arXiv:2406.02243v3

An Explicit Ramsey Bound for de Polignac Numbers

Sayan Goswami

math.NTmath.CO37A4505D1011E2511T30

Paper metadata

Original submission: June 4, 2024
Latest revision: August 17, 2026 · version 3
Claimed MSC: 37A45, 05D10, 11E25, 11T30
arXiv categories: math.NT, math.CO
Paper license: CC BY 4.0 ↗

Abstract

A number $m$ is called a de Polignac number( $\mathbf{POL}$ in short) if it can be expressed as the difference of infinitely many pairs of consecutive prime numbers. For any given $r\in \mathbb N,$ a set $A$ is said to be $Δ_r^\star$ if for all sets $S$ with $|S|=r$ such that $A\cap \{s-t:s>t\in S\}\neq \emptyset.$ In this article we prove that the set $\mathbf{POL}$ is $Δ_r^\star$ with the specific computable value of $r,$ where $r=exp(\mathcal{O}(50)).$ Then we prove that there exists a set $E$ with $|E|\leq 2^{50\cdot \prod_{i=1}^{49}p_i}$ (where $(p_i)_i$ is the enumeration of primes) such that $\mathbb{N}=\bigcup_{t\in E} t^{-1}\mathbf{POL}.$

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Audit summary

Audited against arXiv v3

Not a correctness certificate. A “Correct” result may include yellow typos or minor formal corrections that do not affect substantive soundness. It means this audit found no unresolved substantive error under the stated criteria; it does not replace expert scrutiny or formal verification.

Current report

Detailed mathematical audit

Generated August 18, 2026

01Statements2 reported findingsCorrect

The set of de Polignac numbers has the asserted explicit finite Ramsey property and is multiplicatively syndetic with the stated finite translating-set bound.

Theorem 1.1Correct

De Polignac numbers form an explicit $\Delta_r^*$ set

Pages 1–3 · Theorem 1.1, Lemma 2.4, and its proof · arXiv:2406.02243v3

With $k=50$ , Lemma 2.4 extracts an admissible $k$ -element subset from every set of size $R(k)=\left\lceil k\prod_{p\leq k}p/(p-1)\right\rceil$ . The Banks–Freiberg–Turnage-Butterbaugh theorem, together with the verified bound $k_2\leq50$ , selects two shifts that are consecutive primes infinitely often. Their positive difference belongs simultaneously to the de Polignac set and to the difference set of the original $R(50)$ -element set. Thus $\mathrm{POL}$ is $\Delta_{R(50)}^*$ , with the displayed computable bound.

Theorem 1.2Correct

Explicit multiplicative syndeticity

Pages 1 and 3–4 · Theorem 1.2 and Lemmas 2.5–2.6 · arXiv:2406.02243v3

Lemma 2.6 constructs an admissible $k$ -set among partial sums of any $M_k$ positive integers and therefore proves that $\mathrm{POL}$ meets every $IP_{M_k}$ set. Lemma 2.5 then applies this dual property to the binary generators $1,2,\ldots,2^{M_k-1}$ , giving $\mathbb N=\bigcup_{t=1}^{2^{M_k}-1}t^{-1}\mathrm{POL}$ . The elementary estimate $M_k\leq k\prod_{i=1}^{k-1}p_i$ with $k=50$ yields the claimed bound $|E|\leq2^{50\prod_{i=1}^{49}p_i}$ .

02Proofs5 reported findingsCorrect

The combinatorial extraction, consecutive-prime input, finite-sums construction, and syndeticity argument are correct. The cited theorem is printed with two endpoint inequalities that exclude the values immediately used; the primary source verifies the unique local corrections, so this is a yellow formal finding only.

Lemma 2.4Correct and complete

Every sufficiently large finite set contains an admissible $k$ -subset

Page 2 · Lemma 2.4 · arXiv:2406.02243v3

For each prime $p\leq k$ , deleting a least-populated residue class retains at least a factor $1-1/p$ of the current set and ensures that one residue class is omitted. The definition of $R(k)$ leaves at least $k$ elements. Any chosen $k$ -subset also omits a class modulo every prime $p>k$ by cardinality, so it is admissible.

Theorems 2.1–2.2Minor formal correction

The endpoint hypotheses should include $m=2$ and $k=k_m$

Page 2 · Theorems 2.1–2.2 and Note 2.3; used on pages 2–4 · arXiv:2406.02243v3

The manuscript prints $m>2$ and $k>k_m$ , but every application uses $m=2$ and $k=k_2=50$ . Replace those inequalities by $m\geq2$ and $k\geq k_m$ . The primary theorem of Banks–Freiberg–Turnage-Butterbaugh states exactly these inclusive hypotheses and records that one may take $k_2=50$ . The repair is therefore verified, local, and valid at every downstream use.

Banks–Freiberg–Turnage-Butterbaugh, Consecutive primes in tuples, Theorem 1 ↗

Theorem 1.1 proof headingTypo

The first main-theorem proof is mislabeled

Page 2 · paragraph immediately following Lemma 2.4 · arXiv:2406.02243v3

The heading reads “Proof of Theorem 1.2,” but the paragraph proves the $\Delta_r^*$ assertion of Theorem 1.1. Replace “1.2” by “1.1.” The theorem being proved is uniquely determined by the conclusion and the later separate proof of Theorem 1.2.

Lemma 2.6 notationTypo

The next prime $p_k$ is used without being introduced

Page 3 · proof of Lemma 2.6 · arXiv:2406.02243v3

The lemma introduces the first $k-1$ primes $p_1,\ldots,p_{k-1}$ and later treats every prime $p\geq p_k$ . Define $p_k$ to be the next prime after $p_{k-1}$ . This is the unique intended meaning, and it makes the two cases cover all primes without altering the argument.

Lemmas 2.5–2.6Correct and complete

Finite-sums duality gives the explicit translating set

Pages 3–4 · Lemmas 2.5–2.6 and the final proof · arXiv:2406.02243v3

The recursive pigeonhole construction produces $k$ distinct partial sums forming an admissible set; every positive difference of two selected partial sums is a finite sum of the original generators. The consecutive-prime theorem therefore places an element of $\mathrm{POL}$ in that finite-sums set. Scaling any fixed finite-sums set by an arbitrary $n$ then proves the covering identity used for multiplicative syndeticity.

03Novelty0 reported findingsNo non-novelty findings

No non-novelty findings.

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Paper: arXiv:2406.02243v3
Authors listed: Sayan Goswami
Audit date: August 18, 2026

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An Explicit Ramsey Bound for de Polignac Numbers

Paper metadata

Abstract

Audit summary

Detailed mathematical audit

De Polignac numbers form an explicit Δr∗\Delta_r^*Δr∗​ set

Explicit multiplicative syndeticity

Every sufficiently large finite set contains an admissible kkk-subset

The endpoint hypotheses should include m=2m=2m=2 and k=kmk=k_mk=km​

The first main-theorem proof is mislabeled

The next prime pkp_kpk​ is used without being introduced

Finite-sums duality gives the explicit translating set

Challenge an audit finding

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De Polignac numbers form an explicit $\Delta_r^*$ set

Every sufficiently large finite set contains an admissible $k$ -subset

The endpoint hypotheses should include $m=2$ and $k=k_m$

The next prime $p_k$ is used without being introduced