

A308168


Numbers m that cannot be represented as a ktuple factorial b!k for any b and k < m1.


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
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OFFSET

1,1


COMMENTS

If k >= m1, 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 pq < q1.
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!(pq). 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*bc > c1, so m = (a*b)!(a*bc).
Moreover, the sequence contains all numbers of that form. Proof: If they could be represented as a multifactorial, then it would be a (pq)tuple factorial. But as the second inequality is true, q(pq) is positive, therefore q(pq) 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 0tuple factorials are not defined.


LINKS

Table of n, a(n) for n=1..67.
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 75 < 51.


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



