ON THE K -POWER PART RESIDUE FUNCTION Yang Hai Research Center for Basic Science, Xi’an Jiaotong University, Xi’an, Shaanxi, P.R.China
Fu Ruiqin School of Science, Xi’an Shiyou University, Xi’an, Shaanxi, P.R.China
Abstract
The main purpose of this paper is using the elementary and analytic methods to study the asymptotic properties of the k-power part residue, and give an interesting asymptotic formula for it.
Keywords:
k-power part residues function; Asymptotic formula.
§1. Introduction and results For any positive integer n, the Smarandache k-th power complements bk (n) is the smallest positive integer such that nbk (n) is a complete k-th power (see probem 29 of [1].) Similar to the Smarandache k-th power complements, the additive k-th power complements ak (n) is defined as the smallest nonnegative integer such that ak (n)+n is a complete k-th power. About this problem, some authors had studied it, and obtained some interesting results. For example, in [4] Xu Z.F. used the elementary method to study the mean value properties of ak (n) and d(ak (n)). in [5] Yi Y. and Liang F.C. used the analytic method to study the mean value properties of d(a2 (n)), and obtained a sharper asymptotic formula for it. Similarly, we will define the k-power part residue function as following: For any positive integer n, it is clear that there exists a positive integer N such that N k ≤ n < (N + 1)k . Let n = N k + r, then fk (n) = r is called the k-power part residue of n. In this paper, we use the elementary and analytic methods to study the asymptotic properties of this sequence, and obtain two interesting asymptotic formulae for it. That is, we shall prove the following: Theorem. For any real number x > 1 and any fixed positive integer m and k, we have the asymptotic formula
X n≤x
δm (fk (n)) =
´ ³ Y p 1 2 k2 x2− k + O x2− k , 2(2k − 1) p|m p + 1
142 where
SCIENTIA MAGNA VOL.1, NO.1 Q
denotes the product over all prime divisors of m, and
p|m
½
δm (n) =
max{d ∈ N
| d|n, (d, m) = 1}, if 0, if
n 6= 0, n = 0.
Especially taking m = 1, and note that δ1 (fk (n)) = fk (n) we may immediately get the following: Corollary . For any real number x > 1 and any fixed positive integer k, we have the asymptotic formula X
fk (n) =
n≤x
³ ´ 1 2 k2 x2− k + O x2− k . 2(2k − 1)
§2. Proof of Theorem In this section, we will complete the proof of Theorem. First we need following Lemma. For any real number x > 1 and positive integer m, we have X
δm (n) =
n≤x
3 x2 Y p + O(x 2 +² ), 2 p|k p + 1
where ² is any positive number. Proof. Let s = σ + it be a complex number and f (s) =
∞ P n=1
δm (n) ns .
Note
that δm (n) ¿ n, so it is clear that f (s) is an absolutely convergent series for Re(s)> 2, by the Euler product formula [2] and the definition of δm (n) we get f (s) =
∞ X δm (n) n=1
ns
=
Y p
=
Y
Ã
!
Ã
δm (p2n ) δm (p) δm (p2 ) 1+ + + · · · + + ··· ps p2s pns
p|m
×
!
δm (p) δm (p2 ) δm (p2n ) 1+ + + · · · + + ··· ps p2s pns
Ã
Y p†m
=
Yµ p|m
×
Ã
p†m
=
p|m
¶
1 1 1 1 + s + 2s + · · · + ns + · · · p p p
Y
Y
!
δm (p) δm (p2 ) δm (p2n ) 1+ + + · · · + + ··· ps p2s pns
Ã
!
p p2 p2n 1 + s + 2s + · · · + ns + · · · p p p
1 1 − p1s
!
Y p†m
Ã
1 1−
1 ps−1
!
143
On the k -power part residue function
= ζ(s − 1)
Y µ ps − p ¶
ps − 1
p|m
where ζ(s) is the Riemann zeta-function and
,
Q p
(1)
denotes the product over all
primes. From (1) and Perron’s formula [3], we have X
1 δm (n) = 2πi n≤x
Z
5 +iT 2 5 −iT 2
ζ(s − 1)
Y µ ps − p ¶ xs
where ² is any positive number. Now we move the integral line in (2) from s = ´ ³ time, the function ζ(s − 1)
Q
ps −p ps −1
p|m
with residue
·
ps − 1
p|m
·
xs s
5 2
s
Ã
ds + O
± iT to s =
5
x 2 +² T 3 2
!
, (2)
± iT . This
has a simple pole point at s = 2
x2 Y p . 2 p|m p + 1
(3)
Hence, we have 1 2πi =
ÃZ
5 −iT 2 3 −iT 2
Z
+
5 +iT 2 5 −iT 2
Z
+
3 +iT 2 5 +iT 2
Z
+
3 −iT 2
!
3 +iT 2
ζ(s − 1)
Y µ ps − p ¶ xs p|m
ps − 1
·
s
x2 Y p . 2 p|m p + 1
(4)
We can easily get the estimate ¯ ¯ ÃZ 5 5 ¯ ¯ µ s ¶ Z 3 +iT ! s −iT Y ¯ 1 ¯ 2 2 x 2 +² p −p x ¯ ¯ , + ζ(s − 1) · ds¯ ¿ ¯ 2πi 3 5 ps − 1 s ¯ T −iT +iT ¯ 2 2 p|m
and
¯ ¯ ¯ Z 3 −iT Y µ ps − p ¶ xs ¯¯ ¯ 1 3 2 ¯ ζ(s − 1) · ds¯¯ ¿ x 2 +² . ¯ 2πi 3 s p −1 s ¯ +iT ¯ 2 p|m
(5)
(6)
Taking T = x, combining (2), (4), (5) and (6) we deduce that X n≤x
ds
δm (n) =
3 x2 Y p + O(x 2 +² ). 2 p|m p + 1
(7)
This completes the proof of Lemma. Now we shall use the above lemma to complete the proof of Theorem. For any real number x ≥ 1, let M be a fixed positive integer such that M k ≤ x < (M + 1)k .
(8)
144
SCIENTIA MAGNA VOL.1, NO.1
Then from (7) and the definition of fk (n), we have X
δm (fk (n))
(9)
n≤x
=
M X
X
t=1 (t−1)k ≤n<tk
=
M −1 X
=
t=1
=
M X
k
(t +
t=1
1)k
−
tk
2
´2
δm (fk (n))
M k ≤n≤x
X
δm (j) + O
j=0
³
X
δm (fk (n)) +
t=1 tk ≤n<(t+1)k M (t+1) X X−t
δm (fk (n))
M k ≤n<x
X
k
X
δm (fk (n)) +
δm (fk (n))
M k ≤n<(M +1)k
³
Y
p + O (t + 1)k − tk p + 1 p|m
Ã
!
Ã
´ 3 +² 2
=
M ³ M ´2 X X 3 1Y p (t + 1)k − tk +O t(k−1)( 2 +²) 2 p|m p + 1 t=1 t=1
=
M M X k2 Y p X t2k−3 t2(k−1) + O 2 p|m p + 1 t=1 t=1
=
³ ´ k 2 M 2k−1 Y p + O M 2k−2 . 2(2k − 1) p|m p + 1
Ã
³ ´ k +O M !
!
(10)
On the other hand, we also have the estimate 0 ≤ x − M k < (M + 1)k − M k ¿ x
k−1 k
.
Now combining (9) and (10) we may immediately obtain the asympotic formula X
δm (fk (n)) =
n≤x
³ ´ Y p 1 2 k2 x2− k + O x2− k . 2(2k − 1) p|m p + 1
This completes the proof of Theorem.
References [1] F. Smarandache, Only Problems, Not Solutions, Chicago: Xiquan Publishing House, 1993. [2] Tom M. Apostol, Introduction to Analytic Number Theory, New York: Springer-Verlag, 1976. [3] Pan Chengdong and Pan Chengbiao, Foundation of Analytic Number Theory, Beijing: Science Press, 1997.