|
%I #9 Aug 24 2019 10:18:57
%S 1,3,9,15,56,102,198,228,234,280,312,528,672,756,1050,1320
%N phi(n) - floor(n^(1/3))*tau(n) = 0.
%C See comment to A062516.
%e For m=1320, Phi[m]-k[m]*Tau[m]=320-10*32=0. 16 terms below 100000 [and most likely at all]. phi(n)-Floor[n^(1/3)]*Tau[n] becomes positive for large n. At n=2520 seems last time negative.
%t Flatten[Position[Table[EulerPhi[w]-Floor[w^(1/3)//N]*DivisorSigma[0, w], {w, 1, 100000}], 0]]
%o (PARI) isok(n) = eulerphi(n) - sqrtnint(n, 3)*numdiv(n) == 0; \\ _Michel Marcus_, Aug 24 2019
%Y Cf. A062516, A000005, A000010.
%K fini,nonn
%O 1,2
%A _Labos Elemer_, Jul 20 2001
|
