cp's OEIS Frontend

A229095 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.

%I A229095 #19 Dec 29 2024 03:44:04
%S A229095 1,8,9,67,72,467,801,1071,5141,7193,25688,68488,97768,111816,381061,
%T A229095 7829505,17079937,25615576,44582211,91110856,639359784,3492789629
%N A229095 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 A229095 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 A229095 with(numtheory); P:=proc(q) local n, t; t:=0;
%p A229095 for n from 1 to q do t:=t+n^tau(n); if t mod n=0 then print(n);
%p A229095 fi; od; end: P(10^6);
%o A229095 (PARI) list(lim) = {my(s = 0, f); for(k = 1, lim, s += k^numdiv(k); if(!(s % k), print1(k, ", ")));} \\ _Amiram Eldar_, Dec 29 2024
%Y A229095 Cf. A000005, A227427, A227429, A227502, A227848.
%K A229095 nonn,more
%O A229095 1,2
%A A229095 _Paolo P. Lava_, Sep 13 2013
%E A229095 a(16)-a(18) from _Jinyuan Wang_, Feb 18 2021
%E A229095 a(19)-a(22) from _Amiram Eldar_, Dec 29 2024