Existence of two-fold Langford sequences
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A two-fold Langford sequence of order l and defect d places each p∈{d,…,d+l−1} in two disjoint pairs of positions at distance p, using 4l positions. Does one exist exactly when 3l≥2d−1?
Why it matters and what counts as progress
Why it matters: An exact criterion would cover every order and defect beyond constructions on selected lines or special cases.
What counts as progress: Necessity and sufficiency for all integers d,l≥1, with an explicit construction or a valid exception to the proposed inequality.
How the problem developed
- 1959 · Classification [1]
Roy O. Davies Characterized ordinary Langford sequences (the one-fold case), establishing the congruence conditions for their existence.
- 2021 · Definition and open question [2]
Jamal Alkasasbeh, Christopher Dyer and Michael Howell Defined two-fold Langford sequences in a Skolem-sequence construction and asked for necessary and sufficient conditions for m-fold sequences when m ≥ 2.
- 2026 · Formalized classification [3]
JD-Jones-ASES; Claude Fable 5.1 and GPT-6 contributed during the project Proved that a two-fold sequence of order l and defect d exists exactly when 3l ≥ 2d−1, for positive d and l.
AI contribution
Lean proof characterizes two-fold Langford sequences
For d,l≥1, a two-fold Langford sequence of order l and defect d exists exactly when 3l≥2d−1. Each number p in [d,d+l−1] occupies endpoints of two disjoint pairs at distance p among 4l positions; the proof uses sixteen small cells, concatenation and signed-permutation families.
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