GPT 6 Astra Ultra attempt on Math-5559

Erdős Problem #293: independent continuation

Date: 23 September 2026. Source versions read in full: 293.tex in gpt_pro_5.4.

1) OBJECTIVE AND SOURCE AUDIT

Let $v(k)$ be the first integer greater than one absent from every strictly increasing $k$-term unit-fraction representation of one. The source's $v(k)\le kF(k)+2$ argument is correct, but that argument and the quoted asymptotic improvement already appear in the official discussion. I do not present them as new. Here is a small finite refinement and a complete exact enumeration procedure, rather than a new asymptotic solution.

2) WORK: SEPARATING SMALL AND LARGE DENOMINATORS

Every representation has a denominator at most $k$: otherwise its sum would be at most $k/(k+1)<1$. Thus each representation contributes at most $k-1$ denominators greater than $k$. Since only $k-1$ possible denominators lie in $[2,k]$, the union $D_k$ satisfies, for $k\ge2$, \[ |D_k|\le(k-1)F(k)+(k-1),\qquad v(k)\le(k-1)F(k)+k+1. \tag{1} \] This improves the simple union count when $F(k)>k-1$, but changes no double-exponential exponent and is not a major asymptotic advance.

For reproducible finite checks, suppose a partial decomposition has remaining rational mass $r>0$, needs $j\ge2$ terms, and its last chosen denominator is $b$. The next denominator $x$ must satisfy \[ \max\{b+1,\lfloor1/r\rfloor+1\}\le x\le\lfloor j/r\rfloor. \] The lower bound follows from $1/x<r$; the upper bound follows because all remaining denominators are at least $x$. Replace $(r,b,j)$ by $(r-1/x,x,j-1)$ recursively. When $j=1$, accept precisely when $r=1/x$ for an integer $x>b$. Each branch is finite and the inequalities retain every possible representation; induction on $j$ proves completeness.

Executed with exact fractions, this yields \[ \begin{array}{c|rrr} k&F(k)&|D_k|&v(k)\\\hline 2&0&0&2\\3&1&3&4\\4&6&15&11\\5&72&77&17 \end{array} \] The companion verification/root_checks.py reproduces these counts. No numerical denominator cutoff is assumed.

3) VERIFICATION AND FINAL

Distinctness is enforced by $x>b$, so repeated-unit-fraction counts are not inadvertently included. For $k=1$ the sole representation is $1=1/1$, and $v(1)=2$ separately. UNRESOLVED. The finite refinement and enumeration do not determine the growth of $v(k)$. Source attribution and the greater-than-one convention were checked at https://www.erdosproblems.com/forum/thread/293.

8) COMPLETION ESTIMATE

COMPLETION: 5% This is a subjective estimate of progress toward the entire stated problem, not a probability that the conjecture or an unverified proof is correct.