The main purpose of this paper is to study the distributive properties of k + 1-power free numbers, and give two interesting asymptotic formulae

Hong, Shen

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Scientia MagnaVol. 1 (2005), No. 2, 145-148Two asymptotic formulae on the k + 1-powerfree numbersShen HongXian Yang vocational and technical collegeXian Yang, Shaanxi, P.R.ChinaAbstract The main purpose of this paper is to study the distributive properties of k + 1-power freenumbers, and give two interesting asymptotic formulae.Keywords k + 1-power free numbers; Asymptotic formula.§1. IntroductionA natural number n is called a k + 1-power free number if it can not be divided by anypk+1, where p is a prime number. One can obtain all k+1-power free numbers by the followingmethod:From the set of natural numbers (except 0 and 1)-take off all multiples of 2k+1(i.e. 2k+1, 2k+2, · · · ).-take off all multiples of 3k+1.-take off all multiples of 5k+1.· · · and so on (take off all multiples of all k + 1-power primes).In reference [1], Professor F. Smarandache asked us to study the properties of the k + 1-power free numbers sequence. Yet we still know very little about it.Now we define two new number-theoretic functions U(n) and V (n) as following,U(1) = 1, U(n) =∏p|np,V (1) = 1, V (n) = V (pα11 ) · · ·U (pαrr ) = (pα1 − 1) · · · (pαr − 1),where n is any natural number with the form n = pα11 · · · pαrr . Obviously they are both multi-plicative functions. In this paper, we shall use the analytic method to study the distributionproperties of this sequence, and obtain two interesting asymptotic formulae. That is, we havethe following two theorems:Theorem 1. Let A denote the set of all k + 1-power free numbers, then for any realnumber x ≥ 1, we have the asymptotic formula∑n∈An≤xU(n) =3x2pi2∏p(1 +p2k−2 − 1p2k+1 + p2k − p2k−1 − p2k−2)+O(x32+ε),146 Shen Hong No. 2where ε denotes any fixed positive number and∏pdenotes the product of all the prime numbers.Theorem 2. For any real number x ≥ 1, we have the asymptotic formula∑n∈An≤xV (n) =x22∏p(1− 1pk+1− p2k+1 + p2k − p− 1p2k+3 + p2k+1)+O(x32+ε).§2. Proof of TheoremsIn this section, we shall complete the proof of Theorems. First we prove Theorem 1, letf(s) = 1 +∑n∈An≤xU(n)ns.From the Euler product formula [2] and the definition of U(n), we may havef(s) =∏p(1 +U(p)ps+U(p2)p2s+ · · ·+ U(pk)pks)=∏p(1 +1ps−1+1p2s−1+ · · ·+ 1pks−1)=∏p(1 +1ps−1+p(k−1)s − 1p2s−1(p(k−1)s − p(k−2)s))=ζ(s− 1)ζ(2(s− 1))∏p(1 +p(k−1)s − 1(p2s−1 + ps)(p(k−1)s − p(k−2)s)),where ζ(s) is the Riemann-zeta function. Obviously, we have the following two inequalities|U(n)| ≤ n,∣∣∣∣∣∞∑n=1U(n)nσ∣∣∣∣∣ < 1σ − 2 ,where σ > 2 is the real part of s. So by Perron formula [3]∑n≤xU(n)ns0=12ipi∫ b+iTb−iTf(s+ s0)xssds+O(xbB(b+ σ0)T)+O(x1−σ0H(2x)min(1,log xT))+O(x−σ0H(N)min(1,x||x|| )),where N is the nearest integer to x, ‖x‖ = |x−N |. Taking s0 = 0, b = 3, T = x 32 , H(x) = x,B(σ) = 1σ−2 , we have∑n≤xU(n) =12ipi∫ 3+iT3−iTζ(s− 1)ζ(2(s− 1))R(s)xssds+O(x32+ε),whereR(s) =∏p(1 +p2k−2 − 1p2k+1 + p2k − p2k−1 − p2k−2).Vol. 1 Two asymptotic formulae on the k + 1-power free numbers 147To estimate the main term12ipi∫ 3+iT3−iTζ(s− 1)ζ(2(s− 1))R(s)xssds,we move the integral line from s = 3± iT to s = 32 ± iT . This time, the functionf(s) =ζ(s− 1)xsζ(2(s− 1))sR(s)has a simple pole point at s = 2 with residue x22ζ(2)R(2). So we have12ipi(∫ 3+iT3−iT+∫ 32+iT3+iT+∫ 32−iT32+iT+∫ 3−iT32−iT)ζ(s− 1)xsζ(2(s− 1))sR(s)ds=x22ζ(2)∏p(1 +p2k−2 − 1p2k+1 + p2k − p2k−1 − p2k−2).We can easily get the estimates∣∣∣∣∣ 12pii(∫ 32+iT3+iT+∫ 3−iT32−iT)ζ(s− 1)xsζ(2(s− 1))sR(s)ds∣∣∣∣∣¿∫ 332∣∣∣∣ ζ(σ − 1 + iT )ζ(2(σ − 1 + iT ))R(s)x3T∣∣∣∣ dσ ¿ x3T = x 32and ∣∣∣∣∣ 12pii∫ 32−iT32+iTζ(s− 1)xsζ(2(s− 2))sR(s)ds∣∣∣∣∣¿∫ T0∣∣∣∣∣ζ(1/2 + it)ζ(1 + 2it) x32t∣∣∣∣∣ dt¿ x 32+ε.Note the fact that ζ(2) = pi26 , then from the above we can obtain∑n∈An≤xU(n) =3x2pi2∏p(1 +p2k−2 − 1p2k+1 + p2k − p2k−1 − p2k−2)+O(x32+ε).This completes the proof of Theorem 1.Now we come to prove Theorem 2. Letg(s) = 1 +∑n∈An≤xV (n)ns.From the Euler product formula [2] and the definition of V (n), we also haveg(s) =∏p(1 +V (p)ps+V (p2)p2s+ · · ·+ V (pk)pks)=∏p(1 +p− 1ps+p2 − 1p2s+ · · ·+ pk − 1pks)=∏p(1− 1p(k+1)(s−1)1− 1ps−1−1− 1pksps − 1)= ζ(s− 1)∏p(1− 1p(k+1)(s−1)− (pks − 1)(ps−1 + 1)(pks − p(k−1)s)p2s−1).148 Shen Hong No. 2Now applying Perron formula [3], and the method of proving Theorem 1, we can also obtainthe result of Theorem 2.This completes the proof of Theorems.References[1]F.Smarandache, Only problems, Not solutions, Chicago, Xiquan Publ. 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.

Two asymptotic formulae on the k + 1-power free numbers

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