1,2

To compute the n-th term for fixed n>1: For each prime p that divides n, find the highest power p^E(p) that divides (n-1)!. Then the n-th term of the sequence is the smallest of the numbers p^(E(p)+1). - Jonathan Sondow (jsondow(AT)alumni.princeton.edu), Mar 03 2004

p^(E(p)+1) is smallest when p = P(n), the largest prime dividing n (since E(p) is approximately p^((n-1)/(p-1)), which decreases as p increases). So a(n) = P(n)^(E(P(n))+1) = A006530(n)^A102048(n) for n>1. - Jonathan Sondow (jsondow(AT)alumni.princeton.edu), Dec 26 2004

R. L. Graham, D. E. Knuth and O. Patashnik, "Factorial Factors" Sect. 4.4 in Concrete Mathematics: A Foundation for Computer Science, 2nd ed. Reading, MA: Addison-Wesley, pp. 111-115, 1994.

Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.

Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.

Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.

Index entries for sequences related to factorial numbers.

a(n) = P^(1+Sum(k=1 to [log(n-1)/log(P)], [(n-1)/P^k])) for n>1, where P = A006530(n) is the largest prime dividing n. E.g. a(6) = 9 because 9 is the least integer m with A002034(m) = 6, that is, m divides 6!, but m does not divide k! for k < 6. - Jonathan Sondow (jsondow(AT)alumni.princeton.edu), Dec 26 2004

With[{p=First[Last[FactorInteger[n, FactorComplete->True]]]}, p^(1+Sum[Floor[(n-1)/p^k], {k, Floor[Log[n-1]/Log[p]]}])] (Sandow)

Cf. A002034, A046022.

See also A006530, A102048.

Adjacent sequences: A046018 A046019 A046020 this_sequence A046022 A046023 A046024

Sequence in context: A068795 A072501 A092975 this_sequence A052270 A069117 A098464

nonn,nice

Eric Weisstein (eric(AT)weisstein.com)

More terms from David W. Wilson (davidwwilson(AT)comcast.net) and Christian G. Bower (bowerc(AT)usa.net), independently.