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 A308168 Numbers m that cannot be represented as a k-tuple factorial b!k for any b and k < m-1. 2
 3, 4, 5, 7, 9, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53, 59, 61, 67, 71, 73, 77, 79, 83, 89, 97, 101, 103, 107, 109, 113, 121, 127, 131, 137, 139, 143, 149, 151, 157, 163, 167, 169, 173, 179, 181, 187, 191, 193, 197, 199, 211, 221, 223, 227, 229, 233, 239, 241, 247, 251, 257, 263 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS If k >= m-1, then every number can be represented as a multifactorial: m = m!k. The sequence contains only primes and numbers of the form p*q, where p and q are both prime and satisfy the inequalities p > q and p-q < q-1. Proof: If m has exactly two prime factors p and q (p > q), but p and q do not satisfy the second inequality, then m = p!(p-q). If, on the other hand, m has at least three factors a, b and c, (a >= b >= c > 1, m = a*b*c), then a*b-c > c-1, so m = (a*b)!(a*b-c). Moreover, the sequence contains all numbers of that form. Proof: If they could be represented as a multifactorial, then it would be a (p-q)-tuple factorial. But as the second inequality is true, q-(p-q) is positive, therefore q-(p-q) should also divide m. But m has only two prime factors p and q, so the assumption is wrong and sequence indeed contains all numbers of that form. 1 and 2 are not in the sequence, because (-1)- and 0-tuple factorials are not defined. LINKS Eric Weisstein's World of Mathematics, Multifactorial EXAMPLE 15 is not in the sequence because 15 = 1*3*5 = 5!! 35 is in the sequence because 35 = 7*5 and 7-5 < 5-1. CROSSREFS Cf. A129116. Sequence in context: A128201 A233514 A096262 * A193339 A049646 A033556 Adjacent sequences:  A308165 A308166 A308167 * A308169 A308170 A308171 KEYWORD nonn,easy AUTHOR Elijah Beregovsky, May 15 2019 STATUS approved

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Last modified November 16 20:13 EST 2019. Contains 329206 sequences. (Running on oeis4.)