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A143207
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Numbers with distinct prime factors 2, 3, and 5.
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32
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30, 60, 90, 120, 150, 180, 240, 270, 300, 360, 450, 480, 540, 600, 720, 750, 810, 900, 960, 1080, 1200, 1350, 1440, 1500, 1620, 1800, 1920, 2160, 2250, 2400, 2430, 2700, 2880, 3000, 3240, 3600, 3750, 3840, 4050, 4320, 4500, 4800, 4860
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OFFSET
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1,1
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COMMENTS
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LINKS
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FORMULA
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a(n) ~ sqrt(30) * exp((6*log(2)*log(3)*log(5)*n)^(1/3)). - Vaclav Kotesovec, Sep 22 2020
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MATHEMATICA
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a = {}; Do[If[EulerPhi[x]/x == 4/15, AppendTo[a, x]], {x, 1, 11664}]; a (* Artur Jasinski, Nov 07 2008 *)
n = 10^4; Table[2^i*3^j*5^k, {i, 1, Log[2, n]}, {j, 1, Log[3, n/2^i]}, {k, 1, Log[5, n/(2^i*3^j)]}] // Flatten // Sort (* Amiram Eldar, Sep 24 2020 *)
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PROG
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(Haskell)
import Data.Set (singleton, deleteFindMin, insert)
a143207 n = a143207_list !! (n-1)
a143207_list = f (singleton (2*3*5)) where
f s = m : f (insert (2*m) $ insert (3*m) $ insert (5*m) s') where
(m, s') = deleteFindMin s
(PARI) list(lim)=my(v=List(), s, t); for(i=1, logint(lim\6, 5), t=5^i; for(j=1, logint(lim\t\2, 3), s=t*3^j; while((s<<=1)<=lim, listput(v, s)))); Set(v) \\ Charles R Greathouse IV, Sep 14 2015
(PARI) is(n) = if(n%30, return(0)); my(f=factor(n, 6)[, 1]); f[#f]<6 \\ David A. Corneth, Sep 22 2020
(Magma) [n: n in [1..5000] | PrimeDivisors(n) eq [2, 3, 5]]; // Bruno Berselli, Sep 14 2015
(Python)
from sympy import integer_log
def bisection(f, kmin=0, kmax=1):
while f(kmax) > kmax: kmax <<= 1
while kmax-kmin > 1:
kmid = kmax+kmin>>1
if f(kmid) <= kmid:
kmax = kmid
else:
kmin = kmid
return kmax
def f(x):
c = n+x
for i in range(integer_log(x, 5)[0]+1):
for j in range(integer_log(m:=x//5**i, 3)[0]+1):
c -= (m//3**j).bit_length()
return c
return bisection(f, n, n)*30 # Chai Wah Wu, Sep 16 2024
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CROSSREFS
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KEYWORD
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nonn,easy,changed
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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