A122379 Numbers n such that S(n)! > n^2 > P(n)!, where S(n)! is the smallest factorial divisible by n and P(n) is the greatest prime factor of n.

4, 9, 16, 18, 25, 27, 32, 50, 54, 64, 75, 81, 96, 98, 100, 108, 125, 128, 135, 147, 150, 160, 162, 175, 189, 192, 196, 200, 216, 225, 243, 245, 250, 256, 270, 294, 300, 324, 343, 350, 375, 378, 392, 400, 405, 432, 441, 450, 486, 490, 500, 512, 525, 540, 567

Offset: 1

Author

Jonathan Sondow, Sep 03 2006

Keywords

• nonn

Comments

It is conjectured that n^2 < P(n)! for almost all n. It is known that S(n) = P(n) for almost all n. Clearly, S(n) >= P(n) for all n > 1.

Examples

			S(9)! = 6! = 720 > 81 = 9^2 > 6 = 3! = P(9)!, so 9 is a member.
		

Links

• David A. Corneth, Table of n, a(n) for n = 1..10000
• J. Sondow, A geometric proof that e is irrational and a new measure of its irrationality, Amer. Math. Monthly 113 (2006) 637-641.
• J. Sondow, A geometric proof that e is irrational and a new measure of its irrationality, arXiv:0704.1282 [math.HO], 2007-2010.
• Index entries for sequences related to factorial numbers.

Crossrefs

Cf. A000290, A002034, A006530, A057109, A092495, A102068, A122378, A122380.

Programs

Mathematica

s[n_] := For[k = 1, True, k++, If[Divisible[k!, n], Return[k]]];
p[n_] := FactorInteger[n][[-1, 1]];
okQ[n_] := s[n]! > n^2 > p[n]!;
Select[Range[1000], okQ] (* Jean-François Alcover, Jan 27 2019 *)