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%I #12 Oct 06 2018 04:38:53
%S 1,2,3,9,11,18,54,74,108,135,426,531,585,1361,3456,6771,7245,7392,
%T 11025,11957,21494,27063,41952,68494,72516,108742,128331
%N Numbers k such that the sum of the digits of k^sigma(k) is divisible by k.
%e The sum of the digits of 531^sigma(531) is 9558 and 9558 is divisible by 531, so 531 is in the sequence.
%t Do[s = n^DivisorSigma[1, n]; k = Plus @@ IntegerDigits[s]; If[Mod[k, n] == 0, Print[n]], {n, 1, 10000}]
%t Select[Range[150000],Divisible[Total[IntegerDigits[#^DivisorSigma[ 1,#]]],#]&] (* _Harvey P. Dale_, Jul 19 2013 *)
%K base,nonn,more
%O 1,2
%A _Ryan Propper_, Aug 06 2005
%E More terms from _Ryan Propper_, Oct 10 2005
%E More terms from _Harvey P. Dale_, Jul 19 2013