Scientia Magna Vol. 5 (2009), No. 1, 124-127
On the Smarandache totient function and the Smarandache power sequence Yanting Yang and Min Fang Department of Mathematics, Northwest University, Xi’an, Shaanxi, P.R.China Abstract For any positive integer n, let SP (n) denotes the Smarandache power sequence. And for any Smarandache sequence a(n), the Smarandache totient function St(n) is defined as ϕ(a(n)), where ϕ(n) is the Euler totient function. The main purpose of this paper is S1 using the elementary and analytic method to study the convergence of the function , where S2 Ã ! 2 ¶2 n n µ X X 1 1 , S2 = , and give an interesting limit Theorem. S1 = St(k) St(k) k=1
k=1
Keywords Smarandache power function, Smarandache totient function, convergence.
§1. Introduction and results 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. That is, Y Y SP (n) = min m : n | mm , m ∈ N, p= p . p|n
p|m
For example, the first few values of SP (n) are: SP (1) = 1, SP (2) = 2, SP (3) = 3, SP (4) = 2, SP (5) = 5, SP (6) = 6, SP (7) = 7, SP (8) = 4, SP (9) = 3, SP (10) = 10, SP (11) = 11, SP (12) = 6, SP (13) = 13, SP (14) = 14, SP (15) = 15, · · · . In reference [1], Professor F.Smarandache asked us to study the properties of SP (n). It is clear that SP (n) is not a multiplicative function. For example, 4, SP (3) = 3, SP (24) = 6 6= SP (3) × SP (8). Y SP (8) =Y But for most n, we have SP (n) = p, where denotes the product over all different prime p|n
p|n
divisors of n. If n = pα , k · pk + 1 ≤ α ≤ (k + 1)pk+1 , then we have SP (n) = pk+1 , where αr 1 α2 0 ≤ k ≤Y α − 1. Let n = pα 1 p2 · · · pr , for all αi (i = 1, 2, · · · , r), if αi ≤ pi , then SP (n) = p. p|n
About other properties of the function SP (n), many authors had studied it, and gave some interesting conclusions. For example, in reference [4], Zhefeng Xu had studied the mean value properties of SP (n), and obtained a sharper asymptotic formula: µ ¶ ´ ³ 3 X 1 Y 1 + O x2 + ² , SP (n) = x2 1− 2 p(p + 1) p n≤x
Vol. 5
On the Smarandache totient function and the Smarandache power sequence
where ² denotes any fixed positive number, and
Y
125
denotes the product over all primes.
p
On the other hand, similar to the famous Euler totient function ϕ(n), Professor F.Russo defined a new arithmetical function called the Smarandache totient function St(n) = ϕ(a(n)), where a(n) is any Smarandache sequence. Then he asked us to study the properties of these functions. At the same time, he proposed the following: S1 Conjecture. For the Smarandache power sequence SP (k), converges to zero as S2 Ã ! 2 µ ¶ n n 2 X X 1 1 n → ∞, where S1 = , S2 = . St(k) St(k) k=1 k=1 In this paper, we shall use the elementary and analytic methods to study this problem, and prove that the conjecture is correct. That is, we shall prove the following: S1 Theorem. For the Smarandache power function SP (k), we have lim = 0, where n→∞ S2 Ã ! 2 ¶2 n µ n X X 1 1 S1 = , S2 = . ϕ(SP (k)) ϕ(SP (k)) k=1
k=1
§2. Some lemmas To complete the proof of the theorem, we need the following two simple Lemmas: Lemma 1. For any given real number ² > 0, there exists a positive integer N (²), such c·n , where c is a constant. that for all n ≥ N (²), we have ϕ(n) ≥ (1 − ²) ln ln n Proof． ． See reference [5]． Lemma 2. For the Euler totient fuction ϕ(n), we have the asymptotic formula µ ¶ X 1 ζ(2)ζ(3) ln n = ln n + A + O , ϕ(k) ζ(6) n k≤n
∞ X
∞
µ (n) X µ2 (n) ln n − is a constant. nϕ(n) n=1 nϕ(n) n=1 Proof． ． See reference [6]．
where A = γ
2
§3. Proof of the theorem In this section, we shall prove our Theorem. We separate all integer k in the interval [1, n] into two subsets A and B as follows: A : the set of all square-free integers. B : the set of other positive integers k such that k ∈ [1, n]\A. So we have X X X 1 1 1 = + . 2 2 (ϕ(SP (k))) (ϕ(SP (k))) (ϕ(SP (k)))2 k≤n
k∈A
k∈B
From the definition of the subset A, we may get X k∈A
∞
X 1 = 2 (ϕ(SP (k)))
X 1 1 µ ¶2 ≤ ¶2 ¿ 1. Yµ Y 1 1 k∈A k 2 k=1 2 1− k 1− p p p|k
p|k
126
Yanting Yang and Min Fang
By Lemma 1, we can easily get
No. 1
X µ2 (k) k = O(ln ln k). Note that = O(1). And if ϕ(k) k2 k≤n
k ∈ B, then we can write k as k = l · m, where l is a square-free integer and m is a square-full X 1 integer. Let S denote , then from the properties of SP (k) and ϕ(k) we have (ϕ(SP (k)))2 k∈B
S≤
X lm≤n l2
Y
p2
p|m
1 Yµ
1−
p|lm
Let U (k) =
Y
p, then
1 p
X
¶2 =
m≤n
X 1 X µ2 (l) l2 m2 2 Y . = O (ln ln n) · 2 2 (lm) 2 2 l ϕ p l≤ n p m≤n m
p|m
X
1 Y
m≤n
p|k
1 Y
p2
=
p|m
X a(k) , where m is a square-full integer and the U 2 (k)
k≤n
p|m
arithmetical function a(k) is defined as follows: 1, if k is a square-full integer; a(k) = 0, otherwise. a(k) is a multiplicative function. According to the Euler product formula (see U 2 (k) reference [3] and [5]), we have
Note that
A(s)
=
∞ X k=1
¶ Yµ a(k) 1 = 1 + . U 2 (k)k s p2+s (ps − 1) p
1 , T ≥ 1, we have ln n µ b ¶ µ µ ¶¶ n ζ(b) ln n a(n) ns + O n min 1, + . A(s) ds + O s T T 2U 2 (n)
From the Perron formulas [5], for b = 1 + Z b+iT X a(k) 1 = U 2 (k) 2πi b−iT
k≤n
Taking T = n, we can get the estimate µ b ¶ µ µ ¶¶ n ζ(b) ln n a(n) O + O n min 1, + = O(ln n). T T 2U 2 (n) ns 1 is analytic in Re s > 0, taking c = , then we have s ln n ! Z b−iT Z c+iT Z c−iT ns ns ns ns A(s) ds + A(s) ds + A(s) ds + A(s) ds = 0. s s s s c−iT b+iT c+iT
Because the function A(s) 1 2πi
ÃZ
b+iT
b−iT
! ÃZ Z b−iT T ns ns dy p Note that A(s) ds = O = O(ln n) and A(s) ds = s s c2 + y 2 c−iT −T ÃZ ! c−iTµ ¶ µ ¶ Z b σ b+iT s X n n 1 a(k) 1 O dσ = O . Similarly, A(s) ds = O . Hence, = 2 (k) T ln n s ln n U c c+iT Z
c+iT
k≤n
O(ln n).
Vol. 5
On the Smarandache totient function and the Smarandache power sequence
127
So X k≤n
Now we come to estimate
1 = O(ln n · (ln ln n)2 ). (ϕ(SP (k)))2 X k≤n
(1)
1 , from the definition of SP (n), we may immediately ϕ(SP (k))
αs 1 α2 get that SP (n) ≤ n. Let n = pα 1 p2 · · · ps denotes the factorization of n into prime powers, β1 β2 then SP (n) = p1 p2 · · · pβs s , where βi ≥ 1. Therefore, we can get that p1β1 −1 (p1 − 1)p2β2 −1 (p2 − 1 −1 2 −1 s −1 1) · · · pβs s −1 (ps − 1) ≤ pα (p1 − 1)pα (p2 − 1) · · · pα (ps − 1), thus ϕ(pβ1 1 pβ2 2 · · · pβs s ) ≤ s 1 2 αs 1 α2 ϕ(pα 1 p2 · · · ps ). That is, ϕ(SP (n)) ≤ ϕ(n), according to Lemma 2, we can easily get µ ¶ X X 1 1 ζ(2)ζ(3) ln n ≥ = ln n + A + O . (2) ϕ(SP (k)) ϕ(k) ζ(6) n k≤n
k≤n
Combining (1) and (2), we obtain n µ X
1 ϕ(SP (k))
¶2
¡ ¢ O ln n · (ln ln n)2 0≤ à !2 ≤ µ µ ¶¶2 −→ 0, as n → ∞. n ζ(2)ζ(3) ln n X 1 ln n + A + O ζ(6) n ϕ(SP (k)) k=1
k=1
This completes the proof of our Theorem.
References [1] F. Smarandache, Only Problems, Not Solutions, Xiquan Publishing House, Chicago, 1993. [2] F. Russo, A set of new Smarandache functions, sequences and conjectures in number theory, American Research Press, USA, 2000. [3] Zhefeng Xu, On the mean value of the Smarandache power function, Acta Mathematica Sinica (Chinese series), 49(2006), No.1, 77-80. [4] Huanqin Zhou, An infinite series involving the Smarandache power function SP (n), Scientia Magna, 2(2006), No.3, 109-112. [5] Tom M. Apostol, Introduction to analytical number theory, Springer-Verlag, New York, 1976. [6] H. L. Montgomery, Primes in arithmetic progressions, Mich. Math. J., 17(1970), 33-39. [7] Pan Chengdong and Pan Chengbiao, Foundation of analytic number theory, Science Press, Beijing, 1997, 98. [8] F. Smarandache, Sequences of numbers involved in unsolved problems, Hexis, 2006. [9] Wenjing Xiong, On a problem of pseudo-Smarandache-squarefree function, Journal of Northwest University, 38(2008), No.2, 192-193. [10] Guohui Chen, An equation involving the Euler function, Pure and Applied Mathematics, 23(2007), No.4, 439-445.