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A122378 Numbers m such that m^2 > S(m)!, where S(m)! is the smallest factorial divisible by m. 5
2, 3, 6, 8, 12, 15, 20, 24, 30, 36, 40, 45, 48, 60, 72, 80, 84, 90, 105, 112, 120, 126, 140, 144, 168, 180, 210, 224, 240, 252, 280, 288, 315, 320, 336, 360, 384, 420, 448, 480, 504, 560, 576, 630, 640, 648, 672, 720, 756, 810, 840, 864, 896, 945, 960, 1008, 1080 (list; graph; refs; listen; history; text; internal format)
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
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 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
Sequence in context: A131723 A198442 A035106 * A181687 A194881 A361922
KEYWORD
nonn
AUTHOR
Jonathan Sondow, Sep 03 2006
STATUS
approved

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Last modified April 19 14:10 EDT 2024. Contains 371792 sequences. (Running on oeis4.)