Letting mu be the Moebius function, and nu be the Mertens function defined as nu(n) = SUM(k = 1..n, mu(k)) Then I notice [1] SUM(k = 1..n, mu(k)*[n/k]) = 1 [2] SUM(k = 1..n, nu([n/k])) = 1 where [x] is the floor of x. I wondered if there was any simple relationship linking [1] and [2], or perhaps some relationship to the Moebius transform a(n) = SUM(d|n; b(d)) <=> b(n) = SUM(d|n; mu(n/d)*a(d)).