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Numbers n such that the sum of the digits of phi(n)^sigma(n) is divisible by n.
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%I #14 Dec 02 2016 09:29:26

%S 1,9,18,35,67,189,310,475,516,684,2493,2527,3465,4701,5157,8321,11403,

%T 11780,14709,18396,47250,124385,131952,280342,288820,304299,525195,

%U 536756,899703

%N Numbers n such that the sum of the digits of phi(n)^sigma(n) is divisible by n.

%e The digits of phi(3465)^sigma(3465) sum to 72765 and 72765 is divisible by 3465, so 3465 is in the sequence.

%t Do[s = EulerPhi[n]^DivisorSigma[1, n]; k = Plus @@ IntegerDigits[s]; If[Mod[k, n] == 0, Print[n]], {n, 1, 10^4}]

%t Select[Range[50000],Divisible[Total[IntegerDigits[EulerPhi[#]^ DivisorSigma[1,#]]],#]&] (* _Harvey P. Dale_, Mar 19 2011 *)

%K base,more,nonn

%O 1,2

%A _Ryan Propper_, Aug 06 2005

%E More terms from _Harvey P. Dale_, Mar 19 2011

%E a(22)-a(29) from _Lars Blomberg_, Dec 02 2016