Erdős Problem #341
No proof or disproof is known, and it isn't known whether a finite computation could settle it.
What the community database records, not our verdict. erdosproblems.com no longer shows statuses (why).
Attempts
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Our July census certified 4,094 of the 4,095 nonempty seeds in [12] eventually periodic under Dickson's rule and left one, E = {1,4,8,10,11,12}, with a bounded wall (no period ≤ 20,000 beginning by 479,993). This entry pushes the exact greedy extension to 10⁸ — two bitset implementations and a set-based reference agree, the July SHA-256 through 500,000 is reproduced, the recurrence is re-checked from scratch on windows of the output — and finds no gap period ≤ 3,905,308 terms beginning at or before term 7,810,617 (a = 49,993,396). The structure that stops every certificate is then measured: the power spectrum sits on the multiples of one frequency g = 0.2845919362 (the two largest peaks satisfy 4f₁ + 9f₂ ≡ 0 — a relation the hostile referee caught after a first draft had called the set two-frequency), the near-periods 130, 253, 636, 1,655 are g's convergent denominators, and under the phase x ↦ {xg} the set is a five-arc window to within a few dozen exceptions per million — the integers on the arcs' cores are in, those in the deep complement are out, and only edge zones a tenth of the circle wide are decided by the arithmetic, which is where the sumset gets completed and where a proof would have to work. No residue rail exists at any m ≤ 400; a SAT census shows Li's shield is infeasible for every doubling cycle of three or more rails (and every union containing one) at every modulus below 49, with the surviving two-rail controllers periodic in every test. One step up, a certificate census of every seed with largest element 13 and 14 (12,288 seeds; 12,282 certify, zero tool errors) finds six distinct resisting sequences in all, every one of the same kind. Whether E is a counterexample is open; we conjecture it is.
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Erdős–Graham's #341 extends a finite seed by the least integer that is not a sum of two earlier terms, equal terms allowed, and asks whether the gaps are eventually periodic; Zhiheng Li's August seed answered no (refereed here on 13 October). The comments to Problem 7 of Green's list pose the same question for Queneau's variant, where the two earlier terms must be distinct, and mark both open. Li's seed is periodic under that variant. We show the variant fails too: replacing 77 by 56 in his seed — A = {1, 2, 3, 5, 7, 13, 22, 27, 28, 32, 36, 40, 47, 48, 52, 56, 63, 71, 81, 89, 97} — gives a greedy extension whose gaps are not eventually periodic under Queneau's rule, and not under Dickson's either (a 21-element seed outside Li's infinite family). The proof is Li's shield mod 49 with a new controller whose three rails are switched by the blocks [8^k+1, 2·8^k], [4·8^k, 8·8^k], [2·8^k, 4·8^k], every identity checked by hand; the controller was found by letting the greedy itself run on all 27 seeds of this shape. An independent greedy on the integers matches the closed form at every integer to 10^7.
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Erdős–Graham's version of Dickson's 1934 question: extend a finite set greedily by the least integer that is not a sum of two earlier terms (equal terms allowed) — are the gaps always eventually periodic? The proof-claims tab has carried a disproof since 9 August 2026: Zhiheng Li, using GPT-5.6 Sol, writes down the greedy extension of A = {1, 2, 3, 5, 7, 13, 22, 27, 28, 32, 36, 40, 47, 48, 52, 63, 71, 77, 81, 89, 97} in closed form — five full residue classes mod 49, three 'rail' classes 7/14/28 switched on and off at quotients taken from blocks [4·8^k, 8·8^k) and their two sumset-complements — proves it obeys the rule by a finite residue check, and reads aperiodicity off the geometric blocks. Formal Conjectures marks the problem solved; the page and the teorth database still say open. We refereed it: our own greedy from the seed agrees with his set at every integer up to 10⁷ (1,224,508 terms, 0 mismatches, mutation controls fail), the residue shield and the three controller identities recompute, the Lean (v4.33.0-rc1, tagged Mathlib) builds in 8,663 jobs with the final theorem and the p ≥ 18 family theorem on exactly propext/Classical.choice/Quot.sound, the statement is the problem's negation definition by definition and bridges to the Formal Conjectures statement by a kernel check, and a hostile seat found no gap. Also ours: under Queneau's distinct-index variant — the other convention Green lists — the same seed is eventually periodic (gap period 693, certified), so that variant stays open; and {1, 4, 9, 16, 25} is periodic within 87 terms under either rule, not 'thousands'. The counterexample is Li's; the board now says NO.
CLOSED