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

Author

Labos Elemer, Nov 27 2002

Keywords

• nonn

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

Crossrefs

Subsequence of A039770. A037896 is a subsequence.

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.

Programs

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 *)

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