ON THE GENERALIZATION OF THE FLOOR OF THE SQUARE ROOT SEQUENCE Yao Weili Research Center for Basic Science, Xi’an Jiaotong University Xi’an, Shaanxi, P.R.China
Abstract
The floor of the square root sequence is the natural sequence, where each number is repeated 2n+1 times. In this paper, we use analytic method to study the mean value properties of its generalization, and give an interesting asymptotic formula.
Keywords:
the floor of the square root sequence; mean value; asymptotic formula.
§1. Introduction The floor of the square roots of the natural numbers are: 0,1,1,1,2,2,2,2,2,3,3,3,3,3,3,3,4,4,4,4,4,4,4,4,4, 5,5,5,5,5,5,5,5,5,5,5,6,6, 6,6,6,6,6,6,6,6,6,6,6, 7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,8, · · · This sequence is the natural sequence, where each number is repeated 2n+1 times. In reference [1], Professor F.Smarandache asked us to study the properties of this sequence. We denote the generalization of the sequence as b(n), in which each number is repeated kn + 1 times, and express it as b(n) = [n1/k ]. In reference [2], He Xiaolin and Guo Jinbao studied the mean value of d(b(n)), and obtain an asymptotic formula for it. In this paper, as a generalization of [2], we use analytic method to study the mean value properties of σα (b(n)), and give a general asymptotic formula for σα (b(n)). That is, we shall prove the following : Theorem. For any real number x > 1 and integer n ≥ 1, we have
X
σα (b(n)) =
n≤x
where σα (n) =
P
β+k−1 kζ(α+1) α+k k + O(x k ), α+k x 1 k x log x + O(x), k+ε−1
ζ(2)x + 0(x k ), δ+k−1 ζ(1 − α)x + O(x k ),
if α > 0; if α = 0; if α = −1; if α < 0 and α 6= −1
dα be the divisor function, ζ(n) be the Riemann Zeta func-
d|n
tion, β = max{1, α}, δ = max{0, 1 + α} and ε > 0 be an arbitrary real number.
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SCIENTIA MAGNA VOL.1, NO.1
§2. Proof of the Theorem In this section, we shall complete the proof of the theorem. Firstly, we need following: Lemma 1. Let α > 0 be a fixed real number. Then for x > 1, we have X
σα (n) =
n≤x
and
½
X
σ−α (n) =
n≤x
ζ(α + 1) α+1 x + O(xβ ), α+1
ζ(α + 1)x + O(xδ ), if α 6= 1; ζ(2)x + O(log x), if α = 1.
where β = max{1, α}, δ = max{0, 1 − α}, ζ(n) denotes the Riemann zetafunction. Proof See reference [3]. Lemma 2. Let n be a positive integer, and b(n) = [n1/k ], d(n) be the divisor function, then X
σ0 (b(n)) =
X
³
´
d [n1/k ] =
n≤x
n≤x
1 x log x + O(x). k
Proof. See reference [2]. Now we use the above Lemmas to complete the proof of Theorem. We separate α into three cases respectively. Case 1 , when α > 0, we have X
σα (b(n)) =
n≤x
³
X
=
σα [n
1k ≤n<2k
+
X n≤x
1/k
X
N k ≤n<(N +1)k
X³
=
σα ([n1/k ])
´ ³
³
X
] +
´
σα [n1/k ] + · · ·
2k ≤n<3k
´
σα [n1/k ] + O(N β ) ´
(j + 1)k − j k σα (j) + O(N β ).
j≤N
Let A(n) =
P n≤x
σα (j) and f (j) =
P ³
´
(j + 1)k − j k , applying Abel’s
j≤N
identity and Lemma 1, we have X
σα (b(n))
n≤x
= A(N )f (N ) − A(1)f (1) − =
Z
N
1
kζ(α + 1) α+k N − k(k − 1) α+1
Z 1
0
A(t)f (t)dt + O(N β ) N
ζ(α + 1) α+k+1 t dt α+k
185
On the generalization of the floor of the square root sequence
+O(N β+k−1 ) kζ(α + 1) α+k N + O(N β+k−1 ) α+k β+k−1 kζ(α + 1) α+k x k + O(x k ). α+k
= =
Case 2, when α = −1, we have X n≤x
σ−1 [n
1k ≤n<2k
+
1/k
´ ³
X
=
³
X
] +
´
σ−1 [n1/k ] + · · ·
2k ≤n<3k
´
σ−1 [n1/k ] + O(N ε )
N k ≤n<(N +1)k
X³
σ−1 ([n1/k ])
n≤x
³
X
=
X
σ−1 (b(n)) =
´
(j + 1)k − j k σ−1 (j) + O(N ε ).
j≤N
P
Let A(n) =
j≤N
X
P ³
σ−1 (j) and f (j) = X
σ−1 (b(n)) =
n≤x
j≤N
σ−1 [n1/k ]
n≤x
Z
= A(N )f (N ) − A(1)f (1) − k
= ζ(2)N + O(N = ζ(2)x + O(x
k+ε−1
k+ε−1 k
´
(j + 1)k − j k , we have
N
0
A(t)f (t)dt + O(N ε )
1
)
),
where ε > 0 be an arbitrary real number. Case 3, when α < 0 and α 6= −1, we have X
σα (b(n)) =
n≤x
³
X
=
n≤x
´
³
X
=
³
X
´
σα [n1/k ] + · · ·
2k ≤n<3k
´
σα [n1/k ] + O(N δ )
N k ≤n<(N +1)k
X³
σα ([n1/k ])
σα [n1/k ] +
1k ≤n<2k
+
X
´
(j + 1)k − j k σα (j) + O(N δ ).
j≤N
Let A(n) =
P j≤N
σα (j) and f (j) =
X n≤x
σα (b(n)) =
X n≤x
P ³
´
(j + 1)k − j k , we have
j≤N
σα ([n1/k ])
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SCIENTIA MAGNA VOL.1, NO.1 Z
= A(N )f (N ) − A(1)f (1) − k
1
N
0
A(t)f (t)dt + O(N δ )
= kζ(1 − α)N + O(N ) − (k − 1)ζ(1 − α)N k = ζ(1 − α)N k + O(N k+δ−1 ) = ζ(1 − α)x + O(x
k+δ−1
δ+k−1 k
).
Combining the above result, we have X
σα (b(n)) =
n≤x
β+k−1 kζ(α+1) α+k k + O(x k ), α+k x 1 k x log x + O(x), k+ε−1
ζ(2)x + 0(x k ), δ+k−1 ζ(1 − α)x + O(x k ),
if α > 0; if α = 0; if α = −1; if α < 0 and α 6= −1
This completes the proof of Theorem.
References [1] F.Smarandache, Only Problems, Not Solutions, Chicago, Xiquan Publ. House, 1993. [2] He Xiaolin, On the 80-th problem of F.Smarandache (II), Smarandache Notions Journal, 14 (2004), 74–79. [3] T.M.Apostol, Introduction to Analytic Number Theory, New York, SpringerVerlag, 1976.