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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 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

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