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
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Harvey P. Dale)
FORMULA
a(n) = 510510 * A080681(n). - Amiram Eldar, Mar 10 2020
Sum_{n>=1} 1/a(n) = 1/92160. - Amiram Eldar, Nov 12 2020
MATHEMATICA
a = {}; Do[If[EulerPhi[x 510510] == 92160 x, AppendTo[a, 510510 x]], {x, 1, 100}]; a
sdpdQ[n_]:=Module[{f=FactorInteger[n][[All, 1]]}, Length[f]==7&&Max[f]==17]; Select[Range[510510, 138*10^5, 510510], sdpdQ] (* Harvey P. Dale, Aug 03 2019 *)
PROG
(Python)
from sympy import integer_log
from oeis_sequences.OEISsequences import bisection
def A147574(n):
ptuple = (2, 3, 5, 7, 11, 13, 17)
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, 6), 510510*n, 510510*n) # Chai Wah Wu, Mar 18 2026
CROSSREFS
KEYWORD
nonn
AUTHOR
Artur Jasinski, Nov 07 2008
STATUS
approved
