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A122380 Numbers n such that n^2 > P(n)!, where P(n) is the greatest prime factor of n. 3
2, 3, 4, 6, 8, 9, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36, 40, 45, 48, 50, 54, 60, 64, 72, 75, 80, 81, 84, 90, 96, 98, 100, 105, 108, 112, 120, 125, 126, 128, 135, 140, 144, 147, 150, 160, 162, 168, 175, 180, 189, 192, 196, 200, 210, 216, 224, 225, 240, 243, 245 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
It is conjectured that n^2 < P(n)! for almost all n.
LINKS
Charles R Greathouse IV, 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 and E. W. Weisstein, MathWorld: Smarandache Function
EXAMPLE
15^2 = 225 > 120 = 5! = P(15)!, so 15 is a member.
MATHEMATICA
Reap[For[n = 2, n <= 250, n++, If[n^2 > FactorInteger[n][[-1, 1]]!, Print[n]; Sow[n]]]][[2, 1]] (* Jean-François Alcover, Feb 04 2019 *)
PROG
(PARI) smooth(P:vec, lim)=my(v=List([1]), nxt=vector(#P, i, 1), indx, t); while(1, t=vecmin(vector(#P, i, v[nxt[i]]*P[i]), &indx); if(t>lim, break); if(t>v[#v], listput(v, t)); nxt[indx]++); Vec(v)
list(lim)=my(v=List([2]), u, lower, upper=2, p=2); while(1, lower=upper+1; p=nextprime(p+1); upper=min(sqrtint(p!), lim); if(lower>lim, break); u=select(q->q>=lower, smooth(primes([2, p-1]), upper)); for(i=1, #u, listput(v, u[i]))); Vec(v) \\ Charles R Greathouse IV, Nov 09 2021
CROSSREFS
Sequence in context: A374243 A145807 A278962 * A033501 A336504 A331827
KEYWORD
nonn
AUTHOR
Jonathan Sondow, Sep 03 2006
STATUS
approved

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Last modified July 29 18:21 EDT 2024. Contains 374734 sequences. (Running on oeis4.)