On The Distribution Of {x/p}

The distribution of {p|x} is studied in the first part of this paper.We have the following definition: for sequence of numbers {a_n} ,0 ≤ α < β < 1 are constants, and T(α,β,N) denotes the number offactors in set {1 ≤ n ≤ N : α ≤ {a_n} < β}, {u} is the fractional part ofreal number u. ifthen we say that {a_n} is uniformly distributed modulo 1. Note that asequence of number is uniformly distributed modulo 1 or not has no relationto the values of α and β. By this, {αp} and {αn} are uniformly distributedmodulo 1 when α is an irrational number;{n^α} and {p^α} are uniformlydistributed modulo 1 if α is not an integer;and {n|x} is uniformly distributedmodulo 1 when n ≤ x^1-ε .Uniformly distributed modulo 1 problems play an important part innumber theory. Many classical problems in number theory are uniformlydistributed modulo 1 problems. It is known that the error term of Dirichletdivisor problem iswhich interests people to study the distribution of {nx}.Suppose x is a fixed positive number, R and α are real numberssatisfying 0 ≤ α < 1 and 1 ≤ R < x, N(x,R,α) denote the number ofnatural numbers that satisfyα ≤ {nx} < α + Rn.Wang wei [11] proved that when R ≥ x 2λλ++κ1,N(x,R,α) Rxuniformly for α, where (κ,λ) is an exponential pair. And in the same way,he proved thatN(x,1,α) x13? 1516+uniformly for α. In fact Varbanec [9] proved thatN(x,1,α) (1 + (αx)12)x ,where α = pq is a rational number. Zhai Wenguang [12] made a prove inthis problem. He got:(1)Suppose R < (1 ? α)x, R = ?(x), x 2λ++2κκ log x < R < (1 ? α)2 loxgx,One has asymptotic formula:N(x,R,α) = R log (1 ?R α )x+ d(α)R + O(x 2λ++2κκ log x) + O( xR(12 ?lo gα x)2).where (κ,λ) is any exponential pair,d(α) = γ + ∞l=11 ? α(l + 1)(l + α),γ is the Euler constant.(2)Supose R > (1 ? α)x, (1 ? α)x > x 2λ++2κκ log x, one has formula:N(x,R,α) = x ∞l=11 ? α(l + 1)(l + α) + O(x 2λ++2κκ log x),(κ,λ) is any exponential pair.(3)For any exponential pair (κ,λ), let R < x 2λ++2κκ,N(x,R,α) x 2λ++2κκ+uniformly for α.(4)N(x,1,α) x14+uniformly for α.On the other hand, in 1977, Sa?ari and Vaughan [8] studied the sumθx,y = y?1p≤y(log p)cα(xp),where cα(u) = 1 if 0 ≤ {u} < α;otherwise cα(u) = 0. In fact, it is theasymptotic formula with weighted form of number of solutions satisfyingthe inequality 0 ≤ {xp} < α。By zero-density theorem they got the : Suppose that ?ε > 0, andx 161+ε < y ≤ x, thenθx,y(α) = F(α, xy) + O(exp(?c(ε)( lolgo lgo xg x)13)),where c(ε) is a positive number depending at most on ε. Remark thatsuppose the density hypothesis concerning the distribution of the Riemannzeta function is true, the 161 could be replaced by 12. But the bound of y isstill very large.There are two main results in this paper:Theorem 1.1 Let N(x,R,α) denotes the weighted form of numberof solutions satisfying the inequalityα ≤ {xp} < α + Rp ,where R,α are constants satisfying 1 < R < x1? ,0 ≤ α < 1. Then ifx4372 < R < x1? , we have asymptotic formula:N(x,R,α) =n≤xα≤{nx}<α+RnΛ(n) = R log x(1 R? α)+c(α)R+O(R exp(?Cη(R))),where C is absolut constant,c(α) = γ + 1 ? ∞l=1αl(l + α),η(x) = (log x)35(log log x)?51.定理 1.2 Sa?ari and Vaughan's theorem holds when x y < x.Dirichlet divisor problem in additive square complements is studiedin the second part.Suppose n is natural number. Define {a(n)} := min{k|n + k =m2,k ≥ 0,m ∈ N+}, which is similar to {a2(n)} defined in problem 27of F.Smarandache[3]. a(n) is called additive square complements.Xu Zhefeng [4] studied the sum n≤x d(a(n)) and got it's asymptoticformula, where d(n) is Dirichlet divisor function. Yi Yuan [5] studied thesum n≤x d(n + a(n)) and provedn≤xd(n + a(n)) = 43π2xlog2 x + A1xlog x + A2x + O(x34+ε),Where A1,A2 are constants, ε is any positive number.定理 2.1 We have asymptotic formulan≤xd(n + a(n)) = 43π2xlog2 x + A1xlog x + A2x + O(x43exp(?Cη(x))),where C > 0 is absolute constant, η(x) := (log x)53(log log x)?51.The error term above is the best in present. One must learn more onthe distribution of zeros of Riemann zeta function. For example, 3/4 couldbe proved if RH holds. However, we can prove that the following result onshort interval without any supposition:定理 2.2 Suppose 1/3 < θ < 1/2 is the constant in Lemma 2.2.4,by Kolesnik [2] we can take θ = 43/96 + . Let M(x) denotes the mainterm in Th 2.1, if x1+2 θ+ε ≤ y ≤ x, thenx≤n≤x+yd(n + a(n)) = M(x + y) ? M(x) + O(yx?2ε).