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%I #15 Feb 18 2021 02:37:07
%S 1,8,9,67,72,467,801,1071,5141,7193,25688,68488,97768,111816,381061,
%T 7829505,17079937,25615576
%N Numbers k such that Sum_{i=1..k} i^tau(i) == 0 (mod k), where tau(i) = A000005(i), the number of divisors of i.
%e 1^tau(1) + 2^tau(2) + ... + 8^tau(8) + 9^tau(9) = 1^1 + 2^2 + 3^2 + 4^3 + 5^2 + 6^4 + 7^2 + 8^4 + 9^3 = 6273 and 6273 / 9 = 697.
%p with(numtheory); P:=proc(q) local n, t; t:=0;
%p for n from 1 to q do t:=t+n^tau(n); if t mod n=0 then print(n);
%p fi; od; end: P(10^6);
%Y Cf. A000005, A227427, A227429, A227502, A227848.
%K nonn,more
%O 1,2
%A _Paolo P. Lava_, Sep 13 2013
%E a(16)-a(18) from _Jinyuan Wang_, Feb 18 2021
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