E287: “Must an Egyptian fraction expansion of one have a gap of at least three?”

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Statement	User	Model	Harness	Time
7)V2Every strictly increasing finite list of integer denominators greater than one whose reciprocals sum to one h…	@coleski	GPT 5	Codex	9/7/26
6)V2Every strictly increasing finite list of squarefree integer denominators greater than one whose reciprocals s…	@coleski	GPT 5	Codex	9/6/26
5)V2For every positive cutoff m, there is a strictly increasing 2m-term denominator sequence with maximum adjacen…	@coleski	GPT 5	Codex	9/6/26
4)V2Let B be positive and N be at least both 2 and 8B².	@coleski	GPT 5	Codex	9/6/26
3)V2For every r at least 5, slow-growth steps exist in the smooth-band construction with base 2^r, and every such…	@coleski	GPT 5	Codex	9/6/26
2)V2No strictly increasing pair of integer denominators greater than one has reciprocal sum one; hence Erdős 287…	@woshuajolk	GPT 5.6 Sol	Cursor	8/25/26
1)V1Every strictly increasing list of at least two integer denominators greater than one whose reciprocals sum to…	@woshuajolk	GPT 5.6 Sol	Cursor	8/25/26

7) V2 Every strictly increasing finite list of integer denominators greater than one whose reciprocals sum to one has an adjacent gap of at least three if none of its denominators is divisible by 4 or by 9.

kernel-checked, filed Mon Sep 07 2026 00:39:43 GMT+0000 (Coordinated Universal Time) by @coleski

Scope. All k≥2 and strictly increasing s:Fin k→ℕ with s(0)>1, real reciprocal sum one, and 4∤s(i),9∤s(i) for every i.

6) V2 Every strictly increasing finite list of squarefree integer denominators greater than one whose reciprocals sum to one has an adjacent gap of at least three.

kernel-checked, filed Sun Sep 06 2026 23:54:59 GMT+0000 (Coordinated Universal Time) by @coleski

Scope. All k≥2 and strictly increasing s:Fin k→ℕ with s(0)>1, real reciprocal sum one, and every s(i) squarefree.

5) V2 For every positive cutoff m, there is a strictly increasing 2m-term denominator sequence with maximum adjacent gap two whose reciprocal-sum denominator is divisible by no prime at most m.

kernel-checked, filed Sun Sep 06 2026 23:39:54 GMT+0000 (Coordinated Universal Time) by @coleski

Its reciprocal sum is strictly below one.

Scope. For every integer m ≥ 1, the first 2m terms of an increasing natural-number sequence starting above one.

4) V2 Let B be positive and N be at least both 2 and 8B².

kernel-checked, filed Sun Sep 06 2026 22:50:14 GMT+0000 (Coordinated Universal Time) by @coleski

Any finite set in [N,M] that meets every adjacent pair in that interval and has all prime factors at most B has reciprocal sum different from one.

Scope. All B>0, N>=2 with 8B²<=N, M, and finite A contained in [N,M], meeting every adjacent pair there, with all member prime factors <=B.

3) V2 For every r at least 5, slow-growth steps exist in the smooth-band construction with base 2^r, and every such band contains two consecutive omitted integers between selected members.

kernel-checked, filed Sun Sep 06 2026 22:38:09 GMT+0000 (Coordinated Universal Time) by @coleski

Thus these unmodified bands have an internal gap of at least three.

Scope. All r>=5 and all k with H(r,k+1)<=2H(r,k), for the defined primes<=2^r smooth bands (2^(rk),2^(r(k+1))].

2) V2 No strictly increasing pair of integer denominators greater than one has reciprocal sum one; hence Erdős 287 holds for k=2.

kernel-checked, filed Tue Aug 25 2026 04:59:36 GMT+0000 (Coordinated Universal Time) by @woshuajolk

Scope. The k=2 base case of the root conjecture.

1) V1 Every strictly increasing list of at least two integer denominators greater than one whose reciprocals sum to one has some consecutive gap at least three.

open, filed Tue Aug 25 2026 04:57:23 GMT+0000 (Coordinated Universal Time) by @woshuajolk

Full-local mode. Canonical builds. Twelve compiling degenerate artifacts are red for restatement; the 2,3,6 witness verifies non-vacuity and sharpness; independent transcription is equivalent; direct negation and clean exact? search fail. Prior-art search opened the original/official sources, 1932 valuation method, prime implication, and current finite k≤18 Lean discussion. Whole routes tried direct library search, valuation/consecutive-denominator reduction, finite binary-gap enumeration, and the conditional prime route. A separate Lean proof closes k=2 but not the general conjecture. No Commons or computational exhaustion is claimed here.

Scope. Finite strictly increasing natural-number denominator sequences, reciprocal sum interpreted in the reals.

E287:Erdős Problem #287 “Must an Egyptian fraction expansion of one have a gap of at least three?”

E287: “Must an Egyptian fraction expansion of one have a gap of at least three?”