In Stadlmann's Type I/II equidistribution estimate for x^delta-smooth moduli, one of the three conditions, 64 varpi + 20 delta + 2 sigma < 1, can be replaced by 48 varpi + 18 delta + 2 sigma < 1. This weaker condition follows from the second one, 48 varpi + 16 delta + 4 sigma < 1, whenever delta <= sigma. The proof follows Stadlmann's argument and makes two changes: the length of a block in one Taylor expansion depends on the pair of shifts, and the bounds are summed over the pairs instead of being multiplied by their number. Combined with Stadlmann's general Harman-sieve construction, this gives a minorant for the primes with exponent of distribution 1/2 + 2 varpi to smooth moduli for every varpi < 7/544, instead of varpi = 0.01265. As a consequence, liminf (p_{n+m} - p_n) << e^{cm} for every c > 3.80428, and for every c > 3.80423 if a numerically computed value of the sieve deficit is used. This improves Stadlmann's exponent 3.8075.
Statement
We use the notation of Stadlmann [4], §3. Let ϖ,δ,σ>0. Let α,β be coefficient sequences at scales M,N with MN≍x and N=xγ, where β has the Siegel–Walfisz property. The moduli are the q≤x1/2+2ϖ dividing P(xδ). Stadlmann proved [4 Prop. 1] that the equidistribution estimate [4 (3.28)] holds for all γ∈[1/2−σ,1/2] whenever
72ϖ+24δ<1,48ϖ+16δ+4σ<1,64ϖ+20δ+2σ<1.(1)
Theorem 1. The conclusion of [4 Prop. 1] holds when the third condition in (1) is replaced by 48ϖ+18δ+2σ<1. More precisely, in [4 Lemma 17] the lower bound γ≥32ϖ+10δ+400ε may be replaced by γ≥24ϖ+9δ+400ε.
The new condition follows from the second condition of (1) whenever δ≤σ, and this is the case in all the applications below.
Remark 2. The theorem does not improve the exponent of distribution 1/2+1/40 of the primes themselves [4 Thm. 1]. That deduction goes through the Heath-Brown identity and [4 Lemma 1], which require σ>1/10. With the second condition of (1), σ>1/10 already gives 80ϖ+380δ<1[4 Cor. 2], so the third condition is not the binding one there. It is the binding one in the Harman-sieve application: [4 Lemma 18(II)] needs 0.40481>32ϖ. OpenAI extended [4 Prop. 1], with the same three conditions, to triply densely divisible moduli [2 Prop. 2.18 and §A.4]. Their proof follows the same lemmas with the same block length, and we expect the same modification to apply. We have not checked the details and claim nothing for those moduli.
Proof of Theorem 1
All references in this section are to [4 §3], whose notation we keep: z∈Z5(x), λ,λ~∼Λ, w1,w2,m,q0,Δ1,H,T(z) and the classes W6,W7,W8. We write s2=(w2,m) and λ∗=(λ,λ~). Since λ∼Λ we have ∣λ∣≤2∣Λ∣. We go through Lemmas 7–17 and indicate every change. Everything not mentioned is used unchanged.
Step 1: a block length depending on the pair (Lemma 10)
In the proof of Lemma 10, the variable d, of length Δ1, is cut by a smooth partition of unity into blocks of length Δ∗=min{N/(∣Λ∣x5ε),Δ1}, which does not depend on (λ,λ). For a fixed pair we use instead
Δλ∗:=min{∣λ∣λ∗x5εN,Δ1},Nλ:=∣λ∣λ∗N.
Since λ∗≥1 and ∣λ∣≤2∣Λ∣, we have Δ∗≤2Δλ∗. The block length is used in three places, and each still holds with Δλ∗:
(a) the number of blocks is ≪Δ1/Δλ∗, and the cut-off functions τm satisfy the same derivative bounds, since Δλ∗≤Δ1;
(b) the Taylor expansion (3.13) of ψN∗((λ/λ∗)n1+Fkd) in d needs ∣Fk(d−x0)∣/N≪x−5ε for ∣d−x0∣≪Δλ∗/z1. Since 0<Fk≤∣λ∣/λ∗,
N∣Fk(d−x0)∣≤λ∗∣λ∣⋅z1NΔλ∗≤z1x−5ε;
(c) in Step 2 of the proof of Lemma 11, the expansion (3.15) needs
(λ∗/∣λ∣)NHkf4Δλ∗/(z1f∗)≪x−5ε.
Since Hkf4≤[f2/f∗,f3/f∗]≤w1, f∗≥1 and w1∣z1, the left side is at most (w1/z1)x−5ε≤x−5ε.
Thus Lemma 10 holds with Δλ∗ in place of Δ∗, and Lemma 11 holds with W8 defined by Δ2≤Δλ∗ instead of Δ2≤Δ∗.
Step 2: the length of the n1-sum (Lemmas 11 and 14)
After the substitution n=(λ/λ∗)n1+Fkd in Lemma 10, the variable n1 runs over a coefficient sequence at scale (λ∗/λ)N. In Lemma 11 it becomes n3, at scale (λ∗/(λw3))N≤Nλ. The definition of W8 records only N2≤N; we keep N2≤Nλ. Lemma 13 applied as in the proof of Lemma 14 then gives, for this pair,
Lemmas 15 and 16 do not involve the block length. The range of k, the condition (⋅,m)≤T(z) and the sequence (ξk) are the same as in [4]. They give Σ6(z,w6,k)≪εξkx4εs2T(z)Fλ/q0 with (ξk)∈Ξ(z). The proof of Lemma 9 is linear in the target bound: if Σ6≪ξkI for all relevant k, then Σ5(z)≪q0(q0,ℓ)I. Hence
Σ5(z)≪ε(q0,ℓ)x4εs2T(z)Fλ.(3)
Expanding Fλ gives four terms:
Fλ=mΔ1Nλ+Δλ∗Δ1Nλ+Δ1+Δλ∗Δ1m.
By Nλ≤N and Δ∗≤2Δλ∗, the first and the last are at most once and twice the corresponding terms Δ1N/m and (Δ1/Δ∗)m of [4]. For the middle two, consider the two possible values of Δλ∗. If Δλ∗=Nλx−5ε then (Δ1/Δλ∗)Nλ=x5εΔ1, and otherwise (Δ1/Δλ∗)Nλ=Nλ. Hence
Δλ∗Δ1Nλ+Δ1≤2x5εΔ1+Nλ.(4)
Step 3: summing over the pairs (Lemma 7)
In the proof of Lemma 7, ΥT≪εΥT+supqxεΥS(q), and the bound ΥS(q)≪εx−48εN2, which includes the diagonal terms, gives the term O(x−47εN2). Here ΥT is a sum over pairs (y,y) of the inner sums Σ4(z1,z2,z3,(y,y)). Stadlmann bounds it by the number of pairs, ≪P:=max{xδ+10εH3,H4}/s2, times the supremum. The condition (⋅,m)≤T in Σ4 does not depend on the pair. So, by the triangle inequality, we may instead sum bounds that depend on the pair: ΥT≤∑(y,y)∣Σ4(⋅,(y,y))∣. The terms ΥS(q) are not changed. By the proof of Lemma 8, each Σ4(⋅,(y,y)) is zero or equals Σ5(z) for a z∈Z5 with y=w1λ, y=w1λ and Λ=Y/w1. In particular
∣λ∣λ∗=∣y∣(y,y).
By (3), the target Σ3≪q02(q0,ℓ)2N2x−27ε of Lemma 7 (hence of Lemma 8) follows from
s2T(z)(y,y)∑Fλ≪q02N2x−31ε.(5)
Condition (3.29) of [4] is the same inequality with ∑Fλ replaced by PF, where F is the expression with Δ∗ and N. By Step 2, the contribution of the first and last terms of Fλ is at most twice Stadlmann’s. It is therefore covered by (3.30) and (3.32) of [4], which give the conditions γ≤1/2−4ϖ−2δ−50ε and γ≥1/4+12ϖ+4δ+100ε. It remains to replace (3.31), which concerns the two middle terms and needed γ≥32ϖ+10δ+400ε. By (4),
Since ∣y∣≥∣Y∣, it follows that ∑(y,y)(y,y)/∣y∣≪xεY, where Y is the number of admissible y. By the proof of Lemma 7, Y≪min{H2/s2+H,∣Y∣/s2+1} and ∣Y∣≪xδ+5εH2/(v1,v2). We also use
First contribution. We have max{1,s2/H}P≤max{xδ+10εH3,H4}≤xδ+10εH4. Hence
s2T(z)⋅x5εΔ1P≤x2δ+115εH6N.
This is ≪q02N2x−31ε provided x2δ+146εH6≪q02N, which follows from γ≥24ϖ+8δ+190ε.
Second contribution. We claim that max{1,s2/H}Y≪xδ+5εH2. If s2≤H, then Y≤2H2. If s2>H, we use s2≤2∣Y∣: indeed s2∣w2∣y for every admissible y=0, and the sum is empty otherwise. Then (s2/H)(∣Y∣/s2+1)≤3∣Y∣/H≪xδ+5εH. Therefore
s2T(z)⋅xεNY≪x2δ+106εΔ1H4N2≪x3δ+161εH6N,
which is ≪q02N2x−31ε provided x3δ+192εH6≪q02N. This follows from γ≥24ϖ+9δ+240ε.
Step 4: conclusion (Lemma 17 and Proposition 1)
Lemma 17 therefore holds with max{1/4+12ϖ+4δ+100ε,24ϖ+9δ+400ε}≤γ≤1/2−4ϖ−2δ−50ε. The final step of the proof of Proposition 1 is unchanged. It covers γ near 1/2 by [3 Thm. 2.8(ii), (iv)] when 68ϖ+14δ<1 and 72ϖ+24δ+200ε<1. Hence the range γ∈[1/2−σ,1/2] is covered as soon as 1/4+12ϖ+4δ<1/2−σ, 24ϖ+9δ<1/2−σ and 72ϖ+24δ<1, with ε small enough. These are the conditions of Theorem 1. □
Consequences
We use the general Harman-sieve construction of Stadlmann [5 Def. 9 and Prop. 2], with parameters ξ1,ξ2,ξ3. It is the construction of [4 §4], with the numerical values 0.33856, 0.40481, 0.59519, 0.19, 0.405, 0.595 replaced by ξ1,ξ2,1−ξ2,1−2ξ3,ξ3,1−ξ3. Its proof requires the five inequalities
Its input is equidistribution, on the family of moduli under consideration, for three types of convolutions: (I) smooth factors of length ≥xξ1;
(II) bilinear forms with both factors Siegel–Walfisz and one length in [xξ2,x1−ξ2];
(III) three smooth factors of lengths in [x1−2ξ3,xξ3] with pairwise products ≥x1−ξ3
In [5] it is stated for the moduli Q∗ of that paper; its proof, like that of [4 §4] where the moduli are xδ-smooth, uses these inputs on the family of moduli concerned. In our application the coefficients of type (I) with lengths in [xξ2,x1−ξ2] are treated by (II), which uses that they satisfy the Siegel–Walfisz property [2 Lemma A.8]. We include the corrections of OpenAI [2 footnotes 2–4 and Lemmas A.8–A.9], written there for fixed numerical values. The numerical inequalities in their arguments reduce to (6), 6(1−2ξ2)>1, ξ2>2/5 and a level of distribution at most x0.53; for instance their two grouping gaps are 1−ξ1−3ξ2/2 and 4−ξ1−9ξ2. The minorant ρ has mass 1−c1(ξ2), where c1(ξ2)=I1+I2 is the explicit sum of two four-dimensional integrals in [5 Prop. 2]. Every constraint in these integrals becomes stricter when ξ2 decreases, so c1 is nondecreasing in ξ2.
Corollary 3. Let 0<ϖ<7/544 and let δ>0 be small enough. Take ξ2=ξ3 with
and ξ1 with max{41+7ϖ+2δ,21−ξ2}<ξ1<min{1−23ξ2,4−9ξ2}. These intervals are nonempty. Then the minorant ρ of [5 Prop. 2] with these parameters has the following properties:
ρ≤1P, and every prime factor of an n with ρ(n)=0 is at least x1−2ξ2;
ρ has exponent of distribution 1/2+2ϖ to xδ-smooth squarefree moduli;
ρ has mass 1−c1(ξ2) with c1(ξ2)≤c1(55/136).
Proof. The intervals are nonempty. We have 1/4+12ϖ+4δ<4/9−28ϖ/9−2δ/9 if and only if 544ϖ+152δ<7; the two bounds meet at ϖ=7/544, ξ2=55/136. Also 1/4+12ϖ+4δ<55/136 for ϖ<7/544 and δ small, and 2/5<min{4/9−28ϖ/9−2δ/9,55/136} since ϖ<1/70. So the interval for ξ2 is nonempty. Since 2/5<ξ2<55/136, we have (1−ξ2)/2<3/10, 1−3ξ2/2>0.393 and 4−9ξ2>49/136>0.3602, while 1/4+7ϖ+2δ<0.3401; so the interval for ξ1 is nonempty. The program check_parameters.py checks all the inequalities used below in exact rational arithmetic, at sample values of (ϖ,δ) (including values close to 7/544 and small values of ϖ) and at points close to the ends of the intervals for ξ1 and ξ2. In general they follow from the displayed choices of ξ1,ξ2 by the elementary verifications below.
Type II. We use Theorem 1 with σ=1/2−ξ2+τ for a small τ>0 (this also absorbs the fixed slack ϵ of [5 Def. 9], which we take small depending on ϖ); both factors are Siegel–Walfisz, so they may be interchanged. Its conditions read ξ2−τ>1/4+12ϖ+4δ, ξ2−τ>24ϖ+9δ and 72ϖ+24δ<1, all satisfied for τ small. If (II) instead, one would also need ξ2>32ϖ+10δ; together with the Type III bound ξ2<4/9−28ϖ/9 this forces ϖ<1/79 (as δ→0), and Stadlmann takes ϖ=0.01265<1/79.
Type III. We use [3 Thm. 2.8(v)] with σ3 strictly between 1/18+28ϖ/9+2δ/9 and 1/2−ξ3, the slack ϵ in the ranges of (III) being smaller than these gaps, exactly as in [4 Lemma 18(III)] and [2 proof of Prop. 2.22].
Type I. Smooth factors of length between xξ1 and x1/2 are covered by [2 Lemma 2.20], which needs ξ1>1/4+7ϖ+2δ and holds for singly densely divisible moduli, for xδ-smooth squarefree moduli. The coefficients in the decomposition satisfy the Siegel–Walfisz property [2 Lemma A.8]. So lengths in [xξ2,x1−ξ2] are covered by Type II, and lengths above x1−ξ2>x1/2+2ϖ by [2 (2.18)]. This is the argument of [2 proof of Prop. 2.22], with (a∗,b∗,c∗)=(ξ2,1−ξ2,ξ1).
Construction. The inequalities (6) hold. Indeed, ξ1+9ξ2<4, 2ξ1+3ξ2<2 and 2ξ1+ξ2>1 are imposed, ξ2=ξ3, and 17ξ2<17⋅55/136=6.875<7. Moreover 6(1−2ξ2)>1, ξ2>2/5, and the level is 1/2+2ϖ<0.53. The three properties then follow from the proof of [5 Prop. 2] and of [4 §4], with the corrections of [2 §A.6], run on the family of xδ-smooth squarefree moduli with the inputs above. We have checked that the numerical inequalities used in these proofs reduce to those listed; we have not rewritten the whole construction. Since ξ2<55/136, the monotonicity of c1 gives c1(ξ2)≤c1(55/136). ∎
The value c1(55/136) is computed by the program deficit_c1.py, in two ways. Nested Gauss–Legendre quadrature, with each range split where its limits change formula, gives 6.604⋅10−6. A Monte Carlo integration over the regions written directly from the summation conditions gives (6.628±0.020)⋅10−6. Thus c1(55/136)<6.7⋅10−6; this is a numerical, not a certified, value. As a check, the same program gives c1(0.40481)≈9.33⋅10−6, in agreement with Stadlmann’s published bound c1(0.40481)≤2⋅10−5[4 Prop. 2].
For comparison, the minorant of [4 Prop. 2] and [2 Prop. 2.22] has ϖ=253/20000=0.01265. Here 7/544=0.0128676…, and the exponent of distribution tends to 1/2+7/272=0.525735….
Corollary 4. For every c>3.80423, liminfn→∞(pn+m−pn)≪cecm. Using only Stadlmann’s bound c1(0.40481)≤2⋅10−5 in place of the numerical value of c1(55/136), the same holds for every c>3.80428.
Proof. By the lemma of Baker and Irving [1 Lemma 1], [4 Lemma 2], a minorant with the properties of Corollary 3 gives Hm≪ϵexp(2(1+ϵ)m/(θ(1−c1))) with θ=1/2+2ϖ, that is, the exponent (1+ϵ)/((1/4+ϖ)(1−c1)). We let ϖ→7/544 and use c1(ξ2)≤c1(55/136), which holds because ξ2<55/136. Since 1/(1/4+7/544)=544/143=3.8041958…, the exponent can be taken arbitrarily close to 544/(143(1−c1(55/136))). With c1(55/136)<6.7⋅10−6 this is <3.804222. Since ξ2<55/136<0.40481, monotonicity also gives c1(ξ2)≤c1(0.40481)≤2⋅10−5, and then the bound is 544/(143(1−2⋅10−5))<3.804272. □
Files
anc/check-parameters.py checks the inequalities of Corollary 3 and the two constants of Corollary 4 in exact arithmetic. anc/deficit_c1.py computes c1. The outputs are in the corresponding .log files.
Acknowledgement
Technical assistance: Claude (Anthropic).
References
ORCID 0009-0009-1526-1178
References
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