1, 1, 2, 1, 2, 1, 4, 1, 2, 5, 5, 1, 10, 2, 2, 1, 10, 1, 16, 2, 18, 17, 16, 1, 22, 21, 20, 19, 18, 1, 28, 1, 30, 29, 2, 1, 26, 37, 36, 35, 34, 41, 40, 43, 42, 41, 40, 1, 46, 45, 44, 43, 42, 1, 52, 51, 50, 49, 48, 1, 58, 61, 4, 1, 58, 57, 56, 67, 66, 65, 64, 1, 70, 73, 72, 71, 70, 69, 68, 2
Offset: 1
Consider the products of consecutive integers, (m+9)!/9!, m >= 1. First, 10 is divisible by 2 and 5, but there is a prime gap since 3 is missing from the factorization. 10*11 is divisible by 2, 5, and 11, but 3 and 7 are missing. 10*11*12 is divisible by 2, 3, 5, and 11, but 7 is missing. 10*11*12*13 is divisible by all primes up to 13, except 7. But 10*11*12*13*14 is indeed divisible by every prime from 2 to 13. So a(10) = 5 because 5 consecutive numbers are multiplied together.
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