|
|
A114063
|
|
Numbers k such that phi(k) = tau(k)^4, where tau(k) = A000005(k).
|
|
5
|
|
|
1, 17, 514, 8738, 32301, 33003, 36351, 41504, 42292, 43852, 51860, 62226, 549117, 561051, 571311, 599067, 617967, 629811, 634005, 657495, 673184, 674505, 683168, 701024, 705568, 718964, 722684, 732628, 745484, 759772, 774368
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
For all large enough k, we have tau(k) < k^(1/5) and phi(k) > k^(4/5). Hence, tau(k)^4 < k^(4/5) < phi(k), implying that this sequence is finite. - Max Alekseyev, Mar 10 2016
|
|
LINKS
|
|
|
EXAMPLE
|
phi(33003) = 20736. tau(33003) = 12, 20736 = 12^4.
|
|
PROG
|
(PARI) isok(n) = eulerphi(n) == numdiv(n)^4; \\ Michel Marcus, Jan 22 2014
|
|
CROSSREFS
|
|
|
KEYWORD
|
fini,nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|