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Claim. For the sequence of Problem 460, in both formulations of the greedy rule (the site's, $a_0=0$ with coprimality over $0\le i<k$, and the monograph's, $a_0=n$ with $1\le i<k$), write $S_{<n}(n)=\sum_{a_k<n}1/a_k$ for the truncated sum, $P^-(m)$ for the least prime factor of $|m|$, and

$$f(n)=\sum_{\substack{1\le a<n\\ P^-(n-a)>a}}\frac1a.$$

The note Erdős Problem #460: a trivial divergence and a nontrivial truncated lower bound by Przemyslaw Chojecki (ulam.ai, dated 13 January 2026 and posted in its revised form on 14 January 2026) asserts two results. Its Proposition 7: every $a<n$ with $P^-(n-a)>a$ occurs among the terms $a_k<n$, in either formulation, so $S_{<n}(n)\ge f(n)$ for every $n\ge2$. The proof is that a prime dividing both $n-a$ and $n-a'$ with $a'<a$ divides $a-a'<a$, so such an $a$ is admissible at every stage once available and the greedy rule cannot skip it. Its Theorem 8:

$$\frac1N\sum_{n\le N}f(n)\gg\log\log N,$$

hence $\limsup_{n\to\infty}f(n)=\infty$ and $\limsup_{n\to\infty}S_{<n}(n)=\infty$ in both formulations. The proof changes variables to $m=n-a$, counts the $a$-rough integers up to $N-a$ by Buchstab's asymptotic for $a\in[N^{1/3},N^{1/2}]$, and sums $1/(a\log a)$ over that range. The note's Remark 11 states that $S_{<n}(n)\to\infty$ would follow from $f(n)\to\infty$, which it does not prove; its Conjecture 13 would give that. The author reports using GPT-5.2 for most of the work.

Submission note. Posted to the site's forum by Przemyslaw Chojecki on 14 January 2026:

You're right, in either formulation, if one sums over all $k$ without any upper cut-off on $a_k$, then the series is automatically divergent: for every prime $p>n$ the term $a=n+p$ must occur in the greedy sequence, because before reaching $n+p$ all previously chosen translates satisfy $n-a_i\in[-p+1,n-1]$, so none is divisible by $p$ (the only multiple of $p$ in that interval is $0$, which never arises as $n-a_i$ in these constructions); hence $\gcd(n-(n+p),\,n-a_i)=\gcd(p,\,n-a_i)=1$ for all earlier $i$, making $n+p$ admissible at the moment it becomes available and therefore impossible for the greedy rule to skip. Comparing $\sum_{p>n}1/(n+p)$ with $\sum_p 1/p$ gives divergence for every fixed $n\ge2$.

If instead one imposes the natural ''nontrivial'' restriction $a_k<n$ (or $a_k\le n$), then the above argument disappears and one can at least prove an averaged divergence statement from the forced ''rough'' contribution. Writing $P^-(m)$ for the least prime factor of $|m|$ (with $P^-(\pm1)=\infty$), set

$$> f(n):=\sum_{a=1}^{n-1}\frac1a\,\mathbf 1_{\{P^-(n-a)>a\}},$$

so that (by the

forced-rough lemma) the truncated greedy sum $S_{<n}(n):=\sum_{a_k\le n-1}1/a_k$ satisfies $S_{<n}(n)\ge f(n)$ for every $n$ in both formulations. Averaging over $n\le N$ and changing variables $m=n-a$ yields $\sum_{n\le N}f(n)=\sum_{a\le N-1}\frac1a,\Phi(N-a,a)$, where $\Phi(x,y)$ counts $y$-rough integers $\le x$; Buchstab’s asymptotic for $\Phi(x,y)$ (uniform for $a\in[N^{1/3},N^{1/2}]$, where $u=\log(N-a)/\log a\in[2,3]$ and hence $\omega(u)\gg1$) gives $\Phi(N-a,a)\gg N/\log a$ on that range, and thus

$$> \frac1N\sum_{n\le N}f(n)\ \gg\ \sum_{N^{1/3}\le a\le N^{1/2}}\frac1{a\log a}\ > \asymp\ \log\log N.$$

In particular $\limsup_{n\to\infty}f(n)=\infty$ and

hence $\limsup_{n\to\infty}S_{<n}(n)=\infty$; what remains open is whether $f(n)\to\infty$ (and thus $S_{<n}(n)\to\infty$) along all $n$.

I've re-written the note to discuss both cases and give the full proofs. I think this solves the problem. See here: https://www.ulam.ai/research/erdos-460-v2.pdf

Posted to the site's forum by Przemyslaw Chojecki on 14 January 2026:

You are right. Also I went deeper into trying to prove it, and it comes down to a conjecture which looks adjacent to recent results by Gafni-Tao, but is currently out of reach. I've updated https://www.ulam.ai/research/erdos-460-v2.pdf with the discussion around this dyadic strategy and a plausible conjecture that would imply this pointwise divergence.

At least we have a clear picture what is interesting/hard about the problem.

Covers. In both formulations, $S_{<n}(n)\ge f(n)$ for every $n\ge2$, and $\limsup_{n\to\infty}S_{<n}(n)=\infty$. Whether $S_{<n}(n)\to\infty$ is not settled, and neither are the two restricted sums of the statement.

Depends on. No page of this wiki.

Standing. Claimed. The result is asserted in a dated note posted to the site's discussion thread; no referee, outside reviewer or formalization is recorded. The site's commentary records the reduction to $f(n)\to\infty$ and says that standard estimates on rough numbers give the averaged bound, but the site labels the problem OPEN, so the commentary is not acceptance. The full claim the same notes first made is recorded as withdrawn on its own page.