ON TWO NEW ARITHMETIC FUNCTIONS AND THE K -POWER COMPLEMENT NUMBER SEQUENCES ∗ Xu Zhefeng Department of Mathematics, Northwest University, Xi’an, P.R.China zfxu@nwu.edu.cn
Abstract
The main purpose of this paper is to study the asymptotic property of the kpower complement numbers (where k ≥ 2 is a fixed integer), and obtain some interesting asymptotic formulas.
Keywords:
k-power complement number; Asymptotic formula; Arithmetic function.
§1. Introduction Let k ≥ 2 is a fixed integer, for each integer n, let C(n) denotes the smallest integer such that n × C(n) is a perfect k-power, C(n) is called k-power complement number of n. In problem 29 of reference [1], Professor F. Smarandache asked us to study the properties of the k-power complement number sequences. Let n = pα1 1 pα2 2 · · · pαs s , we define two arithmetic function D(n) and I(n) similar to the derivative and integral function in mathematical analysis as follows: D(n) = D (pα1 1 ) D (pα2 2 ) · · · D (pαs s ) ,
D (pα ) = αpα−1
and I(n) = I (pα1 1 ) I (pα2 2 ) · · · I (pαs s ) ,
I (pα ) =
1 pα+1 . α+1
In this paper, we use the analytic method to study the asymptotic properties of the functions D(n) and I(n) for the k-power complement number sequences, and obtain some interesting asymptotic formulas. That is, we shall prove the following conclusions: ∗ This
work is supported by the N.S.F.(60472068) and the P.N.S.F.of P.R.China .
44
SCIENTIA MAGNA VOL.1, NO.1 Theorem 1. For any real number x ≥ 1, we have the asymptotic formula X
1 D(C(n)) n≤x ³
k k−1 π2
6(k − 1)ζ
=
µ
+O x
´
2k−1 +ε 2k(k−1)
1
· x k−1 Y
Ã
p
¶
!
X p k−2 1 1+ i p + 1 i=1 (k − i)pk−1−i · p k−1
,
where ε denotes any fixed positive number. Theorem 2. For any real number x ≥ 1, we have the asymptotic formula X
I(C(n))d(C(n))
n≤x
Ã
Ã
k X p pk+1−i 6ζ(k(k + 1)) · xk+1 Y 1 1 + − k(k+1) 2 (k+1)i (k + 1)π p + 1 i=2 p p p
=
³
1
!!
´
+O xk+ 2 +ε , where d(n) =
X
1 is the divisor function.
d|n
§2. Proof of the theorems In this section, we shall complete the proof of the theorems. Let f (s) =
∞ X
1 . D(C(n))ns n=1
Because D(n) and C(n) are all multiplicative function, so from the Euler product formula [2] and the definition of D(n) and C(n) we have f (s) =
Yµ p
=
Y
¶
1+
Ã
1+
p
= ζ(ks)
Y p
=
1 1 + + ··· s D(C(p))p D(C(p2 ))p2s Ãk−1 X Ã
i=1
1+
1 (k − i)pk−1−i · pis 1 p(k−1)s Ã
+
k−2 X i=1
1 + ks p
!µ
¶!
1 1 1 + ks + 2ks + · · · p p
1
!
(k − i)pk−1−i · pis
X ζ(ks)ζ((k − 1)s) Y p(k−1)s k−2 1 1 + (k−1)s k−1−i ζ((2k − 2)s) · pis p + 1 i=1 (k − i)p p
!
,
On two new arithmetic functions and the k -power complement numbersequences 1 45
where ζ(s) is Riemann zeta function. Obviously, we have the inequality ¯∞ ¯ ¯X ¯ 1 1 ¯ ¯ ¯ ¯< 1 , σ ¯ D(C(n))n ¯ σ − k−1 n=1
1 | | ≤ 1, D(C(n)) where σ >
1 k−1
is the real part of s. So by Perron formula [3]
X a(n) n≤x
ns0 Z
Ã
!
xs xb B(b + σ0 ) = f (s + s0 ) ds + O s T b−iT µ ¶ µ ¶ log x x 1−σ0 −σ0 +O x H(2x) min(1, ) + O x H(N ) min(1, ) , T ||x|| 1 2iπ
b+iT
where N is the nearest integer to x, kxk = |x − N |. Taking s0 = 0, b = 1+
1 k−1 ,
1 1+ 2k(k−1)
T =x
, H(x) = 1, B(σ) =
1 1 σ− k−1
, we have
X
1 D(C(n)) n≤x =
1 2iπ
Z
1 1+ k−1 +iT
1 1+ k−1 −iT
µ
+O x where R(s) =
Y
2k−1 +ε 2k(k−1)
xs ζ(ks)ζ((k − 1)s) R(s) ds ζ((2k − 2)s) s
¶
, !
Ã
X 1 p(k−1)s k−2 1 + (k−1)s k−1−i · pis p + 1 i=1 (k − i)p
p
.
To calculate the main term 1 2iπ
Z
1 1+ k−1 +iT
1 1+ k−1 −iT
ζ(ks)ζ((k − 1)s)xs R(s)ds, ζ((2k − 2)s)s
we move the integral line from s = 1 + This time, the function f1 (s) =
1 k−1
± iT to s =
1 k
+
1 2k(k−1)
± iT .
ζ(ks)ζ((k − 1)s) xs R(s) ζ((2k − 2)s) s 1
have a simple pole point at s = So we have 1 2iπ
ÃZ
1 1+ k−1 +iT
1 −iT 1+ k−1
Z
+
1 k−1
with residue
1 1 + 2k(k−1) +iT k
1 +iT 1+ k−1
Z
+
k (k−1)ζ ( k−1 )·x k−1 ζ(2)
1 1 + 2k(k−1) −iT k 1 1 + 2k(k−1) +iT k
Z
+
1 ). R( k−1
1 1+ k−1 −iT 1 1 + 2k(k−1) −iT k
!
46
SCIENTIA MAGNA VOL.1, NO.1 ζ(ks)ζ((k − 1)s)xs R(s)ds ζ((2k − 2)s)s
³
=
(k − 1)ζ
k k−1
´
1
· x k−1 Y
ζ(2)
Ã
X p k−2 1 1+ i p + 1 i=1 (k − i)pk−1−i · p k−1
p
!
.
We can easy get the estimate
¿
¯ ¯ ÃZ 1 1 Z 1+ 1 −iT ! ¯ 1 ¯ + 2k(k−1) +iT k k−1 ζ(ks)ζ((k − 1)s)xs ¯ ¯ + R(s)ds¯ ¯ 1 1 1 ¯ 2πi ¯ ζ((2k − 2)s)s 1+ k−1 +iT + −iT k 2k(k−1) ¯ ¯ 1 ¯ ¯ Z 1+ 1 ¯ ζ(k(σ + iT ))ζ((k − 1)(σ + iT )) k−1 x1+ k−1 ¯¯ ¯ R(s) dσ ¯ 1 1 ζ((2k − 2)(σ + iT )) T ¯¯ + 2k(k−1) ¯ k 1
¿
1 1 x1+ k−1 + = x k 2k(k−1) T
and ¯ ¯ 1 ¯ 1 Z k1 + 2k(k−1) ¯ −iT ζ(ks)ζ((k − 1)s)xs ¯ ¯ R(s)ds¯ ¯ ¯ 2πi 1 + 1 +iT ¯ ζ((2k − 2)s)s k 2k(k−1) ¯ ¯ 1 ¯ Z T ¯ ζ(1 + 1 + ikt)ζ( 2k−1 + i(k − 1)t) k1 + 2k(k−1) ¯ ¯ x 2k 2(k−1) ¯ ¯ dt ¿ ¯ ¯ t ζ( 2k−1 + i(2k − 2)t) 0 ¯ ¯ k 1
¿ xk
1 + 2k(k−1) +ε
Note that ζ(2) = X
1 D(C(n)) n≤x
π2 6 ,
.
we have 1
=
k 6(k−1)ζ ( k−1 )·x k−1 2 π
Ã
Q p
1+ µ
p p+1 2k−1
Pk−2
+O x 2k(k−1)
i=1
+ε
¶
! 1
i
(k−i)pk−1−i ·p k−1
.
This completes the proof of Theorem 1. Let g(s) =
∞ X
I(C(n))d(C(n))
n=1
from the definition of I(n) and C(n), we can also have g(s) =
Y p
=
Y p
!
Ã
I(C(p))d(C(p)) I(C(p2 ))d(C(p2 )) + + ··· 1+ ps p2s
Ã
1+
Ã
pk pk−1 p + + · · · + ks s 2s p p p
!µ
¶!
1 1 1 + ks + 2ks + · · · p p
On two new arithmetic functions and the k -power complement numbersequences 2 47
= ζ(ks)
Y p
=
Ã
1 pk pk−1 p 1 − ks + s + 2s + · · · + ks p p p p Ã
Ã
!
k X ζ(ks)ζ(s − k) Y ps−k pk+1−i 1 1 + s−k − ks is ζ(2s − 2k) p p + 1 i=2 p p
!!
.
Now by Perron formula [3] and the method of proving Theorem 1, we can also obtain Theorem 2.
Reference [1] F. Smarndache, Only Problems, Not Solution, Xiquan Publishing House, Chicago, 1993, pp. 26. [2] Tom M. Apostol, Introduction to Analytic Number Theory, SpringerVerlag, New York, 1976. [3] Pan Chengdong and Pan Chengbiao, Foundation of Analytic Number Theory, Science Press, Beijing, 1997, pp. 98.