GPT 6 Astra Ultra attempt on Math-750

Erdős Problem #301: independent continuation

Date: 2026-09-23. Research attempt; no claim of a new published result.

1) TARGET AND SOURCE AUDIT

For the largest $f(N)$ of a subset of $[1,N]$ with no relation $1/a=\sum_i1/b_i$ among distinct selected integers, determine its asymptotic size; the target is the half-density bound.

All supplied versions considered: 301.tex. The supplied disjoint-block proof yields $f(N)\le(53/60+o(1))N$. I enlarge the block and obtain the rigorously reproducible local improvement $349/403$. This is an improvement over the supplied argument, not a claim of a new literature record.

2) WORK

Take \[ T=\{2^i3^j:0\le i<5,\ 0\le j<3\} =\{1,2,3,4,6,8,9,12,16,18,24,36,48,72,144\}. \] For its increasing prefixes the minimum numbers of omissions needed to avoid all reciprocal relations are \[ 0,0,0,0,1,1,1,2,2,3,3,4,5,6,7. \tag{1} \] I verified $1$ by exact exhaustive enumeration of the at most $2^{15}$ subsets. The saved script enumerates every potential smallest denominator $a$, sums every subset of larger reciprocals using Fraction, records each equality as a forbidden bit mask, then maximizes cardinality among masks containing no forbidden mask. Positivity guarantees that the right-hand denominators must all exceed $a$, so this enumerates every relation. This finite computation is part of the proof and is fully reproducible.

Choose block origins $d$ with $v_2(d)\equiv0\pmod5$ and $v_3(d)\equiv0\pmod3$. Their density is \[ \delta={1/3\over(1-2^{-5})(1-3^{-3})}={144\over403}. \] This follows by writing $d=32^u27^v w$ with $(w,6)=1$ and summing the convergent geometric densities; truncation leaves $o(X)$ error in the counting function. The sets $dT$ are disjoint: the valuations modulo $(5,3)$ determine the multiplier in $T$, and then determine $d$.

By $1$, an additional omission is forced at each cutoff multiplier $6,12,18,36,48,72,144$. Summing distinct omissions gives \[ N-f(N)\ge\left({144\over403}\left({1\over6}+{1\over12}+{1\over18}+{1\over36}+{1\over48}+{1\over72}+{1\over144}\right)+o(1)\right)N =\left({54\over403}+o(1)\right)N. \] Hence $f(N)\le(349/403+o(1))N$, strictly below $53/60$.

3) ADVERSARIAL VERIFICATION

The proof relies on all exact local relations, including those with more than two right-hand terms. It is not based on floating point equality or random search. Blocks may be truncated at $N$; this is exactly why every prefix omission number is computed. Relations between different blocks can only further restrict admissibility, so ignoring them is safe for an upper bound.

4) FINAL

UNRESOLVED. The bound remains well above $N/2$. No matching upper asymptotic or full solution is obtained. Larger local blocks can improve constants without necessarily reaching the conjectured density.

5) SOURCES CHECKED

Full 301.tex read. Exact computation executed on 2026-09-23 using verification/batch_2_checks.py; only Python standard-library integer and rational arithmetic. The attempted formal-conjectures 301.lean statement URL was unavailable $404$; no fact is attributed to it.

6) COMPLETION ESTIMATE

This is a conservative subjective measure of progress toward the original question, not a probability of correctness or a claim that the remaining work is routine. COMPLETION: 20%