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Research article

High-dimensional Lehmer problem on Beatty sequences

  • Received: 25 December 2022 Revised: 18 February 2023 Accepted: 09 March 2023 Published: 07 April 2023
  • MSC : 11B83, 11L05, 11N69

  • Let q be a positive integer. For each integer a with 1a<q and (a,q)=1, it is clear that there exists one and only one ˉa with 1ˉa<q such that aˉa1(q). Let k be any fixed integer with k2,0<δi1,i=1,2,,k. rn(δ1,δ2,,δk,α,β,c;q) denotes the number of all k-tuples with positive integer coordinates (x1,x2,,xk) such that 1xiδiq,(xi,q)=1,x1x2xkc(q), and x1,x2,,xk1Bα,β. In this paper, we consider the high-dimensional Lehmer problem related to Beatty sequences over incomplete intervals and give an asymptotic formula by the properties of Beatty sequences and the estimates for hyper Kloosterman sums.

    Citation: Xiaoqing Zhao, Yuan Yi. High-dimensional Lehmer problem on Beatty sequences[J]. AIMS Mathematics, 2023, 8(6): 13492-13502. doi: 10.3934/math.2023684

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  • Let q be a positive integer. For each integer a with 1a<q and (a,q)=1, it is clear that there exists one and only one ˉa with 1ˉa<q such that aˉa1(q). Let k be any fixed integer with k2,0<δi1,i=1,2,,k. rn(δ1,δ2,,δk,α,β,c;q) denotes the number of all k-tuples with positive integer coordinates (x1,x2,,xk) such that 1xiδiq,(xi,q)=1,x1x2xkc(q), and x1,x2,,xk1Bα,β. In this paper, we consider the high-dimensional Lehmer problem related to Beatty sequences over incomplete intervals and give an asymptotic formula by the properties of Beatty sequences and the estimates for hyper Kloosterman sums.



    Let q be a positive integer. For each integer a with 1a<q,(a,q)=1, we know that there exists one and only one ˉa with 1ˉa<q such that aˉa1(q). Let r(q) be the number of integers a with 1a<q for which a and ˉa are of opposite parity.

    D. H. Lehmer (see [1]) posed the problem to investigate a nontrivial estimation for r(q) when q is an odd prime. Zhang [2,3] gave some asymptotic formulas for r(q), one of which reads as follows:

    r(q)=12ϕ(q)+O(q12d2(q)log2q).

    Zhang [4] generalized the problem over short intervals and proved that

    aNaR(q)1=12Nϕ(q)q1+O(q12d2(q)log2q),

    where

    R(q):={a:1aq,(a,q)=1,2a+ˉa}.

    Let n2 be a fixed positive integer, q3 and c be two integers with (n,q)=(c,q)=1. Let 0<δ1,δ21. Lu and Yi [5] studied the Lehmer problem in the sense of short intervals as

    rn(δ1,δ2,c;q):=aδ1qˉaδ2qaˉacmodqna+ˉa1,

    and obtained an interesting asymptotic formula,

    rn(δ1,δ2,c;q)=(1n1)δ1δ2ϕ(q)+O(q12d6(q)log2q).

    Liu and Zhang [6] r-th residues and roots, and obtained two interesting mean value formulas. Guo and Yi [7] found the Lehmer problem also has good distribution properties on Beatty sequences. For fixed real numbers α and β, the associated non-homogeneous Beatty sequence is the sequence of integers defined by

    Bα,β:=(αn+β)n=1,

    where t denotes the integer part of any tR. Such sequences are also called generalized arithmetic progressions. If α is irrational, it follows from a classical exponential sum estimate of Vinogradov [8] that Bα,β contains infinitely many prime numbers; in fact, one has the asymptotic estimate

    #{ prime px:pBα,β}α1π(x) as x

    where π(x) is the prime counting function.

    We define type τ=τ(α) for any irrational number α by the following definition:

    τ:=sup{tR:lim infnntαn=0}.

    Based on the results obtained, we consider the high-dimensional Lehmer problem related to Beatty sequences over incomplete intervals in this paper. That is,

    rn(δ1,δ2,,δk,c,α,β;q):=x1δ1qxkδkqx1xkcmodqx1,xk1Bα,βnx1++xk1,(0<δ1,δ2,,δk1),

    and where k = 2, we get the result of [7].

    By using the properties of Beatty sequences and the estimates for hyper Kloosterman sums, we obtain the following result.

    Theorem 1.1. Let k2 be a fixed positive integer, qn3 and c be two integers with (n,q)=(c,q)=1, and δ1,δ2,,δk be real numbers satisfying 0<δ1,δ2,,δk1. Let α>1 be an irrational number of finite type. Then, we have the following asymptotic formula:

    rn(δ1,δ2,,δk,c,α,β;q)=(1n1)α(k1)δ1δ2δkϕk1(q)+O(qk11τ+1+ε),

    where ϕ() is the Euler function, ε is a sufficiently small positive number, and the implied constant only depends on n.

    Notation. In this paper, we denote by t and {t} the integral part and the fractional part of t, respectively. As is customary, we put

    e(t):=e2πit and {t}:=tt.

    The notation t is used to denote the distance from the real number t to the nearest integer; that is,

    t:=minnZ|tn|.

    Let χ0 be the principal character modulo q. The letter p always denotes a prime. Throughout the paper, ε always denotes an arbitrarily small positive constant, which may not be the same at different occurrences; the implied constants in symbols O, and may depend (where obvious) on the parameters α,n,ε but are absolute otherwise. For given functions F and G, the notations FG, GF and F=O(G) are all equivalent to the statement that the inequality |F|C|G| holds with some constant C>0.

    To complete the proof of the theorem, we need the following several definitions and lemmas.

    Definition 2.1. For an arbitrary set S, we use 1S to denote its indicator function:

    1S(n):={1ifnS,0ifnS.

    We use 1α,β to denote the characteristic function of numbers in a Beatty sequence:

    1α,β(n):={1ifnBα,β,0ifnBα,β.

    Lemma 2.2. Let a,q be integers, δ(0,1) be a real number, θ be a rational number. Let α be an irrational number of finite type τ and H=qε>0. We have

    aδqaBα,β1=α1δϕ(q)+O((ϕ(q))ττ+1+ε),

    and

    aδqaBα,βe(θa)=α1aδ1qe(θa)+O(θ1qε+qε).

    Taking

    H=θ1τ+1+ε,

    we have

    aδqaBα,βe(θa)=α1aδ1qe(θa)+O(θ(ττ+1+ε)).

    Proof. This is Lemma 2.4 and Lemma 2.5 of [7].

    Lemma 2.3. Let

    Kl(r1,r2,,rk;q)=x1q1xk1q1e(r1x1++rk1xk1+rk¯x1xk1p).

    Then

    Kl(r1,r2,,rk;q)qk12kω(q)(r1,rk,q)12(rk1,rk,q)12

    where (a,b,c) is the greatest common divisor of a,b and c.

    Proof. See [9].

    Lemma 2.4. Assume that U is a positive real number, K is a positive integer and that a and b are two real numbers. If

    a=sr+θr2,(r,s)=1,r1,|θ|1,

    then

    kKmin(U,1ak+b)(Kr+1)(U+rlogr).

    Proof. The proof is given in [10].

    We begin by the definition

    rn(δ1,δ2,,δk,c,α,β;q)=S1S2,

    where

    S1:=x1δ1qxkδkqx1xkcmodqx1,xk1Bα,β1,

    and

    S2:=x1δ1qxkδkqx1xkcmodqx1,xk1Bα,βnx1++xk1.

    By the Definition 2.1, Lemma 2.2 and congruence properties, we have

    S1=x1δ1qxkδkqx1xkcmodq1α,β(x1)1α,β(xk1)=1ϕ(q)x1δ1qxkδkqχmodqχ(x1)χ(xk)χ(¯c)1α,β(x1)1α,β(xk1)=S11+S12,

    where

    S11:=1ϕ(q)x1δ1qxkδkq1α,β(x1)1α,β(xk1),

    and

    S12:=1ϕ(q)χmodqχχ0χ(¯c)(x1δ1qxkδkqχ(x1)χ(xk)1α,β(x1)1α,β(xk1)).

    For S2, it follows that

    S2=1ϕ(q)x1δ1qxkδkqnx1++xkχmodqχ(x1)χ(xk)χ(¯c)1α,β(x1)1α,β(xk1)=S21+S22,

    where

    S21:=1ϕ(q)x1δ1qxkδkqnx1++xk1α,β(x1)1α,β(xk1),

    and

    S22:=1ϕ(q)χmodqχχ0χ(¯c)x1δ1qxkδkqnx1++xkχ(x1)χ(xk1)1α,β(x1)1α,β(xk1).

    From the classical bound

    aδq1=δϕ(q)+O(d(q))

    and Lemma 2.2, we have

    S11=1ϕ(q)(x1δ1q1α,β(x1))(xk1δk1q1α,β(xk1))(xkδkq1)=(δk+O(d(q)ϕ(q)))k1i=1(α1δiϕ(q)+O((ϕ(q))ττ+1+ε))=α(k1)ϕk1(q)k1i=1δi+O(qk11τ+1+ε). (3.1)

    From Lemma 2.2, we obtain

    S21=1ϕ(q)(x1δ1q1α,β(x1))(xk1δk1q1α,β(xk1))(xkδkqnxk+(x1++xk1)1)=1ϕ(q)(x1δ1q1α,β(x1))(xk1δk1q1α,β(xk1))(xkδkqxk(x1++xk1)modnd(xk,q)μ(d))=1ϕ(q)(x1δ1q1α,β(x1))(xk1δk1q1α,β(xk1))(dqμ(d)xkδkqdxkxk(x1++xk1)modn1)=1ϕ(q)(x1δ1q1α,β(x1))(xk1δk1q1α,β(xk1))(dqμ(d)(δkqnd+O(1)))=1ϕ(q)(δkϕ(q)n+O(d(q)))k1i=1(α1δiϕ(q)+O((ϕ(q))ττ+1+ε))=α(k1)n1ϕk1(q)k1i=1δi+O(qk11τ+1+ε). (3.2)

    By the properties of exponential sums,

    S22=1nϕ(q)χmodqχχ0χ(¯c)(x1δ1qxkδk1qχ(x1)χ(xk)1α,β(x1)1α,β(xk1))×(nl=1e(x1++xknl))=1nϕ(q)χmodqχχ0χ(¯c)nl=1k1i=1(xiδiq1α,β(xi)χ(xi)e(xinl))(xkδkqχ(xk)e(xknl)). (3.3)

    Let

    G(r,χ):=qh=1χ(h)e(rhq)

    be the Gauss sum, and we know that for χχ0,

    χ(xi)=1qqr=1G(r,χ)e(xirq)=1qq1r=1G(r,χ)e(xirq),

    and

    lnrq0

    for 1ln,1rq1 and (n,q)=1.

    Therefore,

    xkδkqχ(xk)e(xknl)=1qq1rk=1G(rk,χ)f(δk,l,rk;n,q)e(rkqlh)1, (3.4)

    where

    f(δ,l,r;n,p):=1e((lnrq)δq)

    and

    |f(δk,l,rk;n,q)|2.

    For xi(1ik1), using Lemma 2.2, we also have

    xiδiq1α,β(xi)χ(xi)e(xinl)=1qxiδiq1α,β(xi)q1ri=1G(ri,χ)e((lnriq)xi)=1qq1ri=1G(ri,χ)xiδiq1α,β(xi)e((lnriq)xi)=1qq1ri=1G(ri,χ)(α1aδiqe((lnriq)xi)+O(qεlnriq+qε))=1qαq1ri=1G(ri,χ)(f(δi,l,ri;n,q)e(riqln)1+O(qεlnriq+qε)). (3.5)

    Let

    S23=1nϕ(q)χmodqχχ0χ(¯c)nl=1k1i=1(1qαq1ri=1G(ri,χ)f(δi,l,ri;n,q)e(riqln)1)(1qq1rk=1G(rk,χ)f(δk,l,rk;n,q)e(rkqln)1)=1nϕ(q)qkαk1nl=1q1r1=1q1rk=1f(δ1,l,r1;n,q)f(δk,l,rk;n,q)(e(r1qln)1)(e(rkqln)1)×χmodqχχ0χ(¯c)G(r1,χ)G(rk,χ). (3.6)

    From the definition of Gauss sum and Lemma 2.3, we know that

    χmodqχ(¯c)G(r1,χ)G(rk,χ)=q1h1=1q1hk=1χmodqχ(¯c)χ(h1)χ(hk)e(r1h1++rkhkq)=ϕ(q)q1h1=1q1hk=1h1hkcmodqe(r1h1++rkhkq)=ϕ(q)q1h1=1q1hk=1e(r1h1+rk1hk1+rkc¯h1hk1q)=ϕ(q)Kl(r1,r2,,rkc;q)ϕ(q)qk12kω(q)(r1,rkc,q)12(rk1,rkc,q)12ϕ(q)qk12kω(q)(r1,q)(rk,q). (3.7)

    By Mobius inversion, we get

    G(r,χ0)=qh=1e(rhq)=μ(q(r,q))φ(q)φ(q/(r,q))(r,q),

    and

    χ0(¯c)G(r1,χ0)G(rk,χ0)(r1,q)(rk,q).

    Hence,

    χmodqχχ0χ(¯c)G(r1,χ)G(rk,χ)=χmodqχ(¯c)G(r1,χ)G(rk,χ)χ0(¯c)G(r1,χ0)G(rk,χ0)ϕ(q)qk12kω(q)(r1,q)(rk,q). (3.8)

    From (3.8) we may deduce the following result:

    S23kω(q)nqk+12αk1nl=1(q1r=1(r,q)|e(rqln)1|)kkω(q)nqk+12αk1nl=1(q1r=1(r,q)|sinπ(rqln)|)kkω(q)nqk+12αk1nl=1(q1r=1(r,q)rqln)k=kω(q)nqk+12αk1nl=1(dqd<qrq1(r,q)=ddrqln)k=kω(q)nqk+12αk1nl=1(dqd<qdmq1d(m,q)=11mdqln)k=kω(q)nqk+12αk1nl=1(dqd<qdkqμ(k)mq1kd1mkdqln)k.

    It is easy to see

    mkdqln=mknl(q/d)(q/d)n1(q/d)n,

    and we obtain

    S23kω(q)nϕ(q)qk+12αk1nl=1(dqd<qdkqmq1kdmin(qnd,1mkdqln))k.

    Let kd/q=h0/q0, where q01,(h0,q0)=1, and we will easily obtain q/(kd)q0q/d. By using Lemma 2.4, we have

    S23kω(q)nqk+12αk1nl=1(dqd<qdkq((q1)/(kd)q0+1)(qnd+q0logq0))kkω(q)nqk+12αk1nl=1(dqd<qdkq((q1)/(kd)q/(kd)+1)(qnd+qdlogqd))kkω(q)qk12αk1(dqd<qkqn+logq)kqk12d2k(q)(logq+n)k.

    Let

    S24:=q(k1)(ε)nϕ(q)χmodqχχ0χ(¯c)nl=1k1i=1(1qαq1ri=1G(ri,χ)1lnriq)(1qq1rk=1G(rk,χ)f(δk,l,rk;n,q)e(rkqln)1)

    and

    S25:=q(k1)(ε)nϕ(q)χmodqχχ0χ(¯c)nl=1k1i=1(1qαq1ri=1G(ri,χ))(1qq1rk=1G(rk,χ)f(δk,l,rk;n,q)e(rkqln)1).

    By the same argument of S23, it follows that

    S24qk12εd2k(q)(logq+n)k,
    S25qk32+ε(logq+n).

    Since nq13, we have

    S25S24S23qk12+εnkqk2+ε. (3.9)

    Taking n=1, we get

    S12qk12+ε. (3.10)

    With (3.1), (3.2), (3.9) and (3.10), the proof is complete.

    This paper considers the high-dimensional Lehmer problem related to Beatty sequences over incomplete intervals. And we give an asymptotic formula by the properties of Beatty sequences and the estimates for hyper Kloosterman sums.

    This work is supported by Natural Science Foundation No. 12271422 of China. The authors would like to express their gratitude to the referee for very helpful and detailed comments.

    The authors declare that there are no conflicts of interest regarding the publication of this paper.



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