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The Least Prime Number in a Beatty Sequence THE LEAST PRIME NUMBER IN A BEATTY SEQUENCE JÖRN STEUDING AND MARC TECHNAU Abstract. We prove an upper bound for the least prime in an irra- tional Beatty sequence. This result may be compared with Linnik’s theorem on the least prime in an arithmetic progression. 1. Introduction and Statement of the Main Results In 1944, Yuri Linnik [11, 12] showed that for every sufficiently large q and coprime a, there exists a constant ℓ such that the least prime p in the prime residue class a mod q satisfies p cqℓ, where c is another absolute ≤ constant; the best bound so far for the appearing constant in the exponent is ℓ 5 due to Triantafyllos Xylouris [21]. This question dates back to ≤ Savardaman Chowla [5] who conjectured in 1934 that one may even take ℓ =1+ ǫ. Two years later Pál Turán [15] proved that this is true for almost all q and that it holds in general under the assumption of the generalized Riemann hypothesis. In this note we shall investigate the corresponding question about the least prime in a Beatty sequence. Given a positive real number α and a non-negative real β, the associated (generalized) Beatty sequence is defined by (α, β)= nα + β : n N , B {⌊ ⌋ ∈ } where x denotes the largest integer less than or equal to x. If α is ra- ⌊ ⌋ tional, then (α, β) is a union of residue classes and, if at least one of B them is a prime residue class, we may apply Linnik’s theorem to bound the least prime (see the concluding remarks, § 5). Otherwise, if α is irrational, (α, β) does not contain an entire residue class. It follows from a classi- B arXiv:1512.08382v2 [math.NT] 30 May 2016 cal exponential sum estimate due to Ivan M. Vinogradov [17] that there exist infinitely many prime numbers in such a Beatty sequence (details in § 2), hence, in particular there exists a least prime number. However, the problem of estimating the size of the least prime is clearly different as compared to the rational case. In fact, for an integer m 2 consider ≥ Date: November 2015, revised May 2016. 2010 Mathematics Subject Classification. 11N13, 11B83, 11K60. Key words and phrases. Prime number, Beatty sequence, Linnik’s theorem. 1 2 JÖRN STEUDING AND MARC TECHNAU α =4+ √2/m. Since nα =4n for n =1, 2,..., m/√2 =: M, there m ⌊ m⌋ ⌊ ⌋ is no prime amongst the first M elements of (α , 0). It turns out that B m the diophantine character of α has an impact on a non-trivial bound for the least prime in the associated Beatty sequence: Theorem 1. For every positive ǫ there exists a computable positive integer ℓ such that for every irrational α> 1 the least prime p in the Beatty sequence (α, β) satisfies the inequality B 35 16ǫ 2(1 ǫ) 1+ǫ (1) p − α − Bp , ≤L m+ℓ where B = max 1, β , = log(2αB), p denotes the numerator of the n-th { } L n convergent to the regular continued fraction expansion of α = [a0, a1,...] and m is the unique integer such that (2) p 16α2 <p . m ≤L m+1 Since the sequence of numerators pn is recursively given by (3) p 1 =1, p0 = a0 = α , and pn+1 = an+1pn + pn 1 − ⌊ ⌋ − with partial quotients a N for n N, the p ’s are strictly increasing n ∈ ∈ n integers and the integer m satisfying (2) is uniquely determined. In view of the explicit exponents and absolute constants m and ℓ, Theorem 1 may be regarded as the analogue of Linnik’s theorem on the least prime in an arithmetic progression. A careful analysis of the reasoning allows to assign the quantity k from Theorem 1 an explicit (possibly not optimal) value: 44 Theorem 2. On the hypothesis of Theorem 1, and assuming ǫ< 2025 , the least prime p in the Beatty sequence (α, β) satisfies the inequality (1), B where for ℓ one may take the least positive integer satisfying 1 1 3 (4) ℓ 3+9ǫ− 41 + log 1+ ǫ− + log 3711+ 2 17− M 5 , ≥ · 4 ǫ where Mǫ is a constant such that (13) holds. 2. Proof of Theorem 1 In the process of proving the ternary Goldbach conjecture for sufficiently large odd integers Vinogradov [17] obtained the estimate 1 q m m4 1/2 exp(2πimap) N 1+ǫ + + + , ≪ǫ N 1/2 N q q2 p N X≤ THE LEAST PRIME NUMBER IN A BEATTY SEQUENCE 3 where m, q, N are positive integers, the exponential sum is taken over all prime numbers p less than or equal to N, and q is related to a by the existence of an integer a such that a 1 (5) a < . − q q2 For irrational a Vinogradov’s bound is o(π(N)), where π(N) denotes the number of primes p N. Letting a = 1 , this implies that the sequence of ≤ α numbers αp, where p runs through the prime numbers in ascending order, is uniformly distributed modulo one, answering a question of the young Paul Erdős (cf. Turán [15], p. 227). This means that the proportion of fractional parts αp := αp αp which fall in an interval [a, b) [0, 1) is equal { } − ⌊ ⌋ ⊂ to b a (the length of the interval) as follows from a well-known criterion − due to Hermann Weyl [20] stating that a sequence of real numbers xn is uniformly distributed modulo one if and only if, for every integer h =0, 6 1 lim exp(2πihxn)=0. N N →∞ n N X≤ Notice that an integer m lies in (α, β) if and only if m = nα + β for B ⌊ ⌋ some n N. Equivalently, the inequalities ∈ (6) nα + β 1 < m nα + β − ≤ hold. In order to find a prime number p in (α, β) we thus need B p β 1 β (7) − , mod 1, and p>α + β 1. α ∈ α α − Here the right hand-side is to be interpreted modulo one, and if there lies an β 1 β integer in ( −α , α ) the set consists of two disjoint intervals; in any case the 1 Lebesgue measure of the set on the right equals α . In view of Vinogradov’s aforementioned uniform distribution result the number π (α,β)(N) of primes B p (α, β) with p N satisfies ∈ B ≤ 1 π (α,β)(x) π(x). B ∼ α Already Vinogradov provided an error term estimate here. However, for our purpose we shall use the following theorem of Robert C. Vaughan [16]: Let a R and suppose that a and q are coprime integers satisfying (5). ∈ 1 Moreover, for 0 <δ< 2 , define 1 if δ < θ δ, (8) χδ(θ)= − ≤ 0 if either 1 θ δ or δ < θ 1 , − 2 ≤ ≤ − ≤ 2 4 JÖRN STEUDING AND MARC TECHNAU and to be periodic with period 1. Then, for arbitrary real b, every positive integer N, and any real ǫ> 0, ǫ N 3 1 2 4 Nq a b L 8 4 2 5 5 Λ(n)(χδ(n + ) 2δ) ǫ 1 + N +(Nqδ) + δ N − ≪ q 2 δ n N X≤ L Nq with := log( δ ). Here Λ(n) is the von Mangoldt-function counting prime ν a 1 powers p with weight log p. In view of (7) we shall use this with = α , b 2β 1 1 = 2−α and δ = 2α . This leads to 1 (9) log p + log p = log p + (N, q), α Eα pν N pν <α+β 1 pν N pν X≤(α,β) pν (α,βX −α+β ) X≤ ∈B ∈B −⌊ ⌋ where 1 4 N 3 Nq 2 N 5 L 8 4 ǫ (10) α(N, q) c 1 + N + + 2 (2Nqα) |E | ≤ 2 2α 5 q (2α) ! with L = log(2Nqα) and appropriate absolute constant c depending only on ǫ, but not on α. The second sum on the left hand side of (10) may be estimated using a classical inequality due to John B. Rosser & Lowell Schoenfeld [13] (see Lemma 5 below). The number of prime powers pν N 1 ≤ with ν 2 is less than or equal to π(N 2 ), hence we may replace (9) by ≥ 1 1 log p log p+ (N, q) 1.04(α+β 1)+ 1 log p. ≥ α Eα − − α − p N p N pν N p X≤(α,β) X≤ Xν≤2 ∈B ≥ Notice that the last term is negative; it is obviously bounded by 1 1 1 3 1 1 log p< 1 π(N 2 ) log N < 1+ N 2 , − α − α log N pν N Xν≤2 ≥ where we have used a classical inequality for the prime counting function π(x) valid for all x also due to Rosser & Schoenfeld [13]. Using another one of their explicit inequalities, namely N log p > N for N 41, − log N ≥ p N X≤ we thus find a prime p N in (α, β) if we can show that ≤ B N 1 3 1 1 > (N, q) + 1+ N 2 +1.04(α + β 1), α − log N |Eα | log N − THE LEAST PRIME NUMBER IN A BEATTY SEQUENCE 5 which we may also replace by N 1 0.73 > (N, q) +1.81N 2 +1.04(α + β 1). α |Eα | − By (10) this inequality is satisfied if α α2 + αβ α 0.73 >1.81 1 +1.04 − N 2 N 1 3 5 L 8 α α qα 2 α ǫ + c 1 + 1 + + 1 (2Nqα) .
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