The Mean Values Of The K-free Exponential Functions

Let n>1be an integer of canonical form n=p1α1P2α2…p(?)αr.The integer n is called a k-full number if n=p1α1P2α2…prαr,where α1≥l,α2≥k,…,αr,≥k. Let fk(n)be the characteristic function of k-full integers,i.e.We call an integer n=Πir=1piαi exponentially k-free if all the exponents αi(1≤i≤r)are k-free,i.e.are not divisible by the k-th power of any prime (k≥2).Let qk(e)(n)denote the characteristic function of exponentially k-free integers.From L.Toth[2],we know,the function qk(e)(n) is multiplicative,and for every prime power p∝,there are: qk(e)(p)=qk(e)(p2)=qk(e)(p3)=…=qk(e)(p2k-1)=1,qk(e)(p2k)=0.Many authors have studied the properties of the k-free exponential function qk(e)(n).L.Toth[2] proved the following result: where which A being a positive constant,qk(n) denoting the characteristie function of k-free integers.In this paper,we shall study the mean value of the k-free exponential function qk(e)(n)over k-full integers,that is where fk(n) is the characteristic function of k-full integers, i.e.When k=2, we have the following:Theorem1For some D>0, is absolutely convergent for (?)>1/7+∈. In addition, we also improved the above result under the assumption of the Riemann hypothesis(RH).Theorem2If RH is true, then where is absolutely convergent for (?)s>1/7+∈." Then, we shall prove the mean value of the k-free exponential function over cube-full number. We have the following:Theorem3For some D>0, where is absolutely convergent for (?)s>1/9+∈Finally, we also improved the above result under the assumption of the Rie-mann hypothesis(RH).Theorem4If RH is true, then where G(s)=Π(1-1/(p9s)+…)is absolutely convergent for (?)s>1/9+∈.