A078164 Numbers k such that phi(k) is a perfect biquadrate.

1, 2, 17, 32, 34, 40, 48, 60, 257, 512, 514, 544, 640, 680, 768, 816, 960, 1020, 1297, 1387, 1417, 1729, 1971, 2109, 2223, 2289, 2331, 2445, 2457, 2565, 2594, 2608, 2774, 2812, 2834, 2835, 3052, 3260, 3458, 3888, 3912, 3924, 3942, 3996, 4104, 4212, 4218

Offset

1,2

Comments

Corresponding values of phi include 1, 16, 256, 1296, 4096, ... and these arise several times each.

a(3) = A053576(4).

A013776 is a subsequence since phi(2^(4*n+1)) = (2^n)^4. - Bernard Schott, Sep 22 2022

Subsequence of primes is A037896 since in this case: phi(k^4+1) = k^4. - Bernard Schott, Mar 05 2023

Links

Amiram Eldar, Table of n, a(n) for n = 1..10000

Mathematica

k=4; Do[s=EulerPhi[n]^(1/k); If[IntegerQ[s], Print[n]], {n, 1, 5000}]
Select[Range[5000], IntegerQ[Surd[EulerPhi[#], 4]]&] (* Harvey P. Dale, Apr 30 2015 *)

Prog

(PARI) is(n)=ispower(eulerphi(n), 4) \\ Charles R Greathouse IV, Apr 24 2020
(Python)
from itertools import count, islice
from sympy import totient, integer_nthroot
def A078164_gen(startvalue=1): # generator of terms >= startvalue
    return filter(lambda n:integer_nthroot(totient(n), 4)[1], count(max(1, startvalue)))
A078164_list = list(islice(A078164_gen(), 20)) # Chai Wah Wu, Feb 28 2023

Crossrefs

Sequences where phi(k) is a perfect power: A039770 (square), A039771 (cube), this sequence (4th), A078165 (5th), A078166 (6th), A078167 (7th), A078168 (8th), A078169 (9th), A078170 (10th).

Cf. A001317, A053576, A037896, A045544, A000010, A013776.

Sequence in context: A267540 A141068 A162622 * A060387 A003336 A344187

Adjacent sequences: A078161 A078162 A078163 * A078165 A078166 A078167

Keyword

nonn

Author

Labos Elemer, Nov 27 2002

Status

approved