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Erdős problem 337

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Let A⊆NA\subseteq \mathbb{N} be an additive basis (of any finite order) such that ∣A∩{1,…,N}∣=o(N)\lvert A\cap \{1,\ldots,N\}\rvert=o(N). Is it true that lim⁡N→∞∣(A+A)∩{1,…,N}∣∣A∩{1,…,N}∣=∞? \lim_{N\to \infty}\frac{\lvert (A+A)\cap \{1,\ldots,N\}\rvert} {\lvert A\cap \{1,\ldots,N\}\rvert}=\infty?

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FormalConjectures/ErdosProblems/

337.lean

Retained formal statement4 of 4

Ruzsa and Turjányi do prove (under the same hypotheses) that lim⁡N→∞∣(A+A+A)∩{1,…,3N}∣∣A∩{1,…,N}∣=∞, \lim_{N\to \infty}\frac{\lvert (A+A+A)\cap \{1,\ldots,3N\}\rvert} {\lvert A\cap \{1,\ldots,N\}\rvert}=\infty,

FormalConjectures/ErdosProblems/337.leanErdos337.erdos_337.variants.three_fold5 linesExact file
∀ (A : Set ℕ),  A.IsAsymptoticAddBasis →    ((fun N => ↑(A ∩ Set.Icc 1 N).ncard) =o[Filter.atTop] fun N => ↑N) →      Filter.Tendsto (fun N => ↑((A + A + A) ∩ Set.Icc 1 (3 * N)).ncard / ↑(A ∩ Set.Icc 1 N).ncard) Filter.atTop        Filter.atTop
SolvedStatement only, no proof