OFFSET
1,1
COMMENTS
Successive numbers k such that EulerPhi(x)/x = m:
( Family of sequences for successive n primes )
m=1/2 numbers with exactly 1 distinct prime divisor {2} see A000079
m=1/3 numbers with exactly 2 distinct prime divisors {2,3} see A033845
m=4/15 numbers with exactly 3 distinct prime divisors {2,3,5} see A143207
m=8/35 numbers with exactly 4 distinct prime divisors {2,3,5,7} see A147571
m=16/77 numbers with exactly 5 distinct prime divisors {2,3,5,7,11} see A147572
m=192/1001 numbers with exactly 6 distinct prime divisors {2,3,5,7,11,13} see A147573
m=3072/17017 numbers with exactly 7 distinct prime divisors {2,3,5,7,11,13,17} see A147574
m=55296/323323 numbers with exactly 8 distinct prime divisors {2,3,5,7,11,13,17,19} see A147575
Although 39270 has exactly 6 distinct prime divisors (39270=2*3*5*7*11*17), it is not in this sequence because the 6 distinct prime divisors may only comprise 2, 3, 5, 7, 11, and 13. - Harvey P. Dale, Oct 11 2014
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000
FORMULA
a(n) = 30030 * A080197(n). - Charles R Greathouse IV, Sep 14 2015
Sum_{n>=1} 1/a(n) = 1/5760. - Amiram Eldar, Nov 12 2020
MAPLE
N:= 10^7: # for terms <= N
P:= [2, 3, 5, 7, 11, 13]:
r:= convert(P, `*`):
R:= NULL:
for i1 from 1 do
r1:= r*2^(i1-1);
if r1 > N then break fi;
for i2 from 1 do
r2:= r1*3^(i2-1);
if r2 > N then break fi;
for i3 from 1 do
r3:= r2*5^(i3-1);
if r3 > N then break fi;
for i4 from 1 do
r4:= r3*7^(i4-1);
if r4 > N then break fi;
for i5 from 1 do
r5:= r4*11^(i5-1);
if r5 > N then break fi;
q:= ilog[13](N/r5);
R:= R, seq(r5*13^i6, i6 = 0 .. q)
od od od od od:
sort([R]); # Robert Israel, Mar 18 2026
MATHEMATICA
a = {}; Do[If[EulerPhi[x]/x == 192/1001, AppendTo[a, x]], {x, 1, 100000}]; a
PROG
(PARI) is(n)=if(n%30030, return(0)); my(g=30030); while(g>1, n/=g; g=gcd(n, 30030)); n==1 \\ Charles R Greathouse IV, Sep 14 2015
(Python)
from sympy import integer_log
from oeis_sequences.OEISsequences import bisection
def A147573(n):
ptuple = (2, 3, 5, 7, 11, 13)
def g(x, m): return sum(g(x//(ptuple[m]**i), m-1) for i in range(1, integer_log(x, ptuple[m])[0]+1)) if m else x.bit_length()-1
return bisection(lambda x: n+x-g(x, 5), 30030*n, 30030*n) # Chai Wah Wu, Mar 18 2026
CROSSREFS
KEYWORD
nonn
AUTHOR
Artur Jasinski, Nov 07 2008
EXTENSIONS
More terms from Amiram Eldar, Mar 10 2020
STATUS
approved