On a note of the Smarandache power function by Don Hass

Scientia Magna Vol. 6 (2010), No. 3, 93-98

On a note of the Smarandache power function

Wei Huang† and Jiaolian Zhao‡ † Department of Basis, Baoji Vocational and Technical College, Baoji 721013, China ‡ Department of Mathematics, Weinan Teacher’s University, Weinan 714000, China E-mail: wphuangwei@163.com Abstract For any positive integer n, the Smarandache power function SP (n) is defined as the smallest positive integer m such that n|mm , where m and n have the same prime divisors. The main purpose of this paper is to study the distribution properties of the k − th power of SP (n) P P by analytic methods, obtain three asymptotic formulas of (SP (n))k , ϕ((SP (n))k ) and n≤x n≤x P d(SP (n))k (k > 1), and supplement the relate conclusions in some references. n≤x

Keywords Smarandache power function, the k − th power, mean value, asymptotic formula.

§1. Introduction and results For any positive integer n, we define the Smarandache power function SP (n) as the smallest positive integer m such that n|mm , where n and m have the same prime divisors. That is, ½ Y Y ¾ SP (n) = min m : n|mm , m ∈ N+ , p= p . p|m

p|n

If n runs through all natural numbers, then we can get the Smarandache power function seαk 1 α2 quence SP (n): 1, 2, 3, 2, 5, 6, 7, 4, 3, 10, 11, 6, 13, 14, 15, 4, 17, 6, 19, 10, · · · , Let n = pα 1 p2 · · · pk , denotes the factorization of n into prime powers. If αi < pi , for all αi (i = 1, 2, · · · , r), then Q Q we have SP (n) = U (n), where U (n) = p, denotes the product over all different prime p|n

p|n

divisors of n. It is clear that SP (n) is not a multiplicative function. In reference [1], Professor F. Smarandache asked us to study the properties of the sequence SP (n). He has done the preliminary research about this question literature [2] − [4], has obtained some important conclusions. And literature [2] has studied an average value, obtained the asymptotic formula: ¶ µ ¶ µ X 3 1 Y 1 SP (n) = x2 1− + O x 2 +ε . 2 p(1 + p) p n≤x

1 This

paper was partially supported by National Natural Science Foundation of China (10671155), The Natural Sciences Funding of Project Shaanxi Province (SJ08A28) and The Educational Sciences Funding Project of Shaanxi Province (2010JK541).

Wei Huang and Jiaolian Zhao

No. 3

Literature [3] has studied the infinite sequence astringency, has given the identical equation:  2s + 1   , k = 1, 2;   (2s − 1)ζ(s)  ∞  s s µ(n) X 2 +1 2 −1 (−1) − , k = 3; = s − 1)ζ(s) k ))s (2 4s  (SP (n  n=1 s s s  2 +1  2 −1 3 −1   s − + , k = 4, 5. (2 − 1)ζ(s) 4s 9s And literature [4] has studied the equation SP (nk ) = φ(n), k = 1, 2, 3 solubility (φ(n) is the Euler function), and has given all positive integer solution. Namely the equation SP (n) = φ(n) only has 4 positive integer solutions n = 1, 4, 8, 18; Equation SP (n3 ) = φ(n) to have and only has 3 positive integer solutions n = 1, 16, 18. In this paper, we shall use the analysis P (SP (n))k , method to study the distribution properties of the k − th power of SP (n), gave n≤x P P d(SP (n))k (k > 1), some interesting asymptotic formula, has proϕ((SP (n))k ) and n≤x

n≤x

moted the literature [2] conclusion. Specifically as follows: Theorem 1.1. For any random real number x ≥ 3 and given real number k (k > 0), we have the asymptotic formula: µ ¶ X 1 ζ(k + 1) k+1 Y 1 (SP (n))k = x 1− k + O(xk+ 2 +ε ); (k + 1)ζ(2) p (p + 1) p n≤x

µ ¶ X (SP (n))k 1 ζ(k + 1) k Y 1 = x 1− k + O(xk+ 2 +ε ), n kζ(2) p (p + 1) p

n≤x

where ζ(k) is the Riemann zeta-function, ε denotes any fixed positive number, and

denotes

the product over all primes. Corollary 1.1. For any random real number x ≥ 3 and given real number k 0 > 0 we have the asymptotic formula: 0 µ ¶ ¶ µ 0 X 0 Y 6k 0 ζ( 1+k k +2 1 1 +ε k0 ) 1+k k0 2k0 (SP (n)) k0 = 0 x 1 − . + O x 1 (k + 1)π 2 (1 + p)p k0 p n≤x Specifically, we have µ ¶ 4ζ( 32 ) 3 Y 1 2 (SP (n)) = x 1− + O(x1+ε ); 1 π2 2 (1 + p)p p n≤x X

1 2

(SP (n)) 3 =

n≤x

¶ µ 9ζ( 43 ) 4 Y 5 1 3 x + O(x 6 +ε ). 1 − 1 2π 2 3 (1 + p)p p

Corollary 1.2. For any random real number x ≥ 3, and k = 1, 2, 3. We have the asymptotic formula: µ ¶ X 3 1 Y 1 (SP (n)) = x2 1− + O(x 2 +ε ); 2 p(1 + p) p n≤x µ ¶ X 5 1 6ζ(3) 3 Y x + O(x 2 +ε ); 1 − (SP (n))2 = 2 2 3π p (1 + p) p n≤x

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On a note of the Smarandache power function

µ ¶ 7 π2 4 Y 1 (SP (n)) = x 1− 3 + O(x 2 +ε ). 60 p (1 + p) p 3

n≤x

Theorem 1.2. For any random real number x ≥ 3, we have the asymptotic formula: µ ¶ X ¡ ¢ 1 1 ζ(k + 1) k+1 Y k ϕ (SP (n)) = x 1− + O(xk+ 2 +ε ), k (k + 1)ζ(2) (1 + p)p p n≤x

where ϕ(n) is the Euler function Theorem 1.3. For any random real number x ≥ 3, we have the asymptotic formula: X ¡ ¢ 1 d (SP (n))k = B0 x lnk x + B1 x lnk−1 x + B2 x lnk−2 x + · · · + Bk−1 x ln x + Bk x + O(x 2 +ε ). n≤x

where d(n) is the Dirichlet divisor function and B0 , B1 , B2 , · · · , Bk−1 , Bk is computable constant.

§2. Lemmas and proofs αk 1 α2 Suppose s = σ + it and let n = pα 1 p2 · · · pk , U (n) =

p. Before the proofs of the

p|n

theorem, the following Lemmas will be useful. Lemma 2.1. For any random real number x ≥ 3 and given real number k ≥ 1 , we have the asymptotic formula: ¶ µ X 1 1 ζ(k + 1) k+1 Y x + O(xk+ 2 +ε ). 1− (U (n))k = k (k + 1)ζ(2) (1 + p)p p n≤x

Proof. Let Dirichlet’s series A(s) =

∞ X (U (n))k , ns n=1

for any real number s > 1, it is clear that A(s) is absolutely convergent. Because U (n) is the multiplicative function, if σ > k + 1, so from the Euler’s product formula [5] we have A(s)

∞ X (U (n))k ns n=1 ¶ ∞ Yµ X (U (pm ))k = pms p m=0 ¶ Yµ pk pk = 1 + s + 2s + · · · p p p µ ¶ ζ(s)ζ(s − k) Y 1 = 1− k , ζ(2s − 2k) p p (1 + ps−k )

where ζ(s) is the Riemann zeta-function. Letting R(k) = 1, |U (n)| ≤ n, |

∞ P n=1

Q p

(U (n))k | nσ

< ζ(σ − k).

µ 1−

1 pk (1+ps−k )

. If σ > k +

Wei Huang and Jiaolian Zhao

No. 3 1

Therefore by Perron’s formula [5] with a(n) = (U (n))k , s0 = 0, b = k + 32 , T = xk+ 2 , H(x) = x, B(σ) = ζ(σ − k), then we have X

1 (U (n)) = 2πi

k+ 12 −iT

n≤x

where h(k) =

1 ζ(s)ζ(s − k) xs h(s) ds + O(xk+ 2 +ε ), ζ(2s − 2k) s

¶ 1−

k+ 32 +iT

1 pk (1+p)

To estimate the main term 1 2πi

k+ 32 +iT

k+ 12 −iT

we move the integral line from s = k +

3 2

ζ(s)ζ(s − k) xs h(s) ds, ζ(2s − 2k) s

± iT to k +

1 2

± iT , then the function

ζ(s)ζ(s − k) xs h(s) ζ(2s − 2k) s have a first-order pole point at s = k + 1 with residue ¶ µ ζ(s)ζ(s − k) h(s) L(x) = Res s=k+1 ζ(2s − 2k) µ ¶ ζ(s)ζ(s − k) xs = lim (s − k − 1) h(s) s→k+1 ζ(2s − 2k) s =

ζ(k + 1) k+1 x h(k). (k + 1)ζ(s)

Taking T = xk+ 2 , we can easily get the estimate ¯ ¯ ¯ 1 µ Z k+ 32 +iT Z k+ 32 +iT ¶ ζ(s)ζ(s − k) 1 xs ¯¯ x2k+1 ¯ + h(s) ds¯ ¿ = xk+ 2 , ¯ ¯ 2πi ζ(2s − 2k) s ¯ T k+ 12 +iT k+ 21 −iT ¯ ¯ ¯ 1 Z k+ 12 +iT ζ(s)ζ(s − k) 1 xs ¯¯ ¯ h(s) ds¯ ¿ xk+ 2 +ε . ¯ ¯ 2πi k+ 1 −iT ζ(2s − 2k) s ¯ 2 We may immediately obtain the asymptotic formula µ ¶ X 1 ζ(k + 1) k+1 Y 1 (U (n))k = x 1− + O(xk+ 2 +ε ), k (k + 1)ζ(2) (1 + p)p p n≤x

this completes the proof of the Lemma 2.1. Lemma 2.2. For any random real number x ≥ 3 and given real number k ≥ 1, and positive integer α, then we have X (αp)k ¿ ln2k+2 x. pα ≤x α>p

Proof. Because α > p, so pp < pα ≤ x, then p<

ln x ln x < ln x, α ≤ , ln p ln p

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On a note of the Smarandache power function

also,

nk =

n≤x

xk+1 + O(xk ). Thus, k+1 X

(αp)k =

pα ≤x α>p

p≤ln x

Considering π(x) =

αk ¿ lnk+1 x

x α≤ ln ln p

lnk+1 p p≤ln x

¿ lnk+1 x

pk .

p≤ln x

µ ¶ x x +O . we can get from the Able ln x ln2 x Z x X pk = π(x)xk − k π(t)tk−1 dt.

1, by virtue of [5], π(x) =

p≤x

Therefore X

pk =

p≤ln x

lnk x + O(lnk−1 x) − k (k + 1)

ln x

tk dt + O ln t

µZ

ln x

tk dt ln2 t

¶ =

lnk x + O(lnk−1 x). k+1

Thus X

(αp)k =

pα ≤x α>p

p≤ln x

αk ¿ lnk+1 x

x α≤ ln ln p

pk k+1

ln p≤ln x

¿ lnk+1 x

pk ¿ ln2k+2 x.

p≤ln x

This completes the proof of the Lemma 2.2.

§3. Proof of the theorem In this section, we shall complete the proof of the theorem. © ª αk 1 α2 Proof of Theorem 1.1. Let A = n|n = pα 1 p2 · · · pk , αi ≤ pi , i = 1, 2, · · · , r . When n ∈ A : SP (n) = U (n); When n ∈ N+ : SP (n) ≥ U (n), thus X X X£ X ¤ (SP (n))k − (U (n))k = (SP (n))k − (U (n))k ¿ (SP (n))k . n≤x

n≤x

n≤x SP (n)>U (n)

By the [2] known, there is integer α and prime numbers p, so SP (n) < αp, then we can get according to Lemma 2.2 X X X X (SP (n))k < (αp)k ¿ ¿ x ln2k+2 x. n≤x SP (n)>U (n)

Therefore

n≤x SP (n)>U (n)

(SP (n))k −

n≤x

pα <x α>p

(U (n))k ¿ x ln2k+2 x.

n≤x

From the Lemma 2.1 we have X 1 1 ζ(k + 1) k+1 Y (SP (n))k = x (1 − k ) + O(xk+ 2 +ε ) + O(x ln2k+1 x) (k + 1)ζ(2) p (1 + p) p n≤x

1 ζ(k + 1) k+1 Y 1 x ) + O(xk+ 2 +ε ). (1 − k (k + 1)ζ(2) p (1 + p) p

Wei Huang and Jiaolian Zhao

No. 3

This proves Theorem 1.1. Proof of Corollary. According to Theorem 1.1, taking k = k10 the Corollary 1.1 can be π2 π4 obtained. Take k = 1, 2, 3, and ζ(2) = , ζ(4) = , we can achieve Corollary 1.2. Obviously 6 90 [2] so is theorem . Using the similar method to complete the proofs of Theorem 1.2 and Theorem 1.3.

Acknowledgments The authors express many thanks to the anonymous referees for valuable suggestions and comments.

References [1] F. Smarandache, Collected papers, Bucharest, Tempus Publ. Hse., 1998. [2] Zhefeng Xu, On the mean value of the Smarandache power function, Acta Mathematics Sinica (Chinese series), 49(2006), No. 1, 77-80. [3] Huanqin Zhou, An infinite series involving the Smarandache power function SP (n), Scientia Magna, 2(2006), 109-112. [4] Chengliang Tian and Xiaoyan Li, On the Smarandache power function and Euler totient function, Scientia Magna, 4(2008), No. 1, 35-38. [5] Chengdong Pan and Chengbiao Pan, The Elementary Number Theory, Beijing University Press, Beijing, 2003. [6] Tom M. Apostol, Introduction to Analytic Number Theory, New York, Springer-Verlag, 1976.