Some identities on k-power complement

Scientia Magna Vol. 2 (2006), No. 2, 60-63

Some identities on k-power complement Pei Zhang Department of Mathematics, Northwest University Xi’an, Shaanxi, P.R.China Abstract

The main purpose of this paper is to calculate the value of the series +∞ X n=1

(−1)n , nα · aβk (n)

where ak (n) is the k-power complement number of any positive number n, and α, β are two complex numbers with Re(α) ≥ 1, Re(β) ≥ 1. Several interesting identities are given.

Keywords

k-power complement number, identities, Riemann zeta-function.

§1. Introduction For any given natural number k ≥ 2 and any positive integer n, we call ak (n) as a kpower complement number if ak (n) denotes the smallest positive integer such that n · ak (n) is a perfect k-power. Especially, we call a2 (n), a3 (n), a4 (n) as the square complement number, cubic complement number, quartic complement number respectively. In reference [1], Professor F.Smarandache asked us to study the properties of the k-power complement number sequence.

About this problem, there are many authors had studied it, and obtained many results. For example, in reference [2], Professor Wenpeng Zhang calculated the value of the series +∞ X

1 , s (n · a k (n)) n=1 where s is a complex number with Re(α) ≥ 1, k=2, 3, 4. Maohua Le [3] discussed the convergence of the series +∞ X 1 s1 = m (n) a n=1 2 and

+∞ X (−1)n , s2 = a (n) n=2 2

where m ≤ 1 is a positive number, and proved that they are both divergence.

But about the properties of the k-power complement number, we still know very little at present. This paper, as a note of [2], we shall give a general calculate formula for +∞ X

(−1)n

n=1

nα · aβk (n)

.

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Some identities on k-power complement

That is, we shall prove the following: Theorem 1.

For any complex numbers α, β with Re(α) ≥ 1, Re(β) ≥ 1, we have +∞ X

1

n=1

nα · aβk (n)

= ζ(kα)

Y

à 1+

p

where ζ(α) is the Riemann zeta-function,

Y

1 p(k−1)α+(k−1)2 β pα+(k−1)β − 1

1−

! ,

denotes the product over all prime p.

p

Theorem 2. +∞ X

(−1)n

n=1

nα · aβk (n)

For any complex numbers α, β with Re(α) ≥ 1, Re(β) ≥ 1, we have µ

=

2(2kα − 1)(2α+(k+1)β − 1) 1 − (k+1)α+(k−1)β 2 − 2α−(k−1)2 β 4

2

Note that ζ(2) = π6 , ζ(4) = π90 and ζ(8) = obtain the following two corollaries: Corollary 1.

π8 9450 .

¶ ζ(kα)

Y

à 1+

p

Taking α = β, k = 2 in above Theorems, then we have 1 ζ 2 (2α) = ; α (n · a2 (n)) ζ(4α) n=1 +∞ X

1 ζ 2 (2α) 4α − 1 · ; = α (n · a2 (n)) ζ(4α) 4α + 1 n=1 2-n

+∞ X

(−1)n ζ 2 (2α) 3 − 4α = · . α (n · a2 (n)) ζ(4α) 1 + 4α n=1 Taking α = β = 1, 2, k = 2 in Corollary 1, we have +∞ X

1 5 = , n · a2 (n) 2 n=1

+∞ X

1 7 = ; 2 (n · a2 (n)) 6 n=1

+∞ X

1 3 = , n · a2 (n) 2 n=1

35 1 = ; 2 (n · a2 (n)) 34 n=1

2-n

2-n

+∞ X (−1)n 1 =− , n · a (n) 2 2 n=1

+∞ X

+∞ X

! .

From our Theorems we may immediately

+∞ X

Corollary 2.

1 p(k−1)α+(k−1)2 β pα+(k−1)β − 1

1−

91 (−1)n =− .

2 (n · a (n)) 102 2 n=1

62

Pei Zhang

No. 2

§2. Proof of the theorem In this section, we will complete the proof of the theorems. For any positive integer n, we can write it as n = mk · l, where l is a k-free number, then from the definition of ak (n) we have X µ(d) +∞ +∞ X +∞ X X 1 dk |l = β kα α (k−1)β α n=1 n · ak (n) m=1 l=1 m l l X µ(d) +∞ X dk |l = ζ(kα) lα+(k−1)β l=1 µ ¶ Y 1 1 1 = ζ(kα) 1 + α+(k−1)β + 2(α+(k−1)β) + · · · + (k−1)(α+(k−1)β) p p p p ! à 1 Y 1 − p(k−1)(α+(k−1)β) 1 = ζ(kα) 1 + α+(k−1)β 1 p 1 − pα+(k−1)β p à !

1 1 − p(k−1)α+(k−1) Y 2β = ζ(kα) 1+ , pα+(k−1)β − 1 p where µ(n) denotes the M¨ obius function. This completes the proof of Theorem 1.

Now we come to prove Theorem 2. First we shall prove the following identity X µ(d) +∞ +∞ +∞ X X X 1 dk |l = β α mkα lα l(k−1) n=1 n · ak (n) m=1 l=1 2-mk l

2-n

X

µ(d) +∞ 1 X dk |l = mkα lα+(k−1) m=1 l=1 +∞ X

2-m

2-l

à !

1 1 − p(k−1)α+(k−1) Y 2β ζ(kα)(2α+(k−1)β − 1) 2kα − 1 · α+(k−1)β 1+ 2kα 2 − 2(k−1)(α+(k−1)β) p pα+(k−1)β − 1 Ã !

1 α 1 − p(k−1)α+(k−1) 2β ζ(kα)(2k − 1)(2α+(k−1)β ) Y 1+ .

2(k+1)α+(k−1)β − 2α−(k−1)2 β p pα+(k−1)β − 1

=

=

Then use this identity and Theorem 1 we have +∞ X

=

n=1 +∞ X n=1

µ =

(−1)n nα · aβk (n) 1 nα

·

aβk (n)

−2

+∞ X n=1 2-n

1 nα

· aβk (n)

2(2kα − 1)(2α+(k−1)β − 1) 1 − (k+1)α+(k−1)β 2 − 2(k−1)2 β−α

¶ ζ(kα)

Y p

à 1+

1 p(k−1)α+(k−1)2 β pα+(k−1)β − 1

1−

! .

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Some identities on k-power complement

63

This completes the proof of Theorem 2.

References [1] F.Smarandache, Only problems, Not solutions, Xiquan Publishing House, Chicago, 1993.

[2] Zhang Wengpeng, Research on Smarandache Problems in Number Theory, Hexis, 2004, 60-64.

[3] Maohua Le, Some Problems Concerning the Smarandache Square Complementary Function, Smarandache Notions Journal, 14(2004), 220-222.