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Short intervals with a given number of primes
Journal of Number Theory 163 (2016) 159–171
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Journal of Number Theory www.elsevier.com/locate/jnt
Short intervals with a given number of primes Tristan Freiberg Department of Pure Mathematics, University of Waterloo, Waterloo, ON, Canada
a r t i c l e
i n f o
Article history: Received 14 September 2015 Received in revised form 27 November 2015 Accepted 29 November 2015 Available online 8 January 2016 Communicated by Steven J. Miller
a b s t r a c t A well-known conjecture asserts that, for any given positive real number λ and nonnegative integer m, the proportion of positive integers n x for which the interval (n, n + λ log n] contains exactly m primes is asymptotically equal to λm e−λ /m! as x tends to infinity. We show that the number of such n is at least x1−o(1) . © 2016 Elsevier Inc. All rights reserved.
Keywords: Primes in intervals Maynard’s theorem Cramér’s model
1. Introduction Let π(x) ..= #{p x : p prime} denote the prime counting function. One form of the prime number theorem states that, for any given positive real number λ, 1 (π(n + λ log n) − π(n)) ∼ λ x
(x → ∞),
nx
i.e. on average over n x, the interval (n, n + λ log n] contains approximately λ primes. As to the finer questions pertaining to the distribution of primes, we have little more E-mail address: [email protected]. http://dx.doi.org/10.1016/j.jnt.2015.11.009 0022-314X/© 2016 Elsevier Inc. All rights reserved.
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than conjecture in the way of answers. Heuristics based on Cramér’s model suggest that, for any given positive real number λ and nonnegative integer m, #{n x : π(n + λ log n) − π(n) = m} ∼ xλm e−λ /m!
(x → ∞).
(1.1)
However, before the groundbreaking work [5] of Goldston, Pintz and Yıldırım (GPY), it had not even been established that π(n + λ log n) − π(n) m
(1.2)
holds for infinitely many n when λ = 1/5 (for instance) and m = 2. What GPY showed is that, for arbitrarily small λ and m = 2, (1.2) holds for infinitely many n. Only very recently has the breakthrough of Maynard [7] on bounded gaps between primes shown that, for every choice of λ and m, (1.2) holds for infinitely many n. This statement does not preclude the possibility that there are choices of λ and m for which π(n + λ log n) − π(n) = m for at most finitely many n. The purpose of this note is to establish the following. Theorem 1.1. Fix any positive real number λ and any nonnegative integer m. If x is sufficiently large in terms of λ and m, then #{n x : π(n + λ log n) − π(n) = m} x1−ε(x) ,
(1.3)
where ε(x) is a certain function that tends to zero as x tends to infinity. Notation. Throughout, P denotes the set of all primes, 1P : N → {0, 1} the indicator function of P ⊆ N ..= {1, 2, . . .} and p a prime. Given a, q ∈ Z, a (q) denotes the residue class {a +qb : b ∈ Z} (thus, n ≡ a (q) if and only if n ∈ a (q)). Given a large real number x, log2 x ..= log log x, log3 x ..= log log log x and so on. By o(1) we mean a quantity that tends to 0 as x tends to infinity. Expressions of the form A = O(B), A B and B A denote that |A| c|B|, where c is some positive constant (absolute unless stated otherwise); A B is and abbreviation for A B A. Further notation is introduced in situ. 2 2. Background According to Cramér’s model,1 the sequence (1P (n))nx , when x is large, behaves roughly like a sequence (Xn )nx of Bernoulli random variables for which Xn = 1 with probability 1/ log x and Xn = 0 with probability 1 − 1/ log x. Thus, x−1 #{n x : π(x + h) − π(x) = m} is to be thought of as the probability that X1 + · · · + Xh = m. Letting x and h tend to infinity in such a way that h/ log x ∼ λ, we get (in the limit) a Poisson distribution for the sum X1 +· · ·+Xh . Hence the conjectured asymptotic (1.1), 1
For details, we highly recommend the insightful expository article [9] of Soundararajan.
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which, as was shown by Gallagher [4, Theorem 1], would in fact follow from a certain uniform version of the Hardy–Littlewood prime tuples conjecture. We similarly expect the normalized spacings between consecutive primes to follow an exponential distribution, i.e. if dn ..= pn+1 −pn , where pn denotes the nth smallest prime, we have the well-known conjecture 1 #{n x : dn / log n ∈ (a, b]} ∼ x
b
e−t dt
(x → ∞).
a
(Another form of the prime number theorem states that x−1 nx dn / log n ∼ 1 as x tends to infinity, i.e. dn / log n ≈ 1 on average over n x.) However, we do not even know of any specific limit points of the sequence (dn / log n), except for 0 and ∞ (the former following from the aforementioned result of GPY, the latter from an old result of Westzynthius [10]). Nevertheless, it has recently been shown2 [1, Theorem 1.1] that, in a certain sense, 12.5% of positive real numbers are limit points of (dn / log n). This note may be regarded as a continuation of [1], the results of which are utilized here in combination with the very general and powerful quantitative work of Maynard [6]. 3. Proof of Theorem 1.1 We shall consider linear functions L given by L(n) = gn + h, where g, h ∈ Z (it is to be assumed that g = 0). A finite set {L1 , . . . , Lk } of linear functions is admissible if the set of solutions modulo p to L1 (n) · · · Lk (n) ≡ 0 (p) does not form a complete residue system modulo p, for any prime p. (It is to be assumed that 1i
n∈[x,2x) n≡a (q)
n∈[x,2x)
(3.1) Theorem 3.1 (Maynard). Let L = {L1 , . . . , Lk } be an admissible set of k linear functions. Let B be a positive integer, let x be a large real number and let 0 < θ < 1. Let α > 0. 2 Benatar [2] has since claimed that in fact 25% of positive real numbers are limit points of the sequence (dn / log n).
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Suppose that the coefficients of Li (n) ..= gi n + hi ∈ L satisfy 1 gi , hi xα for i = 1, . . . , k, that k (log x)α and that 1 B xα . There is a positive constant C, depending only on θ and α, such that the following holds. If k C, if L, B, x, θ satisfy Hypothesis 1 and if δ > (log k)−1 is such that 1 ϕ(B) ϕ(gi ) k B gi L∈L
1P (L(n)) δ
n∈[x,2x)
x , log x
(3.2)
then # n ∈ [x, 2x) : #({L1 (n), . . . , Lk (n)} ∩ P) C −1 δ log k
x . (eC log x)k
(3.3)
(The implicit constant in (3.1) may depend only on θ and α, and that in (3.3) depends at most on θ and α.) A level of distribution result. We need to show that Hypothesis 1 holds for certain choices of L, B, x, θ. We defer proof of the following result to §4. Lemma 3.2. There is an absolute constant c ∈ (0, 1) such that the following holds. Fix any positive integer k and let x be a real number that is sufficiently large in terms of k. There is a positive integer B, which is either a prime satisfying log2 xη B x2η , where η ..= c/(500k2 ), or is equal to 1, for which we have the following. Let g be any positive integer that is coprime to B and divides plog xη p. Let L = {L1 , . . . , Lk } be any admissible set of k linear functions for which the coefficients of Li (n) ..= gi n + hi satisfy gi = g and 1 hi x for i = 1, . . . , k. For each L ∈ L, ϕ(B) ϕ(g) B g
1P (L(n)) >
n∈[x,2x)
x 2 log x
(3.4)
and qx1/8 (q,B)=1
max (L(a),q)=1
n∈[x,2x) n≡a (q)
1 1P (L(n)) − ϕL (q)
n∈[x,2x)
n∈[x,2x) 1P (L(n)) 1P (L(n)) . (log x)100k2 (3.5)
An Erdős–Rankin type construction. We quote [1, Lemma 5.2]. First, some more notation and terminology: a finite set {h1 , . . . , hk } of integers is admissible if the set {L1 , . . . , Lk } of linear forms given by Li (n) = n + hi , i = 1, . . . , k, is admissible. Given a real number z 1 we let [z] ..= {1, . . . , z}, where z denotes the greatest integer less than or equal to z.
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Lemma 3.3. Fix a positive integer k and k nonnegative real numbers β1 , . . . , βk . Suppose that β1 · · · βk . There is a constant C , depending only on k and β1 , . . . , βk , such that the following holds. Let v, y, z be real numbers satisfying v 1, y C and 2y(1 + (1 + βk )v) 2z y(log2 y)/ log3 y.
(3.6)
Let B be any positive integer such that for all prime divisors l of B (if any), p|B, pl
1/p 1/l 1/ log y.
(3.7)
There exists an admissible set {h1 , . . . , hk } and a sequence (ap (p))py, p B of residue classes such that 1/4 hi = y + βi vy + O ye−(log y) for i = 1, . . . , k and {h1 , . . . , hk } = [z] \
py, p B
ap (p).
Deduction of Theorem 1.1. Fix a positive real number λ and a nonnegative integer m. Let C be the constant of Theorem 3.1, which depends on θ and α. We will apply Theorem 3.1 with θ ..= 1/8 and α ..= 1, so C may be regarded as absolute. We will also apply the theorem with δ ..= 1/2. Let k be the smallest positive integer satisfying k C, k e2 and k e2Cm (i.e. k C, δ (log k)−1 and C −1 δ log k m). Let βi = 2i−k λ, i = 1, . . . , k and let C be the constant of Lemma 3.3, which depends on k (hence m) and λ. Let x be a large real number and define v ..=
log3 x 1 , 3(1 + 3λ) log4 x
y ..= 3(1 + 3λ) log x
log4 x , log3 x
z = (1 + 3λ) log x.
(3.8)
We think of x as tending to infinity, and we tacitly assume throughout that x is “sufficiently large” in terms of any specified fixed quantity, hence ultimately in terms of λ and m. Thus, for instance, as is straightforward to verify, we have v 1, y C and (3.6). Let η and B be as in Lemma 3.2, i.e. η ..= c/(500k2 ) for a certain absolute constant c ∈ (0, 1), and either B = 1 or B is a prime satisfying log2 xη B x2η . As log xη > y, (3.7) is satisfied. Each hypothesis of Lemma 3.3 thus accounted for, we conclude that there exists an admissible set {h1 , . . . , hk } and a sequence (ap (p))py, p B of residue classes such that 1/4 hi = y + 2i−k λ log x + O ye−(log y) for i = 1, . . . , k (we have vy = log x), and {h1 , . . . , hk } = [(1 + 3λ) log x] \
py, p B
ap (p).
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We work with such an admissible set and sequence of residue classes. Note that hk − h1 < −1 + λ log x
(3.9)
1 < h1 < · · · < hk < −1 + 2λ log x
(3.10)
and
We let g ..= py, pB p and h (g) be the residue class modulo g such that h ≡ −ap (p) for each prime p dividing g. Let us suppose that 0 h < g. We let L ..= {L1 , . . . , Lk } be the set of linear functions in which Li (n) ..= gn + h + hi for i = 1, . . . , k. It is straightforward to verify that L is admissible, and that for all positive integers n, (gn + h, gn + h + (1 + 3λ) log x] ∩ P = {L1 (n), . . . , Lk (n)} ∩ P.
(3.11)
We have (g, B) = 1 by definition, and as already noted, log xη > y, so g divides plog xη p. In fact, by Chebyshev’s bounds and since η is very small, we certainly have 0 < g, h + hi < x for i = 1, . . . , k. Therefore, by Lemma 3.2, L, B, x and θ = 1/8 satisfy Hypothesis 1, and (3.2) holds with δ = 1/2 for each L ∈ L. We now invoke Theorem 3.1 with θ = 1/8, α = 1 (we have k log x) and δ = 1/2. We’ve chosen k so that C −1 δ log k m, so we infer that
# {n ∈ [x, 2x) : #({L1 (n), . . . , Lk (n)} ∩ P) m}
x . (eC log x)k
(3.12)
Choose n ∈ [x, 2x) such that #({L1 (n), . . . , Lk (n)} ∩ P) m. Consider the intervals Ij ..= (Nj , Nj + λ log Nj ],
Nj ..= gn + h + j,
j = 0, . . . , 2λ log N0 .
Now, N0 = x1+o(1) , so for j in the given range we have Ij ⊆ (gn + h, gn + h + (1 + 3λ) log x], so by (3.11), Ij ∩ P = (Ij ∩ {L1 (n), . . . , Lk (n)}) ∩ P.
(3.13)
By (3.9), L1 (n) < · · · < Lk (n) < L1 (n) − 1 + λ log x < L1 (n) − 1 + λ log N0 . Thus, if j = h1 − 1 (so that Nj = L1 (n) − 1), then Ij ∩ {L1 (n), . . . , Lk (n)} = {L1 (n), . . . , Lk (n)}.
(3.14)
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By (3.10), Lk (n) = N0 + hk < N0 − 1 + 2λ log x < N0 − 1 + 2λ log N0 . Thus, if j = 2λ log N0 , then Ij ∩ {L1 (n), . . . , Lk (n)} = ∅.
(3.15)
Therefore, by (3.13) and (3.14), #(Ih1 −1 ∩ P) = #({L1 (n), . . . , Lk (n)} ∩ P) m, while on the other hand, by (3.13) and (3.15), #(I2λ log N0 ∩ P) = 0. Now, for any j, if #(Ij+1 ∩ P) < #(Ij ∩ P) then #(Ij+1 ∩ P) = #(Ij ∩ P) − 1. We must conclude that there is some j in the range h1 − 1 j 2λ log N0 for which π(Nj + λ log Nj ) − π(Nj ) = m. By the prime number theorem and the definition of g, B and y, we have g = e(1+o(1))y > (1 + 3λ) log x. Since g(n + 1) + h > gn + h + (1 + 3λ) log x, no two values of n can give rise to the same Nj in this way. We deduce from (3.12) that, with X ..= 4gx, #{N X : π(N + λ log N ) − π(N ) = m}
x . (eC log x)k
It follows that the left-hand side exceeds X 1−ε(X) , where ε(X) ..= (log4 X)2 / log3 X. (Recall that log g = (1 + o(1))y by the prime number theorem, and that, by (3.8), y = 3(1 + 3λ) log x log4 x/ log3 x.) 2 4. Proof of Lemma 3.2 Lemma 3.2 is similar to [1, Theorem 4.2]. We must nevertheless verify the details. First, some more notation: given a positive integer q, χ mod q, or simply χ if q is clear in context, denotes a Dirichlet character to the modulus q, L(s, χ) denotes the L-function associated with it and χ ¯ its complex conjugate. Also, P + (q) denotes the greatest prime + . divisor of q (P (1) .= 1 by convention). We quote [1, Lemma 4.1]. Lemma 4.1. Let T 3 and let P T 1/ log2 T . Among all primitive Dirichlet characters χ mod to moduli satisfying T and P + ( ) P , there is at most one for which the
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associated L-function L(s, χ) has a zero in the region |(s)| exp log P/ log T ,
(s) > 1 − c/ log P,
(4.1)
where c > 0 is a certain (small) absolute constant. If such a character χ mod exists, then χ is real and L(s, χ) has just one zero in the region (4.1), which is real and simple, and P + ( ) log log2 T.
(4.2)
Definition 4.2. For T 3, let (T ) ..= if the “exceptional” character χ mod , as described in Lemma 4.1, exists; let (T ) ..= 1 otherwise. Proof of Lemma 3.2. Let c be the constant of Lemma 4.1. We may assume that c < 1. Fix a positive integer k and let η ..= c/(500k2 ). Let x be a large real number, and let g be a positive integer that divides plog xη p. Note that, by Chebyshev’s bounds, g < x1/22 , say. (This is the reason for taking c < 1.) Let B ..= (x2η ), as in Definition 4.2, and suppose (g, B) = 1. Let h be an integer satisfying 1 h x. Suppose (g, h) = 1 and let L denote the linear function given by L(n) ..= gn + h. Let IL (x) ..= [gx + h, 2gx + h). Let q denote a positive integer and a an integer for which (L(a), q) = 1, noting that this implies (L(a), gq) = 1. We have
1P (L(n)) =
n∈[x,2x) n≡a (q)
1P (n) =
n∈IL (x) n≡L(a) (gq)
1 ϕ(gq)
1P (n) + ΔL (x; q, a),
n∈IL (x)
where
ΔL (x; q, a) ..=
1P (n) −
n∈IL (x) n≡L(a) (gq)
1 ϕ(gq)
1P (n).
n∈IL (x)
We will show that if x is large enough in terms of k, then
max
qx1/8 (q,B)=1
(L(a),q)=1
|ΔL (x; q, a)|
gx . ϕ(g)(log x)2+100k2
(4.3)
Let us show how this implies Lemma 3.2. First, |ΔL (x; 1, 1)| is certainly majorized by the left-hand side of (4.3), so if x is large enough in terms of k, n∈[x,2x)
1P (L(n)) =
1 ϕ(g)
n∈IL (x)
1P (n) + ΔL (x; 1, 1) =
gx (1 + O (1/ log x)) ϕ(g) log(gx)
(4.4)
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by (4.3) and the prime number theorem. If B = 1 then, by (4.2), B is a prime satisfying B log2 xη . In any case, ϕ(B)/B = 1 + O(1/ log2 xη ). As already noted, we have g < x1/22 , i.e. log(gx) < 23 22 log x. Thus, if x is large enough in terms of k we have ϕ(B) ϕ(g) B g
1P (L(n))
n∈[x,2x)
22x (1 + O (1/ log2 xη )) , 23 log x
whence (3.4). Second, we verify that
1P (L(n)) −
n∈[x,2x) n≡a (q)
1 ϕL (q)
1P (L(n)) = ΔL (x; q, a) −
n∈[x,2x)
ϕ(g) ΔL (x; 1, 1). ϕ(gq)
(4.5)
Third, again using (4.3) to bound |ΔL (x; 1, 1)|, then using ϕ(gq) ϕ(g)ϕ(q) and qx 1/ϕ(q) log x, we obtain qx1/8 (q,B)=1
gx ϕ(g) gx 1 |ΔL (x; 1, 1)| . 2 ϕ(gq) ϕ(gq) (log x)2+100k ϕ(g)(log x)1+100k2 1/8
(4.6)
qx
Fourth, we combine (4.3), (4.5) and (4.6) (applying the triangle inequality to the righthand side of (4.5)), obtaining qx1/8 (q,B)=1
max (L(a),q)=1
n∈[x,2x) n≡a (q)
1 1P (L(n)) − ϕL (q)
1P (L(n))
n∈[x,2x)
gx . ϕ(g)(log x)1+100k2 (4.7)
Finally, combining (4.4) with (4.7) yields (3.5). We now establish (4.3) by paraphrasing the proof of [1, Theorem 4.2]. Suppose 1 q x1/8 . By orthogonality of Dirichlet characters we have
1P (L(n)) =
n∈[x,2x) n≡a (q)
1P (n) =
n∈IL (x) n≡L(a) (gq)
1 ϕ(gq)
χ(L(a)) ¯
χ mod gq
χ(n)1P (n).
n∈IL (x)
Letting χ∗ denote the primitive character that induces χ, we have (for characters χ to the modulus gq), (χ(n) − χ(n))1P (n) n∈IL (x)
n∈IL (x) (n,gq)>1
1P (n)
p|gq
1 log(gq) log x,
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whence
1P (L(n)) =
n∈[x,2x) n≡a (q)
1 ϕ(gq)
χ(L(a)) ¯
χ mod gq
χ∗ (n)1P (n) + O(log x).
n∈IL (x)
For the principal character χ0 mod gq we have χ∗0 ≡ 1, and so we deduce that max
(L(a),q)=1
|ΔL (x; q, a)|
1 ϕ(gq)
χ mod gq χ=χ0
∗ + O(log x) χ (n)1 (n) P
(4.8)
n∈IL (x)
It follows from the explicit formula [3, §19, (13)–(14)] that, for nonprincipal characters √ χ mod gq, 2 T N and T N , with Λ denoting the von Mangoldt function, N (ρ) √ + N (log(gqN ))2 , χ(n)Λ(n) |ρ| √ |ρ|< N
nN
where zeros of L(s, χ) having real part at least 1/2. Since the sum is over nontrivial √ χ(n)Λ(n)(1 (n) − 1) N log N , the same bound holds if Λ is replaced by P nN √ √ Λ1P and, via partial summation, 1P . Thus, since gq xO(1) and gx gx + h, ∗ χ (n)1P (n) n∈IL (x)
√ |ρ|< gx
(gx) (ρ) √ + gx(log x)2 . |ρ|
Combining this with (4.8) gives max
(L(a),q)=1
|ΔL (x; q, a)|
=
1 ϕ(gq)
∗ (gx) (ρ) √ + gx(log x)2 |ρ| √
χ mod gq |ρ|< gx χ=χ0
1 ∗ ϕ(gq)
√ d|gq χ mod d |ρ|< gx
(gx) (ρ) √ + gx(log x)2 , |ρ|
(4.9)
∗ where, in the first line, denotes summation over nontrivial zeros of L(s, χ∗ ) having ∗ real part at least 1/2, χ being the primitive character that induces χ, and in the second ∗ line, denotes summation over primitive characters and the innermost sum is over nontrivial zeros of L(s, χ) having real part at least 1/2. Applying (4.9) and changing order of summation, recalling that (g, B) = 1, we find that qx1/8 (q,B)=1
max
(L(a),q)=1
|ΔL (x; q, a)|
T. Freiberg / Journal of Number Theory 163 (2016) 159–171
∗
(gx) (ρ) |ρ|
√ dgx1/8 χ mod d |ρ|< gx (d,B)=1
qx1/8 (q,B)=1 d|gq
169
1 √ + x1/8 gx(log x)2 . ϕ(gq)
Writing d = ab with a = (d, g), for d | gq we have gq = gbc for some integer c. Note that (b, g) divides a. We have ϕ(gq) = ϕ(gbc) ϕ(g)ϕ(b)ϕ(c), and as cx1/8 1/ϕ(c) log x, it follows that max |ΔL (x; q, a)| qx1/8 (q,B)=1
(L(a),q)=1
log x ϕ(g)
a|g bgx1/8 /a (b,g)|a (b,B)=1
1 ϕ(b)
∗
√ χ mod ab |ρ|< gx
(gx) (ρ) √ + x1/8 gx(log x)2 . |ρ|
(4.10)
If a | g and b gx1/8 /a, then a ∈ [R, 2R) and b ∈ [S, 2S) for some pair (R, S) of powers of 2 satisfying 1 R < g and 1 RS < gx1/8 . The number of such pairs is O((log x)2 ). Note that for b ∈ [S, 2S) and S < gx1/8 we have 1/ϕ(b) (log2 b)/b (log x)/S. If √ |ρ| < gx and 1/2 (ρ) 1 then (ρ) ∈ Im ..= [1/2 + m/ log(gx), 1/2 + (m + 1)/ log(gx)) for some integer m satisfying 0 m <
1 2
log(gx), and
|(ρ)| ∈ Jn ..= [n − 1, 2n − 1) √ with n being some power of 2 satisfying 1 n < gx. The number of such pairs (m, n) is O((log x)2 ). Note that for ρ with (ρ) ∈ Im and |(ρ)| ∈ Jn we have (gx) (ρ) /|ρ| √ gx em /n. Thus, log x ϕ(g)
a|g bgx1/8 /a (b,g)|a (b,B)=1
√
1 ϕ(b)
gx(log x)6 ϕ(g)
∗
√ χ mod ab |ρ|< gx
sup R
where N ∗ (R, S, σ, T ) ..=
(gx) (ρ) |ρ|
e−m ∗ N R, S, 1/2 + m/ log(gx), n − 1 , nS
∗
1,
Ra<2R Sb<2S χ mod ab (ρ)σ a|g (b,g)|a | (ρ)|T (b,B)=1
and the innermost sum is over zeros ρ of L(s, χ) in the given region.
(4.11)
(4.12)
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Note that, as g is squarefree, every positive integer may be decomposed uniquely as a product ab of positive integers a | g and (b, g) | a, whence N ∗ (R, S, σ, T )
∗
d<4RS χ mod d
1.
(ρ)σ | (ρ)|T
A result [8, Theorem 12.2] of Montgomery therefore implies that N ∗ (R, S, σ, T ) ((RS)2 T )3(1−σ)/(2−σ) (log(RST ))14
(4.13)
for T 2 and 1/2 σ 1. On the right-hand side of (4.11), we partition the supremum set according as (i) 0 m 12 log(gx) − (44 + 200k2 ) log2 (gx) or (ii) 12 log(gx) − (44 + 200k 2 ) log2 (gx) < m < 12 log(gx). For case (i), we note that for 1/2 σ 1, the following inequalities hold: 1/(2−σ) 1, 6(1 − σ)/(2 − σ) 1 + 2(1 − σ) and 3(1 − σ)/(2 − σ) 1. Thus, (R2 )3(1−σ)/(2−σ) (R6 )(1−σ) , (S 2 )3(1−σ)/(2−σ) S(S 2 )(1−σ) and T 3(1−σ)/(2−σ) T . Hence (4.13) implies N ∗ (R, S, σ, T ) (R6 S 2 )1−σ ST (log(RST ))14 . Recalling that g < x1/22 , note that if √ R < g and RS < gx1/8 then R6 S 2 < gx. It follows that √
gx em ∗ m/2 1/4 14 N R, S, 1/2 + m/ log(gx), n − 1 e (gx) (log x) . nS (log x)8+100k2 (4.14) 2
We divide case (ii) into two sub-cases: (iia) n3/4 S 1/2 (log(gx))22+100k or (iib) 2 3/4 1/2 n S < (log(gx))22+100k . For (iia), we note that if σ = 1/2 + m/ log(gx) then σ 10/11 (provided x is large enough in terms of k, as we assume), hence 3(1 − σ)/(2 − σ) 1/4. We have (R2 )3(1−σ)/(2−σ) (R6 )1−σ < (gx)(1−σ)/2 (for R < g) as before, and (S 2 T )3(1−σ)/(2−σ) S 1/2 T 1/4 . By (4.13) we therefore have √
gx em ∗ em/2 (gx)1/4 (log x)14 N R, S, 1/2 + m/ log(gx), n − 1 2 . 8+100k 3/4 1/2 nS (log x) n S (4.15) For (iib) we apply Lemma 4.1 to the right-hand side of (4.12). Note that in this case we 2 have S < (log(gx))44+200k . Also note that, since g | plog xη p, we have g x(1+o(1))η by the prime number theorem. Recall that η = c/(500k2 ). Thus, for a | g and b < 2S 2η we have ab < x2η and P + (ab) < x2η/ log2 x (if x is large enough in terms of k, as we assume). We find that 1−c
m log2 x2η 1 < + 2η log x 2 log(gx)
and n < exp
log x2η / log2 x2η 2
when m > 12 log(gx) − (44 + 200k2 ) log2 (gx) and n3/4 < (log(gx))22+100k . Therefore, in case (iib), N ∗ (R, S, σ, T ) is at most the number of zeros of L(s, χ) for primitive characters χ mod ab, ab < x2η =.. T , P + (ab) < T 1/ log2 T =.. P ,
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(s) 1 − c/ log P,
171
|(s)| exp log P/ log T
and (ab, B) = 1 (recall that a | g and (g, B) = (b, B) = 1). But by Lemma 4.1 and our choice of B, there are no such zeros, i.e.
N ∗ R, S, 1/2 + m/ log(gx), n − 1 = 0. Combining (4.10), (4.11), (4.14), (4.15) and (4.16) gives (4.3).
(4.16) 2
5. Concluding remarks In §3, we did not quote (the special case of) Maynard’s theorem [6, Theorem 3.1] in its entirety. It continues as follows.3 If, moreover, k (log x)1/5 and all L ∈ L have the form gn + hi with |hi | υ log x (log x)/(k log k) and g 1, then # n ∈ [x, 2x) : #({L1 (n), . . . , Lk (n)} ∩ P) C −1 δ log k
x eCk2 (log x)k
,
(5.1)
where on the left-hand side, [gn, gn + υ log x] ∩ P = {L1 (n), . . . , Lk (n)} ∩ P. This would lead to an improvement of Theorem 1.1 prima facie only for certain λ and m — note here that υ 1/(k log k), so if υ λ and C −1 δ log k m, there is an interdependence −1 between λ, m and δ, viz. λCmδ −1 eCmδ 1. Perhaps the right-hand side of (1.3) can be improved to something of a quality similar to that of (5.1), for all λ and m, via a less ad hoc proof, i.e. a proof that uses Maynard’s sieve alone, and does not involve the Erdős–Rankin construction. References [1] W.D. Banks, T. Freiberg, J. Maynard, On limit points of the sequence of normalized prime gaps, preprint, arXiv:1404.5094, 2014. [2] J. Benatar, Goldbach versus de Polignac numbers, preprint, arXiv:1505.03104, 2015. [3] H. Davenport, Multiplicative Number Theory, 3rd edn., Grad. Texts in Math., vol. 74, SpringerVerlag, New York, 2000, revised and with a preface by H. L. Montgomery. [4] P.X. Gallagher, On the distribution of primes in short intervals, Mathematika 23 (1) (1976) 4–9. [5] D.A. Goldston, J. Pintz, C.Y. Yıldırım, Primes in tuples I, Ann. of Math. (2) 170 (2) (2009) 819–862. [6] J. Maynard, Dense clusters of primes in subsets, preprint, arXiv:1405.2593, 2014. [7] J. Maynard, Small gaps between primes, Ann. of Math. (2) 181 (1) (2015) 383–413. [8] H.L. Montgomery, Topics in Multiplicative Number Theory, Lecture Notes in Math., vol. 227, Springer-Verlag, Berlin–New York, 1971. [9] K. Soundararajan, The distribution of prime numbers, in: A. Granville, Z. Rudnick (Eds.), Equidistribution in Number Theory, an Introduction, in: NATO Sci. Ser. II Math. Phys. Chem., vol. 237, Springer, Dordrech, 2007, pp. 59–83. [10] E. Westzynthius, Über die Verteilung der Zahlen, die zu den n ersten Primzahlen teilerfremd sind, Commentat. Phys.-Math. 5 (25) (1931) 1–37.
3
We are not quoting [6, Theorem 3.1] exactly here. Rather, we are inspecting its proof [6, (6.13) et seq.].
Contents lists available at ScienceDirect
Journal of Number Theory www.elsevier.com/locate/jnt
Short intervals with a given number of primes Tristan Freiberg Department of Pure Mathematics, University of Waterloo, Waterloo, ON, Canada
a r t i c l e
i n f o
Article history: Received 14 September 2015 Received in revised form 27 November 2015 Accepted 29 November 2015 Available online 8 January 2016 Communicated by Steven J. Miller
a b s t r a c t A well-known conjecture asserts that, for any given positive real number λ and nonnegative integer m, the proportion of positive integers n x for which the interval (n, n + λ log n] contains exactly m primes is asymptotically equal to λm e−λ /m! as x tends to infinity. We show that the number of such n is at least x1−o(1) . © 2016 Elsevier Inc. All rights reserved.
Keywords: Primes in intervals Maynard’s theorem Cramér’s model
1. Introduction Let π(x) ..= #{p x : p prime} denote the prime counting function. One form of the prime number theorem states that, for any given positive real number λ, 1 (π(n + λ log n) − π(n)) ∼ λ x
(x → ∞),
nx
i.e. on average over n x, the interval (n, n + λ log n] contains approximately λ primes. As to the finer questions pertaining to the distribution of primes, we have little more E-mail address: [email protected]. http://dx.doi.org/10.1016/j.jnt.2015.11.009 0022-314X/© 2016 Elsevier Inc. All rights reserved.
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than conjecture in the way of answers. Heuristics based on Cramér’s model suggest that, for any given positive real number λ and nonnegative integer m, #{n x : π(n + λ log n) − π(n) = m} ∼ xλm e−λ /m!
(x → ∞).
(1.1)
However, before the groundbreaking work [5] of Goldston, Pintz and Yıldırım (GPY), it had not even been established that π(n + λ log n) − π(n) m
(1.2)
holds for infinitely many n when λ = 1/5 (for instance) and m = 2. What GPY showed is that, for arbitrarily small λ and m = 2, (1.2) holds for infinitely many n. Only very recently has the breakthrough of Maynard [7] on bounded gaps between primes shown that, for every choice of λ and m, (1.2) holds for infinitely many n. This statement does not preclude the possibility that there are choices of λ and m for which π(n + λ log n) − π(n) = m for at most finitely many n. The purpose of this note is to establish the following. Theorem 1.1. Fix any positive real number λ and any nonnegative integer m. If x is sufficiently large in terms of λ and m, then #{n x : π(n + λ log n) − π(n) = m} x1−ε(x) ,
(1.3)
where ε(x) is a certain function that tends to zero as x tends to infinity. Notation. Throughout, P denotes the set of all primes, 1P : N → {0, 1} the indicator function of P ⊆ N ..= {1, 2, . . .} and p a prime. Given a, q ∈ Z, a (q) denotes the residue class {a +qb : b ∈ Z} (thus, n ≡ a (q) if and only if n ∈ a (q)). Given a large real number x, log2 x ..= log log x, log3 x ..= log log log x and so on. By o(1) we mean a quantity that tends to 0 as x tends to infinity. Expressions of the form A = O(B), A B and B A denote that |A| c|B|, where c is some positive constant (absolute unless stated otherwise); A B is and abbreviation for A B A. Further notation is introduced in situ. 2 2. Background According to Cramér’s model,1 the sequence (1P (n))nx , when x is large, behaves roughly like a sequence (Xn )nx of Bernoulli random variables for which Xn = 1 with probability 1/ log x and Xn = 0 with probability 1 − 1/ log x. Thus, x−1 #{n x : π(x + h) − π(x) = m} is to be thought of as the probability that X1 + · · · + Xh = m. Letting x and h tend to infinity in such a way that h/ log x ∼ λ, we get (in the limit) a Poisson distribution for the sum X1 +· · ·+Xh . Hence the conjectured asymptotic (1.1), 1
For details, we highly recommend the insightful expository article [9] of Soundararajan.
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which, as was shown by Gallagher [4, Theorem 1], would in fact follow from a certain uniform version of the Hardy–Littlewood prime tuples conjecture. We similarly expect the normalized spacings between consecutive primes to follow an exponential distribution, i.e. if dn ..= pn+1 −pn , where pn denotes the nth smallest prime, we have the well-known conjecture 1 #{n x : dn / log n ∈ (a, b]} ∼ x
b
e−t dt
(x → ∞).
a
(Another form of the prime number theorem states that x−1 nx dn / log n ∼ 1 as x tends to infinity, i.e. dn / log n ≈ 1 on average over n x.) However, we do not even know of any specific limit points of the sequence (dn / log n), except for 0 and ∞ (the former following from the aforementioned result of GPY, the latter from an old result of Westzynthius [10]). Nevertheless, it has recently been shown2 [1, Theorem 1.1] that, in a certain sense, 12.5% of positive real numbers are limit points of (dn / log n). This note may be regarded as a continuation of [1], the results of which are utilized here in combination with the very general and powerful quantitative work of Maynard [6]. 3. Proof of Theorem 1.1 We shall consider linear functions L given by L(n) = gn + h, where g, h ∈ Z (it is to be assumed that g = 0). A finite set {L1 , . . . , Lk } of linear functions is admissible if the set of solutions modulo p to L1 (n) · · · Lk (n) ≡ 0 (p) does not form a complete residue system modulo p, for any prime p. (It is to be assumed that 1i
n∈[x,2x) n≡a (q)
n∈[x,2x)
(3.1) Theorem 3.1 (Maynard). Let L = {L1 , . . . , Lk } be an admissible set of k linear functions. Let B be a positive integer, let x be a large real number and let 0 < θ < 1. Let α > 0. 2 Benatar [2] has since claimed that in fact 25% of positive real numbers are limit points of the sequence (dn / log n).
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Suppose that the coefficients of Li (n) ..= gi n + hi ∈ L satisfy 1 gi , hi xα for i = 1, . . . , k, that k (log x)α and that 1 B xα . There is a positive constant C, depending only on θ and α, such that the following holds. If k C, if L, B, x, θ satisfy Hypothesis 1 and if δ > (log k)−1 is such that 1 ϕ(B) ϕ(gi ) k B gi L∈L
1P (L(n)) δ
n∈[x,2x)
x , log x
(3.2)
then # n ∈ [x, 2x) : #({L1 (n), . . . , Lk (n)} ∩ P) C −1 δ log k
x . (eC log x)k
(3.3)
(The implicit constant in (3.1) may depend only on θ and α, and that in (3.3) depends at most on θ and α.) A level of distribution result. We need to show that Hypothesis 1 holds for certain choices of L, B, x, θ. We defer proof of the following result to §4. Lemma 3.2. There is an absolute constant c ∈ (0, 1) such that the following holds. Fix any positive integer k and let x be a real number that is sufficiently large in terms of k. There is a positive integer B, which is either a prime satisfying log2 xη B x2η , where η ..= c/(500k2 ), or is equal to 1, for which we have the following. Let g be any positive integer that is coprime to B and divides plog xη p. Let L = {L1 , . . . , Lk } be any admissible set of k linear functions for which the coefficients of Li (n) ..= gi n + hi satisfy gi = g and 1 hi x for i = 1, . . . , k. For each L ∈ L, ϕ(B) ϕ(g) B g
1P (L(n)) >
n∈[x,2x)
x 2 log x
(3.4)
and qx1/8 (q,B)=1
max (L(a),q)=1
n∈[x,2x) n≡a (q)
1 1P (L(n)) − ϕL (q)
n∈[x,2x)
n∈[x,2x) 1P (L(n)) 1P (L(n)) . (log x)100k2 (3.5)
An Erdős–Rankin type construction. We quote [1, Lemma 5.2]. First, some more notation and terminology: a finite set {h1 , . . . , hk } of integers is admissible if the set {L1 , . . . , Lk } of linear forms given by Li (n) = n + hi , i = 1, . . . , k, is admissible. Given a real number z 1 we let [z] ..= {1, . . . , z}, where z denotes the greatest integer less than or equal to z.
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Lemma 3.3. Fix a positive integer k and k nonnegative real numbers β1 , . . . , βk . Suppose that β1 · · · βk . There is a constant C , depending only on k and β1 , . . . , βk , such that the following holds. Let v, y, z be real numbers satisfying v 1, y C and 2y(1 + (1 + βk )v) 2z y(log2 y)/ log3 y.
(3.6)
Let B be any positive integer such that for all prime divisors l of B (if any), p|B, pl
1/p 1/l 1/ log y.
(3.7)
There exists an admissible set {h1 , . . . , hk } and a sequence (ap (p))py, p B of residue classes such that 1/4 hi = y + βi vy + O ye−(log y) for i = 1, . . . , k and {h1 , . . . , hk } = [z] \
py, p B
ap (p).
Deduction of Theorem 1.1. Fix a positive real number λ and a nonnegative integer m. Let C be the constant of Theorem 3.1, which depends on θ and α. We will apply Theorem 3.1 with θ ..= 1/8 and α ..= 1, so C may be regarded as absolute. We will also apply the theorem with δ ..= 1/2. Let k be the smallest positive integer satisfying k C, k e2 and k e2Cm (i.e. k C, δ (log k)−1 and C −1 δ log k m). Let βi = 2i−k λ, i = 1, . . . , k and let C be the constant of Lemma 3.3, which depends on k (hence m) and λ. Let x be a large real number and define v ..=
log3 x 1 , 3(1 + 3λ) log4 x
y ..= 3(1 + 3λ) log x
log4 x , log3 x
z = (1 + 3λ) log x.
(3.8)
We think of x as tending to infinity, and we tacitly assume throughout that x is “sufficiently large” in terms of any specified fixed quantity, hence ultimately in terms of λ and m. Thus, for instance, as is straightforward to verify, we have v 1, y C and (3.6). Let η and B be as in Lemma 3.2, i.e. η ..= c/(500k2 ) for a certain absolute constant c ∈ (0, 1), and either B = 1 or B is a prime satisfying log2 xη B x2η . As log xη > y, (3.7) is satisfied. Each hypothesis of Lemma 3.3 thus accounted for, we conclude that there exists an admissible set {h1 , . . . , hk } and a sequence (ap (p))py, p B of residue classes such that 1/4 hi = y + 2i−k λ log x + O ye−(log y) for i = 1, . . . , k (we have vy = log x), and {h1 , . . . , hk } = [(1 + 3λ) log x] \
py, p B
ap (p).
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We work with such an admissible set and sequence of residue classes. Note that hk − h1 < −1 + λ log x
(3.9)
1 < h1 < · · · < hk < −1 + 2λ log x
(3.10)
and
We let g ..= py, pB p and h (g) be the residue class modulo g such that h ≡ −ap (p) for each prime p dividing g. Let us suppose that 0 h < g. We let L ..= {L1 , . . . , Lk } be the set of linear functions in which Li (n) ..= gn + h + hi for i = 1, . . . , k. It is straightforward to verify that L is admissible, and that for all positive integers n, (gn + h, gn + h + (1 + 3λ) log x] ∩ P = {L1 (n), . . . , Lk (n)} ∩ P.
(3.11)
We have (g, B) = 1 by definition, and as already noted, log xη > y, so g divides plog xη p. In fact, by Chebyshev’s bounds and since η is very small, we certainly have 0 < g, h + hi < x for i = 1, . . . , k. Therefore, by Lemma 3.2, L, B, x and θ = 1/8 satisfy Hypothesis 1, and (3.2) holds with δ = 1/2 for each L ∈ L. We now invoke Theorem 3.1 with θ = 1/8, α = 1 (we have k log x) and δ = 1/2. We’ve chosen k so that C −1 δ log k m, so we infer that
# {n ∈ [x, 2x) : #({L1 (n), . . . , Lk (n)} ∩ P) m}
x . (eC log x)k
(3.12)
Choose n ∈ [x, 2x) such that #({L1 (n), . . . , Lk (n)} ∩ P) m. Consider the intervals Ij ..= (Nj , Nj + λ log Nj ],
Nj ..= gn + h + j,
j = 0, . . . , 2λ log N0 .
Now, N0 = x1+o(1) , so for j in the given range we have Ij ⊆ (gn + h, gn + h + (1 + 3λ) log x], so by (3.11), Ij ∩ P = (Ij ∩ {L1 (n), . . . , Lk (n)}) ∩ P.
(3.13)
By (3.9), L1 (n) < · · · < Lk (n) < L1 (n) − 1 + λ log x < L1 (n) − 1 + λ log N0 . Thus, if j = h1 − 1 (so that Nj = L1 (n) − 1), then Ij ∩ {L1 (n), . . . , Lk (n)} = {L1 (n), . . . , Lk (n)}.
(3.14)
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By (3.10), Lk (n) = N0 + hk < N0 − 1 + 2λ log x < N0 − 1 + 2λ log N0 . Thus, if j = 2λ log N0 , then Ij ∩ {L1 (n), . . . , Lk (n)} = ∅.
(3.15)
Therefore, by (3.13) and (3.14), #(Ih1 −1 ∩ P) = #({L1 (n), . . . , Lk (n)} ∩ P) m, while on the other hand, by (3.13) and (3.15), #(I2λ log N0 ∩ P) = 0. Now, for any j, if #(Ij+1 ∩ P) < #(Ij ∩ P) then #(Ij+1 ∩ P) = #(Ij ∩ P) − 1. We must conclude that there is some j in the range h1 − 1 j 2λ log N0 for which π(Nj + λ log Nj ) − π(Nj ) = m. By the prime number theorem and the definition of g, B and y, we have g = e(1+o(1))y > (1 + 3λ) log x. Since g(n + 1) + h > gn + h + (1 + 3λ) log x, no two values of n can give rise to the same Nj in this way. We deduce from (3.12) that, with X ..= 4gx, #{N X : π(N + λ log N ) − π(N ) = m}
x . (eC log x)k
It follows that the left-hand side exceeds X 1−ε(X) , where ε(X) ..= (log4 X)2 / log3 X. (Recall that log g = (1 + o(1))y by the prime number theorem, and that, by (3.8), y = 3(1 + 3λ) log x log4 x/ log3 x.) 2 4. Proof of Lemma 3.2 Lemma 3.2 is similar to [1, Theorem 4.2]. We must nevertheless verify the details. First, some more notation: given a positive integer q, χ mod q, or simply χ if q is clear in context, denotes a Dirichlet character to the modulus q, L(s, χ) denotes the L-function associated with it and χ ¯ its complex conjugate. Also, P + (q) denotes the greatest prime + . divisor of q (P (1) .= 1 by convention). We quote [1, Lemma 4.1]. Lemma 4.1. Let T 3 and let P T 1/ log2 T . Among all primitive Dirichlet characters χ mod to moduli satisfying T and P + ( ) P , there is at most one for which the
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associated L-function L(s, χ) has a zero in the region |(s)| exp log P/ log T ,
(s) > 1 − c/ log P,
(4.1)
where c > 0 is a certain (small) absolute constant. If such a character χ mod exists, then χ is real and L(s, χ) has just one zero in the region (4.1), which is real and simple, and P + ( ) log log2 T.
(4.2)
Definition 4.2. For T 3, let (T ) ..= if the “exceptional” character χ mod , as described in Lemma 4.1, exists; let (T ) ..= 1 otherwise. Proof of Lemma 3.2. Let c be the constant of Lemma 4.1. We may assume that c < 1. Fix a positive integer k and let η ..= c/(500k2 ). Let x be a large real number, and let g be a positive integer that divides plog xη p. Note that, by Chebyshev’s bounds, g < x1/22 , say. (This is the reason for taking c < 1.) Let B ..= (x2η ), as in Definition 4.2, and suppose (g, B) = 1. Let h be an integer satisfying 1 h x. Suppose (g, h) = 1 and let L denote the linear function given by L(n) ..= gn + h. Let IL (x) ..= [gx + h, 2gx + h). Let q denote a positive integer and a an integer for which (L(a), q) = 1, noting that this implies (L(a), gq) = 1. We have
1P (L(n)) =
n∈[x,2x) n≡a (q)
1P (n) =
n∈IL (x) n≡L(a) (gq)
1 ϕ(gq)
1P (n) + ΔL (x; q, a),
n∈IL (x)
where
ΔL (x; q, a) ..=
1P (n) −
n∈IL (x) n≡L(a) (gq)
1 ϕ(gq)
1P (n).
n∈IL (x)
We will show that if x is large enough in terms of k, then
max
qx1/8 (q,B)=1
(L(a),q)=1
|ΔL (x; q, a)|
gx . ϕ(g)(log x)2+100k2
(4.3)
Let us show how this implies Lemma 3.2. First, |ΔL (x; 1, 1)| is certainly majorized by the left-hand side of (4.3), so if x is large enough in terms of k, n∈[x,2x)
1P (L(n)) =
1 ϕ(g)
n∈IL (x)
1P (n) + ΔL (x; 1, 1) =
gx (1 + O (1/ log x)) ϕ(g) log(gx)
(4.4)
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by (4.3) and the prime number theorem. If B = 1 then, by (4.2), B is a prime satisfying B log2 xη . In any case, ϕ(B)/B = 1 + O(1/ log2 xη ). As already noted, we have g < x1/22 , i.e. log(gx) < 23 22 log x. Thus, if x is large enough in terms of k we have ϕ(B) ϕ(g) B g
1P (L(n))
n∈[x,2x)
22x (1 + O (1/ log2 xη )) , 23 log x
whence (3.4). Second, we verify that
1P (L(n)) −
n∈[x,2x) n≡a (q)
1 ϕL (q)
1P (L(n)) = ΔL (x; q, a) −
n∈[x,2x)
ϕ(g) ΔL (x; 1, 1). ϕ(gq)
(4.5)
Third, again using (4.3) to bound |ΔL (x; 1, 1)|, then using ϕ(gq) ϕ(g)ϕ(q) and qx 1/ϕ(q) log x, we obtain qx1/8 (q,B)=1
gx ϕ(g) gx 1 |ΔL (x; 1, 1)| . 2 ϕ(gq) ϕ(gq) (log x)2+100k ϕ(g)(log x)1+100k2 1/8
(4.6)
qx
Fourth, we combine (4.3), (4.5) and (4.6) (applying the triangle inequality to the righthand side of (4.5)), obtaining qx1/8 (q,B)=1
max (L(a),q)=1
n∈[x,2x) n≡a (q)
1 1P (L(n)) − ϕL (q)
1P (L(n))
n∈[x,2x)
gx . ϕ(g)(log x)1+100k2 (4.7)
Finally, combining (4.4) with (4.7) yields (3.5). We now establish (4.3) by paraphrasing the proof of [1, Theorem 4.2]. Suppose 1 q x1/8 . By orthogonality of Dirichlet characters we have
1P (L(n)) =
n∈[x,2x) n≡a (q)
1P (n) =
n∈IL (x) n≡L(a) (gq)
1 ϕ(gq)
χ(L(a)) ¯
χ mod gq
χ(n)1P (n).
n∈IL (x)
Letting χ∗ denote the primitive character that induces χ, we have (for characters χ to the modulus gq), (χ(n) − χ(n))1P (n) n∈IL (x)
n∈IL (x) (n,gq)>1
1P (n)
p|gq
1 log(gq) log x,
T. Freiberg / Journal of Number Theory 163 (2016) 159–171
168
whence
1P (L(n)) =
n∈[x,2x) n≡a (q)
1 ϕ(gq)
χ(L(a)) ¯
χ mod gq
χ∗ (n)1P (n) + O(log x).
n∈IL (x)
For the principal character χ0 mod gq we have χ∗0 ≡ 1, and so we deduce that max
(L(a),q)=1
|ΔL (x; q, a)|
1 ϕ(gq)
χ mod gq χ=χ0
∗ + O(log x) χ (n)1 (n) P
(4.8)
n∈IL (x)
It follows from the explicit formula [3, §19, (13)–(14)] that, for nonprincipal characters √ χ mod gq, 2 T N and T N , with Λ denoting the von Mangoldt function, N (ρ) √ + N (log(gqN ))2 , χ(n)Λ(n) |ρ| √ |ρ|< N
nN
where zeros of L(s, χ) having real part at least 1/2. Since the sum is over nontrivial √ χ(n)Λ(n)(1 (n) − 1) N log N , the same bound holds if Λ is replaced by P nN √ √ Λ1P and, via partial summation, 1P . Thus, since gq xO(1) and gx gx + h, ∗ χ (n)1P (n) n∈IL (x)
√ |ρ|< gx
(gx) (ρ) √ + gx(log x)2 . |ρ|
Combining this with (4.8) gives max
(L(a),q)=1
|ΔL (x; q, a)|
=
1 ϕ(gq)
∗ (gx) (ρ) √ + gx(log x)2 |ρ| √
χ mod gq |ρ|< gx χ=χ0
1 ∗ ϕ(gq)
√ d|gq χ mod d |ρ|< gx
(gx) (ρ) √ + gx(log x)2 , |ρ|
(4.9)
∗ where, in the first line, denotes summation over nontrivial zeros of L(s, χ∗ ) having ∗ real part at least 1/2, χ being the primitive character that induces χ, and in the second ∗ line, denotes summation over primitive characters and the innermost sum is over nontrivial zeros of L(s, χ) having real part at least 1/2. Applying (4.9) and changing order of summation, recalling that (g, B) = 1, we find that qx1/8 (q,B)=1
max
(L(a),q)=1
|ΔL (x; q, a)|
T. Freiberg / Journal of Number Theory 163 (2016) 159–171
∗
(gx) (ρ) |ρ|
√ dgx1/8 χ mod d |ρ|< gx (d,B)=1
qx1/8 (q,B)=1 d|gq
169
1 √ + x1/8 gx(log x)2 . ϕ(gq)
Writing d = ab with a = (d, g), for d | gq we have gq = gbc for some integer c. Note that (b, g) divides a. We have ϕ(gq) = ϕ(gbc) ϕ(g)ϕ(b)ϕ(c), and as cx1/8 1/ϕ(c) log x, it follows that max |ΔL (x; q, a)| qx1/8 (q,B)=1
(L(a),q)=1
log x ϕ(g)
a|g bgx1/8 /a (b,g)|a (b,B)=1
1 ϕ(b)
∗
√ χ mod ab |ρ|< gx
(gx) (ρ) √ + x1/8 gx(log x)2 . |ρ|
(4.10)
If a | g and b gx1/8 /a, then a ∈ [R, 2R) and b ∈ [S, 2S) for some pair (R, S) of powers of 2 satisfying 1 R < g and 1 RS < gx1/8 . The number of such pairs is O((log x)2 ). Note that for b ∈ [S, 2S) and S < gx1/8 we have 1/ϕ(b) (log2 b)/b (log x)/S. If √ |ρ| < gx and 1/2 (ρ) 1 then (ρ) ∈ Im ..= [1/2 + m/ log(gx), 1/2 + (m + 1)/ log(gx)) for some integer m satisfying 0 m <
1 2
log(gx), and
|(ρ)| ∈ Jn ..= [n − 1, 2n − 1) √ with n being some power of 2 satisfying 1 n < gx. The number of such pairs (m, n) is O((log x)2 ). Note that for ρ with (ρ) ∈ Im and |(ρ)| ∈ Jn we have (gx) (ρ) /|ρ| √ gx em /n. Thus, log x ϕ(g)
a|g bgx1/8 /a (b,g)|a (b,B)=1
√
1 ϕ(b)
gx(log x)6 ϕ(g)
∗
√ χ mod ab |ρ|< gx
sup R
where N ∗ (R, S, σ, T ) ..=
(gx) (ρ) |ρ|
e−m ∗ N R, S, 1/2 + m/ log(gx), n − 1 , nS
∗
1,
Ra<2R Sb<2S χ mod ab (ρ)σ a|g (b,g)|a | (ρ)|T (b,B)=1
and the innermost sum is over zeros ρ of L(s, χ) in the given region.
(4.11)
(4.12)
T. Freiberg / Journal of Number Theory 163 (2016) 159–171
170
Note that, as g is squarefree, every positive integer may be decomposed uniquely as a product ab of positive integers a | g and (b, g) | a, whence N ∗ (R, S, σ, T )
∗
d<4RS χ mod d
1.
(ρ)σ | (ρ)|T
A result [8, Theorem 12.2] of Montgomery therefore implies that N ∗ (R, S, σ, T ) ((RS)2 T )3(1−σ)/(2−σ) (log(RST ))14
(4.13)
for T 2 and 1/2 σ 1. On the right-hand side of (4.11), we partition the supremum set according as (i) 0 m 12 log(gx) − (44 + 200k2 ) log2 (gx) or (ii) 12 log(gx) − (44 + 200k 2 ) log2 (gx) < m < 12 log(gx). For case (i), we note that for 1/2 σ 1, the following inequalities hold: 1/(2−σ) 1, 6(1 − σ)/(2 − σ) 1 + 2(1 − σ) and 3(1 − σ)/(2 − σ) 1. Thus, (R2 )3(1−σ)/(2−σ) (R6 )(1−σ) , (S 2 )3(1−σ)/(2−σ) S(S 2 )(1−σ) and T 3(1−σ)/(2−σ) T . Hence (4.13) implies N ∗ (R, S, σ, T ) (R6 S 2 )1−σ ST (log(RST ))14 . Recalling that g < x1/22 , note that if √ R < g and RS < gx1/8 then R6 S 2 < gx. It follows that √
gx em ∗ m/2 1/4 14 N R, S, 1/2 + m/ log(gx), n − 1 e (gx) (log x) . nS (log x)8+100k2 (4.14) 2
We divide case (ii) into two sub-cases: (iia) n3/4 S 1/2 (log(gx))22+100k or (iib) 2 3/4 1/2 n S < (log(gx))22+100k . For (iia), we note that if σ = 1/2 + m/ log(gx) then σ 10/11 (provided x is large enough in terms of k, as we assume), hence 3(1 − σ)/(2 − σ) 1/4. We have (R2 )3(1−σ)/(2−σ) (R6 )1−σ < (gx)(1−σ)/2 (for R < g) as before, and (S 2 T )3(1−σ)/(2−σ) S 1/2 T 1/4 . By (4.13) we therefore have √
gx em ∗ em/2 (gx)1/4 (log x)14 N R, S, 1/2 + m/ log(gx), n − 1 2 . 8+100k 3/4 1/2 nS (log x) n S (4.15) For (iib) we apply Lemma 4.1 to the right-hand side of (4.12). Note that in this case we 2 have S < (log(gx))44+200k . Also note that, since g | plog xη p, we have g x(1+o(1))η by the prime number theorem. Recall that η = c/(500k2 ). Thus, for a | g and b < 2S 2η we have ab < x2η and P + (ab) < x2η/ log2 x (if x is large enough in terms of k, as we assume). We find that 1−c
m log2 x2η 1 < + 2η log x 2 log(gx)
and n < exp
log x2η / log2 x2η 2
when m > 12 log(gx) − (44 + 200k2 ) log2 (gx) and n3/4 < (log(gx))22+100k . Therefore, in case (iib), N ∗ (R, S, σ, T ) is at most the number of zeros of L(s, χ) for primitive characters χ mod ab, ab < x2η =.. T , P + (ab) < T 1/ log2 T =.. P ,
T. Freiberg / Journal of Number Theory 163 (2016) 159–171
(s) 1 − c/ log P,
171
|(s)| exp log P/ log T
and (ab, B) = 1 (recall that a | g and (g, B) = (b, B) = 1). But by Lemma 4.1 and our choice of B, there are no such zeros, i.e.
N ∗ R, S, 1/2 + m/ log(gx), n − 1 = 0. Combining (4.10), (4.11), (4.14), (4.15) and (4.16) gives (4.3).
(4.16) 2
5. Concluding remarks In §3, we did not quote (the special case of) Maynard’s theorem [6, Theorem 3.1] in its entirety. It continues as follows.3 If, moreover, k (log x)1/5 and all L ∈ L have the form gn + hi with |hi | υ log x (log x)/(k log k) and g 1, then # n ∈ [x, 2x) : #({L1 (n), . . . , Lk (n)} ∩ P) C −1 δ log k
x eCk2 (log x)k
,
(5.1)
where on the left-hand side, [gn, gn + υ log x] ∩ P = {L1 (n), . . . , Lk (n)} ∩ P. This would lead to an improvement of Theorem 1.1 prima facie only for certain λ and m — note here that υ 1/(k log k), so if υ λ and C −1 δ log k m, there is an interdependence −1 between λ, m and δ, viz. λCmδ −1 eCmδ 1. Perhaps the right-hand side of (1.3) can be improved to something of a quality similar to that of (5.1), for all λ and m, via a less ad hoc proof, i.e. a proof that uses Maynard’s sieve alone, and does not involve the Erdős–Rankin construction. References [1] W.D. Banks, T. Freiberg, J. Maynard, On limit points of the sequence of normalized prime gaps, preprint, arXiv:1404.5094, 2014. [2] J. Benatar, Goldbach versus de Polignac numbers, preprint, arXiv:1505.03104, 2015. [3] H. Davenport, Multiplicative Number Theory, 3rd edn., Grad. Texts in Math., vol. 74, SpringerVerlag, New York, 2000, revised and with a preface by H. L. Montgomery. [4] P.X. Gallagher, On the distribution of primes in short intervals, Mathematika 23 (1) (1976) 4–9. [5] D.A. Goldston, J. Pintz, C.Y. Yıldırım, Primes in tuples I, Ann. of Math. (2) 170 (2) (2009) 819–862. [6] J. Maynard, Dense clusters of primes in subsets, preprint, arXiv:1405.2593, 2014. [7] J. Maynard, Small gaps between primes, Ann. of Math. (2) 181 (1) (2015) 383–413. [8] H.L. Montgomery, Topics in Multiplicative Number Theory, Lecture Notes in Math., vol. 227, Springer-Verlag, Berlin–New York, 1971. [9] K. Soundararajan, The distribution of prime numbers, in: A. Granville, Z. Rudnick (Eds.), Equidistribution in Number Theory, an Introduction, in: NATO Sci. Ser. II Math. Phys. Chem., vol. 237, Springer, Dordrech, 2007, pp. 59–83. [10] E. Westzynthius, Über die Verteilung der Zahlen, die zu den n ersten Primzahlen teilerfremd sind, Commentat. Phys.-Math. 5 (25) (1931) 1–37.
3
We are not quoting [6, Theorem 3.1] exactly here. Rather, we are inspecting its proof [6, (6.13) et seq.].