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Numbers k such that phi(k) is a perfect biquadrate.
+10
19
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
COMMENTS
Corresponding values of phi include 1, 16, 256, 1296, 4096, ... and these arise several times each.
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
(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)))
Numbers k such that phi(k) is a perfect 5th power.
+10
8
1, 2, 51, 64, 68, 80, 96, 102, 120, 1285, 2048, 2056, 2176, 2560, 2570, 2720, 3072, 3084, 3264, 3840, 4080, 7957, 8227, 8279, 9079, 9139, 9709, 9919, 10355, 10595, 11667, 11673, 11691, 12099, 12393, 12483, 12753, 12987, 13797, 14715, 14763
COMMENTS
Numbers of the form u = 2^(5*k)*3^(5*m + 1), k>=1, m>=0, are terms because phi(u) = 2^(5*k)*3^(5*m) = (2^k*3^m)^5. - Marius A. Burtea, Sep 26 2022
EXAMPLE
phi of the sequence includes 1, 32, 1024, 7776, ...; powers arise several times; a(3) = A053576(5) = 51.
MATHEMATICA
k=5; Do[s=EulerPhi[n]^(1/k); If[IntegerQ[s], Print[n]], {n, 1, 5000}]
Select[Range[15000], IntegerQ[Surd[EulerPhi[#], 5]]&] (* Harvey P. Dale, Jul 26 2019 *)
CROSSREFS
Cf. A039770 (square), A039771 (cube), A078164 (4th), A078165 (5th, this sequence), A078166 (6th), A078167 (7th), A078168 (8th), A078169 (9th), A078170 (10th power), A001317, A053576, A045544, A000010.
Numbers k such that phi(k) is a perfect sixth power.
+10
8
1, 2, 85, 128, 136, 160, 170, 192, 204, 240, 4369, 8192, 8224, 8704, 8738, 10240, 10280, 10880, 12288, 12336, 13056, 15360, 15420, 16320, 47197, 47239, 47989, 49267, 49589, 50557, 51319, 52429, 52649, 55699, 57589, 57953, 59495, 63973
EXAMPLE
phi of the sequence includes 1, 64, 4096, 46656,..; powers arise several times; a(3)= A053576(6) = 85; in sequence relatively large jumps are observable when power of new numbers appear.
MATHEMATICA
k=6; Select[Range[65000], IntegerQ[EulerPhi[#]^(1/k)]&] (* Harvey P. Dale, Feb 20 2011 *)
Numbers k such that phi(k) is a perfect 7th power.
+10
8
1, 2, 255, 256, 272, 320, 340, 384, 408, 480, 510, 21845, 32768, 32896, 34816, 34952, 40960, 41120, 43520, 43690, 49152, 49344, 52224, 52428, 61440, 61680, 65280, 280999, 281587, 282637, 282949, 283897, 294409, 297449, 300409, 302039, 304399
EXAMPLE
phi of the sequence includes 1, 128, 16384, 279936, etc..; powers arise several times; a(3) = A053576(7) = 255; in sequence rather large jumps arise when power of new numbers appear.
MATHEMATICA
k=7; Do[s=EulerPhi[n]^(1/k); If[IntegerQ[s], Print[n]], {n, 1, 1000000}]
CROSSREFS
Cf. A039770 (square), A039771 (cube), A078164 (4th), A078165 (5th), A078166 (6th), A078167 (7th, this sequence), A078168 (8th), A078169 (9th), A078170 (10th power), A001317, A053576, A045544, A000010.
Numbers k such that phi(k) is a perfect 9th power.
+10
8
1, 2, 771, 1024, 1028, 1088, 1280, 1360, 1536, 1542, 1632, 1920, 2040, 327685, 524288, 524296, 526336, 557056, 559232, 655360, 655370, 657920, 696320, 699040, 786432, 786444, 789504, 835584, 838848, 983040, 986880, 1044480, 1048560
EXAMPLE
phi of the sequence includes 1, 512, 262144,.. etc.; powers arise several times; a(3) = A053576(9) = 771; in sequence smoother ranges and quite large jumps arise when power of new numbers appear as phi-values.
MATHEMATICA
k=9; Do[s=EulerPhi[n]^(1/k); If[IntegerQ[s], Print[n]], {n, 1, 10000000}]
CROSSREFS
Cf. A039770 (square), A039771 (cube), A078164 (4th), A078165 (5th), A078166 (6th), A078167 (7th), A078168 (8th), A078169 (9th, this sequence), A078170 (10th power), A001317, A053576, A045544, A000010.
Numbers k such that phi(k) is a perfect tenth power.
+10
8
1, 2, 1285, 2048, 2056, 2176, 2560, 2570, 2720, 3072, 3084, 3264, 3840, 4080, 1114129, 2097152, 2097184, 2105344, 2228224, 2228258, 2236928, 2621440, 2621480, 2631680, 2785280, 2796160, 3145728, 3145776, 3158016, 3342336
EXAMPLE
phi of the sequence includes 1, 1024, 1048576,.. etc.; powers emerge several times; a(3) = A053576(10) = 1285; in sequence smoother ranges and quite large jumps alternate when power of new numbers appear as phi-values.
MATHEMATICA
k=10; Do[s=EulerPhi[n]^(1/k); If[IntegerQ[s], Print[n]], {n, 1, 10000000}]
CROSSREFS
Cf. A039770 (square), A039771 (cube), A078164 (4th), A078165 (5th), A078166 (6th), A078167 (7th), A078168 (8th), A078169 (9th), A078170 (10th power, this sequence), A001317, A053576, A045544, A000010.
Numbers k such that phi(k) is a perfect 11th power.
+10
1
1, 2, 3855, 4096, 4112, 4352, 5120, 5140, 5440, 6144, 6168, 6528, 7680, 7710, 8160, 5570645, 8388608, 8388736, 8421376, 8912896, 8913032, 8947712, 10485760, 10485920, 10526720, 11141120, 11141290, 11184640, 12582912, 12583104
COMMENTS
Given the fact that phi(n) > sqrt(n) for all n except n=2 and n=6 we can see that every 11th power does appear as value only a finite number of times. What bounds on the density of this sequence can be proved?
MATHEMATICA
For[n = 1, n < 100000, n++, If[EulerPhi[n]^(1/11) == Floor[EulerPhi[n]^(1/11)], Print[n]]]
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