A143207 Numbers with distinct prime factors 2, 3, and 5.

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

Offset: 1

Author

Reinhard Zumkeller, Aug 12 2008

Keywords

• nonn
• easy

Comments

Numbers of the form 2^i * 3^j * 5^k with i, j, k > 0. - Reinhard Zumkeller, Sep 13 2011

Integers k such that phi(k)/k = 4/15. - Artur Jasinski, Nov 07 2008

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 1..10000 (terms 1..1000 from T. D. Noe)

Crossrefs

Cf. A069819.

Subsequence of A143204 and of A051037.

Programs

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
-- Reinhard Zumkeller, Sep 13 2011

Magma

[n: n in [1..5000] | PrimeDivisors(n) eq [2,3,5]]; // Bruno Berselli, Sep 14 2015

Mathematica

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 *)

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

Python

from sympy import integer_log
def A143207(n):
    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

Formula

A001221(a(n)) = 3; A020639(a(n)) = 2; A006530(a(n)) = 5; A143201(a(n)) = 6.

a(n) = 30*A051037(n); A007947(a(n)) = A010869(n). - Reinhard Zumkeller, Sep 13 2011

a(n) ~ sqrt(30) * exp((6*log(2)*log(3)*log(5)*n)^(1/3)). - Vaclav Kotesovec, Sep 22 2020

Sum_{n>=1} 1/a(n) = 1/8. - Amiram Eldar, Sep 24 2020

Extensions

New name from Charles R Greathouse IV, Sep 14 2015