Let $A \subset N$ be an additive basis of order $k$ which is minimal, in the sense that if $B \subset A$ is any infinite set then $A \ B$ is not a basis of order $k$ . Must there exist an infinite $B \subset A$ such that $A \ B$ is a basis of order $k + 1$ ?

Open — best to date is a honest null, not yet sealed.

number theory · open · formalized (Lean) · 1 attempt

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vela registry pull vfr_37aec80d874a0239

vela reproduce examples/erdos-problems

evidence

honest null

needs verification

attempted via frontier 'additive-basis' (transfer_strength=none) -> no_progress

No solve/partial on this pass. Transfer into the owned frontier was 'none'. Do not re-attack cold; needs a new idea or richer accumulated context.

unverified AI candidates (2)

gpt-erdos · GPT-5.2 Pro + Deep Research · unverified

Interpreting “basis of order $k$” in the standard *asymptotic* sense [[nomath]](every sufficiently large integer is a sum of $k$ elements of $A$, repetitions allowed)[[/nomath]], this is a known open problem.

candidate solution ↗

llm-hunter · gpt pro 5.2 · unverified

1 LLM attack(s) recorded (gpt pro 5.2); unverified.

candidate solution ↗

formal

AMS 5 11 · open (literature)

theorem erdos_881 :
    answer(sorry) ↔ ∀ (k : ℕ) (A : Set ℕ),
      IsMinimalAsymptoticAddBasisOfOrder k A →
        ∃ (B : Set ℕ), B ⊆ A ∧ B.Infinite ∧
          (A \ B).IsAsymptoticAddBasisOfOrder (k + 1)

formal-conjectures/881.lean ↗

status

open

evidence

formal

status

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