are no any primepand positive integermcoprime withp, such thatφ(pm)=S(pk) andS(pk)≥S(mk).

2) For any positive integerk, if there are some primepand positive integermcoprime withp, such thatφ(p2m)=S(p2k) andS(p2k)≥S(mk). Thenp=2k+1 or 2≤p≤k. Furthermore,

(I) if 2k+1=p, then

(II) otherwise, i.e., 2≤p≤k, thenk≥3 and

3) For any positive integerk, if there are some primepand positive integermcoprime withp, such thatφ(pαm)=S(pαk) andS(pαk)≥S(mk). Thenαk+1>pα-3(p2-1) and 1≤φ(m)≤q, where

4) For any positive integerk, there exist some primepand positive integermcoprime withp, such thatφ(p3m)=S(p3k) andS(p3k)≥S(mk), namely,m=1,2.

2)Letpbeanoddprime, α≥1andn=2αp.

3)If2r-1isaprimeandn=22r-1(2r-1),then

Remark For convenience, throughout the paper we denote [·] to be the Gauss function.

2 The Proofs for Our Main Results

Before proving our main results, the following Lemmas are necessary.

2) For any primepand positive integerkwithk≤p, we have

Lemma 2.2 For any positive integerαand primep, we haveS(pα)≤(α-kα)p, wherekα(p+1)≤α<(kα+1)(p+1).

Proof For 0<α

Now forα=m≥p+1, ifS(pm)=(m-km)pwith

then

Thus forα=m+1, we know that

Hence we have two cases as following.

(I) If

thenkm+1=km. By the definition ofS(n), we haveS(pm+1)≤S(pm)+p, and so

therefore in this case Lemma 2.2 is true.

(II) Otherwise, we havem+1=(km+1)(p+1), and thenkm+1=km+1 andm-km=(km+1)p, where

Note that

therefore

This means that Lemma 2.2 is true.

By the definition ofS(n), we immediately have the following.

Lemma 2.3 Letpbe a prime andmbe a positive integer. Then

The Proof for Theorem 1.1 1) Sincepis a prime, and so

Thus, by the definition ofS(n), we haveS(ppr)=pr+1-pr+p, and then (1) of Theorem 1.1 is proved.

2) Since

andpr‖(pr+1-pr), and so for any positive integerrandα=pr-twitht∈[1,r], we haveS(pα)=pr+1-pr, thus (2) of Theorem 1.1 is true.

3) Forα=pr-twithr+1

(3)

In fact, form=1, i.e.,α=pr-r-1, we have

And then by the definition ofS(n), we can obtain

which means that (3) is true form=1. Now suppose that (3) is true for anym=k(≥1), i.e.,

Then form=k+1, by Lemma 2.3, we have

(A)

or

(B)

For the case (A), by Lemma 2.3, we have

and then

which means that (3) is true.

whichmeansthattheidentity(3)issatisfied.

Fromtheabove,theidentity(3)istrue.

Nowweprove(3)ofTheorem1.1.

1)Supposethatforanypositiveintegerk1andm=pk1such thatα=pr-r-pk1. Fromr+m∈[r+1,pr-pr-1], we haver+pk1∈[r+1,pr-pr-1], thus by the identity (3) and (1) of Theorem 1.1, we can obtain

2) Suppose that for any positive integerk1,s∈[1,k1] andm=pk1-s, such thatα=pr-r-(pk1-s). Fromr+m∈[r+1,pr-pr-1], i.e.,r+pk1-s∈[r+1,pr-pr-1], (3) and (2) of Theorem 1.1, we have

3) Suppose that there is some positive integerk1ande∈[k1+1,pk1-pk1-1], such thatm=pk1-e, namely,α=pr-r-(pk1-e). Fromr+m∈[r+1,pr-pr-1] we haver+pk1-e∈[r+1,pr-pr-1]. Now set

then

Similar to the previous discussions, we have the following three cases.

1′) If there is some positive integerk2such thatm1=pk2, i.e.,

and

Thus by (3) and (1) of Theorem 1.1, we have

which satisfies (1) of Theorem 1.1.

2′) Suppose that there is some positive integerk2andt1∈[1,k2], such thatm1=pk2-t1, i.e.,

and so

Thus by (3) and (2) of Theorem 1.1, we have

which satisfies (2) of Theorem 1.1.

3′) Suppose that there is some positive integerk2andt1∈[k2+1,pk2-pk2-1], such thatm1=pk2-t1, i.e.,

Now set

then

and so

Similar to the previous discussions, we know thatα∈[pr-1,pr] is a positive integer. Thus, one can repeat the above discussions 1)-3).

From the above discussions, Theorem 1.1 is proved.

The Proof for Corollary 1.2 For any positive integerski(1≤i≤n) with 1≤k1

(**)

Note that for anykm(1≤m≤n-1), we have

Thus from (**) we can get

Hence

Thus Corollary 1.2 is proved.

The Proof for Theorem 1.3 1) If there are some primepand positive integermcoprime withp, such thatS(pk)=φ(pm) andS(pk)≥S(mk). Then forp=2, we have

Byφ(2m)≡ 0(mod 2) we havem≥3. While byS(2k)≥S(mk), we havem=1, this is a contradiction. And sop≥3, thus from the definition ofS(n) and the assumption thatpis coprime withm, we have

2) Suppose that there exist some positive integerα, primepand positive integermcoprime withp, such thatφ(p2m)=S(p2k) andS(p2k)≥S(mk).

(I) For the case 2k≤p, by (2) of Lemma 2.1, we have

i.e., 2k=(p-1)φ(m). Note thatpis a prime, ifp=2, then by 2k≤p=2 we havek=1, and soφ(m)=2, thusm=3,4,6. Hence from gcd(p,m)=1 andp=2, we can getm=3. In this case,

which means that (p,m)=(2,3) is a solution.

Now forp≥3, by 2k≤pwe have

and soφ(m)=1, i.e.,m=1 or 2 andp=2k+1, hence

(II) For the case 2k>p, suppose thatt1andt2are both nonnegative integers such that

(4)

and

Then byS(p2k)=φ(p2m) and Lemma 2.2, we have

(5)

and

Now from (5), we know that

which means that

(6)

Note that 2k>p, i.e., 2kp>p2, thus we have three cases as following.

1) For the case

which means that

and sop2-1|p+2, i.e.,p+2≥p2-1. While

2) For the case

i.e.,

3) Therefore we must have (p2-1)φ(m)-(p+1)>p2, namely,

By (6), we have

i.e.,

(7)

thus 2p2-(2k+1)p-3≤0, and so

(8)

Note thatpis a prime, and so 2p-(2k+1)≤1. If 2p-(2k+1)=1, then by (8), we know that (p,k)=(2,1),(3,2). From (p,k)=(2,1), we have 2k=p=2, this is a contradiction to 2k>p. Sok=2,p=3 or 2p<2k+1. Byk=2,p=3 and (7), we haveφ(m)=2, and thenm=3,4,6. Note that gcd(p,m)=1 and then forp=3, we havem=4, thus

namely,

(9)

Note that

whichisacontradiction.Hencek≥3,thusweprove(2)ofTheorem1.3.

3)Forα≥3.Ifαk≤p,thenby(2)ofLemma2.1,wehave

thus

henceαk=p=2,whichisacontradictiontotheassumptionα≥3.Andsoαk>p.Nowsupposethatt1andt2arebothnonnegativeintegerssuchthat

(10)

and

(11)

namely,

thus

and so

i.e.,

(12)

Note that for any positive integerm, we haveφ(m)≥1, therefore we must haveαk+1≥pα-3(p2-1).

If

i.e.,φ(m)=1. In this case, forα=3 we have 3k+1=p2-1, i.e.,p2=3k+2, which is impossible. Soα>3, and then

We can conclude that

(13)

Otherwise, fromα-3≥pα-4(p-1)-1, we have

(14)

It is easy to see that forα≥4 there is no any primep>5 satisfying (14). Hencep=2 or 3. Byp=3 and (14) we haveα≥2(3α-4+1). While 2(3α-4+1)>αforα≥5. Therefore from (14) we haveα=4, and then 4k+1=3α-1-3α-3=24, which is a contradiction. Thus we must havep=2.

Now fromp=2 and (14), we haveα>2α-4+2, and soα=4,5,6. Thus byαk+1=pα-1-pα-3andα=4, we haveαk+1=4k+1=23-2=6, which is a contradiction. Forα=5, we have 5k+1=12, which is also a contradiction. Forα=6,6k+1=24, it is also a contradiction. Hence (14) is not true, and soα-3

Now byφ(m)=1, gcd(p,m)=1 andφ(pαm)=S(pαk), we have

this meanspα-3+p=pα-2. Note thatpis a prime, thus we havep=2 andα=4. And so 4k+1=23-2=6, which is a contradiction.

From the above we must haveαk+1>pα-3(p2-1)≥p+1. Without loss of the generality, set

Now by (12), we have

(15)

and so 1≤φ(m)≤q. Thus we prove (3) of Theorem 1.3.

Thusweprove(4)ofTheorem1.3.

FromtheaboveTheorem1.3isproved.

we have

i.e.,

(16)

Thus from

we have

(17)

wecanobtain1≤m≤d.

ThusweproveTheorem1.4.

TheProofforCorollary1.5 1)Ifp=2r+1isaprimeandα=2r,n=22r(2r+1).Then

Ontheotherhand,bythedefinitionofσ(n)and(1)ofTheorem1.1,wealsohave

2)Sincen=2p-1(2p-1)isaperfectnumber,soσ(n)=2p(2p-1).Thusfrom(1)ofLemma2.1and2p-1isaprimenumber,wehave

Notethat

andso

Bythesimilarway,wecanprovepart(3).

ThusweproveCorollary1.5.

3 Some Examples

Inthissection,someexamplesforbothTheorem1.1andCorollary1.2aregiven.

Example3.1Letp=3,α=35=243,thenby(1)ofTheorem1.1wehave

Ontheotherhand,from

163+54+18+6+2+0=243,

and the definition ofS(n), we also haveS(3243)=489.

Example 3.2 Letp=3,α=36-4=725. Namely, be takingr=6,t=4 in (2) of Theorem 1.1, we know that

On the other hand, from

486+162+54+18+6+2+0=728,

485+161+53+17+5+1+0=722,

and the definition ofS(n), we also haveS(3725)=1 458.

Example 3.3 Letp=3,α=5 017, i.e.,

thus from (2) of Theorem 1.1, we have

4×2 187+16×81=10 044.

On the other hand, from

3 348+1 116+372+124+

41+13+4+1=5 019,

and

3 347+1 115+371+123+

41+13+4+1=5 015,

we also haveS(35 017)=10 044.

4 Conclusion

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Smarandache函數的準確計算公式以及相關數論方程的求解

廖群英, 羅文力

(四川師范大學 數學與軟件科學學院, 四川 成都 610066)

Smarandache函數; 歐拉函數; 高斯函數; 完全數

O

A

1001-8395(2017)01-0001-10

2016-01-03

國家自然科學基金(11401408)和四川省科技廳研究項目(2016JY0134)

廖群英(1974—),女,教授,主要從事編碼和密碼學理論的研究,E-mail:qunyingliao@sicnu.edu.cn

Foundation Items: This work is supported by National Natural Science Foundation of China (No.11401408) and Project of Science and Technology

10.3969/j.issn.1001-8395.2017.01.001

(編輯 周 俊)

Received date:2016-01-03

Department of Sichuan Province (No.2016JY0134)

2010 MSC：12E20; 12E30; 11T99