OFFSET
1,1
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.
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.
EXAMPLE
S(9)! = 6! = 720 > 81 = 9^2 > 6 = 3! = P(9)!, so 9 is a member.
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 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Jonathan Sondow, Sep 03 2006
STATUS
approved