GPT 6 Astra Ultra attempt on Math-5564

FORMAL RESTATEMENT

Erdős #299; a density deduction with all quantifiers checked, 2026-09-24. Can there be an infinite increasing sequence of positive integers $a_1<a_2<\cdots$ with bounded successive gaps, but with no finite subset of its terms whose reciprocals sum to $1$?

QUICK LITERATURE AND CONTEXT CHECK

The answer is no, as a consequence of Bloom's positive-density theorem for unit fractions. We state that theorem as an external input and prove the bounded-gap reduction, including the case where the gap bound holds only eventually. The same argument gives infinitely many pairwise disjoint finite unit-sum subsets.

OBJECTIVE AND ATTACK PLAN

Convert the bound on consecutive terms into a positive lower density bound. Apply the density theorem after any finite deletion, obtaining both the requested contradiction and the strengthened disjointness assertion.

WORK

Suppose that for some positive integer $D$ and index $i_0$, \[ a_{i+1}-a_i\leq D\qquad(i\geq i_0). \tag{1} \] The hypothesis $a_{i+1}-a_i=O(1)$ has exactly this consequence after increasing $D$ to an integer. Summing $1$, \[ a_{i_0+j}\leq a_{i_0}+Dj\qquad(j\geq0). \tag{2} \] For every $N\geq a_{i_0}$, all indices $j$ with $0\leq j\leq\lfloor(N-a_{i_0})/D\rfloor$ yield terms at most $N$. Thus, writing $A=\{a_i:i\geq1\}$, \[ |A\cap[1,N]|\geq1+\left\lfloor\frac{N-a_{i_0}}D\right\rfloor, \qquad \liminf_{N\to\infty}\frac{|A\cap[1,N]|}{N}\geq\frac1D>0. \tag{3} \]

Declared external input. Bloom's theorem states that every subset $E\subseteq\mathbb N$ with positive upper density contains a finite subset $S$ satisfying $\sum_{n\in S}1/n=1$. Its stronger upper-density formulation certainly applies under $3$. The technical proof is external; the companion attempt for problem 298 gives a detailed reduction to Bloom's Proposition 1.

Applying the theorem to $A$ gives a finite unit-sum subset, contradicting the proposed avoidance property. This answers the question in the negative.

For the strengthened assertion, apply the theorem repeatedly. After finite sets $S_1,\ldots,S_r$ have been chosen, their union is finite, so \[ \liminf_{N\to\infty}\frac{|(A\setminus\bigcup_{j\leq r}S_j)\cap[1,N]|}{N} =\liminf_{N\to\infty}\frac{|A\cap[1,N]|}{N}\geq1/D. \] The theorem therefore supplies a further finite unit-sum subset disjoint from all predecessors. Induction constructs infinitely many pairwise disjoint such subsets. Alternatively, it gives a unit-sum subset beyond any prescribed cutoff by deleting the corresponding finite initial segment.

VERIFICATION AND SCOPE

No natural-density assumption is needed, and no restriction on the finitely many early gaps is required. The selected denominators are distinct because they form a subset of an increasing integer sequence. This proof does not give an effective bound for the largest denominator of the first selected subset; that is not required by the problem.

SOURCES

• Current statement and the reduction to problem 298: https://www.erdosproblems.com/299.
• T. F. Bloom, On a density conjecture about unit fractions, positive-density theorem: https://arxiv.org/html/2112.03726.

CONCLUSION

Attempt status: solved, using Bloom's stated density theorem. No such bounded-gap sequence exists; every one contains infinitely many disjoint finite unit-sum subsets. Completion rate estimate: 100% of the problem, relative to the external theorem.