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Claim. With $b(a)$ the least $b>a$ at which the lowest-terms denominator $v_{a,b}$ of $\sum_{a\le n\le b}1/n$ satisfies $v_{a,b}<v_{a,b-1}$ (the author's convention, one more than the site's), Theorem 1 of the note states that for almost all $a$

$$b(a)>a+\exp\Bigl(\tfrac12\sqrt{\log a\log\log a}\Bigr),$$

so that for every fixed $C$ the set of $a$ with $b(a)\le a+(\log a)^C$ has density zero. The proof rests on the note's Lemma 2: with $L_n$ the least common multiple of $1,\ldots,n$, $v_{a,b}/v_{a,b-1}\ge b/\gcd(b,L_{b-a})^2$, so a drop at $b$ forces $\gcd(b,L_{b-a})>\sqrt b$; for each shift $n$ up to the exponential the integers $a\in[x,2x)$ with a drop at $a+n$ are counted through the divisors $d>\sqrt x$ of $L_n$ by Rankin's trick and a weak form of the prime number theorem, and the count summed over $n$ is $o(x)$.

Submission note. Posted to erdosproblems.com as a proof claim by Wouter van Doorn (account Woett) on 24 September 2026, giving "GPT-5.6 Sol" as the AI used:

We prove that for almost all $a$ we have

$$> b(a) > a + \exp\left(\frac{1}{2} \sqrt{\log a \log \log a} \right). >$$

In particular, contrary to some earlier speculation, for all $C$ we have that the set of $a$ for which $b(a) \le a + (\log a)^C$ holds has density $0$. I still believe that $b(a) \le a + O_{\varepsilon}(a^{\varepsilon})$ holds for all $\varepsilon > 0$. Let $L_n$ be the least common multiple of the first $n$ positive integers. The proof of the lower bound rests on the observation that, if the denominator of the sum decreases at $b$, then $\gcd(b, L_{b-a}) > \sqrt{b}$. For a large $x$ and a given $n \le \exp\left(\frac{1}{2} \sqrt{\log 2x \log \log 2x} \right)$, we then count the number of possible values of $a \in [x, 2x)$ for which $b := a + n$ has this property that $\gcd(b, L_n) > \sqrt{b}$. Applying Rankin's trick, PNT and then summing over all possible $n$, we find our zero density result after some algebra.

Covers. The size of $b(a)-a$ for a density-one set of $a$. It leaves the existence question to the 2024 paper and the limit inferior to the 2026 preprint, and gives no upper bound; the author conjectures in the note that $b(a)<a+a^{\varepsilon}$ for every $\varepsilon>0$ and all large $a$. The result is a proved lower bound on a density-one set, so the value is proved; it does not determine how $b(a)$ grows.

Standing. A four-page note in the author's GitHub repository Woett/Mathematical-shorts (uploaded 24 September 2026), not posted to arXiv, with no journal record and no independent review; it is not held in the library; its statements are compiled from the posted PDF and its proof is not checked, so the claim is pending. The author filed it on the site's proof-claim tab the same day, without a scope label, naming the system GPT-5.6 Sol; the note's declaration of AI usage credits ChatGPT 5.6-Sol Pro with the proof of the main result and points to the machine write-up, "A density-one lower bound for the first decrease of a harmonic denominator", in the author's repository linked above. A thread comment of the same day states the result as holding for infinitely many $a$, weaker than the note's almost all.