363141domain: NAppears in sequencesNumbers k such that Sum_{j=1..k} tau(j)^j == 0 (mod k), where tau(j) = A000005(j), the number of divisors of j.at n=11A229207G.f. satisfies A(x) = exp( Sum_{k>=1} (A(x^k) + A(w*x^k) + A(w^2*x^k))/3 * x^k/k ), where w = exp(2*Pi*i/3).at n=39A363404