arXiv:1512.08382v2 [math.NT] 30 May 2016 n coprime and aadmnCol 5 h ojcue n13 htoemyev may one that 1934 in conjectured ℓ who [5] Chowla Savardaman osat h etbuds a o h pern osatin constant appearing the for far is so bound best the constant; where by defined is sequence Beatty (generalized) rm eiu class residue prime all usinaottelatpiei etysequence. co Beatty the a investigate in shall prime least we the note about this question In hypothesis. Riemann inl then tional, es rm setecnldn eak,§5.Ohrie if Otherwise, 5). § t remarks, theorem concluding Linnik’s the apply may (see we prime class, least residue prime a is them B scmae otertoa ae nfc,fra integer an clear for is fact, prime In least case. the numbe rational of prime the least size to a the compared sequence exists estimating as Beatty there of particular a problem in such t the hence, [17] in 2), Vinogradov numbers § M. prime in Ivan many to infinitely due exist estimate sum exponential cal H ES RM UBRI ETYSEQUENCE BEATTY A IN NUMBER PRIME LEAST THE ( + 1 = n14,Yr ink[1 2 hwdta o vr sufficientl every for that showed 12] [11, Linnik Yuri 1944, In e od n phrases. and words Key 2010 Date ie oiiera number real positive a Given ,β α, ℓ q ≤ n hti od ngnrludrteasmto ftegener the of assumption the under general in holds it that and 1. ) oebr21,rvsdMy2016. May revised 2015, November : ⌊ ahmtc ujc Classification. Subject Mathematics inlBat eune hsrsl a ecmae ihLin with progression. arithmetic compared an be in prime may least result the This on theorem sequence. Beatty tional Abstract. 5 ǫ x osntcnana niersdecas tflosfo cl a from follows It class. residue entire an contain not does w er ae á uá 1]poe htti stu o al for true is this that proved [15] Turán Pál later years Two . nrdcinadSaeeto h anResults Main the of Statement and Introduction ⌋ u oTinaylsXlui 2] hsqeto ae bac dates question This [21]. Xylouris Triantafyllos to due eoe h ags nee esta reulto equal or than less integer largest the denotes B a hr xssaconstant a exists there , ( ,β α, ÖNSEDN N ACTECHNAU MARC AND STEUDING JÖRN epoea pe on o h es rm na irra- an in prime least the for bound upper an prove We ) a sauino eiu lse n,i tlatoeof one least at if and, classes residue of union a is B mod ( rm ubr etysqec,Lni’ theorem. Linnik’s sequence, Beatty number, Prime ,β α, q = ) satisfies α {⌊ n o-eaiereal non-negative a and nα 1 + p 11,1B3 11K60. 11B83, 11N13, ℓ ≤ β uhta h es prime least the that such ⌋ cq : ℓ n where , ∈ N } , c saohrabsolute another is β m h associated the , α x h exponent the ≥ If . sirrational, is .However, r. on the bound o rresponding ydifferent ly nik’s 2 a there hat large y consider p α ntake en (details nthe in alized sra- is most assi- to k q 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
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) . q 2 N 4 2N N 5 ! Obviously, N needs to be larger than max α4, β and q larger than α2. { } Since L depends on α and β and we would like to eliminate this depen- dency, we shall take both N and q somewhat larger. Therefore, we make the ansatz N = 35α2Bqηǫ and q = 16α2η with some large parameter η, L L to be specified later, and B = max 1, β . (The exponent 35 was chosen { } with hindsight from both this proof and the proof of Theorem 2; from the above inequality we see that it should be at least 16 and the further in- crease is due to the positive contribution from the ǫ-term.) Then, the latter inequality can be rewritten as 51 1 1 1+ǫ 51 3 1 (1+ǫ) 0.73 >1.81 − 2 α− B− 2 η− 2 +1.04 − (α + β 1)α− B− η− L L − 8 8 1 51 1 1+ǫ 1 35 1 1 ǫ (11) + cL − η− 2 + − 4 B− 4 η− 4 +2− 2 − 2 α− 2 B− 2 η− 2 L L L ǫ 51 +67ǫ 1 +7ǫ 1 +ǫ 1+ǫ +(2+ǫ)ǫ +2 − 5 α− 5 B− 5 η− 5 , L L 67 7 2+ǫ 1 where = log(2 α Bη ). Assuming ǫ< 35 , as we may, all exponents L 1 κ belonging to α, B and η are negative. Since L − η− for any κ> 0, L ≪κ the above inequality is fulfilled for all sufficiently large η, say η η . ≥ 0 Since η is intertwined with q a little care needs to be taken. In order to find a suitable η recall that α is irrational. Hence, by Dirichlet’s approxi- a mation theorem, there are infinitely many solutions q to inequality (5); in view of a = 1 we may take for a the reciprocals of the convergents pn to α q qn the continued fraction expansion of α. (For this and further fundamental results about diophantine approximation and continued fractions we refer pm+ℓ to Hardy & Wright [8].) We shall choose ℓ such that η0 , where m ≤ pm is defined by (2), for then the choice q = p will yield an η η . In m+ℓ ≥ 0 fact, if Fn denotes the n-th Fibonacci number (defined by the recursion Fn+1 = Fn + Fn 1 and the initial values F0 = 0 and F1 = 1), we observe − by (3) and induction that
pm+ℓ pm+ℓ 1 + pm+ℓ 2 ... Fj+1pm+ℓ j + Fjpm+ℓ (j+1) ≥ − − ≥ ≥ − − 6 JÖRN STEUDING AND MARC TECHNAU for j m + ℓ 1. In view of Binet’s formula, ≤ − 1 n n F = (G +( G)− ), n √5 − 1 √ where G = 2 ( 5+1) is the golden ratio, we thus find
pm+ℓ 1 ℓ 1 (12) η F G − . ℓ √5 ≥ pm+1 ≥ ≥
For sufficiently large ℓ the right hand side of this inequality will exceed η0 and, hence, we obtain the desired choices to satisfy (11). This completes the proof of Theorem 1.
3. Proof of Theorem 2 In order to make the inequalities from the previous section effective, we require an effective version of Vaughan’s theorem. In order to state this we
first introduce some notation: let dk(x) denote the number of representa- tions of the positve integer x as a product of exactly k positive integers.
When k = 2 we also write d(x) = d2(x). It is well-known that for every ǫ> 0 there is a constant Mǫ > 0 such that (13) d(x) M xǫ. ≤ ǫ Indeed, one may take
1/ǫ 2 e (14) M = ǫ e log 2 (see, e.g., [18, p. 38]). Now we state:
Theorem 3 ([16]). Let a R and suppose that a and q are coprime integers ∈ 1 L satisfying (5). Suppose that 0 <δ< 2 , := log(Nq/δ) and let χδ be given by (8). Then, for any ǫ> 0 and Mǫ satisfying (13), (15)
3 8 1 3 1 Λ(n)(χ (an b) 2δ) 10 L 3422Nq− 2 + 251N 4 + 38(δNq) 2 δ − − ≤ n N X≤ 3 1 3 ǫ 2 4 +ǫ + 11+17− M 5 (Nqδ− ) 4 δ 5 N 5 . 4 ǫ We postpone the proof to the next section and proceed with the proof of Theorem 2. In order to not be too repetitive we only sketch briefly what modifications are to be made. We continue with the notation introduced in the previous section. First observe that the exponents belonging to the last term in (15) have changed slightly. After making the implied changes, THE LEAST PRIME NUMBER IN A BEATTY SEQUENCE 7
44 assuming ǫ < 2025 , and estimating the arising exponents, (15) may be replaced by 51 1 1 1+ǫ 51 3 1 (1+ǫ) 0.73 >1.81 − 2 α− B− 2 η− 2 +1.04 − (α + β 1)α− B− η− L L − 3 8 8 ǫ 3 + 10 L − η− 2 3711+ 2 17− M 5 , L · 4 ǫ which, if satisfied, ensures the existence of a prime p N in (α, β). Since ≤ B 8 8 ǫ 8 1 8 L − η− 4 < 65 1+ ǫ− L this is surely satisfied if ǫ 3 8 1 8 3 η 4 > 2 10 65 1+ ǫ− 3711+ 2 17− M 5 , · · 4 ǫ Recalling (12) we thus find any choice of ℓ N, such that (4) holds, to be ∈ admissible. This concludes the proof of Theorem 2.
4. Proof of Theorem 3 The present section is devoted to a proof of Theorem 3. To this end, we follow Vaughan [16] and replace his estimates with effective inequalities. 4.1. Outline of the argument. The basic idea is to use a finite Fourier expansion of the function χδ given by (8) (see Lemma 6). The main term
2δ n N Λ(n) arises from the zeroth coefficient in this expansion, whereas all other≤ terms are handled as error, which reduces the problem to estimat- P ing sums of the type Λ(n)e(αhn) , h n X X with variables varying in between suitable ranges. This is achieved by applying Vaughan’s identity to Λ(n), yielding sums of the type
axbye(αhxyz) . h x y z X X X X
These are handled by combining variables in a suitable way and applying some estimates on exponential sums. 4.2. The Piltz divisor function. Since combining variables naturally introduces divisor functions to the coefficients in the new sums we shall need explicit bounds for these. Concerning bounds of the type (13), we have already noted that the choice (14) is admissible for every ǫ > 0. However, for most arguments in the present section we need three special instances, namely ǫ 1 , 1 , 1 . Therefore we would like to do somewhat ∈ { 2 4 6 } better in these cases. 8 JÖRN STEUDING AND MARC TECHNAU
νj Indeed, following a well-known argument, if x = j pj with distinct primes p then it holds that j Q d(x) ν +1 = j =: Π Π , say. xǫ · ǫνj 1 · 2 1/ǫ 1/ǫ pj pj e pj 1 1 1 (16) d(x) min 139x 6 , 9x 4 , 2x 2 . ≤ We also need the following n o Lemma 4. Suppose that X 2. Then ≥ d (x)2 3000X(log X)8, 3 ≤ x X X≤ d(x)2 7X(log X)3. ≤ x X X≤ Proof. We follow the approach presented in [10, sec. 1.6]. We start with the inequality 1 d (pν)2 d (x)2 X 1 3 . 3 ≤ − p pν x X p X ν 0 X≤ Y≤ X≥ The sum on the right hand side can be calculated explicitly and, together with p 2, we obtain ≥ 5 9 5 d (pν)2 1 − p2 +4p +1 1 1 3 = 1 1+ 1+ . pν − p p2 ≤ p p2 ν 0 X≥ By [13, inequality (3.20)], 1 1 1 1+ < exp + +0.26149 ... log X, p (log X)2 2 p X Y≤ so that 1 8 1 5 d (x)2 X 1+ exp − exp , 3 ≤ p p2 p2 x X p X p p X≤ Y≤ X X THE LEAST PRIME NUMBER IN A BEATTY SEQUENCE 9 which is 3000X(log X)8 for X 6100. For smaller values the bound can ≤ ≥ be easily verified by a computer. Similarly, 1 1 4 1 1 3 d(x)2 X 1+ 1 X exp − 1 , ≤ p − p ≤ p2 − p x X p X p ! p X X≤ Y≤ X Y≤ which is 7X(log X)3 for X 171. Again, for smaller X the bound is ≤ ≥ verified by a computer. 4.3. Miscellany. Lemma 5 ([13]). It holds that Λ(n) c N, ≤ 0 n N X≤ for some constant c0, where one may take c0 =1.03883. Lemma 6. Let 0 <δ< 1 and let L N. Suppose that χ is given by (8). 2 ∈ δ Then there are coefficients cℓ± such that 1 3 c± min 2δ + , | ℓ | ≤ L +1 2ℓ and 1 χδ(y) ⋚ 2δ + c±e(ℓy). ± L +1 ℓ 0< ℓ L X| |≤ Proof. See [9, Lemma 2.1], resp. [1, pp. 18–21]. 4.4. Bounds on some exponential sums. Lemma 7 ([16]). Suppose that X 1, Y 1 and let β R. Then ≥ ≥ ∈ 1 1 (17) min Y, < 4XY q− +4Y +(X + q) log q, 2 αx β x X X≤ k − k XY 1 1 7 (18) min , < 10XY q− + X + q log(2XY q), x 2 αx 2 x X X≤ k k Proof. The case when X q is treated in [19, Lemma 8a and 8b]. Splitting ≤ the sum in (17) into X/q +1 sums of length at most q and applying ⌊ ⌋ [19, Lemma 8a] yields the first inequality. The second inequality can be obtained by splitting the sum in (18) in essentially the same way as in the proof of [19, Lemma 8b]. 10 JÖRN STEUDING AND MARC TECHNAU Lemma 8 ([16]). Let a1,...,aX , b1,...,bY be complex numbers and write l = log(2XY q). Then 1 7 (19) max axe(αxy) l 10XY q− + X + q max ax . Z XY/x ≤ 2 x X | | x X ≤ y Z ≤ ≤ ≤ X X 1 2 3 2 2 (20) max axbye(αxy) l 2 ax by Z Y ≤ | | | | x X ≤ y Z x X y Y X≤ X≤ X≤ X≤ 1 1 (167XY q− + 70X +6Y + 10q) 2 , · Proof. The inequality (19) follows directly from (18). For the proof of (20) we refer the reader to [16]. There is only one Vinogradov symbol in the proof and it is hiding only a factor 2 and a 1 1 factor in the second argument of min(X, αh − ). The application of (6) 2 k k is to be replaced by an application of (17). Lemma 9. Suppose that 0 δ < 1 , L := log(Nq/δ) and ≤ 2 (21) J J ′ H q N, J ′ < 2J. ≤ ≤ ≤ ≤ Then 3 7 1 3 1 Λ(n)e(αjn) 10 L 560JNq− 2 + 41JN 4 + 86(JNq) 2 ′ ≤ J j 2 1 1 (22) u := min N 5 J − 5 ,q,Nq− n o and use Vaughan’s identity, Λ(n)e(αjn)= S S + S S , 1,j − 2,j 3,j − 4,j n N X≤ THE LEAST PRIME NUMBER IN A BEATTY SEQUENCE 11 where S1,j = Λ(n)e(αjn), n u X≤ S2,j = µ(d)Λ(n)e(αjdrn), d,n u r N/dn XX≤ ≤X S3,j = µ(d)Λ(n)e(αjdrn), d u n N/d r N/dn X≤ X≤ ≤X S4,j = µ(d)Λ(n)e(αjmn), u 2 1 4 2 14 S2,j L 8J ′Nq− + J ′u + q max 1. | | ≤ 5 5 x J′u2 J j 1 3 1 2 4 ǫ 1 2 max 1 min 139q , 3M 5 ǫ(J N) , 11(JNq− ) , x J′u2 ≤ 4 ≤ j x jX|J′ n o ≤ so that 2 1 3 + 3 ǫ 4 +ǫ 1 S2,j L 2224JNq− 2 +3M 5 J 5 4 N 5 + 31(JNq) 2 . | | ≤ 4 ǫ J j S3,j 1 max (log y)e(αxrn) | | ≤ Z J′N/x J j 2 1 S3,j L 20J ′Nq− +2J ′u +7q max 1 | | ≤ x J′u J j 1 2 1 1 1 max 1 min 9q 2 , 11(J N) 10 , 3(JNq− ) 2 . x J′u ≤ ≤ j x jX|J′ n o ≤ For the previous two cases we did combine j with small variables ( u), 1 ≤ which does not immediately work in case k = 4. If m N 2 we combine 1 ≪ j and m for otherwise we have n N 2 and combine j with n instead. In ≪ either case we run into the problem that the range of n is dependent on m. This can be resolved by restricting the range of m to short intervals. Choose complex coefficients c˜j of modulus 1 such that c˜jS4,j R 0. Then ∈ ≥ S = c˜ µ(d) Λ(n)e(αjmn) | 4,j| j J j J′ J j J′ u k where ♦ means that M only takes values of the form 2 u (k N ) and M ∈ 0 P SM := c˜j µ(d) Λ(n)e(αjmn), J j J′ m M u 1 where m M is to be understood as M < m 2M. When u M < N 2 , ∼ ≤ ≤ by (20) and Lemma 4 SM max d3(x)Λ(n)e(αxn) | | ≤ Z N/M x J′M ≤ u 1 3 ♦ 7 480360 S L 535414JNq− 2 + JN 4 | M | ≤ log(4N) u M SM max d(m) c˜jΛ(n) e(αmy) | | ≤ Z J′N/M m M ≤ Ju 1 1 1 ♦ 5 229 S 11L 107JNq− 2 + J 2 Nu− 2 | M | ≤ log(4N) N 1/2 M 1 3 Λ(n)e(αjn) c0N 4 JN 4 ′ ≤ · J j ♦ ♦ Λ(n)e(αjn) S1,k + + SM ′ ≤ ′ | | J j 4.5. Proof of Theorem 3. When N 28 1022 we obtain (15) by the ≤ · trivial estimate 1 3 Λ(n)(χ (αn β) 2δ) 3c N 4 N 4 . δ − − ≤ 0 · n N X≤ Subsequently we shall assume that N > 28 1022. We apply Lemma 6 with 1 · L = Rδ− , R to be specified later, and obtain ⌊ ⌋ δ Λ(n)(χδ(αn β) 2δ) ⋚ c0 N + c±e( ℓβ) Λ(n)e(ℓαn). − − ± R ℓ − n N 0< ℓ L n N X≤ X| |≤ X≤ By Abel’s method of partial summation the sum on the right hand side is found to be (23) 1 3 S(L) L S(u) 2 min 2δ + , Λ(n)e(ℓαn) =3 +3 du, ≤ L +1 2ℓ L ˆ u2 ℓ L n N ̟ X≤ X≤ where 3 3 S(u) := Λ(n)e(αhn) and ̟ := . 4δ + 2 ≥ 4δ h u n N L+1 X≤ X≤ Suppose for the moment that we already had a suitable bound for S(H), 3 H L, e.g., 4δ ≤ ≤ 3 7 1 3 1 (24) S(H) 10 L 1120HNq− 2 + 82HN 4 + 294(HNq) 2 ≤ 6 3 + 3 ǫ 4 +ǫ + 62+10− M 5 H 5 4 N 5 . 4 ǫ Employing (23), taking R = Nq, and using the estimate above gives (15) as well. In order to establish (24) we note that (20) together with Lemma 8 implies that S(H) max Λ(n)e(αhn) ≤ Z N h H ≤ n Z X≤ X≤ 3 1 1 (log(2HNq )) 2 (log N) 2 c 2 ≤ 0 1 1 1 1 √167HNq− 2 + √70HN 2 + √6H 2 N + √10(HNq) 2 . · THE LEAST PRIME NUMBER IN A BEATTY SEQUENCE 15 It then suffices to show (24) under the assumption (21) of Lemma 9. Ap- plying said Lemma to the right hand side of S(H) Λ(n)e(αjn) ≤ +1 0 k 5. Concluding Remarks We begin with an historical remark. Beatty sequences of the form (α, 1 ) B 2 appear first in 1772 in astronomical studies of Johann Bernoulli III. [4]. Elwin Bruno Christoffel [6] studied homogeneous Beatty seqquences (α, 0) B in 1874; those appear as well in the treatise [14] by the physicist and Nobel laureate John William Strutt (Lord Rayleigh), published in 1894. The name, however, is with respect to Samuel Beatty who popularized the topic by a problem he posed in 1926 in the American Mathematical Monthly [2, 3]. The proposed problem was to show that (α, 0) (α′, 0) = N is a B ∪ B 1 1 disjoint union for irrational α and α′ related to one another by α + α′ =1; this is now well-known as both, Beatty’s theorem and Rayleigh’s theorem. The generalization of this statement to generalized Beatty sequences has been established by Aviezri S. Fraenkel [7]; already its formulation is much more complicated than in the homogeneous case β = β′ =0. The case of Beatty sequences (α, β) with α < 1 is trivial. Since then B 0 (n + 1)α + β nα + β < 2 consecutive elements differ by at most ≤ ⌊ ⌋ − ⌊ ⌋ 1, whence (α, β) = N [ α + β , + ). The case of integral α is trivial B ∩ ⌊ ⌋ ∞ as well. However, we shall discuss briefly the case of rational non-integral α> 1. Given α = a with coprime a and q 2, it follows that (α, β) is a q ≥ B disjoint union of residue classes ab (25) ( a , β)= + β + aN . B q q 0 1 b q ≤[≤ a We observe that for some values of q and β there is no prime residue class and even no prime number contained, e.g., ( 15 , 3) = 10, 18 + 15N . It B 2 { } 0 seems to be an interesting question to characterize for which parameters α Q and β real there appears at least one prime residue class in (α, β). ∈ B In the homogeneous case, however, there is always at least one prime residue a 1 class contained in ( , 0) which results from the choice b = a− mod q B q in (25). 16 JÖRN STEUDING AND MARC TECHNAU References [1] R.C. Baker, Diophantine inequalities, London Mathematical Society Monographs, Clarendon Press, 1986. [2] S. Beatty, Problem 3173, Amer. Math. Monthly 33 (1926), 159 [3] S. Beatty, A. Ostrowski, J. Hyslop, A.C. Aitken, Solutions to Problem 3173, Amer. Math. Monthly 34 (1927), 159-160 [4] J. Bernoulli III, Sur une nouvelle espèce de calcul, in: Recueil pour les astronomes, vols. 1, 2, Berlin 1772 [5] S. Chowla, On the least prime in an arithmetical progression, J. Indian Math. Soc., n. Ser., 1 (1934), 1-3 [6] E.B. Christoffel, Observatio Arithmetica, Brioschi Ann. 6 (1873-75), 148-152 [7] A.S. Fraenkel, The bracket function and complementary sets of integers, Canad. J. Math. 21 (1969), 6-27 [8] G.H. Hardy, E.M. Wright, An introduction to the theory of numbers, 6th ed., Oxford University Press 2007 [9] G. Harman, Prime-detecting sieves, Princeton University Press, 2007. [10] H. Iwaniec and E. Kowalski, Analytic number theory, American Mathematical Society (AMS), 2004. [11] Yu. Linnik, On the least prime in an arithmetic progression. I: The basic theorem, Mat. Sb., N. Ser., 15 (1944), 139-178 [12] Yu. Linnik, On the least prime in an arithmetic progression. II: The Deuring- Heilbronn theorem, Mat. Sb., N. Ser., 15 (1944), 347-368 [13] J.B. Rosser, L. Schoenfeld, Approximate formulas for some functions of prime numbers, Ill. J. Math. 6 (1962), 64-94 [14] J.W. Strutt, The theory of sound. Vol. I, 2nd ed., Macmillian, London 1894 [15] P. Turán, Über die Primzahlen der arithmetischen Progression, Acta Litt. Sci. Szeged 8 (1937), 226-235 [16] R.C. Vaughan, On the distribution of αp modulo 1, Mathematika 24 (1977), 135- 141 [17] I.M. Vinogradov, A new estimate of a certain sum containing primes (Russian), Rec. Math. Moscou, n. Ser., 2(44), No. 5, 783-792 (1937); engl. translation: New estimations of trigonometrical sums containing primes, C. R. (Dokl.) Acad. Sci. URSS, n. Ser., 17, 165-166 (1937) [18] I.M. Vinogradov, Selected works, Springer-Verlag, 1985. [19] I.M. Vinogradov, The method of trigonometrical sums in the theory of numbers, Dover Publications, reprint of the 1954 translation edition, 2004. [20] H. Weyl, Sur une application de la théorie des nombres à la mécaniques statistique et la théorie des pertubations, L’Enseign. math 16 (1914), 455-467 [21] T. Xylouris, Über die Nullstellen der Dirichletschen L-Funktionen und die kleinste Primzahl in einer arithmetischen Progression, Bonner Mathematische Schriften 404, Univ. Bonn 2011, (Diss.) THE LEAST PRIME NUMBER IN A BEATTY SEQUENCE 17 Jörn Steuding, Department of Mathematics, University of Würzburg, Campus Hubland Nord, Emil-Fischer-Straße 40, D-97074 Würzburg, Ger- many E-mail address: [email protected] Marc Technau, Department of Mathematics, University of Würzburg, Campus Hubland Nord, Emil-Fischer-Straße 40, D-97074 Würzburg, Ger- many E-mail address: [email protected]