Arithmetic functions on Beatty

Alex G. Abercrombie 3 Middle Row, Golden Hill Pembroke SA71 4TD, UK [email protected] William D. Banks Department of University of Missouri Columbia, MO 65211, USA [email protected] Igor E. Shparlinski Department of Computing Macquarie University Sydney, NSW 2109, Australia [email protected]

Abstract We study sums of the form

Sα(f,x)= f(n), 6 n x, n∈Bα where f is an arbitrary arithmetic function satisfying a mild growth condition, and Bα = (⌊αk⌋)k∈N is the homogeneous Beatty corresponding to a α > 1. We show that for almost all α> 1 the asymptotic formula −1 Sα(f,x) ∼ α f(n) (x →∞) 6 nx holds, and we give a strong bound on the error term. Our results extend and improve the earlier results of several authors.

1 1 Introduction

1.1 Background For a real number α > 1, the homogeneous Beatty sequence corresponding to α is the sequence of natural numbers given by

Bα =(⌊αk⌋)k∈N, where ⌊t⌋ denotes the greatest 6 t. Beatty sequences appear in a variety of contexts and have been extensively explored in the literature. In particular, summatory functions of the form

Sα(f, x)= f(n) (1) n6x, n∈B α have been studied when the arithmetic function f is • a multiplicative or an additive function (see [1, 2, 8, 9, 10, 11]);

• a Dirichlet character (see [2, 3, 5]);

• the characteristic function of primes or smooth numbers (see [4, 6, 7]). For an arbitrary arithmetic function f we denote

S(f, x)= S1(f, x)= f(n). (2) n6x Abercrombie [1] has shown that for the divisor function τ the asymptotic formula −1 5/7+ε Sα(τ, x)= α S(τ, x)+ O(x ) (3) holds for any ε> 0 and almost all α> 1 (with respect to Lebesgue measure), where the implied constant depends only on α and ε. This result has been improved and extended by Zhai [14] as follows. For a fixed integer r > 1, let τr(n) be the number of ways to express n as a product of r natural numbers, expressions with the same factors in a different order being counted as different (in particular, τ2 = τ is the usual divisor function). In [14] it is shown that the asymptotic formula

−1 (r−1)/r+ε Sα(τr, x)= α S(τr, x)+ O(x ) (4)

2 holds for any ε> 0 and almost all α> 1 (in the special case r = 2 a similar result has also been obtained by Begunts [8]). The estimate (4) has been further improved by L¨uand Zhai [11] as follows:

O(x(r−1)/r+ε) if2 6 r 6 4; S (τ , x)= α−1S(τ , x)+ (5) α r r O(x4/5+ε) if r > 5. 1.2 Our result In this paper, we use the methods of [1] to derive an asymptotic formula for Sα(f, x) which holds for almost all α > 1 whenever f satisfies a rather mild growth condition. In particular, we do not stipulate any conditions on the multiplicative or additive properties of f (or on any other properties of f except for the rate of growth). Our general result, when applied to the divisor functions, yields a statement stronger than (3) and an improvement of (5) for all r > 4, and it can be applied to many other number theoretic functions (and to powers and products of such functions), including:

ˆ the M¨obius function µ(n),

ˆ the Euler function ϕ(n),

ˆ the number of prime divisors ω(n),

ˆ the sum σg(n) of the digits of n in a given base g > 2.

On the other hand, we note that although the results of [1, 11, 14] are formulated as bounds which hold for almost all α, the methods of those papers are somewhat more explicit than ours, and the results can be applied to any “individual” numbers α whose rational approximations satisfy certain hypotheses; thus, one can derive variants of (3), (4) and (5) for specific values of α (or over some interesting classes of α, such as the class of algebraic numbers).

1.3 Notation Throughout the paper, implied constants in the symbols O, ≪ and ≫ may depend (where obvious) on the parameters α,ε but are absolute otherwise. We recall that the notations U = O(V ), U ≪ V , and V ≫ U are all

3 equivalent to the assertion that the inequality |U| 6 cV holds with some constant c> 0. We also use t to denote the distance from t ∈ R to the nearest integer.

1.4 Acknowledgements During the preparation of this paper, I. S. was supported in part by ARC grant DP0556431.

2 Main Result

2.1 Formulation We denote −1 ∆α(f, x)= Sα(f, x) − α S(f, x) (6) and M(f, x)=1+max |f(n)| : n 6 x .

Theorem 1. For fixed ε> 0 and almost all real numbers α> 1, the following bound holds: 2/3+ε ∆α(f, x) ≪ x M(f, x).

2.2 Preparations

We follow the arguments of [1]. For any real number x > 1, let ψx be the trigonometric polynomial of Vaaler [13] given by

2πimt ψx(t)= ax(m) e (t ∈ R), 1/2 16|m|6x where for each integer m in the sum we put

πm (1 −|m |) cot(πm )+ |m | m a (m)= − x x x x with m = . (7) x 2πim x x1/2 +1 As in [1, Section 3] we note that the inequality

u(1 − u) cot(πu) 6 1 (0 6 u 6 1) 4 immediately implies the uniform bound 1 a (m) ≪ (1 6 |m| 6 x1/2). (8) x |m|

The function ψx is an exceptionally good approximation to the “sawtooth” function ψ(t) = {t} − 1/2, where {t} denotes the fractional part of t ∈ R. Indeed, by [1, Corollary 2.9] we have

csc2(πt) csc2(πt) ψ(t) − ψ (t) 6 ≪ . (9) x 2(x1/2 + 1)2 x To prove the theorem, we can clearly assume that α> 1 is irrational. In this case, one sees that a natural number n is a term in the Beatty sequence −1 Bα (that is, n = ⌊αk⌋ for some k ∈ N) if and only if α n lies in the

t ∈ R : 1 − α−1 6 {t} < 1 . As the characteristic function ξα of that set satisfies the relation

−1 −1 ξα(t)= α + ψ(t) − ψ(t + α ) for every t ∈ R, it follows that

−1 f(n)= f(n) ξα(α n) 6 n6x n x, n∈Bα = f(n) α−1 + ψ(α−1n) − ψ(α−1(n + 1)) . n6x Taking into account the definitions (1), (2) and (6), we see that

∆α(f, x) 6 Qα(f, x) + |f(n)| Rα(n, x), (10) n6x where −1 −1 Qα(f, x)= f(n) ψx(α n) − ψx(α (n + 1)) , n6x and

−1 −1 −1 −1 Rα(n, x)= ψ(α n) − ψx(α n) + ψ(α (n + 1)) − ψx(α (n + 1)) . 5 2.3 Growth of the function Qα(f,x) We need the following estimate on the finite differences of the function Qα(f, x) which could be of independent interest. Lemma 1. For a fixed irrational α> 1 we have

Qα(f,y) − Qα(f, x) ≪ (y − x) M(f,y) (1 6 x 6 y 6 2x).

Proof. For any t ∈ R we have

ψy(t) − ψx(t)= S1 + S2, where

2πimt 2πimt S1 = (ay(m) − ax(m)) e and S2 = ay(m) e . 1/2 1/2 1/2 16|m|6x x <|m|6y In view of (8) the latter sum is bounded by

1 y1/2 − x1/2 y − x S ≪ ≪ ≪ . 2 |m| x1/2 x 1/2 1/2 x <|m|6y

To bound S1, we put

F (u)= πu(1 −|u|) cot(πu)+ |u|,

1/2 so that ax(m) = −F (mx)/(2πim) in the notation of (7). If 1 6 |m| 6 x then m(x1/2 − y1/2) |m|(y − x) m − m = ≪ , y x (x1/2 + 1)(y1/2 + 1) x3/2 and since F is continuous and piecewise-differentiable on the interval (−1, 1) it follows that F (m ) − F (m ) y − x a (m) − a (m)= − y x ≪ . y x 2πim x3/2 Therefore, y − x S 6 a (m) − a (m) ≪ . 1 y x x 6 6 1/2 1 |m| x 6 Thus, we have established the uniform bound y − x ψ (t) − ψ (t) ≪ (t ∈ R, 1 6 x 6 y 6 2x). (11) y x x Now write Qα(f,y) − Qα(f, x)= S1 + S2 + S3, where −1 −1 S1 = f(n) ψy(α n) − ψx(α n) , n6x −1 −1 S2 = − f(n) ψy(α (n + 1)) − ψx(α (n + 1)) , n6x −1 −1 S3 = f(n) ψy(α n) − ψy(α (n + 1)) . x

Sj ≪ (y − x)M(f, x) (j =1, 2), and clearly, S3 ≪ (y − x)M(f,y). This completes the proof. 2.4 Concluding the proof

−1 Now put λ = α and expand Qα(f, x) as a Fourier series in λ:

2πimnλ 2πim(n+1)λ Qλ−1 (f, x)= f(n) ax(m) e − e n6x 16|m|6x1/2 2πimnλ = g(n) ax(m) e n6x+1 1/2 16|m|6x 2πikλ = e g(n) ax(m), 16|k|6(x+1)x1/2 n6x+1 |m|6x1/2 nm=k where f(n) if n = 1; g(n)= f(n) − f(n − 1) if2 6 n 6 x;   −f(n − 1) if x < n 6 x + 1.  7 By the Parseval identity we have

2 1 2 Qλ−1 (f, x) dλ = g(n) ax(m) . (12) 0 16|k|6(x+1)x1/2 n6x+1 1 2 |m|6x / nm=k The inner sum on the right of (12) is bounded above by 1 g(n) a (m) ≪ M(f, x) x |m| n6x+1 6 6 1/2 1 2 |k|/(x+1) |m| x |m|6x / m| k nm=k x ≪ M(f, x) τ(|k|) min 1, . |k| Thus, the integral on the left of (12) is bounded by

1 2 2 τ(k) −1 2 2 2 Qλ (f, x) dλ ≪ τ(k) + x 2 M(f, x) 0 k k6x x

τ(k)2 ≪ x(log x)3 6 k x together with partial summation (for the second sum). Now put 3 Θ= , 1+3ε and observe that the preceding bound implies

1 Θ 2 Θ 3 Θ 2 Qλ−1 (f, N ) dλ ≪ N (log N) M(f, N ) (N > 1). 0 Then, since

∞ Θ 2 ∞ 1 Q −1 (f, N ) 1 λ dλ ≪ < ∞, N Θ+1(log N)6M(f, N Θ)2 N(log N)3 N=1 0 N=1 8 it follows that the integral

∞ Θ 2 1 Q −1 (f, N ) λ dλ N Θ+1(log N)6M(f, N Θ)2 0 N=1 converges. This implies that the series

∞ Θ 2 Qα(f, N ) N Θ+1(log N)6M(f, N Θ)2 N=1 converges for almost all α > 1. Let α be fixed with that property, and note that Θ (Θ+1)/2 3 Θ Qα(f, N ) ≪ N (log N) M(f, N ) (N > 1). For any given real number x > 1, let N be the unique integer for which N Θ 6 x< (N + 1)Θ. Then,

Θ (Θ+1)/(2Θ) 3 Qα(f, N ) ≪ x (log x) M(f, x) = x2/3+ε/2(log x)3M(f, x) ≪ x2/3+ε M(f, x). By Lemma 1 we also see that

Θ Θ Θ Qα(f, x) − Qα(f, N ) ≪ (N + 1) − N M(f, x) Θ−1 ≪ N M(f, x) ≪ x(Θ−1)/Θ M(f, x)= x2/3−ε M(f, x). Therefore, 2/3+ε Qα(f, x) ≪ x M(f, x) (13) for almost all α. To bound the sum in (10) we put log x L = , 2log2 and for each j =1,...,L we denote by Nj the set of natural numbers n 6 x for which 2−j−1 < min α−1n, α−1(n + 1) 6 2−j.

We also denote by N⋆ the set of natural numbers n6 x such that min α−1n, α−1(n + 1) 6 2−(L+1). 9 If n ∈Nj, then (9) implies that

2 −1 2 −1 −1 Rα(n, x) ≪ csc (πα n) + csc (πα (n + 1)) x −1 −2 −1 −2 −1 2j −1 ≪ α n + α (n + 1) x ≪ 2 x , and the bound |ψ(t) − ψx(t)| 6 1, which follows from [1, Lemma 2.8] (which in turn follows from [13]), implies that Rα(n, x) ≪ 1 holds for all n ∈ N⋆; therefore,

L −1 2j |f(n)| Rα(n, x) ≪ x 2 |f(n)| + |f(n)| n6x j=1 n∈N n∈N j ⋆ L −1 2j 6 x 2 Nj + N⋆ M(f, x). j=1 Using [1, Lemma 2.4 and Corollary 2.7] one sees that for almost all α > 1 and uniformly for x > 1, the upper bounds

−j 3 Nj ≪ 2 x + (log x) (j =1,...,L) and −L 3 N⋆ ≪ 2 x + (log x) hold. Since 2L ≍ x1/2, it follows that 1/2 |f(n)| Rα(n, x) ≪ x M(f, x) (14) n6x for almost all α> 1. Combining (10), (13) and (14), we obtain the stated result.

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