摘要
设Q(x)表示≤x的无四次方因子数的个数,在本文中,我们证明了:在RH(Riemann假设)下有Q(x)=x/ζ(4)+O(x^(2/11+ε))从而改进了Graham和Pintz的结果.
A Positive integer is said to be 4-free if it is divisible by no 4-th-powew of a prime. An elementary argument shows that(x1/4),Let Δ(x) denote the left hand side of the above. It seems plausible that Δ(x) = 0(x1/8+ε), bat this is not known, even assuming the Riemann Hypothesis(RH).Montgomery and Vaughan[2] assumed RH and obtained Δ(x) = 0(x1/5+ε)Li Honghn[3] improved 1/5 to9/48 , and Graham and Pintz[4] have improved9/48 to 7/38 . This paper, we shall prove theorem. Assume RH. For everyε>0,Δ(x)=O(x2/11+ε).
