%I
%S 1,4,5,10,25,50,100,446,1000,9775,10000,100000
%N Numbers n such that the sum of the digits of (n^n  1) is divisible by n.
%C n = 10^k is a member of the sequence, for all k >= 0. Proof: Let n = 10^k for some nonnegative integer k. Then n^n  1 has k*10^k 9's and no other digits, so its digits sum to 9*k*10^k = 9*k*n, a multiple of n.
%e The digits of 9775^9775  1 sum to 175950 and 175950 is divisible by 9775, so 9775 is in the sequence.
%p sumdigs:= n > convert(convert(n,base,10),`+`);
%p select(n > sumdigs(n^n1) mod n = 0, [$1..10^5]); # _Robert Israel_, Dec 03 2014
%t Do[k = n^n  1; s = Plus @@ IntegerDigits[k]; If[Mod[s, n] == 0, Print[n]], {n, 1, 10^5}]
%o (Python)
%o A109675_list = [n for n in range(1,10**4) if not sum([int(d) for d in str(n**n1)]) % n]
%o # _Chai Wah Wu_, Dec 03 2014
%K base,hard,more,nonn
%O 1,2
%A _Ryan Propper_, Aug 06 2005
