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%I #17 Jun 27 2023 07:07:38
%S 1,4,5,10,25,50,100,446,1000,9775,10000,100000,995138,996544,998866,
%T 1000000
%N Numbers k such that the sum of the digits of (k^k - 1) is divisible by k.
%C k = 10^m is a term of the sequence for all m >= 0. Proof: Let k = 10^m for some nonnegative integer m. Then k^k - 1 has m*10^m 9's and no other digits, so its digits sum to 9*m*10^m = 9*m*k, a multiple of k.
%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^n-1) 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**n-1)]) % n]
%o # _Chai Wah Wu_, Dec 03 2014
%Y Cf. A007953, A048861.
%K base,hard,more,nonn
%O 1,2
%A _Ryan Propper_, Aug 06 2005
%E a(13)-a(16) from _Michael S. Branicky_, Jun 25 2023