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kth powers in a generalization of Piatetski-Shapiro sequences

  • The article considers a generalization of Piatetski-Shapiro sequences in the sense of Beatty sequences. The sequence is defined by (αnc+β)n=1, where α1, c>1, and β are real numbers.

    The focus of the study is on solving equations of the form αnc+β=smk, where m and n are positive integers, 1nN, and s is an integer. Bounds for the solutions are obtained for different values of the exponent k, and an average bound is derived over k-free numbers s in a given interval.

    Citation: Yukai Shen. kth powers in a generalization of Piatetski-Shapiro sequences[J]. AIMS Mathematics, 2023, 8(9): 22411-22418. doi: 10.3934/math.20231143

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  • The article considers a generalization of Piatetski-Shapiro sequences in the sense of Beatty sequences. The sequence is defined by (αnc+β)n=1, where α1, c>1, and β are real numbers.

    The focus of the study is on solving equations of the form αnc+β=smk, where m and n are positive integers, 1nN, and s is an integer. Bounds for the solutions are obtained for different values of the exponent k, and an average bound is derived over k-free numbers s in a given interval.



    The Piatetski-Shapiro sequences are defined by N(c)=(nc)n=1, where c>1 and cN. In 1953, Piatetski-Shapiro [8] proved that N(c) contains infinitely many primes if c(1,1211). Moreover, he showed that the prime counting function π(c)(x), which counts the number of primes in N(c) up to x, satisfies the asymptotic relation π(c)(x)x1/clogx as x. Since then, the range of c for which it is known that N(c) contains infinitely many primes has been extended several times, and it is now known that the above formula holds for all c(1,28172426) thanks to the work of Rivat and Sargos [10].

    This study is related to the topic of Beatty sequences. A non-homogeneous Beatty sequence is a sequence of integers obtained from fixed real numbers α>0 and β, defined as Bα,β=(αn+β)n=1, which is sometimes referred to a generalized arithmetic progression. It is well-known that the sequence contains infinitely many prime numbers if α is irrational [4]. Moreover, it is possible to establish an asymptotic relation for the distribution of primes in such sequences, which is the subject of extensive research in number theory. We get that

    #{prime px:pBα,β}α1π(x),x

    holds, where π(x) is the prime counting function.

    The author of this study proposes a generalization of Piatetski-Shapiro sequences in the context of Beatty sequences, which similarly consists of infinitely many prime numbers. Specifically, let α1 and β be real numbers. We investigate the following generalized Piatetski-Shapiro sequences:

    N(c)α,β=(αnc+β)n=1.

    The Piatetski-Shapiro sequences have deep connections to several fundamental concepts in number theory, such as smooth numbers, square-free numbers and so on. Previous research by Liu, Shparlinski, and Zhang [5] focused on the distribution of squares in Piatetski-Shapiro sequences, while Qi, Guo, and Xu [9] investigated an intriguing equation related to these sequences. In this paper, we aim to extend their work by exploring the distribution of k-th powers in a generalization of Piatetski-Shapiro sequences, which can be viewed as an extension of their previous findings. To be precise, we define Qα,βc,k(s;N) as the number of solutions to the equation:

    αnc+β=smk,1nN,m,nZ.

    We mention the trivial bound

    Qα,βc,k(s;N)min(N,s1k(αNc+β)1k).

    We prove the following theorem.

    Theorem 1.1. Let k>1 be an integer. For any exponent pair (κ,λ), we have

    Qα,βc,k(s;N)=γ(kγk+1)1s1kN1c+ck+O(sλk(1+κ)Nkκ+cλk(1+κ)+ε+sκλkNkκ+c(λκ)k+ε+s1kNc+ck).

    We also study Qα,βc,k(s;N) on average over positive k-free integers sS. Recall that γ=c1and define that

    Qα,βc,k(S,N)=sSs is k-free Qα,βc,k(s;N).

    We remark that only the case SαNc+β is meaningful, hence we always assume this. Liu, Shparlinski and Zhang [5] showed that

    Q1,0c,2(S,N)=12γπ2(2γ1)S11kN1c2+O(S15N15+2c5+S58N3c8+S18N14+3c8+SN1c).

    We obtain the following result.

    Theorem 1.2. For any c>1,cN, we have

    Qα,βc,k(S,N)=kζ(k)(k1)(kγk+1)S11kN1c+ck+O(S3545kN15+4c5k+S134kN3c4k+S1234kN14+3c4k+SN1c+S11kNc+ck).

    We remark that the topic is relative to harmonic numbers and their relationships which are interesting mathematical concepts, including number theory, calculus, and physics, as well as the study of degenerate versions or special cases.For further details, refer to [6,7].

    We denote by t and {t} the greatest integer t and the fractional part of t, respectively. We also write e(t)=e2πit for all tR, as usual. We make considerable use of the sawtooth function defined by

    ψ(t)=tt12={t}12(tR).

    In this study, we consider the Piatetski-Shapiro sequence (nc)n=1, where denotes the floor function, and γ is defined as the inverse of the constant c. The set of primes in the natural numbers is denoted by P. We use the notation mM to indicate that m lies in the interval (M,2M].

    In order to state our results, we introduce some notation. Throughout the paper, the symbol ε represents an arbitrarily small positive constant, which may vary from one occurrence to another. The implied constants in the symbols O, , and may depend on the parameters c, ε, α, and β, but are absolute otherwise. For given functions F and G, the notations FG, GF, and F=O(G) are all equivalent to the assertion that the inequality |F|C|G| holds with some constant C>0.

    Lemma 2.1. Let

    L(Q)=Ii=1AiQai+Jj=1BjQbj,

    where Ai,ai,Bj,bj>0. Then, (1) for any Q2Q1>0 there exists Q[Q1,Q2] such that

    L(Q)Ii=1Jj=1(AbjiBaij)1ai+bj+Ii=1AiQai1+Jj=1BjQbj2.

    (2) For any Q1>0 there exists Q(0,Q1] such that

    L(Q)Ii=1Jj=1(AbjiBaij)1ai+bj+Jj=1BjQbj1.

    Proof. See [3, Lemma 2.4]

    Lemma 2.2. For any J>0, there holds

    ψ(x)=1|j|Jaje(jx)+O(|j|Jbje(jx)),

    where

    aj|j|1,bjJ1.

    Proof. This is the result of Vaaler [11].

    Lemma 2.3. (1384+ε,5584+ε) is an exponent pair.

    Proof. See [1, Theorem 6].

    Lemma 2.4. Let α,α1,α2 be real constants such that

    α1andαα1α20.

    Let M,M1,M2,x1 and let

    Φ=(φm)mMandΨ=(ψm1,m2)m1M1,m2M2

    be two sequences of complex numbers supported on mM,m1M1 and m2M2 with |φm|1. Then, for the sum

    SΦ,Ψ(x;M,M1,M2)=mMm1M1m2M2φmψm1,m2e(xmαmα11mα22MαMα11Mα22)

    we have

    SΦ,Ψ(x;M,M1,M2)(x14M12(M1M2)34+M710M1M2+M(M1M2)34+x14M1110M1M2)log2(2MM1M2).

    Proof. See [2, Theorem 3]

    Denote γ=c1 and θ=αγ. A kth power equals αnc+β if and only if

    smkαnc+β<smk+1,

    which is equivalent to

    θ(smkβ)γn<θ(smk+1β)γ.

    Let M=s1k(αNc+β)1k, by a normal construction,

    Qα,βc,k(s;N)=mM(θ(smkβ)γθ(smk+1β)γ)+O(1)=S1+S2+O(1), (3.1)

    where

    S1=mM(θ(smk+1β)γθ(smkβ)γ)=θmM(γ(smkβ)γ1+O((smkβ)γ2))=θmM(γsγ1mk(γ1)+O(sγ2mk(γ2)))=γ(kγk+1)1s1kN1c+ck+O(s1kNc+ck), (3.2)

    and

    S2=mM({θ(smk+1β)γ}{θ(smkβ)γ})=mM(ψ(θ(smk+1β)γ)ψ(θ(smkβ)γ)).

    Consider S2. From Lemma 2.2 we have

    S2=S3+O(S4), (3.3)

    where

    S3=mM1|j|Jaj(e(jθ(smk+1β)γ)e(jθ(smkβ)γ)),

    and

    S4=mM|j|Jbj(e(jθ(smk+1β)γ)+e(jθ(smkβ)γ)),

    for any J1. We begin with S3. Remembering aj|j|1, we have

    S31|j|J|j|1|mMe(jθsγmkγ)|.

    Summing over m, we obtain

    mMe(jθsγmkγ)logMmax1LM|Lm2Le(jθsγmkγ)|.

    Using the exponent pair (κ,λ) we get

    Lm2Le(jθsγmkγ)(jθsγLkγ1)κLλ.

    Then,

    mMe(jθsγmkγ)jκθκsκλkNc(kγκκ+λ)k+ε,

    which yields

    S3JκθκsκλkNc(kγκκ+λ)k+ε.

    We can readily eliminate the contribution of S4. By utilizing the fact that bj is bounded by J1, we obtain the following result:

    S4J1|j|J|mMe(jθsγmkγ)|J1s1kNck+J11|j|J|mMe(jθsγmkγ)|.

    By a similar argument, we have

    S4J1s1kNck+JκsκλkNc(kγκκ+λ)k+ε.

    It yields that

    J=sλκ1k+kκNc(1+κkγκλ)k+kκ.

    Applying Lemma 2.1 to the bounds on terms in (3.3), we have

    S2sλk(1+κ)Nkκ+cλk(1+κ)+ε+sκλkNkκ+c(λκ)k+ε.

    Now the result follows from (3.3) and (3.1). Applying the exponent pair (1384+ε,5584+ε) from Lemma 2.3 by Bourgain [1], people can get the asymptotic formula

    Qα,βc,k(s;N)=γ(kγk+1)1s1kN1c+ck+O(s5597kN1397+55c97k+ε+s12kN1384+c2k+ε).

    This proof is almost identical to the proof given in [5, Theorem 2.3], so we will provide only a brief outline. We start that

    Φk(S)=sSs is k-free s1k.

    Applying a commonly known result, (see [11, p. 181]),

    sSs is k-free 1=Sζ(k)+O(S1k),

    and a partial summation, we obtain

    Φk(S)=kζ(k)(k1)S11k+O(logS). (4.1)

    We proceed as in the proof of Theorem 1.1, so set θ=αγ, T=(αNc+β)γ, and

    Qα,βc,k(S;N)=K0+K1+O(1),

    where

    K0=smkαNc+βsSs is k-free (θ(smk+1β)γθ(smkβ)γ),

    and

    K1=smkαNc+βsSs is k-free (ψ(θ(smk+1β)γ)ψ(θ(smkβ)γ)).

    Using (3.2) and (4.1), we compute K0 directly as follows

    K0=γ(kγk+1)1N1c+ckΦk(S)+O(SN1c+S11kNc+ck).

    By Lemma 2.2

    K1K11+K12,

    where

    K11=smkαNc+βsSs is k-free 0<|j|<Jaj(ψ(θ(smk+1β)γ)ψ(θ(smkβ)γ)),

    and

    K12=smkαNc+βsSs is k-free 0|j|<Jbj(ψ(θ(smk+1β)γ)ψ(θ(smkβ)γ)).

    Using a similar approach in [5, p. 250] we have

    K110<|j|<Jj1|S(R,D,M;j)|,

    where

    S(R,D,M;j)=rR,dD,mMrdkmkαNc+β,rdkSμ(d)e(jθrγdkγmkγ).

    From Lemma 2.4, we have

    S(R,D,M;j)=(jRγDkγMkγ)14R12(DM)34+R710DM+R(DM)34+(jRγDkγMkγ)14R1110DM.

    Noting that γ>12, it can be easily verified that the fourth term can be combined with the third term on the right-hand side. We can obtain

    K11j14N14+3c4kS1234k+NckS7101k+N3c4kS134k.

    Hence,

    |K11|+|K12|J14N14+3c4kS1234k+J1NckS11k+NckS7101k+N3c4kS134k,

    where the term J1NckS11k results from the choice j=0 in the summation on the right-hand side in Lemma 2.2. Now Lemma 2.1 gives

    |K11|+|K12|N15+4c5kS3545k+N3c4kS134k+N14+3c4kS1234k.

    We have the final result.

    We have proved Theorems 1.1 and 1.2.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    The author is supported in part by the Fundamental Research Funds for the Central Universities (No. xzy012021030) and the Shaanxi Fundamental Science Research Project for Mathematics and Physics (No. 22JSY006).

    We declare no conflict of interest.



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