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

Dongmei, Ren

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Scientia MagnaVol. 1 (2005), No. 2, 46-48On the square-free number sequenceRen DongmeiResearch Center for Basic Science, Xi’an Jiaotong UniversityXi’an, Shaanxi, P.R.ChinaAbstract 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 acorollary.Keywords Square-free number sequence; Asymptotic formula.§1. IntroductionA 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. Smarandacheasked us to study the properties of the square-free number sequence. About this problem, itseems 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 number sequence, andobtain two interesting asymptotic formulas for it. That is, let S(x) =∑n≤x,n∈A1, we shall provethe followings:Theorem 1. For any real number x ≥ 1, we have the asymptotic formulalnS(x) =ln 6ln 10× lnx+O(1).Theorem 2. For any real number x ≥ 1, we have the asymptotic formula∑n≤x,n∈B1 = x+O(x2 ln 2ln 10),where B denote the complementary set of those numbers whose all digits are square numbers.Let B′ denote the set of those numbers whose all digits are square numbers. Then we havethe following:Theorem 3. For any real number x ≥ 1,we have the asymptotic formula∑n≤x,n∈B1n= lnx+ γ − C +O(x−ln 52ln 10),where C is a computable constant, γ denotes the Euler’s constant.Let A′ denote the complementary set of A, we have following:Corollary. For any real number x ≥ 1, we have the asymptotic formulaVol. 1 On the Square-Free number sequence 47∑n≤x,n∈A′1n= lnx+ γ −D +O(x−ln 53ln 10),where D is a computable constant.§2. Proof of TheoremsIn this section, we shall complete the proof of Theorems. First we need the following onesimple lemma.Lemma. For any real number x ≥ 1, we have the asymptotic formula∑n≤x,n∈B′1n= C +O(x−ln 52ln 10).Proof. In the interval [10r−1, 10r), (r ≥ 2), there are 3× 4r−1 numbers belong to B′ , andevery number’s reciprocal isn’t greater than 110r−1 ; when r = 1, there are 4 numbers belong toB′ and their reciprocals aren’t greater than 1. Then we have∑n∈B′1n< 3 +∞∑r=13× 4r10r,then∑n∈B′ 1 is convergent to a constant C. So∑n≤x,n∈B′1n=∑n∈B′1n−∑n>x,n∈B′1n= C +O( ∞∑r=k3× 4r10r)= C +O(x−ln 52ln 10).Now we come to prove Theorem 1. First for any real number x ≥ 1, there exists a non-negative integer k, such that 10k ≤ x < 10k+1(k ≥ 1) therefore k ≤ log x < k + 1. If a numberbelongs 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∑n≤x,n∈A1 ≤k+1∑r=16r =65× (6k+1 − 1) < 6k+25<625× x ln 6ln 10 ,and∑n≤x,n∈A1 ≥k∑r=16r =65× (6k − 1) ≥ 6k > 16× x ln 6ln 10 .So we have16× x ln 6ln 10 <∑n≤x,n∈A1 <625× x ln 6ln 10 .Taking the logarithm computation on both sides of the above, we get48 Ren Dongmei No. 2ln(xln 6ln 10 ) + (− ln 6) <∑n≤x,n∈A1 < ln(xln 6ln 10)+ (2× ln 6− ln 5).SolnS(x) = ln ∑n≤x,n∈A1 = ln(x ln 6ln 10)+O(1) = ln 6ln 10× lnx+O(1).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 itsdigits are square numbers. So in the interval [10r−1, 10r), (r ≥ 2), there are 3× 4r−1 numbersdon’t belong to B; when r = 1, there are 4 numbers don’t belong to B. Then we have∑n≤x,n∈B1 =∑n≤x1−∑n≤x,n∈B′1= x+O(4 + 3× 4 + 3× 42 + · · ·+ 3× 4k)= x+O(4k+1)= x+O(x2×ln 2ln 10).This completes the proof of the Theorem 2. Now we prove the Theorem 3. In reference [2], weknow the asymptotic formula:∑n≤x1n= lnx+ γ +O(1x),where γ is the Euler’s constant.Then from this asymptotic formula and the above Lemma, we have∑n≤x,n∈B1n=∑n≤x1n−∑n≤x,n∈B′1n= lnx+ γ +O(1x)− C +O(x−ln 52ln 10)= lnx+ γ − C +O(x−ln 52ln 10).This completes the proof of the Theorem 3. Now the Corollary immediately follows from theLemma 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.

On the square-free number sequence

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On the square-free number sequence

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