|
%I #24 Nov 17 2018 21:31:12
%S 1,5,34,63,76,128,136,170,315,364,380,444,640,680,972,1820,1824,1836,
%T 2142,2220,4788,4860,6000,8064,8568,8736,9120,9180,10710,23940,40320,
%U 42840,43680
%N Numbers k such that phi(k) = tau(k)^2.
%C No more terms through 10^9. - _Ryan Propper_, Jun 09 2006
%C This sequence is finite because phi(k) >= sqrt(k) for all k >= 6, and for any e > 0, tau(k) < k^e for k large enough. Choosing e=1/4 gives tau(k)^2 < sqrt(k) <= phi(k). It remains unknown, however, if this sequence is full. - _Nathaniel Johnston_, Apr 28 2011
%t Do[If[EulerPhi[n] == DivisorSigma[0, n]^2, Print[n]], {n, 10^5}] (* _Ryan Propper_, Jun 09 2006 *)
%t Select[Range[10^5], EulerPhi[#] == DivisorSigma[0, #]^2 &] (* _Alonso del Arte_, Aug 25 2011 *)
%Y Cf. A039770.
%K easy,fini,nonn
%O 1,2
%A _Benoit Cloitre_, Mar 25 2002
%E More terms from _Ryan Propper_, Jun 09 2006
|
