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Numbers n such that phi(n) = p^3, where p is product of digits of n.
3

%I #26 Aug 27 2014 13:58:15

%S 1,14611,141152,2831112,113216152,419521121,5342121213,7112216125,

%T 11191442171,11371542112,12192256111,12327134121,12432712113,

%U 13161313512,14132133215,113335111416,127131142512,133412214117,196112145112,311721152134,312118111536,312127213125,322631161151,411111553831,432122133312

%N Numbers n such that phi(n) = p^3, where p is product of digits of n.

%H Hiroaki Yamanouchi, <a href="/A153428/b153428.txt">Table of n, a(n) for n = 1..404</a> (terms a(1)-a(48) from _Max Alekseyev_)

%e phi(113216152) = (1*1*3*2*1*6*1*5*2)^3, so 113216152 is in the sequence.

%t Do[If[Apply[Times,IntegerDigits[n]]^3==EulerPhi[n],Print[n]],{n,300000000}]

%Y Cf. A068572, A153427.

%K base,nonn

%O 1,2

%A _Farideh Firoozbakht_, Jan 02 2009

%E a(6)-a(18) from _Donovan Johnson_, Nov 01 2010

%E a(19)-a(22) from _Donovan Johnson_, Aug 15 2013

%E a(23)-a(48) from _Max Alekseyev_, Aug 18 2013