A359415 Numbers k such that phi(k) is a 5-smooth number where phi is the Euler totient function.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 24, 25, 26, 27, 28, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 44, 45, 48, 50, 51, 52, 54, 55, 56, 57, 60, 61, 62, 63, 64, 65, 66, 68, 70, 72, 73, 74, 75, 76, 77, 78, 80, 81, 82, 84, 85, 88, 90, 91, 93, 95, 96, 97, 99

Offset

1,2

Links

Table of n, a(n) for n=1..79.

Project Euler, Problem 204: Generalised Hamming Numbers.

Example

20 is in the sequence because totient(20) = 8 and divisors of 8 are [1,2,4].

Maple

isA359415 := proc(n)
    numtheory[factorset](numtheory[phi](n)) minus {2, 3, 5} ;
    if  nops(%) =0 then
        true;
    else
        false;
    end if;
end proc:
for n from 1 to 100 do
    if isA359415(n) then
        printf("%d, ", n) ;
    end if;
end do: # R. J. Mathar, Mar 22 2023

Mathematica

Select[Range[100], Max[FactorInteger[EulerPhi[#]][[;; , 1]]] <= 5 &] (* Amiram Eldar, Dec 30 2022 *)

Prog

(Python)
from sympy import totient
def isok(n):
  f = totient(n)
  while f & 1 == 0: f >>= 1
  while f % 3 == 0: f //= 3
  while f % 5 == 0: f //= 5
  return f == 1
(PARI) issm(n) = n<7||vecmax(factor(n, 5)[, 1])<7; \\ A051037
isok(k) = issm(eulerphi(k)); \\ Michel Marcus, Jan 04 2023

Crossrefs

Cf. A000010 (phi), A051037 (5-smooth numbers).

Sequence in context: A214922 A004830 A081330 * A080682 A182049 A038770

Adjacent sequences: A359412 A359413 A359414 * A359416 A359417 A359418

Keyword

nonn,easy

Author

Darío Clavijo, Dec 30 2022

Status

approved