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Claim. The first question of Problem 460 has answer yes: the truncated sum $S_{<n}(n)=\sum_{a_k<n}1/a_k$ tends to infinity with $n$, in both formulations of the greedy rule, the site's ($a_0=0$, coprimality over $0\le i<k$) and the monograph's ($a_0=n$, $1\le i<k$). The claimant is Przemyslaw Chojecki (ulam.ai), who posts on the site as Przemek Chojecki.

Submission note. Posted to the site's forum by Przemyslaw Chojecki on 13 January 2026:

I'm not sure whether $a_0 = 0$ with $i=0$ included in $gcd$ condition is the right one, but both versions should be provable. See here: https://www.ulam.ai/research/erdos-460.pdf - I also write the summary of the proof below.

Posted to the site's forum by Przemyslaw Chojecki on 13 January 2026:

As Kevin Barreto mentioned, we might have $a_0=0$ with $i=0$ included in $gcd$ condition. The full proof of both versions is here: https://www.ulam.ai/research/erdos-460.pdf - here's the sketch:

Fix $n\ge2$ and define a greedy increasing sequence $(a_k)$ by requiring the translates $n-a_k$ to be pairwise coprime: in Case A one enforces $\gcd(n-a_k,n-a_i)=1$ for all $1\le i<k$, while in Case B one includes $i=0$ with $a_0=0$, which is equivalent to adding the standing constraint $\gcd(a_k,n)=1$. To study Erdős' divergence question for $\sum_k 1/a_k$ (and its natural subsums), the note introduces a dichotomy for $1\le a\le n-1$: call $a$ rough if the least prime factor $P^{-}(n-a)$ exceeds $a$ (equivalently, $n-a$ has no prime divisor $\le a$), and proper otherwise. This yields a canonical subsum $S_{\mathrm{rough}}(n)=\sum_{k:,a_k\le n-1,
P^{-}(n-a_k)>a_k}1/a_k$, with complementary ''proper'' subsum capturing the genuinely interactive part of the greedy dynamics.

The main reduction is that the rough part is forced and admits an explicit description independent of the history: for any $a<n$ with $P^{-}(n-a)>a$, one has $\gcd(n-a,n-a')=1$ for every $a'<a$ (since any common prime divisor would divide $a-a'$ and hence be $\le a$), so such an $a$ is admissible at every stage once it becomes available (and in Case B this automatically includes coprimality with $n$). A greedy minimality argument then shows the process cannot skip any rough $a$: if $a$ were the least skipped rough integer, at the first step where $a_{k-1}<a<a_k$ it would be admissible, contradicting the definition of $a_k$ as the least admissible choice. Consequently the set of rough values appearing with $a_k<n$ is exactly $R(n)={1\le a\le n-1: P^{-}(n-a)>a}$, and hence

$$S_{\mathrm{rough}}(n)=\sum_{a\in > R(n)}\frac1a=\sum_{\substack{1\le a\le n-1\\ P^{-}(n-a)>a}}\frac1a,$$

giving

an explicit, history-free lower bound for $\sum_k 1/a_k$ in both cases and reducing the divergence problem to estimating this rough sum plus the remaining proper contribution.

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:

Indeed, this note v1 basically reduced the problem to its core of estimating a subsum. The full solution of the problem (divergence) with discussion is here in a re-written note v2: https://www.ulam.ai/research/erdos-460-v2.pdf

This should settle the problem as is.

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.

The postings. On 13 January 2026 the claimant posted the note Greedy Coprimality Sequences and a Forced "Rough" Subsum, dated that day, to the site's discussion thread as the full proof of both versions, with a sketch: call $a<n$ rough when the least prime factor $P^-(n-a)$ exceeds $a$; a rough $a$ is coprime to $n-a'$ for every $a'<a$, so the greedy rule cannot skip it, and the rough terms below $n$ are exactly the set $R(n)=\{a<n:P^-(n-a)>a\}$ (its Proposition 8). A reply the same evening observed that this reduces the divergence question to estimating the rough sum plus the remaining proper contribution and leaves the essence unresolved. On 14 January the claimant agreed that the first note reduced the problem to the core of estimating a subsum and posted the rewritten note Erdős Problem #460: a trivial divergence and a nontrivial truncated lower bound as solving the problem and settling it as stated. The site's curator, Thomas Bloom, replied at 09:39 that the note does not establish $f(n)\to\infty$ for every $n$, which is what is required, where

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

The claimant answered at 12:48 that this is right, that a proof comes down to a conjecture adjacent to results of Gafni and Tao and out of reach, and that the second note had been updated with that discussion. The files at both URLs were replaced after posting: the first opens by pointing to the second, and the second carries the revised content.

What the revised note proves. Its Lemma 2 and Corollary 3 give the uncut divergence: $a=n+p$ is a term for every prime $p>n$, so without a cutoff the sum is infinite for every $n\ge2$ in either formulation (for $n\ge3$ in the monograph's, whose sequence ends when $n=2$). Its Proposition 7 gives $S_{<n}(n)\ge f(n)$ for every $n\ge2$ in both formulations. Its Theorem 8 gives $\frac1N\sum_{n\le N}f(n)\gg\log\log N$ from Buchstab's asymptotic for rough numbers, hence $\limsup_{n\to\infty}S_{<n}(n)=\infty$. Its Remark 11 states that $S_{<n}(n)\to\infty$ would follow from $f(n)\to\infty$, which the arguments do not give. Its Conjecture 13, a multiscale lower bound on the counts of rough numbers in dyadic intervals ending near $n$, would give $f(n)\gg\log\log n$ and so $S_{<n}(n)\to\infty$. The claimant reports using GPT-5.2 for most of the work, with the direction of the exploration the claimant's own. The site's commentary credits Chojecki with the reduction to $f(n)\to\infty$ and attributes the averaged bound to standard estimates on rough numbers.

Standing. Withdrawn: the claimant retitled the result below a proof, a reduction with an open conjecture, and the revised note asserts only the bound and the unboundedness. That proved part is recorded as the pending partial claim Chojecki's forced rough lower bound. The problem's standing takes nothing from this page.

Depends on. Nothing in this wiki.