A147577 Numbers with exactly 4 distinct odd prime divisors {3,5,7,11}.

1155, 3465, 5775, 8085, 10395, 12705, 17325, 24255, 28875, 31185, 38115, 40425, 51975, 56595, 63525, 72765, 86625, 88935, 93555, 114345, 121275, 139755, 144375, 155925, 169785, 190575, 202125, 218295, 259875, 266805, 280665, 282975, 317625

Offset: 1

Author

Artur Jasinski, Nov 07 2008

Keywords

• nonn

Comments

Numbers k such that phi(k)/k = m

( Family of sequences for successive n odd primes )

m=2/3 numbers with exactly 1 distinct prime divisor {3} see A000244

m=8/15 numbers with exactly 2 distinct prime divisors {3,5} see A033849

m=16/35 numbers with exactly 3 distinct prime divisors {3,5,7} see A147576

m=32/77 numbers with exactly 4 distinct prime divisors {3,5,7,11} see A147577

m=384/1001 numbers with exactly 5 distinct prime divisors {3,5,7,11,13} see A147578

m=6144/17017 numbers with exactly 6 distinct prime divisors {3,5,7,11,13,17} see A147579

m=3072/323323 numbers with exactly 7 distinct prime divisors {3,5,7,11,13,17,19} see A147580

m=110592/323323 numbers with exactly 8 distinct prime divisors {3,5,7,11,13,17,19,23} see A147581

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A060735, A143207, A147571-A147575, A147576-A147580.

Programs

Mathematica

a = {}; Do[If[EulerPhi[x]/x == 32/77, AppendTo[a, x]], {x, 1, 1000000}]; a
Select[Range[350000],EulerPhi[#]/#==32/77&] (* Harvey P. Dale, Mar 25 2016 *)

Python

from sympy import integer_log
def A147577(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 i11 in range(integer_log(x,11)[0]+1):
            for i7 in range(integer_log(x11:=x//11**i11,7)[0]+1):
                for i5 in range(integer_log(x7:=x11//7**i7,5)[0]+1):
                    c -= integer_log(x7//5**i5,3)[0]+1
        return c
    return 1155*bisection(f,n,n) # Chai Wah Wu, Oct 22 2024

Formula

Sum_{n>=1} 1/a(n) = 1/480. - Amiram Eldar, Dec 22 2020