Scientia Magna Vol. 1 (2005), No. 2, 46-48
On the square-free number sequence Ren Dongmei Research Center for Basic Science, Xi’an Jiaotong University Xi’an, Shaanxi, P.R.China Abstract The main purpose of this paper is to study the number of the square-free number sequence, and give two interesting asymptotic formulas for it. At last, give another asymptotic formula and a corollary. Keywords Square-free number sequence; Asymptotic formula.
§1. Introduction A number is called a square-free number if its digits don’t contain the numbers: 0, 1, 4, 9. Let A denote the set of all square-free numbers. In reference [1], Professor F. Smarandache asked us to study the properties of the square-free number sequence. About this problem, it seems that none had studied it, at least we have not seen such a paper before. In this paper, we use the elementary method to study the number of the square-free X number sequence, and obtain two interesting asymptotic formulas for it. That is, let S(x) = 1, we shall prove n≤x,n∈A
the followings: Theorem 1. For any real number x ≥ 1, we have the asymptotic formula ln S(x) =
ln 6 × ln x + O(1). ln 10
Theorem 2. For any real number x ≥ 1, we have the asymptotic formula X
³ 2 ln 2 ´ 1 = x + O x ln 10 ,
n≤x,n∈B
where B denote the complementary set of those numbers whose all digits are square numbers. Let B 0 denote the set of those numbers whose all digits are square numbers. Then we have the following: Theorem 3. For any real number x ≥ 1,we have the asymptotic formula X n≤x,n∈B
¶ µ ln 5 1 2 = ln x + γ − C + O x− ln 10 , n
where C is a computable constant, γ denotes the Euler’s constant. Let A0 denote the complementary set of A, we have following: Corollary. For any real number x ≥ 1, we have the asymptotic formula
Vol. 1
On the Square-Free number sequence
47
µ ¶ ln 5 1 3 − ln 10 = ln x + γ − D + O x , n
X n≤x,n∈A0
where D is a computable constant.
§2. Proof of Theorems In this section, we shall complete the proof of Theorems. First we need the following one simple lemma. Lemma. For any real number x ≥ 1, we have the asymptotic formula X n≤x,n∈B0
µ ¶ ln 5 1 2 − ln 10 =C +O x . n
Proof. In the interval [10r−1 , 10r ), (r ≥ 2), there are 3 × 4r−1 numbers belong to B 0 , and 1 every number’s reciprocal isn’t greater than 10r−1 ; when r = 1, there are 4 numbers belong to 0 B and their reciprocals aren’t greater than 1. Then we have
then
∞ X 1 X 4r <3+ 3× r, n 10 0 r=1
n∈B
P n∈B0
1 is convergent to a constant C. So
X n≤x,n∈B0
X 1 1 = − n n 0 n∈B
X n>x,n∈B0
Ã∞ ! µ ¶ X 3 × 4r ln 5 1 2 − ln 10 =C +O =C +O x . n 10r r=k
Now we come to prove Theorem 1. First for any real number x ≥ 1, there exists a nonnegative integer k, such that 10k ≤ x < 10k+1 (k ≥ 1) therefore k ≤ log x < k + 1. If a number belongs to A, then its digits only contain these six numbers: 2, 3, 5, 6, 7, 8. So in the interval [10r−1 , 10r )(r ≥ 1), there are 6r numbers belong to A. Then we have X
1≤
n≤x,n∈A
k+1 X
6r =
r=1
ln 6 6 6k+2 62 × (6k+1 − 1) < < × x ln 10 , 5 5 5
and X n≤x,n∈A
1≥
k X
6r =
r=1
ln 6 1 6 × (6k − 1) ≥ 6k > × x ln 10 . 5 6
So we have ln 6 1 × x ln 10 < 6
X n≤x,n∈A
1<
ln 6 62 × x ln 10 . 5
Taking the logarithm computation on both sides of the above, we get
48
Ren Dongmei
³ ln 6 ´ 1 < ln x ln 10 + (2 × ln 6 − ln 5).
X
ln 6
ln(x ln 10 ) + (− ln 6) <
No. 2
n≤x,n∈A
So ln S(x) = ln
X n≤x,n∈A
³ ln 6 ´ ln 6 1 = ln x ln 10 + O(1) = × ln x + O(1). ln 10
This proves the Theorem 1. Now we prove Theorem 2. It is clear that if a number doesn’t belong to B, then all of its digits are square numbers. So in the interval [10r−1 , 10r ), (r ≥ 2), there are 3 × 4r−1 numbers don’t belong to B; when r = 1, there are 4 numbers don’t belong to B. Then we have X
1=
n≤x,n∈B
X n≤x
1−
X
1
n≤x,n∈B0
¡ ¢ = x + O 4 + 3 × 4 + 3 × 42 + · · · + 3 × 4k ³ 2×ln 2 ´ ¡ ¢ = x + O 4k+1 = x + O x ln 10 . This completes the proof of the Theorem 2. Now we prove the Theorem 3. In reference [2], we know the asymptotic formula: µ ¶ X 1 1 = ln x + γ + O , n x
n≤x
where γ is the Euler’s constant. Then from this asymptotic formula and the above Lemma, we have X
X 1 X 1 1 = − n n n n≤x,n∈B n≤x n≤x,n∈B0 µ ¶ µ ¶ ln 5 1 2 − ln 10 = ln x + γ + O −C +O x x µ ¶ ln 5 2 − ln 10 = ln x + γ − C + O x . This completes the proof of the Theorem 3. Now the Corollary immediately follows from the Lemma and Theorem 3.
Reference [1] F.Smarandache, Only problems, Not Solutions, Xiquan Publ. House, Chicago, 1993. [2] Tom M.Apostol, Introduction to Analytic Number Theory, Springer-Verlag, New York, 1976.