Erdős problem #282 — live-page audit, reduction, and exact finite check

Access date: 2026-07-26 (UTC)

Claim labels used throughout:

• (a) elementary-rigorous: proved directly here, or a direct observation of the cited source.
• (b) rigorous-modulo-named-theorem: depends on the specifically named published result.
• (c) plausible/structural-unverified: not used as a theorem.
• (d) computational-only: established only by the reproducible exact computation.

Result in one paragraph

(a) The original distinct-denominator problem is equivalent to the usual odd-greedy termination question restricted to inputs below \(2/3\). The distinctness rule can affect an initial finite block, but after the first unconstrained choice \(n\geq 5\), the new residual is \(<1/n\), so all later denominators increase automatically.

(d) Exact exhaustive arithmetic verifies termination for every reduced fraction

(a) I fetched the rendered live problem page and its discussion thread through the Bright Data browser path, rather than relying on the stale tracker or a search snippet.

(a) The live status was OPEN. It listed 0 claimed proofs and Currently working on this problem: None. It listed Vjeko_Kovac as “Interested in collaborating,” which is not a current-worker marker. Therefore the mandatory stop condition did not fire.

Current statement

(a) Verbatim excerpt from the live LaTeX source (short quotation):

> Let \(A\subseteq \mathbb{N}\) be an infinite set and consider the following greedy algorithm for a rational \(x\in (0,1)\): choose the minimal \(n\in A\)

(a) Precise mathematical transcription of the remainder: require \(n\geq 1/x\), subtract \(1/n\), and repeat. Finite termination gives a sum of distinct unit fractions with denominators in \(A\). The concrete question asks whether termination always occurs when the reduced denominator of \(x\) is odd and \(A\) is the odd positive integers; the general question asks which pairs \((x,A)\) terminate. The exact live source is at erdosproblems.com/latex/282.

Known results listed on the page

(b) The page credits Fibonacci with termination for \(A=\mathbb N\), and attributes the odd-denominator question to Stein.

(b) The page cites Graham’s 1964 theorem that \(m/n\) is a sum of distinct unit fractions with all denominators congruent to \(a\bmod d\) exactly when

It then asks whether the associated greedy algorithm always terminates.

(b) The page also cites Graham’s characterization for square denominators: the eligible rationals are those in

It says Erdős and Graham expected the square-denominator greedy algorithm not to terminate in general, perhaps for almost every eligible rational.

All four live comments

(a) The comments were read in full:

1. On 2026-07-07, trajenta pointed out that the displayed rule did not explicitly exclude reused denominators, using \(2/3\) and two copies of \(1/3\) as the example.

2. Thomas Bloom replied the same day that the intended “obvious” rule chooses the largest eligible unit fraction that has not already been used.

3. Vjeko_Kovac (2025-08-11) described a tail-inequality/Cantor-set view of nontermination and suggested a connection to rational points in Cantor-like sets and irrationality questions. The comment expressly said it was undecided whether this helps any concrete question on the page.

4. Dogmachine (2025-08-09) asked about the analogous greedy question for shifted-prime denominators.

(c) The Vjeko_Kovac characterization predates the distinctness clarification and uses the immediately preceding element of \(A\) in its tail inequality. I did not treat it as a theorem for the clarified “unused denominators” algorithm. The page itself warns that comments are unverified.

(a) The paper linked in that comment really is Koizumi, “Irrationality of the reciprocal sum of doubly exponential sequences,” arXiv:2504.05933 (submitted 2025-04-08). Its abstract discusses uniqueness/irrationality for reciprocal sums of doubly exponential sequences and says it gives heuristic insight into an Erdős–Graham problem. I did not use it in the finite verification.

Convention check against the original source

(a) Printed page 30 of Erdős–Graham, Old and New Problems and Results in Combinatorial Number Theory (1980), explicitly defines the greedy choice as the largest eligible unit fraction “not yet used.” It then asks Stein’s odd-denominator question. Thus Bloom’s 2026 clarification agrees with the primary source.

(a) Accordingly, this report uses:

• only odd positive denominators;
• no denominator more than once;
• the largest eligible unused unit fraction at every step;
• termination only when the residual is exactly zero. In particular, a unit-fraction residual does not terminate immediately if its denominator was already used.

For example, the clarified algorithm gives

not \(1/3+1/3\).

Primary-source literature check

(b) The following sources were verified to exist and to say the claimed things:

• R. L. Graham, “On Finite Sums of Unit Fractions,” Proc. London Math. Soc. 14 (1964), 193–207. Its concluding results include the arithmetic-progression denominator criterion cited on the live page.
• R. L. Graham, “On Finite Sums of Reciprocals of Distinct \(n\)th Powers,” Pacific J. Math. 14 (1964), 85–92. Theorem 4 and Corollary 1 give the general power and square-denominator characterizations.
• J. Pihko, “Remarks on the ‘Greedy Odd’ Egyptian Fraction Algorithm,” Fibonacci Quarterly 39 (2001), 221–227, DOI 10.1080/00150517.2001.12428725. It calls termination a well-known open problem and studies possible numerator transitions.
• J. Pihko, “Remarks on the ‘Greedy Odd’ Egyptian Fraction Algorithm II,” Fibonacci Quarterly 48 (2010), 202–208, DOI 10.1080/00150517.2010.12428097. Its Corollary 3.6 produces infinitely many inputs with prescribed numerator run \(a,a+1,\ldots,p-1,1\) for every odd prime \(p>a>1\).
• J. Louwsma and J. Martino, “Rational Numbers with Two-Term Odd Greedy Expansion,” Integers 25 (2025), A46, DOI 10.5281/zenodo.15536292. It states that odd-denominator termination remains open and classifies the two-term case. It explicitly permits repeats, while observing that the convention variants coincide below \(2/3\).

(a) The 1980 source identifies Stein’s item as a personal communication, not a paper.

(a) Searches using the exact phrases “odd greedy expansion,” “greedy odd Egyptian fraction algorithm,” Stein plus odd denominators, and recent 2024–2026 variants found no primary source claiming a uniform termination proof or counterexample. This is a search miss, not a proof that no such work exists.

Elementary reduction: it is enough to consider \(x<2/3\)

Let \(U(r)\) be the least odd \(n\) with \(1/n\leq r\), ignoring used denominators.

(a) Lemma 1. If \(0

Proof. Suppose the current choice is \(n=U(r)\).

• If \(n=3\), then \(r-1/3<1/3=1/n\).
• If \(n\geq5\), minimality gives \(r<1/(n-2)\), hence

\[ r-\frac1n < \frac1{n-2}-\frac1n =\frac{2}{n(n-2)} <\frac1n. \]

Thus the next eligible denominator is greater than \(n\). Induction proves the claim. \(\square\)

(a) Lemma 2. For any rational \(x\in[2/3,1)\), the distinct algorithm either terminates during a finite initial block of consecutive odd denominators, or reaches a residual \(r<2/3\) from which the used-denominator restriction never binds again.

Proof. The first selected denominator is 3. As long as the unconstrained choice has already been used, the clarified algorithm takes the next unused odd, so the used set remains an initial consecutive odd block. The sum \(1/3+1/5+1/7+\cdots\) diverges, so this forced consecutive phase cannot continue forever while keeping a nonnegative residual. It either hits zero or reaches a first unconstrained selected denominator \(n\geq5\). The calculation in Lemma 1 makes the new residual \(<1/n\leq1/5<2/3\), and all subsequent denominators increase automatically. Subtracting odd-denominator fractions preserves an odd reduced denominator. \(\square\)

(a) Corollary. Uniform termination for the original distinct algorithm on all odd-denominator \(x\in(0,1)\) is equivalent to uniform termination for the ordinary odd-greedy algorithm on odd-denominator \(x\in(0,2/3)\). The forward implication is immediate; Lemmas 1–2 give the reverse implication.

This also makes the below-\(2/3\) results in papers that permit repetition relevant to the clarified problem.

Exact transition used by the checker

Let the current reduced residual be \(a/b\), with \(b\) odd, and let \(d\) be the last selected odd denominator (initially \(d=1\)).

(a) Every later selected denominator exceeds \(d\). Indeed, an unused smaller odd denominator was already inadmissible when \(d\) was chosen, and the residual only decreases.

Therefore define

Then \(n\) is exactly the next greedy denominator. Put

If \(c=0\), the process terminates. Otherwise the next reduced residual is

(a) These formulas preserve nonnegativity, reduced form, and odd denominator. They involve only integer addition, multiplication, division with remainder, and gcd.

(a) Once the distinctness constraint is inactive, \(n\) is the least odd integer at least \(b/a\), so

This is only a subdoubling bound, not descent: \(a'\) can exceed \(a\).

(b) Pihko’s 2010 theorem shows that numerator sequences can contain arbitrarily long runs \(a,a+1,\ldots,p-1,1\). For fixed \(a\), its infinitely many integer denominators \(b\) are unbounded, so one may take \(a/b<2/3\); Lemma 1 then makes its convention agree with the distinct convention here. Consequently, no proof based on a fixed number of steps always decreasing the numerator can work.

Reproducible exhaustive computation

Checker: erdos282_wave5p_reverify.py

Run from the repository root:

python runs/erdos282_wave5p_reverify.py --progress

(d) The default run enumerates every pair

d32db44ddbf87a4b7d992b0204a233ec6d104405a44fa4a25db56dfc5316e384

48dbaf586120474923c8389f44921f1e05bd823db9cdd884911e9f8468fa0a41