A114076 -id:A114076 - cp's OEIS frontend

A372670 Numbers k such that k * phi(k) is a fifth power.

1, 8, 256, 500, 864, 8192, 9826, 16000, 27648, 54000, 132651, 209952, 246924, 262144, 314432, 333396, 512000, 884736, 1061208, 1562500, 1728000, 6718464, 7002306, 7294032, 7901568, 8388608, 8541936, 10061824, 10668672, 13122000, 13564278, 15432750, 16384000

Offset: 1

Seiichi Manyama, May 10 2024

nonn

To look for terms it suffices to see if fifth powers have a divisors pair (k, m) such that phi(m) = k. - David A. Corneth, May 21 2024

			8 * phi(8) = 32 = 2^5.

David A. Corneth, Table of n, a(n) for n = 1..8846 (first 38 terms from R. J. Mathar)

Cf. A000584, A002618, A114076.

PARI
```
isok(n) = ispower(n*eulerphi(n), 5);
```

If n is in the sequence and prime p divides n, then p^5*n is in the sequence.

A358051 Squares k such that phi(k) is a cube.

Original entry on oeis.org

1, 16, 1024, 2500, 5184, 50625, 65536, 160000, 331776, 810000, 3779136, 4194304, 4691556, 5345344, 7001316, 10240000, 16867449, 20820969, 21233664, 27060804, 36905625, 39062500, 51840000, 52200625, 228765625, 241864704, 268435456, 269879184, 300259584, 333135504

Offset: 1

Darío Clavijo, Oct 27 2022

nonn

Project Euler, Problem 342. The totient of a square is a cube.

Cf. A000010, A114076.

Select[Range[20000]^2, IntegerQ[Surd[EulerPhi[#], 3]] &] (* Amiram Eldar, Oct 27 2022 *)

isok(k) = issquare(k) && ispower(eulerphi(k), 3); \\ Michel Marcus, Oct 27 2022

from sympy.ntheory.factor_ import totient
from gmpy2 import *
def isok(k):
  if is_square(k):
    j = isqrt(k)
    a,b = iroot(totient(j) * j, 3)
    return b

a(n) = A114076(n)^2. - Amiram Eldar, Oct 27 2022

cp's OEIS Frontend

A372670 Numbers k such that k * phi(k) is a fifth power.

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