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.
Squarefree semiprimes that are in this sequence (35, 77, 143, 187, 209, 221, ...) are all in A259282 and they are the only semiprimes there. (See the Echi and Ghanmi reference for a proof.) - Elijah Beregovsky, Feb 05 2020
LINKS
O. Echi, N. Ghanmi, The Korselt set of pq, International Journal of Number Theory, Vol. 8 (2012), 2, 299-309.
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
KEYWORD
nonn,easy
AUTHOR
Elijah Beregovsky, May 15 2019
STATUS
approved