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A359415 Numbers k such that phi(k) is a 5-smooth number where phi is the Euler totient function. 0
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
LINKS
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
KEYWORD
nonn,easy
AUTHOR
Darío Clavijo, Dec 30 2022
STATUS
approved

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Last modified April 23 02:53 EDT 2024. Contains 371906 sequences. (Running on oeis4.)