OFFSET
1,1
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
FORMULA
Sum_{n>=1} 1/a(n) = 1/480. - Amiram Eldar, Dec 22 2020
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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Artur Jasinski, Nov 07 2008
STATUS
approved