ON THE INTEGER PART OF A POSITIVE INTEGER’S K-TH ROOT

ON THE INTEGER PART OF A POSITIVE INTEGER’S K -TH ROOT 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 method and analytic method to study the asymptotic properties of the integer part of the k-th root positive integer, and give two interesting asymptotic formulae.

Keywords:

k-th root; Integer part; Asymptotic formula.

§1. Introduction And Results For any positive integer n, let s(n) denote the integer part of k-th root of n.

For example, s(1) = 1, s(2) = 1, s(3) = 1, s(4) = 1, · · ·, s(2k ) = 1, s(2k + 1) = 1, · · ·, s(3k ) = 1, · · ·. In problem 80 of [1], Professor F.Smarandache asked us to study the properties of the sequence s(n). About this problem, it seems that none had studied it, at least we have not seen related paper before.

In this paper, we use the elementary method and analytic method to study the asymptotic properties of this sequence, and obtain two interesting asymptotic formulae. That is, we shall prove the following: Theorem 1. For any real number x > 1, we have the asymptotic formula X

µ

¶

x Ω(s(n)) = x ln ln x + (A − log k) x + O , ln x n≤x where Ω(n) denotes the total number of prime divisors of n, A is a constant.

Theorem 2. Let m be a fixed positive integer and ϕ(n) be the Euler totient function, then for any real number x ≥ 1, we have the asymptotic formula X

³

1

´

ϕ((s(n), m)) = h(m)x + (k + 1)h(m) + O x1− 2k +ε ,

n≤x

where (s(n), m) denotes the greatest common divisor of s(n) and m, h(m) = ϕ(m) Q (1 + α − α−1 m p ), and ε is any positive number. pα km

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§2. Some Lemmas To complete the proof of the theorems, we need the following two simple lemmas.

Lemma 1. For any real number x > 1, then we have µ ¶ X x Ω(n) = x log log x + Ax + O , log x n≤x where A = γ +

P

P

p

p

(log(1 − p1 ) + p1 ) +

1 p(p−1) ,

γ is the Euler constant.

Proof. See reference[2].

Lemma 2. Let m be a fixed positive integer and ϕ(n) be the Euler totient function, then for any real number x ≥ 1, we have the asymptotic formula X

³

´

1

ϕ((m, n)) = x · h(m) + O x 2 +ε ,

n≤x

where (m, n) denotes the greatest common divisor of m and n, h(m) = ϕ(m) Q (1 + α − α−1 m p ), and ε is any positive number. pα km

Proof. Let F (s) =

∞ X ϕ((m, n))

ns

n=1

,

then from the Euler Product formula [3] and the multiplicative property of ϕ(m, n), we may get F (s) = =

p

Ã

Y pα km

×

Y

ϕ((m, p)) ϕ((m, p2 )) + + ··· 1+ ps p2s

!

ϕ((m, p)) ϕ((m, pα−1 )) ϕ((m, pα )) 1 ) 1+ + · · · + + ( s αs (α−1)s p p p 1 − p1s

Yµ

p†m

= ζ(s)

!

Ã

Y pα km

¶

1+

1 1 + 2s + · · · s p p

ÃÃ

ϕ((m, p)) ϕ((m, pα−1 )) 1+ + · · · + ps p(α−1)s

!µ

1 1− s p

¶

where ζ(s) is the Riemann zeta-function.

Obviously, we have inequality

¯ ¯ ∞ ¯X K ϕ((m, n)) ¯¯ ¯ , ¯ ¯< σ ¯ ¯ n σ−1 n=1

|ϕ((m, n))| < K,

where σ > 1 is the real part of s. So by Perron formula [4], taking b = 2, T = 1 K x 2 , H(x) = K, B(σ) = σ−1 , then we have X

1 ϕ((m, n)) = 2πi n≤x

Z

2+iT

2−iT

ζ(s)R(s)

³ 1 ´ xs ds + O x 2 +ε , s

!

ϕ((m, pα )) + , pαs

63

On the integer part of a positive integer’s k -th root

where ÃÃ

Y

R(s) =

ϕ((m, p)) ϕ((m, pα−1 )) 1+ + · · · + ps p(α−1)s

pα km

!µ

1 1− s p

¶

!

ϕ((m, pα )) + . pαs

To estimate the main term 1 2πi

Z

2+iT

2−iT

ζ(s)R(s)

xs ds s

we move the integral line from s = 2 ± iT to s = 1/2 ± iT . This time, we have a simple pole point at s = 1 with residue R(1)x. That is 1 2πi

ÃZ

2+iT

2−iT

Z

+

1 +iT 2

2+iT

Z

+

1 −iT 2 1 +iT 2

Z

+

2−iT 1 −iT 2

!

ζ(s)R(s)

xs ds = R(1)x. s

Taking T = x3/2 , we can easily get the estimate ¯ ¯ ÃZ 1 Z 2−iT ! ¯ 1 +iT 2 xs ¯¯ ¯ + ζ(s)R(s) ds¯ ¯ 1 ¯ 2πi s ¯ 2+iT −iT 2 ¯ ¯ Z 2¯ 1 x2 ¯¯ x2 ¯ ¿ = x2 , ¯ζ(σ + iT )R(s) ¯ dσ ¿ 1 ¯ T¯ T 2

and ¯ ¯ Z T ¯ 1 Z 12 −iT xs ¯¯ ¯ ζ(s)R(s) ds¯ ¿ ¯ ¯ 2πi 1 +iT s ¯ 0 2

¯ 1 ¯ ¯ 1 1 x 2 ¯¯ ¯ ¯ζ( + it)R(s) ¯ dt ¿ x 2 +ε . ¯ 2 t ¯

Noting that R(1) =

ÃÃ

Y

ϕ((m, p)) ϕ((m, pα−1 )) 1+ + ··· + p p(α−1)

pα km

=

!µ

1 1− p

¶

ϕ((m, pα )) + pα

ϕ(m) Y α−1 (1 + α − ). m pα km p

So we have the asympotic formula X

³

1

´

ϕ((m, n)) = x · h(m) + O x 2 +ε ,

n≤x

where h(m) = 2.

ϕ(m) m

Q

(1 + α −

pα km

α−1 p ).

This completes the proof of Lemma

!

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§3. Proof of Theorems In this section, we will complete the proof of Theorems. First we prove Theorem 1. For any real number x > 1, let M be a fixed positive integer with M k ≤ x ≤ (M + 1)k , from the definition of s(n) we have X

Ω(s(n)) =

M X

X

X

Ω(s(n)) +

t=1 (t−1)k ≤n<tk

n≤x

=

M −1 X

X

X

Ω(s(n)) +

t=1 tk ≤n<(t+1)k

=

M −1 X

Ω(M )

M k ≤n≤x





X

[(t + 1)k − tk ]Ω(t) + O 

t=1

= k

Ω(s(n))

M k ≤n<x

Ω(M )

M k ≤n<(M +1)k

M X

³

´

tk−1 Ω(t) + O M k−1 log M ,

t=1

where we have used the estimate Ω(M ) ¿ log n.

P Let B(y) = Ω(n), then by Able’s identity and Lemma 1, we can easily n≤y

deduce that M X

Z

t

k−1

Ω(t) = M

k−1

B(M ) − (k − 1)

t=1

M

Z

= M k−1 (M log log M + AM ) − (k − 1) Ã

+O

Mk log M

y k−2 B(y)dy

2 M

2

!

(y k−1 log log y + Ay k−1 )dy

k−1 = M log log M + AM − (M k log log M + AM k ) + O k k

=

k

1 k 1 M log log M + AM + O k k

Ã

Mk log M

!

.

Therefore, we can obtain the asymptotic formula X

à k

Ω(s(n)) = M log log M + AM + O

n≤x

Mk log M

On the other hand, we also have the estimate 0 ≤ x − M k < (M + 1)k − M k ¿ x

k−1 k

.

!

.

Ã

Mk log M

!

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On the integer part of a positive integer’s k -th root

Now combining the above, we may immediately obtain the asymptotic formula µ

X

Ω(s(n)) = x log log x + (A − log k) x + O

n≤x

¶

x . log x

This completes the proof of Theorem 1.

Now we come to prove Theorem 2. For any fixed positive integer m, we have X n≤x

=

X

ϕ((s(n), m)) =

1

ϕ(([n k ], m))

n≤x

X

ϕ(([i ], m)) + · · · +

1≤i<2k

=

X

X

1 k

1

ϕ(([i k ], m)) + O (N ε )

N ≤i<(N +1)k

[(j + 1)k − j k ]ϕ((j, m)) + O (N ε ) .

j≤N

From Lemma 2, we can let A(N ) =

X

³

1

´

ϕ((j, m)) = N · h(m) + O N 2 +ε ,

j≤N

f (j) = [(j + 1)k − j k ], Then by Able’s identity, we can easily obtain X

[(j + 1)k − j k ]ϕ((j, m))

j≤N

Z

= A(N )f (N ) − A(1)f (1) − ³

1

´

1

N

A(t)f 0 (t)dt

= [N · h(m) + O N 2 +ε ][(N + 1)k − N k ] Z

−

1

N

³

´

1

[t · h(m) + O t 2 +ε ] · k[(t + 1)k−1 − tk−1 ]dt ³

1

´

= k · N k h(m) + O k · N k− 2 +ε − (k − 1)h(m)(nk − 1) ³

1

= (k + N k + 1)h(m) + O k · N k− 2 +ε ³

1

´

´

= h(m)x + (k + 1)h(m) + +O x1− 2k +ε .

This completes the proof of Theorem 2.

References [1] F. Smarandache, Only Problems, Not Solutions, Xiquan Publishing House, Chicago, 1993.

[2] G. H. Hardy and S. Ramanujian, The normal number of prime factors of a number n, Quart. J. Math. 48 (1917), 76-92.

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[3] Tom M. Apostol, Introduction to Analytic Number Theory, New York, Springer-Verlag, 1976.

[4] Pan Chengdong and Pan Chengbiao, Foundation of Analytic number Theory, Beijing, Science Press, 1997.