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. Ivi´ c[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. Ivi´ c, the Riemann zeta-function, Wiley-Interscience, New York, 1985.