ON TWO NEW ARITHMETIC FUNCTIONS AND THE K-POWER COMPLEMENT NUMBER SEQUENCES

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.