Some Arithmetic Functions Involving Exponential Divisors

1 2 Journal of Integer Sequences, Vol. 13 (2010), 3 Article 10.3.7 47 6 23 11

Xiaodong Cao Department of Mathematics and Physics Beijing Institute of Petro-Chemical Technology Beijing, 102617 P. R. China [email protected]

Wenguang Zhai School of Mathematical Sciences Shandong Normal University Jinan, 250014 Shandong P. R. China [email protected]

Abstract

In this paper we study several arithmetic functions connected with the exponential divisors of integers. We establish some asymptotic formulas under the Riemann hypothesis, which improve previous results. We also prove some asymptotic lower bounds.

1 Introduction and results

a1 ar Suppose n> 1 is an integer with prime factorization n = p1 ··· pr . An integer d is called an b1 br exponential divisor (e-divisor) of n if d = p1 ··· pr with bj|aj(1 ≤ j ≤ r), which is denoted by the notation d|en. For convenience 1|e1. The properties of the exponential divisors attract the interests of many authors(see, for example, [4, 6, 7, 9, 11, 12, 14, 15, 16, 19, 20, 21, 22]).

1 a1 ar An integer n = p1 ··· pr is called exponentially squarefree (e-squarefree) if all the ex- ponents a1, ··· ,ar are squarefree. The integer 1 is also considered to be an e-squarefree number. a1 ar Suppose f and g are two arithmetic functions and n = p1 ··· pr . Subbarao [17] ﬁrst introduced the exponential convolution (e-convolution) by

b1 br c1 cr (f ⊙ g)(n)= ··· f(p1 ··· pr )g(p1 ··· pr ), b1Xc1=a1 brXcr=ar which is an analogue of the classical Dirichlet convolution. The e-convolution ⊙ is com- mutative, associative and has the identity element µ2, where µ is the Möbius function. Furthermore, a function f has an inverse with respect to ⊙ iff f(1) =6 0 and f(p1 ··· pr) =6 0 for any distinct primes p1, ··· ,pr. The inverse of the constant function f(n) ≡ 1 with respect to ⊙ is called the exponential (e) (e) 2 analogue of the Möbius function and it is denoted by µ . Hence d|en µ (d) = µ (n) for n ≥ 1, µ(e)(1) = 1, and µ(e)(n) = µ(a ) ··· µ(a ) for n = pa1 ··· par > 1. Note that 1 r 1 P r |µ(e)(n)| = 1 or 0, according as n is e-squarefree or not. An integer d is called an exponential squarefree exponential divisor (e-squarefree e- b1 br divisor) of n if d = p1 ··· pr with bj|aj(1 ≤ j ≤ r) and b1, ··· , br are squarefree. Observe that the integer 1 is an e-squarefree and it is not an e-divisor of n > 1. Let t(e) denote the number of e-sqaurefree e-divisors of n. Now we introduce some notation for later use. As usual, let µ(n) and ω(n) denote the Möbius function , and the number of distinct prime factors of n respectively. If t is real, then {t} denotes the fractional part of t,ψ(t) = {t} − 1/2. Throughout this paper, ε is a small fixed positive constant and m ∼ M means that cM ≤ m ≤ CM for some constants

(e) 1/2 ∆ µ (n)= A1x +(x exp(−c1(log x) )), (1) n≤x X where 0 < ∆ < 9/25 and c1 > 0 are constants and

∞ µ(k) − µ(k − 1) A := m(µ(e))= 1+ . 1 pk p ! Y Xk=2 T´oth [20] proved that if the Riemann hypothesis (RH) is true, then

(e) 91/202+ε µ (n)= A1x + O(x ). (2) n≤x X e Subbarao [17] and Wu [22] studied the asymptotic properties of the sum n≤x |µ (n)|. T´oth [20] proved that if RH is true, then P (e) 1/5+ε |µ (n)| = B1x + O(x ), (3) n≤x X 2 where ∞ µ2(a) − µ2(a − 1) B := (1 + ). 1 pa p α=4 Y X For the function t(e)(n), T´oth [20] proved the asymptotic formula

(e) 1/2 1/4+ε t (n)= C1x + C2x + O(x ), (4) n≤x X where ∞ 2ω(k) − 2ω(k−1) C := 1+ , 1 pk p ! Y Xk=2 1 ∞ 2ω(k) − 2ω(k−1) − 2ω(k−2) + 2ω(k−3) C := ζ( ) 1+ . 2 2 pk p ! Y Xk=4 P´etermann [13] proved the formula (4) with a better error term O(x1/4), which is the best unconditional result up to date. In order to further reduce the exponent 1/4, we have to know more information about the distribution of the non-trivial zeros of the Riemann zeta-function. In this short paper we shall prove the following theorem. Theorem 1. If RH is true, then

(e) 37 +ε µ (n) = A1x + O x 94 , (5) n≤x X (e) 1 38 +ε |µ (n)| = B1x + B2x 5 + O(x 193 ), (6) n≤x X (e) 1/2 3728 +ε t (n) = C1x + C2x + O x 15469 , (7) n≤x X where B2 is a computable constant. Remark 1. Numerically, we have 37 91 = 0.39361 ··· , = 0.45049 ··· , 94 202 38 3728 = 0.1968 ··· < 1/5, = 0.240 ··· < 1/4. 193 15469

Let ∆µ(e) (x), ∆|µ(e)|(x), ∆t(e) (x) denote the error terms in (5), (6) and (7) respectively. We have the following result. Theorem 2. We have

1/4 ∆µ(e) (x) =Ω(x ), (8) 1/8 ∆|µ(e)|(x) =Ω(x ), (9) 1/6 ∆t(e) (x) =Ω(x ). (10)

3 2 Some generating functions

In this section we shall study the generating Dirichlet series of the functions µ(e)(n), |µ(e)(n)| and t(e)(n), respectively. We consider only |µ(e)(n)| and t(e)(n), since T´oth [20] already proved the following formula, which is enough for our purpose,

∞ µ(e)(n) ζ(s) = U(s) (ℜs> 1), (11) ns ζ2(2s) n=1 X ∞ u(n) where U(s) := n=1 ns is absolutely convergent for ℜs> 1/5. We ﬁrst consider the function |µ(e)(n)|. The function µ(e) is multiplicative and µ(e)(pa)= µ(a) for every primeP power pa, namely for every prime p, µ(e)(p) = 1,µ(e)(p2)= −1,µ(e)(p3)= −1,µ(e)(p4) = 0,µ(e)(p5)= −1,µ(e)(p6) = 1,µ(e)(p7) = 1,µ(e)(p8) = 0,µ(e)(p9) = 0,µ(e)(p10)= 1,µ(e)(p11)= −1, ··· . Hence by the Euler product we have for ℜs> 1 that

∞ |µ(e)(n)| ∞ |µ(m)| = 1+ . (12) ns pms n=1 p m=1 ! X Y X Applying the product representation of Riemann zeta-function

ζ(s)= (1 + p−s + p−2s + ··· )= (1 − p−s)−1 (ℜs> 1), (13) p p Y Y we have for ℜs> 1

−1 ζ(s)ζ(5s)= (1 − p−s)(1 − p−5s) . (14) p Y Let ∞ m f|µ(e)|(z): =1+ |µ(m)|z (15) m=1 X ∞ =1+ z + z2 + z3 + z5 + z6 + z7 + z10 + z11 + |µ(m)|zm. m=12 X For |z| < 1, it is easy to verify that

5 f|µ(e)|(z)(1 − z)(1 − z ) (16) ∞ = 1+ z + z2 + z3 + z5 + z6 + z7 + z10 + z11 + |µ(m)|zm (1 − z − z5 + z6) m=12 ! ∞ X 4 8 9 (1) m = 1 − z − z + z + cm z , m=12 X where

(1) cm := |µ(m)| − |µ(m − 1)| − |µ(m − 5)| + |µ(m − 6)|. (m ≥ 12)

4 Furthermore, we have

5 4 −1 8 −1 f|µ(e)|(z)(1 − z)(1 − z )(1 − z ) (1 − z ) (17) ∞ 4 8 9 (1) m 4 8 8 16 = 1 − z − z + z + cm z (1 + z + z + ··· )(1 + z + z + ··· ) m=12 ! ∞ X 9 (1) m =1+ z + Cm z . m=12 X We get from (12), (13), (15) and (17) by taking z = p−s that

∞ |µ(e)(n)| ζ(s)ζ(5s) = V (s) (ℜs> 1), (18) ns ζ(4s)ζ(8s) n=1 X ∞ v(n) 1 where V (s) := n=1 ns is absolutely convergent for ℜs> 9 . We now consider the function t(e)(n), which is also multiplicative and t(e)(pa) = 2ω(a) for every prime powerP pa. Therefore for every prime p, t(e)(p) = 1,t(e)(p2)= t(e)(p3)= t(e)(p4)= t(e)(p5) = 2,t(e)(p6) = 4,t(e)(p7)= t(e)(p8)= t(e)(p9) = 2,t(e)(p10) = 4, ··· . By the Euler product we have for ℜs> 1 that

∞ t(e)(n) ∞ 2ω(m) = 1+ . (19) ns pms n=1 p m=1 ! X Y X Let ∞ ω(m) m ft(e) (z):=1+ 2 z (20) m=1 X ∞ =1+ z + 2z2 + 2z3 + 2z4 + 2z5 + 4z6 + 2z7 + 2z8 + 2z9 + 4z10 + 2z11 + 2ω(m)zm. m=12 X For |z| < 1, it is easy to check that

2 6 2 ft(e) (z)(1 − z)(1 − z )(1 − z ) ∞ 4 7 8 9 (2) m = 1 − z − 2z − 2z + 2z + cm z m=10 X and

2 6 2 4 −1 ft(e) (z)(1 − z)(1 − z )(1 − z ) (1 − z ) (21) ∞ 4 7 8 9 (2) m 4 8 = 1 − z − 2z − 2z + 2z + cm z (1 + z + z + ··· ) m=10 ! ∞ X 7 (2) m = 1 − 2z + Cm z . m=8 X 5 Combining (13) and (19)–(21) we get

∞ t(e)(n) ζ(s)ζ(2s)ζ2(6s) = W (s) (ℜs> 1), (22) ns ζ(4s) n=1 X ∞ w(n) 1 where W (s) := n=1 ns is absolutely convergent for ℜs> 7 . P 3 Proof of Theorem 2

(e) ζ(s) We see from (11) that the generating Dirichlet series of the function µ (n) is ζ2(2s) U(s), 1 which has infinitely many poles on the line ℜs = 4 , whence the estimate (8) follows. From the expression (18) we see that the generating Dirichlet series of the function |µ(e)(n)| is ζ(s)ζ(5s) 1 ζ(4s) V (s), which has infinitely many poles on the line ℜs = 8 , whence the estimate (9) follows. Via (22), we get the estimate (10) with the help of Theorem 2 of Küleitner and Nowak [8] (or by Balasubramanian, Ramachandra and Subbarao’s method in [1]).

4 The Proofs of (5) and (7)

We follow the approach of Theorem 1 in Montgomery and Vaughan [10]. Throughout this section, we assume RH. We ﬁrst prove (5). It is well-known that the characteristic function of the set of squarefree integers is

µ2(n)= |µ(n)| = µ(d), (23) 2 Xd |n We write x x ∆(x) := µ2(d) − := D(x) − . (24) ζ(2) ζ(2) n≤x X

Deﬁne the function a1(n) by

2 a1(n)= µ (m)µ(d), (25) 2 mdX=n it is easy to see that

∞ µ2(n) ∞ ζ(s) s a (n) = n=1 n = 1 , ℜs> 1, (26) ζ2(2s) ζ(2s) ns P n=1 X Thus x T (x) := a (n)= µ2(m)µ(d)= µ(d)D( ). (27) 1 1 d2 n≤x md2≤x 1 X X dX≤x 2 6 1/2 Suppose 1

T1(x)= S1(x)+ S2(x), (28) where x S (x)= µ(d)D , (29) 1 d2 Xd≤y and x S (x)= µ(d)D = µ2(m)µ(d). (30) 2 d2 1 2 yy

We ﬁrst evaluate S1(x). From (24) we get

x x S (x) = µ(d) +∆ (31) 1 d2ζ(2) d2 Xd≤y x µ(d) x = + µ(d)∆ . ζ(2) d2 d2 Xd≤y Xd≤y

To treat S2(x), we let µ(d) g (s)= ζ−1(s) − , (s = σ + it), (32) y ds Xd≤y so that for ℜs = σ > 1 µ(d) g (s)= . (33) y ds Xd>y Hence g (2s)ζ(s) ∞ b (n) y = 1 , (34) ζ(2s) ns n=1 X for σ > 1, where

2 b1(n)= µ (m)µ(d) (35)

md2=n Xd>y

1 ε Let 2 <σ< 2, δ = 10 . Assume RH, from 14.25 in Titchmarsh [18] we have

µ(d) −1 1 −σ+δ δ = ζ (s)+ O y 2 (|t| + 1) . (36) ds Xd≤y 7 δ −1 δ 1 In addition, RH implies that ζ(s) ≪ |t| +1 and ζ (s) ≪ (|t| + 1) uniformly for σ > 2 + δ 1 2 +δ and |s − 1| >ε, and n≤y µ(n) ≪ y . From (32), (33) and (36), we have

P − 1 δ 1 g (2s) ≪ y 2 (|t| + 1), (σ ≥ + δ). y 2 Thus

ζ(s) − 1 3δ 1 g (2s) ≪ y 2 (|t| + 1), (σ ≥ + δ, |s − 1| >ε). (37) y ζ(2s) 2 From (30), (34), (35) and Perron’s formula([18], Lemma 3.19), we obtain

1+ε+ix2 1 ζ(s) s −1 δ S2(x)= b1(n)= gy(2s) x s ds + O(x ), (38) 2πi 2 ζ(2s) n≤x 1+ε−ix X Z δ 1 since b(n) ≪ n by a divisor argument. If we move the line of integration to σ = 2 + δ, then by the residue theorem

1+ε+ix2 1 ζ(s) s −1 gy(2s) x s ds (39) 2πi 2 ζ(2s) Z1+ε−ix ζ(s) s −1 = I1 + I2 − I3 + Res gy(2s) x s , s=1 ζ(2s) where

2 1 2 1+ε+ix 2 +δ+ix 1 ζ(s) s −1 1 ζ(s) s −1 I1 = gy(2s) x s ds,I2 = gy(2s) x s ds, 2πi 1 +δ+ix2 ζ(2s) 2πi 1 +δ−ix2 ζ(2s) Z 2 Z 2 1+ε−ix2 1 ζ(s) s −1 I3 = gy(2s) x s ds. 2πi 1 +δ−ix2 ζ(2s) Z 2 From (37) it is not diﬃcult to see that

− 1 1 +8δ Ij ≪ y 2 x 2 , (j = 1, 2, 3). (40) Combining (38)–(40), we get

x µ(d) − 1 1 +ε ε S (x)= + O y 2 x 2 + x . (41) 2 ζ(2) d2 Xd>y Finally, combining (27), (28), (31) and (41) we obtain

x x 1 +ε − 1 ε T (x)= + µ(d)∆ + O x 2 y 2 + x . (42) 1 ζ2(2) d2 Xd≤y 17 +ε In [5], Jia proved the estimate ∆(u) ≪ u 54 . Inserting this estimate into (42) and on 10 taking y = x 57 , we get

x 37 +ε T (x)= + O x 94 . (43) 1 ζ2(2) 8 The asymptotic formula (5) follows form (25)-(27)and(43) by the well-known convolution method. Now we prove the asymptotic formula (7). We deﬁne the functions a2(n) by the following identity

ζ(s)ζ(2s) ∞ a (n) = 2 , ℜs> 1. (44) ζ(4s) ns n=1 X Hence a2(n)= µ(d)d(1, 2; m). 4 dXm=n 1 Thus by the same approach as (42), we easily get that for 1 ≤ y ≤ x 4

T2(x) := a2(n) (45) n≤x X 1 ζ(2) ζ( 2 ) 1 x 1 +ε − 3 ε = x + x 2 + µ(d)∆ 1, 2; + O x 2 y 2 + x . ζ(4) ζ(2) d4 Xd≤y 1057 +ε Graham and Kolesnik [2] proved that ∆(1, 2; u) ≪ u 4785 . Inserting this estimate into 2671 (45) and on taking y = 15469 , we get

1 ζ(2) ζ( 2 ) 1 3728 +ε T (x)= a (n)= x + x 2 + O x 15469 . (46) 2 2 ζ(4) ζ(2) n≤x X The asymptotic formula (7) follows from (22), (44) and (46) by the convolution method.

5 Proof of (6)

Throughout this section we assume RH. Deﬁne the function a3(n) by

ζ(s)ζ(5s) ∞ a (n) = 3 , ℜs> 1. (47) ζ(4s) ns n=1 X 1 Similar to (45) we have for some 1 ≤ y ≤ x 4 that

T3(x) := a3(n) n≤x X 1 ζ(5) ζ( ) 1 x 1 3 1 3 5 5 2 +ε − 2 5 +ε − 10 ε = x + 4 x + µ(d)∆ 1, 5; 4 + O x y + x y + x . ζ(4) ζ( 5 ) d Xd≤y 1 Unfortunately, we can not improve T´oth’s exponent 5 in (3) by the above formula even if we use the conjectural bound ∆(1, 5; t) ≪ t1/12+ε. So we use a diﬀerent approach to prove (6).

9 Let q4(n) denote the characteristic function of the set of 4-free numbers, then

∞ q (n) ζ(s) 4 = , ℜs> 1. ns ζ(4s) n=1 X We write x ∆ (x) := q (n) − . (48) 4 4 ζ(4) n≤1 X From (47) and (48), it follows from the Dirichlet hyperbolic approach that for some 1 1 ≤ y ≤ x 5 ,

B3(x)= a3(n)= q4(m) (49) n≤x 5 X mdX≤x = q4(m)+ q4(m) 1 x x d≤y m≤ m≤ x 1 d5 y5 5 X X X ysummation formula, we get

q4(m) 1 (50) 5 m≤ x m Xy5 x x q4(m) 5 m≤ 5 1 y q4(m) = y + m≤t dt x 1 6 P ( ) 5 5 1 t 5 y5 Z P x t y 1 y5 ζ(4) +∆4(t) = 1 q4(m)+ 6 dt 5 5 5 x m≤ x 1 t Xy5 Z 4 x y 1 x 5 1 y5 ∆4(t) = 1 q4(m)+ 4 − 1 + 6 dt. 5 4ζ(4) y 5 5 x m≤ x ! 1 t Xy5 Z In addition, we have by Euler-Maclaurin formula 1 1 = ζ(5) − − ψ(y)y−5 + O(y−6). (51) d5 4y4 Xd≤y If RH is true, Graham and Pintz [3] showed that

7 +ε ∆4(x)= O(x 38 ), (52)

10 and so x ∞ ∞ y5 ∆4(t) ∆4(t) ∆4(t) 6 dt = 6 dt − 6 dt (53) 1 t 5 1 t 5 x t 5 Z Z Z y5 ∞ ∆4(t) 3 3 − 190 +ε 38 = 6 dt + O x y . 1 t 5 Z Combining (49)–(53) and taking on y = x31/193, we obtain

T3(x) = q4(m) (54) md5≤x X ∞ ζ(5) 1 ∆4(t) 1 1 5 = x + 6 dt − x ζ(4) 5 5 4ζ(4) Z1 t x 1 7 +ε 3 −6 − q (m)ψ ( ) 5 + O x 38 y 38 + xy 4 m m≤ x Xy5 ζ(5) 1/5 7 +ε 3 −5 = x + C x + O(x 38 y 38 + xy ), ζ(4) 3

ζ(5) 1/5 38 +ε = x + C x + O(x 193 ), ζ(4) 3 where we used the trivial estimate x 1 x q (m)ψ ( ) 5 ≪ 4 m y5 m≤ x Xy5 and where ∞ 1 ∆4(t) 1 C3 := 6 dt − . 5 5 4ζ(4) Z1 t Now the asymptotic formula (6) follows from (18), (47), (49) and (54) by the convolution approach.

6 Acknowledgments

The authors express their gratitude to the referee for a careful reading of the manuscript and many valuable suggestions, which highly improve the quality of this paper. This work is supported by National Natural Science Foundation of China (Grant No. 10771127) and Research Award Foundation for Young Scientists of Shandong Province (No. BS2009SF018).

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2000 Mathematics Subject Classiﬁcation: Primary 11N37. Keywords: exponential divisor, error term, asymptotic lower bound.

Received January 29 2010; revised version received March 10 2010. Published in Journal of Integer Sequences, March 12 2010.

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