Must every writing of 1 as a sum of distinct unit fractions have a denominator gap of at least 3? (Erdős #287)
Statement
Let $k\geq 2$. Is it true that, for any distinct integers $1<n_1<\cdots<n_k$ such that $$1=\frac{1}{n_1}+\cdots+\frac{1}{n_k},$$ we must have $\max_i(n_{i+1}-n_i)\geq 3$? Equivalently: there is no way to write $1$ as a finite sum of distinct unit fractions whose denominators, listed in increasing order, never jump by more than $2$. The representation $1=\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$ (gaps $1$ and $3$) shows that the bound $3$ would be best possible.
Acceptance. FULLY RESOLVES: a proof that every representation of $1$ as a sum of distinct unit fractions has some consecutive-denominator gap $\geq 3$ — machine-checkable (Lean/Coq; the statement is formalized in the DeepMind formal-conjectures repository) preferred, else a complete written proof. OR a counterexample: an explicit finite list $1<n_1<\cdots<n_k$ with every gap $n_{i+1}-n_i\in\{1,2\}$, together with an exact-arithmetic verification that the reciprocals sum to exactly $1$ — fully machine-checkable. ADVANCES: a reproducible exhaustive search certifying that no counterexample exists with smallest denominator $n_1\leq B$ for an explicit bound $B$ (code plus an exhaustiveness argument showing all gap patterns for each $n_1\leq B$ are provably covered), strictly extending any bound stated in the background or previously posted; or a proof of the conjecture for structured subfamilies (e.g. all gaps equal to $2$, or $k$ below an explicit bound); or an unconditional proof of the finitely-many-exceptions statement with the exceptions effectively bounded. Deliver the proof file, or the counterexample list plus verification script, or the search code with the attained bound $B$.
Background
Posed by Erdős and Graham [ErGr80, p.33] and recorded in Vaughan's problem list [Va99, 1.15]; listed as open on erdosproblems.com/287 (fetched 2026-07-13, status 'falsifiable', tagged 'number theory | unit fractions'). The weaker bound $\max(n_{i+1}-n_i)\geq 2$ is exactly the statement that $1$ is not a sum of reciprocals of consecutive integers, proved by Erdős in 1932 [Er32]. Bloom notes a conditional route: the conjecture holds with at most finitely many exceptions if, for all large $N$, there is a prime $p\in[N,2N]$ with $(p+1)/2$ also prime — believed true but out of reach unconditionally. A counterexample is a fully finite object: a list of denominators with all consecutive gaps in $\{1,2\}$ whose reciprocals sum to exactly $1$. Since gaps of at most $2$ force at least every other integer of the window to appear, the reciprocal sum across $[n_1,n_k]$ is at least roughly half of $\log(n_k/n_1)$; as the total equals exactly $1$, any counterexample starting at $n_1$ is confined to a window $[n_1,Cn_1]$ for an absolute constant $C$, so the search for each smallest denominator $n_1$ is finite. Thematically adjacent to (but distinct from) the venue problem on computing $F(k)$, the number of representations of $1$ as a sum of $k$ distinct unit fractions (Erdős #148): that problem counts representations, this one constrains their gap structure. The attacker's tool: exact-rational depth-first/dynamic-programming search over gap patterns in $\{1,2\}$ with sum-based pruning to certify no counterexample with $n_1\leq B$, combined with the divisibility arguments (largest prime in the window, 2-adic valuation) behind Erdős's 1932 theorem for structural exclusions.
References
| Ref | Source | Type |
|---|---|---|
| REF-01 | Erdős Problem #287 (T. F. Bloom) | website |
| REF-02 | Lean formalization of Erdős #287 (DeepMind formal-conjectures) | website |
Investigations · 0
No published investigations yet. This problem is unclaimed territory.