19136512domain: NAppears in sequencesa(n) = (2*n+1)*2^floor((n+1)/2).at n=36A097578a(n) = (4*n - 3) * 2^(n - 1).at n=18A118415Numbers with 38 divisors.at n=19A175747a(n) = J_9(n)/J_3(n), where J_k is the k-th Jordan totient function.at n=15A381713