OFFSET
1,1
COMMENTS
It is conjectured that m^2 < S(m)! for almost all m.
For each k > 1, at most tau(k!)/2 = A000005(k!)/2 are in the sequence because of that k. So at most Sum_{k = 1..m} tau(k!)/(2*m!) of the numbers up to m! are terms. This tends to 0 as m tends to infinity. - David A. Corneth, Dec 29 2019
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.
J. Sondow and E. W. Weisstein, MathWorld: Smarandache Function
EXAMPLE
15^2 = 225 > 120 = 5! = S(15)!, so 15 is a member.
MATHEMATICA
nmax = 1100;
Do[m = 1; While[!IntegerQ[m!/n], m++]; S[n] = m, {n, 1, nmax}];
Select[Range[nmax], #^2 > S[#]!&] (* Jean-François Alcover, Dec 04 2018 *)
PROG
(PARI) upto(n) = {my(res = List(), maxf = 1, olddiv, newdiv, n2 = n^2, cf = 1); while(maxf! < n2, maxf++); maxf--; olddiv = divisors(0!); newdiv = divisors(1!); for(i = 2, maxf, olddiv = newdiv; cf*=i; newdiv = divisors(cf); cans = setminus(Set(newdiv), Set(olddiv)); for(j = 1, #cans, if(cans[j]^2 > cf, if(cans[j] <= n, listput(res, cans[j]) , next(2) ); ) ) ); listsort(res); res } \\ David A. Corneth, Dec 29 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
Jonathan Sondow, Sep 03 2006
STATUS
approved