Scientia Magna
Vol. 2 (2006), No. 2, 55-59
On the Product of the Square-free Divisor of a
Natural Number
Yanyan Han
Department of Mathematics, Shandong Normal University
Jinan, Shandong, P.R.China
Abstract In this paper we study the product of the square-free divisor of a natural number
Q
Psd (n) =
d. According to the Dirichlet divisor problem, we turn to study the asymptotic
d|n
µ(d)6=0
P
formula of
log Psd (n). This article uses the hyperbolic summation and the convolution
n≤x
method to obtain a better error term.
Keywords
Square-free number, Dirichlet divisor problem, hyperbolic summation, convolution.
§1. Introduction and main results
F.Smarandache introduced the function Pd (n) :=
Q
d in Problem 25[1] . Now we define a
d|n
similar function Psd (n), which denotes the product of all square-free divisors of n, i.e.,
Y
Psd (n) :=
d.
d|n
µ(d)6=0
In the present paper, we shall prove the following Theorem.
Theorem. We have the asymptotic formula
X
1
3
1
log Psd (n) = A1 x log2 x + A2 x log x + A3 x + O(x 2 exp(−D(log x) 5 (log log x)− 5 ),
n≤x
where A1 , A2 , A3 are constants, D > 0 is an absolute constant.
Rt
Notations. [x] = maxk∈Z {k ≤ x}.ψ(t) = t − [t] − 12 , ψ1 (t) = 0 ψ(u)du. µ(n) is the
M obius function. ε denotes a fixed small positive constant which may be different at each
occurence. γ is the Euler constant. B1 , B2 , B3 , C1 , C2 , D1 , D2 , D3 , D4 are constants.
§2. Some preliminary lemmas
We need the following results:
Lemma 1. Let f (n) be an arithmetical function for which
X
n≤x
f (n) =
l
X
j=1
xaj Pj (log x) + O(xa ),
56
Yanyan Han
X
No. 2
|f (n)| = O(xa1 logr x),
n≤x
1
k
where a1 ≥ a2 ≥ · · · ≥ al > > a ≥ 0, r ≥ 0, P1 (t), · · · , Pl (t) are polynomials in t of degrees
not exceeding r , and k ≥ 1 is a fixed integer. If
X
h(n) =
µi (d)f (n/dk ), i ≥ 1
dk |n
then
X
h(n) =
l
X
xaj Rj (log x) + δ(x),
j=1
n≤x
where R1 (t), · · · , Rl (t) are polynomials in t of degrees not exceeding r. and for some D > 0
1
3
1
δ(x) ¿ x k exp (−D(log x) 5 (log log x)− 5 ).
Proof. See Theorem 14.2 of A. Ivić[3] when l = 1 and the similar proof is used when l > 1
.
Lemma 2. Suppose that f (u) ∈ C 3 [u1 , u2 ], then
Z u2
¯ u2 Z
¯ u2
X
¯
¯
f (n) =
f (u)du − ψ(u)f (u)¯ + ψ1 (u)f 0 (u)¯ −
u1
u1 <n≤u2
u1
u1
u2
ψ1 (u)f 00 (u)du.
u1
Lemma 3. We have
X
1≤m≤y
ψ(y)
1
= log y + γ −
+O
m
y
µ
1
y2
¶
, y ≥ 1.
Lemma 4. We have
X
log m = y log y − y − ψ(y) log y +
1≤m≤y
ψ1 (y)
+ D1 + O
y
µ
1
y2
¶
, y ≥ 1.
Lemma 5. We have
µ
¶
X log m
log y
1
log y
=
−
−
+
D
+
O
, y ≥ 1.
3
m2
y
y
y2
1≤m≤y
Lemma 6. We have
µ 2 ¶
X log2 m
log2 y 2 log y
2
log y
=−
, y ≥ 1.
−
− + D4 + O
m2
y
y
y
y2
1≤m≤y
Lemma 7. We have
X
1
d(n) = y log y + (2γ − 1)y + O(y 3 ).
n≤y
Lemma 8. We have
X
1
1
1
1
d(n)n− 2 = 2y 2 log y + (4γ − 4)y 2 − 2γ + 3 + O(y − 6 ).
n≤y
Vol. 2
57
On the Product of the Square-free Divisor of a Natural Number
Lemma 9. We have
X
1
1
1
1
1
d(n)n− 2 log n = 2y 2 log2 y + (4γ − 8)y 2 log y + (16 − 8γ)y 2 + 8γ − 16 + O(y − 6 log y).
n≤y
Lemma 2 is the Euler-Maclaurin summation formula (see [2]). Lemma 3, 4, 5, 6 follow from
Lemma 2 directly. Lemma 7 is a classical result about the Dirichlet divisor problem. Lemma
8, 9 can be easily obtained by Lemma 2 and Lemma 7.
§3. Proof of the theorem
It is easily seen that
X
log Psd (n) =
log d,
n=dl,µ(d)6=0
which implies that (σ > 1)
∞
X
log Psd (n)
ns
n=1
= ζ(s)
Ã
∞
X
|µ(d)| log d
= ζ(s) −
ds
ds
d=1
µ
¶0
0
0
ζ(s)
1
1
= −ζ(s)
= −ζ(s)ζ (s)
+ 2ζ 2 (s)ζ (2s) 2
ζ(2s)
ζ(2s)
ζ (2s)
∞
∞
X
X
=
h1 (n)n−s + 2
h2 (n)n−s .
where
h1 (n) =
n=1
X
µ(d)f1 (n/d2 ), f1 (n) =
d2 |n
X
!0
d=1
n=1
h2 (n) =
∞
X
|µ(d)|
µ2 (d)f2 (n/d2 ),
X
log m;
m|n
X
f2 (n) =
d(k) log m,
µ2 (d) =
µ(d)µ(k).
n=dk
n=m2 k
d2 |n
X
Our Theorem follows from the following Proposition 1 and Proposition 2.
Proposition 1. We have
X
1
3
1
h1 (n) = B1 x log2 x + B2 x log x + B3 x + O(x 2 exp(−D(log x) 5 (log log x)− 5 ).
n≤x
Proposition 2. We have
X
3
1
1
h2 (n) = C1 x log x + C2 x + O(x 2 exp(−D(log x) 5 (log log x)− 5 ).
n≤x
We only proof Proposition 2. The proof of Proposition 1 is similar and easier. We have
X
f2 (n) =
n≤x
=
X
1
m≤x 3
X X
n≤x m2 k=n
log m
X
n≤ mx2
= S1 + S2 − S3
X
d(k) log m =
d(n) +
X
1
n≤x 3
d(n) log m
m2 n≤x
d(n)
X
log m −
1
x 2
)
m≤( n
X
1
m≤x 3
log m
X
1
n≤x 3
d(n)
58
Yanyan Han
No. 2
By Lemma 7, 5, 6
µ³
¶¶
X µ x
x
x
x ´ 13
S1
=
log 2 + (2γ − 1) 2 + O
log m
m2
m
m
m2
1
m≤x 3


X log m
X log m
 1 X log m 
− 2x
+ O x 3
2 
2
2
m
m
m3
1
1
1
m≤x 3
m≤x 3
m≤x 3
µ
³
´¶
1
2
1 1
= (x log x + (2γ − 1)x) − x− 3 log x − x− 3 + D3 + O x− 3 log x
3
µ
³ 2
´¶
³ 4 ´
1
1 1
2 1
−2x − x− 3 log2 x − x− 3 log x − 2x− 3 + D4 + O x− 3 log x
+ O x 9 +ε
9
3
2
= (x log x + (2γ − 1)x)
By Lemma 4, 8, 9
S2
=
X
1
µ ³ ´1
¶
³
x ³ x ´ 12
x´
1 x 2
log −
+ D1 + O log
d(n)
2 n
n
n
n
n≤x 3
X
1
1
1
1 1
1 1 X
= ( x 2 log x − x 2 )
d(n)n− 2 − x 2
d(n) log n n− 2
2
2
1
1
n≤x 3
n≤x 3

+D1
X
X


d(n) + O log x
d(n)
1
1
n≤x 3
¶µ
n≤x 3
³ 1 ´¶
1
1
2 1
1 1
x 2 log x − x 2
x 6 log x + (4γ − 4)x 6 − 2γ + 3 + O x− 18
2
3
µ
³ 1
´¶
1
1
1 1 2 1
1
− x2
x 6 log2 x + (4γ − 8)x 6 log x + (16 − 8γ)x 6 + 8γ − 16 + O x− 18 log x
2
9
3
³ 1 ´
X
+D1
d(n) + O x 3 +ε
µ
=

1
n≤x 3
By Lemma 4 and Lemma 7,
µ
¶
X
1
1 1
S3
=
d(n)
x 3 log x − x 3 + D1 + O (log x)
3
1
n≤x 3
=
X
X
X
1
1 1
x 3 log x
d(n) − x 3
d(n) + D1
d(n)
3
1
1
1

n≤x 3

n≤x 3
n≤x 3
X


+O log x
d(n)
1
n≤x 3
¶µ
³ 1 ´¶
1
1
1 1
1 1
3
3
3
3
x log x − x
x log x + (2γ − 1)x + O x 9
=
3
3
³ 1 ´
X
+O x 3 +ε + D1
d(n)
µ
1
n≤x 3
Vol. 2
On the Product of the Square-free Divisor of a Natural Number
59
Combining the above estimates we get
X
f2 (n) = D3 x log x + ((2γ − 1)D3 − 2D4 ) x +
n≤x
³ 4 ´
1
3 − 2γ 1
x 2 log x + (5 − 2γ)x 2 + O x 9 +ε ,
2
which gives Proposition 2 immediately by using Lemma 1 .
References
[1] Smarandache F, Only problems, not Solutions, Chicago, Xiquan Publ, House, 1993.
[2] Chengdong Pan and Chengbiao Pan, Foundation of Analytic Number Theory, Science
Press, Beijing, 1997.
[3] A. Ivić, the Riemann zeta-function, Wiley-Interscience, New York, 1985.