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Level A · Machine-checkable Hard Number theory P-oeis-78729

Least k > 0 such that (k+1)(k+2)⋯(k+n) + 1 is prime

(k+1)(k+2)(k+3)(k+4) + 1 = (k^2 + 5k + 5)^2, which is never prime. Hence a(4) = 0. Conjecture: a(n) = 0 if and only if n = 4.

From the catalogue. Imported from The Formal Conjectures Authors (Google DeepMind and contributors) (Apache-2.0) — original. Nobody has started on it here yet: tasks are created as soon as someone asks for one or submits a claim. A Lean proof is checked against the statement below by the Lean kernel; a curator confirms before the problem counts as resolved.

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@misc{cairn-oeis-78729,
  title        = {Least k > 0 such that (k+1)(k+2)⋯(k+n) + 1 is prime},
  author       = {{Cairn Commons contributors}},
  howpublished = {\url{https://cairn-commons.com/problems/oeis-78729}},
  year         = {2026},
  note         = {Open problem on Cairn Commons, CC BY 4.0. Accessed 2026-09-29}
}

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Current state

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The problem

The question

, which is never prime. Hence . Conjecture: if and only if .

The sequence is the least positive integer such that is prime, if such exists; otherwise .

Formal statement (Lean 4)

From Formal Conjectures, module FormalConjectures.OEIS.«78729».

theorem conjecture (n : ℕ) (hn : 0 < n) : a n = 0 ↔ n = 4

What counts as progress

  • A Lean proof of the pinned statement (or of its negation, for a yes/no question) — checked by the Lean kernel against the upstream statement; a curator confirms before the problem is marked resolved.
  • Partial results: special cases, weaker bounds, reductions — as verified claims.
  • Computations and numerical evidence with published code (reproducible).
  • Literature: the problem may have been solved or partly solved already. Report it as a literature claim.
  • A precise flaw in the formal statement (a misformalisation) — report it upstream too.

References

Source and licence

Imported from Formal Conjectures (OEIS), commit e6d1743831c2. Statements and descriptions © The Formal Conjectures Authors, Apache License 2.0; reformatted for this page.